Modelling of a universal motor with speed control by

Henrik Grop Master Thesis

Supervisor: Dr. Juliette Soulard

Royal Institute of Technology Department of Electrical Engineering Electrical Machines and Power Electronics Stockholm 2006

XR-EE-EME 2006:003

Abstract Noise emitted by a universal motor is believed to be strongly related to the fluctuations in the electromagnetic torque produced by the machine. The variations of the current feeding the machine give rise to the fluctuations in the torque. The variations of the current are depending on the power electronics used to control the speed of the machine. This thesis presents a method of modelling a universal motor in a non-linear way based on a linear model and calculations from FEM using Flux version 8.1 by Cedrat. The FEM model is derived starting from drawings of the machine and the winding connections. A non-linear model implemented in Simulink is used to connect a triac, and a PWM. The fluctuations in the torque and the total harmonic distortion of the motor current are compared for the two control methods. It is found that the PWM produces less variation in the torque in the zero to peak sense, but it introduces high frequency ripple in the torque wave form instead. The difference between the triac and the PWM is most significant when the machine is controlled to rotate at low speed. At low speed, the PWM reduces the torque peak with approximately 5 % compared to the triac. Compared to the triac, the PWM controller reduces the THD in the machine current by approximately 24 % at low speed and high torque.

I

II

Acknowledgements This master thesis was written at the Royal Institute of Technology, School of Electrical Engineering, division of Electrical Machines and Power Electronics. The work was carried out in the period of September 2005 - March 2006. First of all, I would like to thank Ankarsrum Industries for initiating the subject of this thesis. A special thanks goes to Mr. Werner Fritz and Mr. Mats Pettersson who provided drawings and measurement data. I would like to thank my supervisor, Dr Juliette Soulard for outstanding assistance and support during the work of this thesis. All staff at the division have been very helpful in many ways, therefore I would like to thank you all. At last I would like to thank Mr. Jan Timmerman for very good times, especially at Roebel’s bar on Friday evenings.

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IV

Table of contents Abstract ......................................................................................................................................I Acknowledgements................................................................................................................ III Table of contents...................................................................................................................... V 1.

Introduction ...................................................................................................................... 1 1.1. Purpose of this work................................................................................................... 1 1.2. Outline of the thesis.................................................................................................... 1

2.

Universal motors .............................................................................................................. 2

3.

The investigated machine ................................................................................................ 3 3.1. Design......................................................................................................................... 3 3.1.1. Stator .................................................................................................................. 3 3.1.2. Rotor................................................................................................................... 4 3.1.3. Commutator........................................................................................................ 4 3.2. Windings .................................................................................................................... 5 3.2.1. Stator winding .................................................................................................... 5 3.2.2. Rotor winding..................................................................................................... 5 3.3. Measured performance............................................................................................... 8

4.

Linear model..................................................................................................................... 9 4.1. Electrical system ...................................................................................................... 10 4.1.1. Back emf .......................................................................................................... 10 4.2. Mechanical system ................................................................................................... 12 4.2.1. Electromagnetic torque produced by the rotor................................................. 13 4.3. Electromechanical system ........................................................................................ 14 4.4. Implementation in Simulink..................................................................................... 14 4.5. Simulation results using the linear model ................................................................ 15

5.

Finite element model ...................................................................................................... 17 5.1. Geometrical simplifications for the Finite Element Method (FEM) modelling....... 17 5.1.1. Stator ................................................................................................................ 17 5.1.2. Rotor................................................................................................................. 17 5.1.3. Rotor shaft ........................................................................................................ 17 5.2. Flux 2D..................................................................................................................... 18 5.3. Full pitch motor model............................................................................................. 18 5.3.1. Geometry.......................................................................................................... 18 5.3.2. Electric circuit .................................................................................................. 18 5.4. Short-pitch motor model .......................................................................................... 19 V

5.4.1. Geometry.......................................................................................................... 19 5.4.2. Mesh ................................................................................................................. 20 5.4.3. Electric circuit .................................................................................................. 21 5.4.4. End winding reactance ..................................................................................... 24 5.5. Results from full-pitch motor................................................................................... 27 5.5.1. Verification of commutation ............................................................................ 27 5.6. Results from short-pitch motor ................................................................................ 28 5.6.1. Machine resistance ........................................................................................... 28 5.6.2. Machine current and torque at 8000 rpm ......................................................... 29 5.6.3. Simulation time ................................................................................................ 31 6.

Non-linear analytical model .......................................................................................... 32 6.1. Air gap flux for the induced voltage ........................................................................ 32 6.2. Inductance of the field windings .............................................................................. 34 6.3. Inductance of the rotor winding ............................................................................... 36 6.4. Results from non-linear analytical model ................................................................ 37 6.4.1. Machine current................................................................................................ 37 6.4.2. Torque-speed characteristics ............................................................................ 38 6.4.3. Current wave forms.......................................................................................... 38 6.4.4. Torque wave forms........................................................................................... 40 6.5. Conclusions .............................................................................................................. 41

7.

Speed control .................................................................................................................. 42 7.1. Triac ......................................................................................................................... 42 7.2. Sinus PWM .............................................................................................................. 49

8.

Conclusions and future work ........................................................................................ 52

References ............................................................................................................................... 53 List of symbols ........................................................................................................................ 54 Appendix ................................................................................................................................. 56

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VII

VIII

Chapter 1 - Introduction

1. Introduction

1.1. Purpose of this work The purpose of this work is to investigate the possibility to model a small universal motor in a non-linear way. The model should be used to compare two different power electronic converters used to control the speed of the machine.

1.2. Outline of the thesis Chapter 2 describes universal motors very shortly. In chapter 3, the machine studied in this thesis is presented thoroughly. The different parts in the machine are described. Drawings of the rotor and the stator are provided. The winding type is described and a drawing of the winding in developed form is included. In chapter 4, the electrical and mechanical equations which describe the machine are derived and a linear model of the machine is presented. Chapter 5 describes the FEM model of the machine. The simplifications that are made to the geometry of the machine are described. A simple type of winding is used at first and then the real machine winding is described. Only the model with the real machine winding is thoroughly described. In chapter 6, a non-linear model of the machine is presented. This model is implemented in Matlab/Simulink. Chapter 7 presents the control of the machine. A triac controller is compared to a sinus PWM controller. In Chapter 8, conclusions are drawn and a discussion of future work is presented.

1

Chapter 2 - Universal motors

2. Universal motors Universal, or series motors, are widely used in fractional horsepower ratings in many domestic appliances such as drilling machines, vacuum cleaners and food mixers. Large universal motors, in the range of 500 hp are used for traction applications. This type of machine is preferable where a high power to weight ratio is required. It can be fed with either DC current or single phase AC current. The single phase AC current is most common to use as supply in most applications [1]. Due to the high starting torque and the possibility to be fed with direct current, this machine is as well used as starter motor for engines. The machine is called a series motor because the excitation winding is connected in series with the rotor winding. Further on, it is called universal motor because it can be operated with both AC and DC supplies. The speed of a universal motor can be very high, typically in the range of 5000-20000 rpm [2]. The speed is entirely dependent on the load and the supply voltage. Without load, a universal motor may reach dangerously high speeds. In fact, the no-load speed is only limited by friction and windage losses. Therefore, a universal motor should never be operated without a load. Since an unloaded universal motor can destroy itself, the load should always be connected directly to the shaft and should not be belt connected [2].

2

Chapter 3 - The investigated machine

3. The investigated machine The different parts of the machine and their purpose are discussed here. The design of the modelled machine is introduced.

3.1. Design Variations of the flux in magnetic materials induce eddy currents. These currents cause unwanted losses in the material. To minimize these losses, the stator and the rotor are made of laminations [2].

3.1.1. Stator The stator lamination is made of an electrical steel of quality M800-65A. This is a cold rolled non grain-oriented electrical steel, which is also called DK70. This material is alloyed with aluminium and silicone which increases the electrical resistance and thereby decreases the eddy currents in the material. Small eddy currents imply small losses in the laminations. The eddy current losses also depend on the thickness of the laminations. Thinner laminations give smaller eddy current losses [3]. A drawing of the cross-section of the stator lamination is found in figure 3.1. The stator consists of two excitation poles and magnetic paths between the poles. The holes in the lamination are used for fixating the sheets together with screws when the stator is assembled. The small rectangular shapes which look like holes are actually flanges, used for guidance when stacking the sheets during assembly. The inner space, to the left and to the right of the field poles, contains the field winding which produces the magnetising flux in the magnetic circuit.

Figure 3.1 – Cross-section of a stator lamination sheet.

3

Chapter 3 - The investigated machine

3.1.2. Rotor The rotor, which supports the armature winding, is laminated in the same way as the stator. The lamination is made of the same steel quality as the stator lamination. The laminations are punched and stacked together on the rotor shaft of the machine. Twelve armature slots hold the coils which conduct the armature current. Twelve rotor teeth complete the magnetic circuit and guide the flux towards the air gap. A drawing of the rotor is presented in figure 3.2.

Figure 3.2 – Drawing of a rotor lamination sheet and a 3D-view of the rotor without the armature coil.

3.1.3. Commutator To feed the rotating armature winding with current, there must be some physical electrical connection between the outer static part of the machine and the inner rotating part. This connection is achieved by using a set of brushes and a commutator. The brushes are fixed and the commutator is attached on the rotor shaft. Moreover, the brush-commutator construction has another purpose which is to reverse the current in the rotor coils when the coil leaves one excitation pole and enters the region of the next pole. The current reversal in each coil must be done in a specific rotor position in order to maintain the direction of the produced electromagnetic torque and thereby the direction of rotation. In order to do this, the commutator is made of several segments, depending on the number of coils and their arrangement in the armature. The segments are isolated from each other with some electrical insulating material. The brushes slide on the commutator surface during rotation and thereby create a connection to the armature coils. The connection of the coils is always made in such a way to make the current in the coils underneath the north pole of the excitation field have an opposite direction compared to the current in the coils underneath the south pole. In other words, with respect to the direction of the magnetizing field, the switching of the commutator ensures a constant current pattern in the rotor [4]. Thus the rotor produces a unidirectional torque which results in a rotating movement. The commutator in the test machine has 24 segments. A drawing showing the principle of the commutator assembled to the rotor shaft is presented in figure 3.3.

4

Chapter 3 - The investigated machine

Figure 3.3 – The commutator assembled on the rotor shaft.

3.2. Windings 3.2.1. Stator winding The stator winding, wound around the field poles produces the magnetizing flux in the machine [3]. Since the field winding is connected in series with the rotor winding, it carries the same current as the rotor winding. Therefore it must be wound with larger conductors than the rotor winding as the current is split in two parallel circuits in the rotor. The number of turns per coil in the field winding is 110 (there are two field coils connected in series). The diameter of the copper wire is 0.90 mm.

3.2.2. Rotor winding The rotor winding carries the rotor current. It produces a flux which creates the electromagnetic torque in the air gap when interacting with the magnetizing flux. The current in the armature winding is fed by a brush-commutator system. The coil ends are connected to the different segments in the commutator. The coil sides are placed in slots in the rotor. For DC-machines, the most common way of placing these coil sides is by letting the distance between the forward coil side and the backward coil side to be one pole pitch. This is called a full pitch winding. The pole pitch is defined as

One pole pitch = 180 o electrical = where p is the total number of stator poles. 5

o

360 mechanical p

Chapter 3 - The investigated machine There are several ways of inserting the coils in the rotor. The two most common ones are the lap winding and the wave winding (see figure 3.4). In the lap winding, the two ends of one coil are connected to segments in the commutator sitting next to each other. This makes the coil lap back on itself. [4] In the wave winding, the coil ends are connected to two commutator segments spread apart. This way of connecting the coil ends makes the coils look like a wave pattern as they are put into the machine [1].

Figure 3.4 – Left image – Coil configuration in a lap winding. Right image – Coil configuration in a wave winding [6].

The rotor winding in the studied universal motor is a short-pitch lap winding. This means the machine is wound in two layers with two coils side by side in each slot, i.e. four different coils in a slot. Each of the four coils in one slot of the rotor has seventeen conductors. The diameter of the conductor wire is 0.50 mm. For a full pole pitch winding in this machine, each coil would span six rotor slots. In the studied machine, the coil span is equal to five slots. The coil pitch can be calculated in electrical degrees [4] by:

coil pitch=180° ×

y 5 =180° × =150° electrical τp 6

where y - is the coil span, in number of slots, and τ p - is the pole pitch, in number of slots The mechanical coil pitch angle is the same as the electrical since the machine has two poles. The winding principle with end connections to the commutator is shown in figure 3.5. Note that only two coils are shown in the first layer in one slot in this figure. The rest of the coils are wound in a similar manner.

6

Chapter 3 - The investigated machine

Figure 3.5 – The principle of winding the coils and connecting the coil ends to the commutator. The figure is seen from the commutator side of the machine [13].

By studying figure 3.5, a drawing of the winding in developed form has been constructed [4]. The first coils which are illustrated in figure 3.5 are first drawn in respective rotor slots in figure 3.6. The ends of the coils are then connected to the correct commutator segment.

Figure 3.6 – The armature winding in developed form with the coils in figure 3.5 displayed.

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Chapter 3 - The investigated machine The remaining coils are then drawn in a similar way by doing a one slot translation in a counter-clockwise direction until the fourth coil in slot 1 and slot 6 are drawn. These are the last coils wound in the rotor. The complete rotor winding in developed form is viewed in figure 3.7. This figure is very important for the definition of the finite element model of the machine (chapter 5). The winding type of this machine makes the geometry in the finite element model complex.

Figure 3.7 – The configuration of the coils in the armature winding with connections to the commutator segments.

3.3. Measured performance The performance of the machine has been tested at Ankarsrum and the test results are presented in figure A1 of the appendix.

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Chapter 4 - Linear model

4. Linear model The system of differential equations that describes the machine is derived from the equivalent circuit of the universal motor. The universal motor is basically a DC-machine with the field winding connected in series with the rotor winding. The equivalent circuit is presented in figure 4.1 [2].

Figure 4.1 – Equivalent circuit of the universal motor

In figure 4.1, the parameters are R a - rotor winding resistance R f - field winding resistance La - rotor winding inductance Lf - field winding inductance u(t) - terminal voltage e(t) - back emf i(t) - current in the machine J - moment of inertia of machine and load D - viscous damping constant T(t) - electromagnetic torque TL - load torque ωm (t) - angular velocity of machine

9

Chapter 4 - Linear model

4.1. Electrical system By applying Kirchoff’s voltage law on the circuit in figure 4.1, the following equation can be derived. u (t ) − ( La + L f

) dtd i(t ) − e(t ) = ( R

a

+ R f ) i (t )

(eq. 4.1)

This differential equation can be solved if an expression for the induced back emf can be determined. This expression will be derived in next section.

4.1.1. Back emf As the rotor rotates in a magnetic field, an electromotive force is induced in the turns of the rotor winding. The emf induced in one turn of a coil is given by equation 4.2. This is the general form of Faraday’s law for a moving conductor in a time varying magnetic field [8]. econductor = ∫ Ea • dl = − ∫ C

S

∂B • ds + ∫ ( v × B ) • dl ∂t C

(eq. 4.2)

where v is the velocity vector of the conductor and B is the magnetic flux density vector in which the conductor is moving. C is the path along the conductor in the magnetic field and S is the surface bounded by C . The first term on the right side of equation 4.2 is the transformer emf due to time variations of the magnetic flux density. If the armature reaction is not considered, the coil short circuited by the brush, i.e the coil undergoing commutation, is the only contribution to the transformer action of equation 4.2. To simplify the analytical model of the machine, this transformer action of the coil undergoing commutation is not considered. The second term to the right of equation 4.2 is the flux cutting emf. A homogenous magnetic field beneath the magnetising poles of the machine is assumed. The flux cutting back emf produced in one turn of a coil according to figure 4.2 together with the simplifications made, is given by equation 4.3 [5]. econductor = ∫ Ea • dl ≈ ∫ ( v × B ) • dl C

C

10

(eq. 4.3)

Chapter 4 - Linear model

Figure 4.2 – Back emf induced in a conductor moving in a homogenous magnetic field [5].

The magnitude of the velocity of the conductors in figure 4.2 is given by: v = ωm r

(eq. 4.4)

where ωm is the angular velocity of the rotor and r is the radius of the rotor. Further on, the magnitude of the electric field along the conductor is Ea = ωm rBn

(eq. 4.5)

Bn is the normal-component of the magnetic flux density, i.e. the component perpendicular to the velocity of the conductor. The electric field Ea is in the opposite direction in conductor 2 compared to conductor 1.

By integrating the electric field along the length of the conductor moving in the magnetic field, the total emf voltage produced in the conductor is given. Assuming a constant electric field along the conductor, the emf induced in one conductor is: econductor = E ⋅ lconductor = ωm rBn ⋅ lconductor

(eq. 4.6)

lconductor is the length of the conductor moving in the magnetic field, i.e. the active length of the machine. Since the electric fields in the two conductors are of opposite directions, the total induced emf voltage in one turn is: eturn = 2 E ⋅ lconductor = 2ωm rBn ⋅ lconductor

(eq. 4.7)

The flux density seen by the conductors beneath each pole in the machine is calculated according to:

11

Chapter 4 - Linear model

Bn =

ψn Apole

=

ψn

2π rpolel pole p

= ⋅c

ψn p

2π rpolel pole c

(eq. 4.8)

Where p is the number of poles in the machine and c is the pole coverage factor. rpole is the radius of the pole which approximately is the same as the radius of the rotor i.e: rpole ≈ r

(eq. 4.9)

Further on, l pole is the same length as lconductor , i.e. l pole = lconductor

(eq. 4.10)

Again, index n stands for the normal component, and ψ n is the normal component of the flux beneath a pole. Substituting equation 4.8-4.10 in 4.7 gives the expression of the induced emf in one turn of a coil. eturn = ωm

p ψ πc n

(eq. 4.11)

This is the induced emf in one turn of a coil due to a conductor moving in a magnetic field. The total emf induced in the rotor winding consisting of N turns is the single turn voltage multiplied by the total number of turns. If the N turns are connected in parallel paths, the total emf induced is divided by the number of parallel paths a . Therefore, the total induced emf in the rotor winding is: e=

N p ωmψ n = K aωmψ n [V] a πc

(eq. 4.12)

N p is called the armature constant (or the rotor constant). This constant contains a πc information on the configuration of the rotor winding. Ka =

Assuming non-saturated operating conditions, the flux ψ n is directly proportional to the current in the field winding. Since the field winding is connected in series with the rotor winding, the flux is proportional to the rotor current according to. e = K a Kψ ωm I a [V]

(eq. 4.13)

Where Kψ is the flux constant.

4.2. Mechanical system The differential equation, describing the mechanical behaviour of the motor is given by:

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Chapter 4 - Linear model

Te (t ) = TL + Dωm (t ) + J

Where D is the friction constant.

d ωm (t ) dt

(eq. 4.14)

The machine speed and current 1 are connecting the electrical and mechanical differential equations, which leads to an electromechanical system. In next section, an expression for the developed electromagnetic torque is derived.

4.2.1. Electromagnetic torque produced by the rotor Assuming the coils are placed in the air gap, an expression for the developed electromagnetic torque in a universal motor is achieved by summing the forces acting on each conductor in the rotor winding. The force acting on one conductor in the rotor is given by [5]. Fe ,conductor = ∫ (I × B ) dl

(eq. 4.15)

C

Where C is the path along the current carrying conductor located in the magnetic field B . Assuming a constant magnetic field along C , the magnitude of the force acting on one conductor is equal to: Fe ,conductor = IBn l conductor

(eq. 4.16)

Once again, Bn is the normal component of the flux density, i.e. the component perpendicular to the current and the rotor periphery as shown in figure 4.3. lconductor is the length of the conductor located in the magnetic field which is the same as the active length of the machine l .

Figure 4.3 – A conductor carrying a current in a homogenous magnetic field

The electromagnetic torque produced by one conductor is then given by multiplying the force acting on the conductor, by the radius of the rotor. Te ,conductor = IBn lr

(eq. 4.17)

The total number of conductors in the rotor winding is 2 N , where N is the total number of turns. As stated in section 4.1.1, the number of parallel circuits in the rotor is a , and the I current in each conductor is a This means that the total electromagnetic torque acting on a the rotor is given by: 1

See section 4.3

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Chapter 4 - Linear model

Te =

2 NI a Bn lr a

(eq. 4.18)

With Bn replaced by equation 4.8, the torque is obtained by: Te =

Np I aψ n = K a I aψ n [ Nm ] πac

(eq. 4.19)

Again, with the flux ψ n directly proportional to the machine current, equation 4.19 can be written as [5].

Te = K a Kψ I a2

(eq. 4.20)

According to equation 4.20, the torque pulsates at twice the supply frequency.

4.3. Electromechanical system The machine is now described by the following set of differential equations forming an electromechanical system according to equation 4.21. d ⎧ ⎪⎪u (t ) − ( La + L f ) dt i (t ) − K a Kψ i (t )ωm (t ) = ( Ra + R f ) i(t ) ⎨ ⎪ K K i 2 (t ) = T + Dω (t ) + J d ω (t ) L m m ⎪⎩ a ψ dt

(eq. 4.21)

The back emf and the developed torque have been substituted with the expressions derived in section 4.1.1 and 4.2.1 respectively.

4.4. Implementation in Simulink The system of differential equations given by equation 4.21 is implemented in Simulink as a block diagram. The input voltage u (t ) and the load torque TL are input signals. The current and rotor speed are outputs. The linear simulink model of the machine is presented in figure 4.4.

Figure 4.4 – Block diagram modelling the universal motor.

14

Chapter 4 - Linear model

Further on, the block denoted Universal Motor in figure 4.4 consists of a block representing the electrical system in equation 4.21 and another block representing the mechanical system as shown in figure 4.5. 1 Machi ne current

2

Load Torque

Load T orque

i

i

w

2

w

Rotor speed Flux

1

Flux

Voltage Ka

Input voltage

Ka

Mechani cal system

El ectrical system

Figure 4.5 – The electrical and mechanical systems of the universal motor.

The contents of the electrical and the mechanical system are shown in figure A2 and figure A3 in appendix.

4.5. Simulation results using the linear model Even though this model does not take saturation into account, simulations are run in order to verify the behaviour of the model. The machine constant K a is calculated according to equation 4.12. The unknown parameters included in the model are at this stage only guessed. The parameters used in the simulations are: Known parameters Ra = 1.651 Ω

Guessed parameters La = 10 mH

R f = 1.164 Ω

L f = 10 mH

u = 120 V / 60 Hz

K a Kψ = 0.21 J = 0.0001 kg ⋅ m2 D = 0 Nm ⋅ s

Figure 4.6 shows the rotor speed when a constant load torque TL = 0.5 Nm is applied. The maximum speed reached is approximately 5300 rpm with a load torque of 0.5 Nm.

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Chapter 4 - Linear model

Figure 4.6 – Simulated rotor speed vs time with TL =0.5 Nm .

The speed given by the simulations is far lower than the speed of the machine measured at this load torque which would be approximately 8000 rpm at this load torque. This is due to the fact that the saturation is not taken into account in this linear model. The current and torque wave form are of perfect sine shape which are shown in figure 4.7 and figure 4.8. The torque pulsates with twice the frequency of the current as expected.

Figure 4.7 – Machine current as function of time

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Figure 4.8 – Produced electromagnetic torque as function of time.

Chapter 5 - Finite element model

5. Finite element model A finite element model of a full pitch machine is introduced in this chapter and a FEM model of the short pitch machine described in chapter 3 is thoroughly described.

5.1. Geometrical simplifications for the Finite Element Method (FEM) modelling The simplifications of the geometry are presented in this section. These simplifications are made to • •

Obtain a better mesh of the geometry Simplify the drawing of the geometry

The results from FEM simulations are more trustworthy if the triangles of the mesh are equilateral. If sharp edges and small radii’s are present in the geometry, the number of elements in those regions has to be very high in order to get a satisfactory mesh. Therefore it is desirable to have as few small radii’s and sharp corners as possible.

5.1.1. Stator In the finite element model, some holes in the stator geometry were not included. This simplification of the geometry can be made because the positions of the holes are not critical. This is the case for the holes situated in regions where the area of the magnetic path is large enough to ensure that the iron in these regions is not saturated. Further on, small smoothening radii’s in the stator were not taken into account in the finite element model. Besides simplifying the drawing of the geometry in the finite element program, this also makes the meshing of the geometry simpler to implement.

5.1.2. Rotor Some simplifications are also made to the smoothening radii’s of the rotor in the finite element model.

5.1.3. Rotor shaft The rotor shaft is the part of the machine that mechanically transfers the electromagnetic torque produced by the rotor to a connected load. The material properties of the shaft are critical in a mechanical point of view, but since the shaft is made of a solid non-magnetic steel material, it is not necessary to include the real properties of the shaft material in an electromagnetic model. The only crucial thing that must be correct is the diameter of the shaft. 17

Chapter 5 - Finite element model In the finite element model, the shaft is treated as a hole filled with vacuum in the centre of the rotor.

5.2. Flux 2D The software used in the FEM modelling in this project is the 2D 8.1 version of Flux which is a dedicated tool for making FEM models of electrical machines. This is due to the possibility to assign rotating properties to the air gap, and external connections with electrical circuits [7]. These are two features which are used in the simulation of this machine.

5.3. Full pitch motor model A simplified finite element model is obtained for a machine with two full-pitch rotor coils in each slot. This model is used to verify that the commutation works properly. This simplified motor is also used to test how to implement the non-linear model in Simulink. The model is only briefly presented here because the procedure used to create it is very similar to the one for the short-pitch motor.

5.3.1. Geometry The geometry of the full-pitch model is almost the same as for the test motor model. The only difference is that instead of having four regions in each rotor slot, the full-pitch machine only has two. The geometry of the full-pitch machine is presented in figure A9 in appendix. The rotor slots are divided in two regions which contain the rotor coils. The region coils are then linked to the electric circuit.

5.3.2. Electric circuit The rotor coils in the electric circuit in figure A10 in appendix have twice as many turns as for the short pitch machine presented in section 5.4. This is because the total number of coils in the full pitch winding is half the number of coils in the short pitch winding and the number of ampere turns in the rotor must be kept constant in order to have similar properties as the real machine.

18

Chapter 5 - Finite element model

5.4. Short-pitch motor model

5.4.1. Geometry Making the geometrical simplifications stated in the previous sections, the geometry is drawn in Flux2D. The final drawing of the universal machine is presented in figure 5.1. The rotor, the stator and the air are sectioned into regions which are assigned the respective material properties. Each rotor slot is separated into four regions so that the rotor coils in the electric circuit can be associated with the right region. The black slot corresponds to rotor slot 1a at rotor position zero, i.e.

θ mec = 0 The winding description given by Ankarsrum is shown in figure 3.5. It is drawn to give a clockwise rotation at the commutator side of the machine. In Flux, a counter clockwise rotation is defined positive. Instead of redrawing the drawing of the winding and in this way give the machine a counter clockwise rotation, a negative speed of the rotor is chosen in Flux. The coils in the rotor are at first inserted in the bottom of the slots, then in the outer region closer to the air gap. This removes the symmetry in the machine and the entire geometry of the machine must therefore be drawn in Flux. This implies that totally 48 rotor coil regions must be described. The large number of coils and regions give rise to very time- consuming simulations. The simulation time is especially high when the machine is fed with a current source of high amplitude. This is because the iron becomes highly saturated and the solver has difficulties to find a solution of the problem.

Figure 5.1 – The finite element model of the short pitch machine seen from the commutator side. The black slot corresponds to rotor slot 1a at rotor position zero degrees.

19

Chapter 5 - Finite element model

5.4.2. Mesh The geometry has to be meshed into a number of triangular surface elements. The most sensitive regions need a very fine mesh in order to get good results. The air gap is divided into three separate regions, one closest to the rotor, one in the middle and one closest to the stator. This is usually done when the variations in the torque are to be studied. The mesh can be seen in figure 5.2 which also includes a view of the fine mesh in the air gap region. The middle gap is remeshed at each timestep, the rotor air gap moving as the rotor and the stator air gap being fixed. To eliminate so called mesh noise, the mesh has to be implemented in such a way that the mesh in the air gap appears to be constant in time [9]. When the simulation time is increased by one time step the mesh in the air gap should appear to be exactly the same as in the previous time step.

Figure 5.2 – The mesh of the short pitch model. The right image shows an enlargement of the air gap region with its three-layer mesh.

20

Chapter 5 - Finite element model

5.4.3. Electric circuit Figure 5.3 shows a sketch of the position of the commutator segments relative to the rotor slots and relative to the brushes at position θ mec = 0 . The figure also contains information about how the machine is wound. This can be seen under Winding Arrangement in the right side of the figure. The brushes are on a fictional horizontal axis.

Figure 5.3 – The rotor slots with its coils and their relative position to the commutator at rotor angle zero degrees.

Figure 5.4 shows one commutator segment and one brush. The commutator segment is approaching the brush. The connection is established when the edge of the commutator segment reaches the edge of the brush. The brush starts to conduct current through the commutator segment. The connection is maintained until the other edge of the segment leaves the brush at the other edge of the brush.

Figure 5.4 – The commutation process.

A virtual set of brushes and a commutator are attached to the rotor in the FEM model as shown in figure 5.3. It is important to notice that no physical commutator or brush is actually included in the FEM model. These parts are solely modelled by variable resistances called 21

Chapter 5 - Finite element model brush-segment components in the electrical circuit. The instantaneous resistance of the brushsegment is governed by the rotor angle θ mec . The virtual commutator–brush approach of the problem simplifies the determination of the angular parameters of the switches. The brush-segment component is characterized by five parameters [9]: • •

The on state resistance Ron in ohms. The off state resistance Roff in ohms.

•

The angular position, θ pos in degrees, of the commutator segment with respect to the brush position. The brush opening angle, θbru in degrees. The opening angle of the commutator segment, θbar in degrees.

• •

The brush-segment component is in full conduction whenever the mechanical angle of the rotor θ mec is opposite to the angular position of the commutator segment with respect to the brush θ pos [9].

Figure 5.5 – Brush-segment conduction characteristics [9]

22

Chapter 5 - Finite element model The drive circuit presented in figure A5 in appendix is fed by a 120 V / 60Hz voltage source in series with a small inductance representing the end windings of the machine. The 48 square shaped brush-commutator segments in the upper and lower section of the drawing represent the brush-commutator connections. Figure 5.6 shows θ pos for commutator segment number 22 with respect to the right brush and to the left brush at θ mec = 0 . In figure A5 in appendix, this gliding contact to the right brush is handled by brush-segment number G22 and brushsegment G46 handles the contact to the left brush. A table showing the numbering of the brush-segments and their corresponding segment and brush is found in appendix. The angular parameters of the brush-segment components for the modelled machine at position zero, i.e. θ mec = 0 , are presented in figure 5.7.

Figure 5.6 – The commutator and its segments with positions relative to the right and left brush. These angles are used to set the switch parameter θ pos .

23

Chapter 5 - Finite element model

Brush-segment number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Brush-segment number 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Pos. r.r.b.

-7.5 7.5 22.5 37.5 52.5 67.5 82.5 97.5 112.5 127.5 142.5 157.5 172.5 -172.5 -157.5 -142.5 -127.5 -112.5 -97.5 -82.5 -67.5 -52.5 -37.5 -22.5

Figure 5.7 – The position of the segments relative the brushes,

Pos. r.l.b.

172.5 -172.5 -157.5 -142.5 -127.5 -112.5 -97.5 -82.5 -67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5 82.5 97.5 112.5 127.5 142.5 157.5

θ pos at θ mec = 0 . Pos r.r.b: Position relative

right brush, Pos r.l.b: Position relative left brush.

5.4.4. End winding reactance The effect of the end windings is not included in a two dimensional FE problem such as the one that is used. Therefore the end-winding reactance should be added in the circuit. In [10], an empirical method to determine the end-winding reactance is proposed. A simplified drawing of the end winding is viewed in figure 5.8.

Figure 5.8 – The simplified shape of a coil end [10].

The angle γ is the angle between the end winding and the radial rotor plane. ly is the length of the end winding. α is calculated approximately by measuring the length of the end winding and the nearest length between the forward and the backward coil, which is the distance τ. 24

Chapter 5 - Finite element model

cos γ =

τ

(eq. 5.1)

ly

[7] states that inductance of a coil end Ly is equal to:

⎛ b ' sin γ Ly = 0.2nl2l y ln ⎜ kz ⎜ 2g 1,1 ⎝

⎞ −6 ⎟⎟ ⋅10 H ⎠

(eq. 5.2)

where nl - Total number of conductors in a coil end bkz' - Ideal reduced commutation zone, the length along the anchor where the current in the winding is reversed g1,1 - A coil ends average geometrical distance from itself

g1,1 = 0.2235 ( b1 + h1 )

(eq. 5.3)

b1 is the width of the coil and h1 is the height of the coil showed in figure 5.9.

Figure 5.9 - Cross section of a coil end [10].

For simplicity, the reduced commutation zone length bkz' , is substituted with the non-reduced length bkz . This gives a somewhat larger inductance.

⎛ β − 1 + fλ ⎞ bkz = τ s ⎜1 + ⎟ uc ⎝ ⎠ where uc is the number of coils, side by side in a slot .

25

(eq. 5.4)

Chapter 5 - Finite element model

β=

bc − bi

λτ

(eq. 5.5)

with bc - the width of the brush bi - the thickness of the insulation between two adjacent commutator segments

λτ - the distance between two adjacent commutator segments λτ =

π dk nk

(eq. 5.6)

Where d k is the diameter of the commutator and nk is the number of segments in the commutator.

τ s is the distance between two adjacent rotor slots τs =

2π r Q

(eq. 5.7)

D is the outer diameter of the rotor and Q is the total number of rotor slots. f λ is the reduction from full pole pitch expressed in number of commutator segments. fλ =

nk − ucτ p p

(eq. 5.8)

τ p is the pole pitch in number of rotor slots. With expressions 5.1-5.8 and some geometrical data from drawings and approximations, a first value of the end winding inductance can be calculated. It is found to be approximately. Ly ≈ 5μ H

(eq. 5.9)

At 60 Hz this inductance corresponds to a reactance equal to

X Ly = ω Ly ≈ 2mΩ

(eq. 5.10)

There are one end winding per coil on both sides of the machine and there are a = 2 number of parallel paths in the rotor. With 4 coils per rotor slot, the total end winding reactance can be written. 1 X Ltoty = 2 4Qω Ly = 48 ⋅ 2mΩ = 96mΩ a

(eq. 5.11)

At maximum current in the machine, this reactance results in a total reactive voltage drop of less than 1V. Therefore the effect of the end windings could be neglected. 26

Chapter 5 - Finite element model

5.5. Results from full-pitch motor The FEM model of the full-pitch motor is primarily used to verify that commutation works properly i.e. that the brush-segments in the electric circuit is described properly.

5.5.1. Verification of commutation The method used to describe the brush-commutator connection was explained in detail in section 5.4 which deals with the FEM model of the short-pitch machine. The model of the full-pitch machine is fed with a constant current and the commutation process is studied. To speed up the simulation time, no stator current is present. To further decrease the simulation time, the coils in the rotor have less turns than they should have (reduced level of magnetization). If the commutation is described properly, the coils situated in the region under one stator pole should always have opposite current direction compared to the coils situated in the region under the other stator pole. The coils in rotor slots 1-6 should have the same current direction at θ mec = 0 . Every time the rotor has rotated to a position where a commutator segment enters the brush, the associated coils should start to commutate. The first time step corresponds to θ mec = 10o , which means that the coils in slot 1 and slot 7 are fully commutated. When the rotor has rotated to position θ mec = 180o , the coils in slot 1 and slot 7 should have reversed their current direction. The commutation order of the coils in slot 1-3 and their corresponding coils on the opposite side of the rotor, i.e. coil 7-9, are shown in figure 5.10 and figure 5.11 respectively. All the curves are as expected, so the commutator was well described.

Figure 5.11 – Current in conductors in slot 7-9

Figure 5.10 – Current in conductors in slot 1-3

27

Chapter 5 - Finite element model

5.6. Results from short-pitch motor 5.6.1. Machine resistance The winding resistance is represented by the resistors connected in series with the coils. Their values depend on the loading of the machine since losses increase the temperature of the winding and therefore its resistance. This heating results in a linearly increasing resistance for the coils. It was assumed that the rotor and the stator resistances provided by Ankarsrum have been measured at room temperature. In the electric circuit associated to the FEM model, the resistance at the unknown working temperature should be used. To identify the value of the machine resistance at a load corresponding to 8000 rpm, the FEM model is simulated. The current and the torque are compared to the measured values. This was done for different values of the machine resistance. Using the results for the current, a linear approximation is done and an approximate value of the resistance is determined. Finally, the machine is once again simulated with the identified resistance and the simulated current is compared to the measured current. It was intended to do this procedure for several speeds, i.e. different loads, but the simulation time turned out to be long. Therefore this simulation was only performed at 8000 rpm. Figure 5.12 shows the relationship between the total machine resistance and current together with the linear approximation at 8000 rpm. It was found that a total machine resistance of 5.7 Ω corresponds to a winding temperature of approximately 280 οC which is non-realistic. The high resistance can however be explained by the contact between the brush and the commutator that is not so well characterized.

Figure 5.12 – Relationship between the current and the total machine resistance at 8000 rpm.

28

Chapter 5 - Finite element model

5.6.2. Machine current and torque at 8000 rpm The most relevant results from the FEM simulations are presented here. The machine speed is fixed at 8000 rpm and the applied voltage is 120V / 60 Hz. The value of the sum of the stator and rotor resistances is 5.7 Ω.

Figure 5.13 – The voltage applied to the machine in the simulation.

Figure 5.14 shows the current drawn by the machine. The steady-state current seems to be reached almost instantaneously. Therefore the simulation time can be set to two periods. This amount of time ensures that the steady-state is obtained in the second period of the current wave form. The RMS value of the current is 6.21 A which can be compared to the measured value 6.29 A. It can be seen that the current is distorted due to saturation of the magnetic circuit. This implies that a non-linear model is necessary to predict the performance of this machine. Since the current is adjusted with the total resistance of the machine, the produced electromagnetic torque should also be analyzed to verify the model.

Figure 5.14 - The machine current wave form at 8000 rpm, 120V / 60Hz and total machine resistance 5.7Ω.

29

Chapter 5 - Finite element model

Figure 5.15 – Produced electromagnetic torque

In figure 5.15, the electromagnetic torque is shown. It pulsates with twice the supply frequency and also presents some high frequency ripple at the torque peaks. The average torque from the simulations is 0.496 Nm. This value corresponds very well to the measured value which is 0.498 Nm. The simulated current is approximately 0.1 A smaller than the measured value. Increasing the simulated current by reducing the total resistance would lead to an increased average torque, but the results for the torque would still be very close to measurements.

Figure 5.16 – The input power to the machine

Another parameter that can be used to test the validity of the model is the input power, which is shown in figure 5.16. The average power consumed by the motor in the simulation is 697.8 W. This is 9.5 % higher than the measured value which is 637 W.

30

Chapter 5 - Finite element model

5.6.3. Simulation time The simulation time for this machine with so many coil regions in the geometry is very high. It should be mentioned that the results from one working point of the short-pitch motor model presented in this chapter took more than a weekend to produce. The simulations were carried out on a Pentium 4, 2.40 GHz with 512 Mb of RAM. Due to the high simulation time, it was decided to use the full pitch motor model in the following FEM simulations.

31

Chapter 6 - Non-linear analytical model

6. Non-linear analytical model As the short-pitch machine turned out to be very time consuming to simulate, the full-pitch machine will instead be investigated. Unfortunately, the full pitch machine also requires long simulation time. Therefore only four loads/speeds have been simulated. To further decrease the simulation time, the time stepping procedure in FEM is terminated after one period of the electromagnetic torque, i.e. half the period of the current wave form. This is believed to give results accurate enough according to the results obtained for the 8000 rpm simulation of the short-pitch machine. This simulation showed no transient behaviour at the start. It is therefore assumed that the full-pitch machine does not show transients either. To begin with, two ways of making a non-linear analytical model in Simulink have been considered. These are:

• •

Combined Simulink-Flux model Pure Simulink model

In the combined Simulink-Flux model, a block representing the FE model of the machine would be implemented in Simulink. This method would probably yield more accurate result compared to the pure Simulink model. The risk of serious difficulties which may occur when the programs are connected is considered high. Further on, the simulation time is expected to be even higher than with FEM only. These are the reasons why the pure Simulink model, was chosen. As the iron in the machine becomes saturated, the air gap flux will no longer show a linear behaviour with the machine current. Therefore, a method to calculate the flux in the air gap for different currents is needed in order to determine the induced back emf in equation 4.13 and the electromagnetic torque in equation 4.20. The air gap flux calculation procedure is presented in section 6.1. Further on, the inductance of the field winding will also become saturated. The calculation of the non-linear field winding inductance is presented in section 6.2. No method has been investigated to calculate the rotor winding inductance but it is believed to be small compared to the inductance of the field winding.

6.1. Air gap flux for the induced voltage The normal component of the flux density under one pole is calculated using Flux. The flux density is not homogenous in the air gap beneath a field pole due to armature reaction. Therefore an average value of the normal component of the flux density is calculated according to

Bn,avg =

1 Bn dl lC C∫

Where C is the integration path shown in figure 6.1. lC is the length of C .

32

(eq. 6.1)

Chapter 6 - Non-linear analytical model

Figure 6.1 – The path where the flux density is calculated

An increasing current is applied to the full-pitch machine in the FEM simulation and equation 6.1 is applied for the different current levels. This gives the average value of the flux density which generates the back emf in the rotor winding. The average flux density in the air gap beneath a pole as function of the machine current is shown in figure 6.2. With Bn available, it is straight forward to calculate the average flux cut by the conductors in the rotor winding using the width of the excitation pole. A lookup table is added to the linear Simulink model in section 4. This table has the machine current as input and the flux, which is used in equation 4.12 and 4.19 as output. With the air gap flux known for any current, the back-emf can be calculated.

Figure 6.2 – The average flux density in the air gap calculated by FEM.

33

Chapter 6 - Non-linear analytical model

6.2. Inductance of the field windings In the linear analytical model, it is assumed that the flux produced by the field windings is directly proportional to the machine current. Unfortunately, this is not true when the machine operates with high currents at high loads, i.e. the iron in the machine is highly saturated. To deal with the non-linearity in the inductance of the field windings, the following procedure is used. The objective is to calculate the field winding inductance of the machine for different rates of saturation. This is achieved by simulating the flux linkage seen by the stator- and the rotor winding for different current levels with FEM. The method to obtain the flux linkage from magneto-static FEM simulations is thoroughly described in [11] and is shortly presented here. The method is based on the magnetic vector potential. This quantity reflects how much flux per unit length is circulating around a certain point. Starting with one of Maxwell’s equations

∇• B = 0

(eq. 6.2)

Equation 6.2 implies that the curl of another vector field A can be used to express B such as: B = ∇× A

(eq. 6.3)

where A is the magnetic vector potential. The total flux seen by a coil is given by integrating the flux density over the area of the coil.

φ=

∫∫ B • ds = ∫∫ (∇ × A ) • ds

Acoil

(eq. 6.4)

Acoil

By applying Stoke’s theorem, equation 6.4 becomes:

φ=

∫∫ B • ds = ∫ A • dl

Acoil

(eq. 6.5)

C

where C is the countour bounding the area Acoil . In a 2D model A , has only one component in the axial direction (A z ) . Equation 6.5 can therefore be written as:

φ = ∫ A • dl = Leff ⋅ ( A1 − A2 ) + Lend ⋅ ( Aend 1 − Aend 2 )

(eq. 6.6)

C

where Leff is the length of the coil, i.e the effective length of the machine. A1 is the averaged vector potential on one side of the winding and A 2 is the averaged vector potential on the other side of the winding. Neglecting the end windings, equation 6.6 becomes: 34

Chapter 6 - Non-linear analytical model

φ = ∫ A • dl ≈ Leff ⋅ ( A1 − A2 )

(eq. 6.7)

C

This method is applied to the field winding of the full-pitch motor model, with A1 the averaged vector magnetic potential in the right slot holding the stator winding and A 2 the averaged vector magnetic potential in the left stator winding slot. The total flux seen by the field winding is:

ψ field = N field Leff ⋅

(∑

p

Ar ,i − ∑ i =1 Al ,i p

i =1

)

(eq. 6.8)

N field is the number of turns in the field winding, Ar ,i and Al ,i are the averaged vector potential in the right and left region holding field coil i .

For a non-linear magnetic circuit, two inductances can be defined. The absolute inductance is defined as the proportionality factor between the flux and the current i.e. the usual way in a linear case. L f ,abs =

φ Ia

(eq. 6.9)

An incremental inductance can also be defined as: L f ,inc =

dφ dI a

(eq. 6.10)

Figure 6.3 illustrates the values of the absolute and incremental inductance for a level of current where the circuit is saturated. As can be seen, the value of the incremental inductance is much lower than the absolute inductance at saturation. They are equal in the linear region. For the Simulink model of the universal motor, it is the incremental inductance that should be used as the inductive voltage drop in the field winding can be expressed as: ef =

dφ f dt

=

dφ f dia di ⋅ = L f ,inc (ia ) ⋅ a dia dt dt

(eq. 6.11)

The saturated flux linkage seen by the stator widning is shown in figure 6.3. The two different interpretations of the inductance concept are also shown in the same figure.

35

Chapter 6 - Non-linear analytical model

Flux linkage

0.25 L

= dΨ/di

inc

Flux linkageΨ [Wb]

0.2 Instantaneous operating point 0.15 L

= Ψ/i

abs

0.1

0.05

0

0

5

10

15 I [A]

20

25

30

Figure 6.3 – The flux linkage seen by the stator winding and the two ways of interpreting the inductance.

As for the air gap flux used to model the back emf, the incremental self inductance of the field winding L f , is implemented in the non-linear Simulink model as a lookup table. With the machine current I a as input, the lookup table returns the incremental field winding inductance, saturated or non-saturated depending on the level of I a . The incremental inductance of the field winding as function of the machine current is presented in figure 6.4. The electrical system of the non-linear analytical model is shown in figure A4 in appendix.

Field winding inductance 100 90 80

Lf = dΨ/dI [mH]

70 60 50 40 30 20 10 0

0

5

10

15 I [A]

20

25

30

Figure 6.4 – The field winding inductance calculated based on equation 6.10.

6.3. Inductance of the rotor winding The inductance of the rotor winding is chosen based on trial and error. It is considered constant and independent of the load. The inductance of the rotor winding is set to 6 mH.

36

Chapter 6 - Non-linear analytical model

6.4. Results from non-linear analytical model The interesting results from the non-linear analytical model are presented here. These results are compared to the FEM results for the full-pitch machine.

6.4.1. Machine current The RMS value of the current drawn by the machine is plotted versus the produced electromagnetic torque in figure 6.5. It is found out that the current obtained from the non-linear analytical model compares well to the current in the FEM model. The FEM value of the current is between 5-7 % (depending on the torque) less than with the Simulink model at a given torque.

Electromagnetic torque vs current 0.8

FEM Simulink

0.7 0.6

Te [Nm]

0.5 0.4 0.3 0.2 0.1 0

3

4

5

6 Irms [A]

7

8

9

Figure 6.5 – Electromagnetic torque vs. current.

37

Chapter 6 - Non-linear analytical model

6.4.2. Torque-speed characteristics The torque available at different speeds is presented in figure 6.6. The torque from the non-linear analytical model is overestimated by approximately 7 % for 10000 rpm. This might be explained by the inductance of the rotor winding which may show non-linear behaviour in reality. However, no method of modelling this inductance in a non-linear way has been investigated. Further on, the commutation is not modelled in Simulink. This factor also influences the torque/speed characteristics of the machine. In FEM time-stepping simulations with a voltage source and the rotor forced to rotate at a certain speed, the time steps are coarse in order to keep the simulations reasonably long. The large time steps generate solutions which are somewhat noisy. This is a source of error when the rms value of the current and the average produced electromagnetic torque are calculated from FEM. The FEM simulations for the two highest loads are only simulated for half the current period. If the entire period were investigated, the current would increase. This was seen by comparing calculations from the third highest torque (8000 rpm in figure 6.6). The current increase led to better results for this point. Electromagnetic torque vs speed 0.8

FEM Simulink

0.7 0.6

Te [Nm]

0.5 0.4 0.3 0.2 0.1 0 5000

6000

7000

8000 9000 speed [Rpm]

10000

11000

Figure 6.6 – Electromagnetic torque vs. speed.

6.4.3. Current wave forms Some current wave forms generated by the non-linear analytical model at different loads and speeds are presented in this section. It is interesting to note that the wave form at 8000 rpm for the simulated full-pitch machine is of similar shape as for the short-pitch machine at 8000 rpm which is presented in figure 5.14, in section 5.6.2. The current wave forms for 6500, 7000, 8000 and 10000 rpm and their related load torque are presented in figure 6.7. For comparison, the current from the 8000 rpm FEM simulation is provided in figure 6.8.

38

Chapter 6 - Non-linear analytical model

Current vs time at 6500 rpm, 0.69 Nm from non-linear analytical model

20

Current vs time at 7000 rpm, 0.631 Nm from non-linear analytical model

15

Machine current Ia

Machine current Ia

15 10 10 5 Ia [A]

Ia [A]

5 0

0

-5 -5 -10 -10

-15 -20

0

0.005

0.01

0.015 0.02 Time [s]

0.025

-15

0.03

0

Current vs time at 8000 rpm, 0.524 Nm from non-linear analytical model

10

5

5

-5

-10

-10

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.025

0.03

-15

0.03

Machine current Ia

0

-5

-15

0.015 0.02 Time [s]

Current vs time at 10000 rpm, 0.321 Nm from non-linear analytical model

10

0

0.01

15

Machine current Ia

Ia [A]

Ia [A]

15

0.005

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.03

Figure 6.7 – The current wave forms simulated at different loads and speeds by the non-linear analytical Simulink model.

Figure 6.8 – The current wave form simulated with FEM at 8000 rpm, 0.52 Nm load torque.

39

Chapter 6 - Non-linear analytical model

6.4.4. Torque wave forms The electromagnetic torque corresponding to the currents presented in figure 6.7 in section 6.4.3 is plotted in figure 6.9. The torque simulated by FEM is plotted in figure 6.10 as a reference. Electromagnetic torque vs time at 6500 rpm, 0.69 Nm from non-linear analytical model

Electromagnetic torque vs time at 7000 rpm, 0.631 Nm from non-linear analytical model

2

Electromagnetic torque

1.8

1.8

1.6

1.6

1.4

1.4 Torque [Nm]

Torque [Nm]

2

1.2 1 0.8

1 0.8 0.6

0.4

0.4

0

0.2 0

0.005

0.01

0.015 0.02 Time [s]

0.025

2

0

0.03

Electromagnetic torque vs time at 8000 rpm, 0.524 Nm from non-linear analytical model

1.8

1.6

1.6

1.4

1.4

1 0.8

0.4

0.2

0.2 0.01

0.015 0.02 Time [s]

0.025

0.025

0.03

Electromagnetic torque

1

0.6

0.005

0.015 0.02 Time [s]

0.8

0.4

0

0.01

1.2

0.6

0

0.005

2

1.8

1.2

0

Electromagnetic torque vs time at 10000 rpm, 0.321 Nm from non-linear analytical model

Electromagnetic torque

Torque [Nm]

Torque [Nm]

1.2

0.6

0.2

Electromagnetic torque

0

0.03

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.03

Figure 6.9 – The torque wave forms simulated at different loads and speeds with the non-linear analytical Simulink model.

Figure 6.10 – The torque wave form simulated with FEM at 8000 rpm, 0.52 Nm load torque.

40

Chapter 6 - Non-linear analytical model

6.5. Conclusions The current waveforms clearly show that a non-linear model is required to predict the performance of such universal motors (full-pitch and/or short-pitch). A way to obtain the required non-linear functions from FEM has been presented. Good correlation has been obtained for the current and torque (wave forms, RMS value and mean value respectively). At high speed, the commutation might be added in the model to help obtaining better correlation. Further on, the non-linearity of the rotor winding inductance should be investigated. The good correlation between FEM and the non-linear analytical model offers to use the latter for investigating control strategies.

41

Chapter 7 - Speed control

7. Speed control In this chapter, two speed control strategies are implemented in Matlab/Simulink together with the non-linear analytical model of the full-pitch motor. These two strategies are implemented with triacs or controlled switches. The two circuits are shown in figure 7.1 and figure 7.2. The influence of the torque and current waveforms are compared.

Figure 7.1 – Triac motor controller block.

Figure 7.2 – PWM motor controller block.

7.1. Triac To control the speed of the machine, an open loop triac controller that controls the effective applied machine terminal voltage, is implemented in Simulink. It is believed that the switching of the machine voltage caused by the triac adds harmonics to the supply current and affects the torque wave form of the machine in a negative way.

42

Chapter 7 - Speed control The triac block is shown in figure 7.1. The control angle block is high whenever the control angle has passed in the electrical period of the voltage. If the triac is feeding a pure resistive load, the zero current detection block can be removed. The machine is an inductive load. Therefore, for each half period, the triac can only turn off the applied terminal voltage once the current has reached zero. The sr-latch in figure 7.1 holds the on-state of the switch until the zero current detection block detects a zero crossing in the current wave form so that the behaviour of a triac is described. Some signals in the triac block are plotted in figure. This plot corresponds to a control angle of 45ο and a current that lags the voltage by 30ο. For simplicity, the amplitude of the load current and the supply voltage are both set to unity. The triac is turned on as the control angle has elapsed in each half period. For a resistive load, the triac is always turned off as the control angle signal turns low. However, if the load is inductive (as in figure 7.3), it is seen that the current is not zero as the control angle turns low. This means that the triac can not turn off. As the current crosses zero, the current state (Zero current detection block in figure 7.1) turns high for a very short time. This spike is sensed by the reset input in the flip-flop and the output voltage is set to zero by the switch. The triac is then turned on once the control angle turns high in the next half cycle.

Signals in the triac block

1

Supply voltage

0 -1 10

100

200

300

400

500

600

700

Load current

0 Amplitude

-1 1 0.5 0 1 0.5 0

0

100

200

300

400

500

600

700

Control angle

0

100

200

300

400

500

600

700

Current state

10

100

200

300

400

500

600

700

Output voltage

0 -1 0

100

200 300 400 500 Electrical angle [degrees]

600

700

Figure 7.3 – The signals in the triac controller block in figure 7.1.

For control angles α higher than the current lag ϕ , the operation of the triac is perfectly stable. However, with control angles lower than the current lag, the triac is unstable. This is due to the fact that the current need to cross zero during the period when the control angle signal is low. If this is not the case, the triac is constantly on. The phenomena is presented in figure 7.5 where the current lag is ϕ = 40 o and the control angle is α = 30 o . However, it is assumed that the torque and current wave forms are mostly affected at high control angles. Therefore this problem is not investigated further as only values of alpha higher than the current lag will be chosen from now on.

43

Chapter 7 - Speed control To verify that the triac controller works properly, the output voltage at different control angles and pure resistive load are plotted in figure 7.4. The output voltages with inductive load at different control angles are plotted in figure 7.5. Note that the upper left plot for the inductive load shows unstable operation.

44

Chapter 7 - Speed control

Figure 7.4 – Triac output voltage at different control angles with pure resistive load

Figure 7.5 - Triac output voltage at different control angles with inductive load

45

ϕ = 40o

Chapter 7 - Speed control It is assumed that the torque wave form is mostly affected at high torque and high control angles α . Therefore, the motor and its controller are simulated at 0.69 Nm which is the highest load simulated in chapter 6. The terminal voltage, machine current and the produced electromagnetic torque are shown in figure 7.6-7.8. The speed of the machine varies between 5000 rpm down to 2000 rpm in steps of 1000 rpm. The control angles corresponding to the simulated speeds are α = 56 o − 78.5 o . Applied terminal voltage, α =56o

Applied terminal voltage, α =65o

200

Terminal voltage

150

150

100

100

50

50

Voltage [V]

Voltage [V]

200

0 -50

0 -50

-100

-100

-150

-150

-200

0

0.005

0.01

0.015 0.02 Time [s]

0.025

-200

0.03

Terminal voltage

0

0.005

Applied terminal voltage, α =72o

150

100

100

50

50

0 -50

-150

-150

0.01

0.015 0.02 Time [s]

0.025

-200

0.03

Terminal voltage

-50 -100

0.005

0.03

0

-100

0

0.025

Applied terminal voltage, α =78.5o

150

-200

0.015 0.02 Time [s]

200

Terminal voltage

Voltage [V]

Voltage [V]

200

0.01

0

0.005

0.01

0.015 0.02 Time [s]

Figure 7.6 – The terminal voltage at different control angles

46

0.025

0.03

Chapter 7 - Speed control

Current vs time at 5000 rpm, 0.69 Nm, α =56o

Current vs time at 4000 rpm, 0.69 Nm, α =65o

20

Machine current Ia

15

15

10

10

5

5 Ia [A]

Ia [A]

20

0

0

-5

-5

-10

-10

-15

-15

-20

0

0.005

0.01

0.015 0.02 Time [s]

0.025

-20

0.03

Machine current Ia

0

Current vs time at 3000 rpm, 0.69 Nm, α =72o

15

10

10

5

5

0

-5

-10

-10

-15

-15

0.005

0.01

0.015 0.02 Time [s]

0.025

0.025

0.03

-20

0.03

Machine current Ia

0

-5

0

0.015 0.02 Time [s]

Current vs time at 2000 rpm, 0.69 Nm, α =78.5o

15

-20

0.01

20

Machine current Ia

Ia [A]

Ia [A]

20

0.005

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.03

Figure 7.7 – The machine current at different control angles

Electromagnetic torque vs time at 5000 rpm, 0.69 Nm, α =56o

Electromagnetic torque vs time at 4000 rpm, 0.69 Nm, α =65o

2

Electromagnetic torque

1.8

1.8

1.6

1.6

1.4

1.4 Torque [Nm]

Torque [Nm]

2

1.2 1 0.8

1.2 1 0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0

0.03

Electromagnetic torque

0

Electromagnetic torque vs time at 3000 rpm, 0.69 Nm, α =72o

1.8

1.6

1.6

1.4

1.4

1.2 1 0.8

0.4

0.2

0.2 0.015 0.02 Time [s]

0.025

0

0.03

Electromagnetic torque

1

0.6

0.01

0.03

0.8

0.4

0.005

0.025

1.2

0.6

0

0.015 0.02 Time [s]

Electromagnetic torque vs time at 2000 rpm, 0.69 Nm, α =78.5o

1.8

0

0.01

2

Electromagnetic torque

Torque [Nm]

Torque [Nm]

2

0.005

0

0.005

0.01

0.015 0.02 Time [s]

0.025

Figure 7.8 – The produced torque at different control angles

47

0.03

Chapter 7 - Speed control The total harmonic distortion (THD) are calculated for the current wave forms according to [12].

I THD

⎛ I (n) = 100 ⋅ ∑n = 2 ⎜ ⎜I ⎝ (1) ∞

⎞ ⎟ ⎟ ⎠

2

(eq. 7.1)

[%]

where I (1) is the fundamental component and I (n ) are the harmonics. This calculation is performed in Simulink. The triac controller definitely adds harmonics to the machine current as shown in table 7.1 where the THD for different control angles are shown. Table 7.1 – The total harmonic distortion of the current at different speeds at 0.69 Nm load torque with triac control.

Control angle (degrees)

Speed (rpm)

THD (%)

56

5000

38.13

65

4000

40.79

72

3000

42.76

78.5

2000

44.46

As a direct consequence to the current distortion, the torque wave form is also affected due to the triac controller. This is seen by comparing figure 7.8, with figure 6.9 which shows the torque wave form without speed control. The peak of the produced electromagnetic torque when the machine is speed controlled with a triac is higher while the rise time of the torque seems to be shorter. The machine without triac speed control has a smoother variation of torque with time. With triac speed control, the current drops to zero under a certain period. Therefore the machine does not produce any torque during that phase. At a certain time, the triac starts to conduct the current and the torque increases rapidly. This is most likely to produce noise as well as affect the wearing of gears and transmissions connected to the machine.

48

Chapter 7 - Speed control

7.2. Sinus PWM Instead of having a triac controller controlling the speed of the machine, a PWM controller is tested in the model. The PWM block is shown in figure 7.2. A triangle wave of high frequency is compared to a reference voltage. Depending on which signal is the highest, the switch in figure 7.2 is high, + Vd or low, − Vd . The fundamental of the output voltage can be calculated as [12] ∧

V 1 = m a ⋅ Vd

(eq. 7.2)

Where m a is the modulation index, which is defined as [12] ∧

ma =

V ref

(eq 7.3)

∧

V tri The switching frequency used is f s = 3 kHz . The time step that is used for the simulations is 10 μs i.e, the sample frequency is 100 kHz. The current and torque wave forms are shown in figure 7.9 and figure 7.10 respectively.

Current vs time at 5000 rpm, 0.69 Nm, ma=0.93

Current vs time at 4000 rpm, 0.69 Nm, ma=0.88

20

Machine current Ia

15

15

10

10

5

5 Ia [A]

Ia [A]

20

0

0

-5

-5

-10

-10

-15

-15

-20

0

0.005

0.01

0.015 0.02 Time [s]

0.025

-20

0.03

Machine current Ia

0

10

10

5

5 Ia [A]

Ia [A]

15

0

-5

-10

-10

-15

-15

0.005

0.01

0.015 0.02 Time [s]

0.025

0.025

0.03

0.03

Machine current Ia

0

-5

0

0.015 0.02 Time [s]

20

Machine current Ia

15

-20

0.01

Current vs time at 2000 rpm, 0.69 Nm, ma=0.78

Current vs time at 3000 rpm, 0.69 Nm, ma=0.825

20

0.005

-20

0

0.005

0.01

0.015 0.02 Time [s]

0.025

Figure 7.9 – The current wave form when sinus PWM-control is used.

49

0.03

Chapter 7 - Speed control

Electromagnetic torque vs time at 5000 rpm, 0.69 Nm, ma=0.93

Electromagnetic torque vs time at 4000 rpm, 0.69 Nm, ma=0.88

2

Electromagnetic torque

1.8

1.8

1.6

1.6

1.4

1.4 Torque [Nm]

Torque [Nm]

2

1.2 1 0.8

1 0.8 0.6

0.4

0.4

0.2

0.2 0

0.005

0.01

0.015 0.02 Time [s]

0.025

0

0.03

0

1.8

1.6

1.6

1.4

1.4 Torque [Nm]

1.8

1 0.8

0.4

0.2

0.2 0 0.01

0.015 0.02 Time [s]

0.025

0.03

0.03

Electromagnetic torque

1

0.4

0.005

0.025

0.8 0.6

0

0.015 0.02 Time [s]

1.2

0.6

0

0.01

2

Electromagnetic torque

1.2

0.005

Electromagnetic torque vs time at 2000 rpm, 0.69 Nm, ma=0.78

Electromagnetic torque vs time at 3000 rpm, 0.69 Nm, ma=0.825

2

Torque [Nm]

1.2

0.6

0

Electromagnetic torque

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.03

Figure 7.10 – The torque wave form when sinus PWM-control is used.

The distortion in the current wave form is calculated in the same way as for the triac controller. The results are gathered in table 7.2. Table 7.2 – The total harmonic distortion of the current at different speeds at 0.705 Nm load torque with PWM-control.

Modulation index

Speed (rpm)

THD (%)

0.93

5000

33.68

0.88

4000

33.66

0.825

3000

33.70

0.78

2000

33.80

The current wave form is now less distorted compared to the triac control even though it contains high frequency ripple due to the switching. The reduction of the current distortion is most significant at low speeds. At 2000 rpm, the THD is approximately 24 % lower with PWM control. Figure 7.11 shows the harmonic content in the current at 2000 rpm, 0.69 Nm, for the two different control strategies. The third harmonic is significantly reduced when PWM control is used. On the other hand, a switching frequency at 3000 Hz is introduced. 50

Chapter 7 - Speed control

Current harmonics, 2000 rpm 0.69 Nm, Sinus-PWM controlled

Current harmonics, 2000 rpm 0.69 Nm, Triac controlled 15

15

Ia(n)

Ia(n)

10 Amplitude [A]

Amplitude [A]

10

5

5

0

0

100

200

300

400

500

600

700

800

900

0

1000

Frequency [Hz]

0

500

1000

1500 2000 2500 Frequency [Hz]

3000

3500

4000

Figure 7.11 – The harmonic content in the current at 2000 rpm, 0.69 Nm, for the two different control strategies. Triac control to the left and sinus-PWM to the right.

The produced electromagnetic torque for the machine with a PWM controller consists of high frequency ripple. The long period where the machine produces no torque introduced by the triac is not present for the PWM. The high frequency ripple in the current can most likely be reduced by the use of filters. This would lead to less ripple in the torque wave form as well. The peak torque is approximately 5 % lower with PWM-control compared to triac at 2000 rpm, 0.69 Nm. This would probably reduce the 120 Hz noise produced by the machine. Further on, with PWM, the torque wave forms are similar independent of the machine speed. This is not the case for the triac where the zero current period prevents the machine from producing a torque during a certain period of time. This dead time has to be compensated with a high peak in the torque in order to keep the speed. It would be interesting to simulate at least one point with FEM together with a triac controller, in order to look at the variations of the normal forces acting on the rotor teeth and excitation poles.

51

Chapter 8 - Conclusions and future work

8. Conclusions and future work Two FEM models of a full-pitch and a short-pitch universal motor have been developed. Comparisons with measurements showed that these models are giving realistic results. This project has shown how to derive a non-linear analytical model of a universal motor. Although the simulation times in the FEM simulations are really long for the short pitch machine, it is assumed that a model of the short pitch machine can be created in a similar manner as the non linear Simulink model of the full pitch machine presented in this report. A general conclusion drawn is that nonlinearities due to saturation are difficult to model. To decrease the simulation time for the FEM simulations, only half or even one fourth of the geometry could be used for the full-pitch machine. This would lead to fewer elements and a less complex problem to solve. However, this would not be possible for the short-pitch machine because there is no symmetry with respect to the coils in this machine. The non-linearity of the rotor winding inductance should be analyzed. The errors in the torque/speed characteristics of the machine might be reduced by taking it into account. To further improve the model of the machine, the effects of commutation should be described. A transformer induced emf in the rotor was found in a FEM simulation. The influence of this should be investigated. If the machine is speed controlled with a triac, the wave form of the produced electromagnetic torque is higher in amplitude at low speeds compared to the PWM controller. The level of 120 Hz noise emitted by the machine may therefore be reduced if the machine is controlled by PWM. The higher torque peaks caused by the triac controller also most likely increase wearing of gears connected to the machine. More load levels should be analyzed for the triac and the PWM. A closed loop control should be implemented in Simulink for the triac control as well as for the PWM control. It would be interesting to connect Simulink with Flux and perform the control simulations on this model and compare the results to the simulation results provided in chapter 7. In this way, the effect of slotting and armature reaction would be fully taken into account and the forces acting on the different parts of the machine could be investigated. The FEM simulation used to calculate the back emf and the field inductance should be performed at more rotor angles to investigate how these quantities varies as the position of the rotor is changed.

52

References

References [1] P.C. Sen – “Principles of electrical machines and power electronics”, second edition, ISBN 0-471-02295-0, 1997. [2] Peter F. Ryff, David Platnick, Joseph A. Karnas – “Electrical machines and transformers, Principles and applications”, ISBN 0-13-247222-8, New Yersey 1987. [3] http://www.sura.se – Information of stator- and rotor laminations [4] MG Say EO Taylor - “Direct current machines”, ISBN 0-273-01219-3, Bath, Great Britain, 1980. [5] http://www.ece.msstate.edu – Back emf calculation [6] http://www.reliance.com – Information of winding types in DC – machines [7] http://www.cedrat.com – Flux features [8] David K. Cheng – “Field and wave electromagnetics”, Second edition - United States, 1992, ISBN 0-201-12819-5. [9] Flux Help [10] Emil Alm 1944 – ”D.T.V Elektroteknik III, Elektromaskinlara II D”. [11] P.Thelin, J. Soulard, H.-P. Nee and C. Sadarangani, “Comparison between Different Ways to Calculate the Induced No-Load Voltage of PM Synchronous Motors using Finite Element Methods”, Presented at PEDS’01, Bali, Indonesia, Oct. 2001. [12] Hans-Peter Nee – “Kompendium i Eleffektsystem”, KTH, Stockholm 2003. [13] – Drawings Ankarsrum.

53

List of symbols

List of symbols Number of stator poles Coil pitch Pole pitch

p y τp

[− ] [−] [−]

Active length of machine Instantaneous current Instantaneous voltage Instantaneous emf Instantaneous rotor speed Time Supply frequency Field winding resistance Rotor winding resistance

L eff i u e ωm t f Rf Ra

[m] [A ] [V ] [V ]

[s −1 ] [s] [ Hz] [Ω ] [Ω ]

Field winding inductance Rotor winding inductance

Lf La

[H ] [H ]

Moment of inertia Viscous damping constant Instantaneous electromagnetic torque

[kg ⋅ m 2 ] [ Nm ⋅ s] [ Nm ]

Load torque Rotor constant

J D Te TL Ka

Flux constant

Kψ

[ Wb ⋅ A −1 ]

Electromagnetic force Pole coverage factor Radius of rotor Radius of stator pole Number of parallel paths in the rotor Number of turns in the rotor Length of rotor conductor in magnetic field Length of stator pole Flux linkage Flux Flux density Flux density, normal component Position of rotor Brush resistance at on state

Fe c r rpole a N l conductor l pole Ψ φ B Bn θ mec R on

[ N] [−] [m] [m]

54

[ Nm ] [−]

[−] [−] [m] [m] [ Wb] [ Wb] [T ] [T ] [− ] [Ω ]

List of symbols Brush resistance at off state Position of commutator segment

R off θ pos

[Ω ]

Opening angle of a brush Opening angle of a commutator segment In plane length between forward and backward coil in rotor Length of end winding

θ bru θ bar τ ly

[− ]

Inductance of one coil end Average value of simultaneously commutating coils Reactance of one coil end

Ly β X Ly

[H ]

Number of conductors in a coil end Reduced commutation zone Angle between end winding and plane of rotor lamination A coil ends average geometrical distance from itself

nl b 'kz γ g 1,1

[− ]

Height of a coil end Width of a coil end Distance between two adjacent rotor slots Distance between two adjacent commutator segments

[m] [m] [m]

Width of insulation between commutator segments Number of coils side by side in a rotor slot

h1 b1 τs λτ dk Q nk fλ bc bi uc

Vector magnetic potential Vector differential operator

A ∇

Conductance

G

Diameter of commutator Total number of rotor slots Total number of commutator segments Reduction from full pitch expressed in number of segments Width of brush

55

[− ] [− ] [m] [m] [− ] [Ω ] [m] [− ] [m]

[m] [m] [− ] [− ] [− ] [m] [m] [− ]

[ Wb ⋅ m −1 ] [− ] [ Ω −1 ]

Appendix

Appendix

56

Appendix

Performance test I (A*1000) Cos phi

P1 (W*10) Eta

25,000

1.00

22,500

0.90

20,000

0.80

17,500

0.70

15,000

0.60

12,500

0.50

10,000

0.40

7,500

0.30

5,000

0.20

2,500

0.10

0.00

0 0

10

20

30

40

50

60

70

Ncm

Motortype: KS4242/110 94119202 Armature winding: 4x17t/s Ø 0.50mm Field winding: 110t/c Ø 0.90mm Testvoltage: 120V 60Hz Rotation:

File: PT433.xls Author: ASUL/AH,LC Date: 2004-05-28

Remark: TP20040104 KIA Mörkrotorplåt 02746

Figure A1 – Results from the performance test of the short pitch machine.

57

80

Eta, Cos phi

n (rpm*1) P2 (W*10)

Lf

58

Figure A2 – The electrical system for the linear Simulink model.

Rotor winding resistance

Ra

Field winding resistance

Rf

Voltage

2

Rotor constant

k

Rotor winding inductance

La

Field winding inductance

Ka

3

w

1

f lux

-K-

Gain

flux

2

Product1

E

Product

(Rf +Ra)*i

Vc-(Rf +Ra)*i

Vc-(Rf +Ra)*i-E

Product2

(Vc-(Rf +Ra)*i-E)/(Lf +La)

Integrator

1 s

i Machine current

i

1

Appendix

59

i

Ka

4

Flux

3

2

Product3

1 Load Torque

Ka*phi*i=Te

To Workspace4

Te

Viscous damping

D

J

Product7

Dw

Te-TL

Moment of inertia

Te-TL-Dw

Product8

(Te-TL-Dw)/J

Integrator1

1 s

To Workspace1

w

w

1

Appendix

Figure A3 – The mechanical system for the linear and non-linear Simulink models.

60

Figure A4 – The electrical system for the non- linear Simulink model.

Constant2

Ra

Constant1

Rf

Voltage

2

Constant5

k

La

6e-3

k

3

w

1

Flux

2

Product1

Product

E

f lux

Lf (i)

(Rf +Ra)*i

Vc-(Rf +Ra)*i

Look-Up Table1

Lf +La

Product2

|u|

Abs

Look-Up Table

(Vc-(Rf +Ra)*i-E)/(Lf +La)

abs(i)

Integrator

1 s

i To Workspace

i

1

Appendix

Appendix

Figure A5 – The circuit connected to the short pitch machine in the finite element model.

61

Appendix

Figure A6 – The coils corresponding to the rotor coils in the short pitch machine. This is an enlargement of figure A5.

62

Appendix

Component V1 R1, R2 R3- R26 L1

Component descr i pt i on Vol t age sour ce St at or coi l r esi st ances Rot or coi l r esi st ances End wi ndi ng i nduct ance

Figure A7 – Description of the components in the circuit in figure A5 and A6.

63

Appendix

Coi l B1p B1n B1a B1b B1c B1d B2a B2b B2c B2d B3a B3b B3c B3d B4a B4b B4c B4d B5a B5b B5c B5d B6a B6b B6c B6d

Descr i pt i on Exci t at i on coi l Exci t at i on coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l

in in in in in in in in in in in in in in in in in in in in in in in in

r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or

sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot

Coi l

1a 1b 1c 1d 2a 2b 2c 2d 3a 3b 3c 3d 4a 4b 4c 4d 5a 5b 5c 5d 6a 6b 6c 6d

B7a B7b B7c B7d B8a B8b B8c B8d B9a B9b B9c B9d B10a B10b B10c B10d B11a B11b B11c B11d B12a B12b B12c B12d

Descr i pt i on

Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l

i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or

sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot

7a 7b 7c 7d 8a 8b 8c 8d 9a 9b 9c 9d 10a 10b 10c 10d 11a 11b 11c 11d 12a 12b 12c 12d

Figure A8 – Description of the components in the circuit in figure A5 and A6.

64

Appendix

Figure A9 – The FEM geometry of the full pitch machine.

65

Appendix

Figure A10 – The circuit connected to the full pitch machine in the finite element model.

66

Henrik Grop Master Thesis

Supervisor: Dr. Juliette Soulard

Royal Institute of Technology Department of Electrical Engineering Electrical Machines and Power Electronics Stockholm 2006

XR-EE-EME 2006:003

Abstract Noise emitted by a universal motor is believed to be strongly related to the fluctuations in the electromagnetic torque produced by the machine. The variations of the current feeding the machine give rise to the fluctuations in the torque. The variations of the current are depending on the power electronics used to control the speed of the machine. This thesis presents a method of modelling a universal motor in a non-linear way based on a linear model and calculations from FEM using Flux version 8.1 by Cedrat. The FEM model is derived starting from drawings of the machine and the winding connections. A non-linear model implemented in Simulink is used to connect a triac, and a PWM. The fluctuations in the torque and the total harmonic distortion of the motor current are compared for the two control methods. It is found that the PWM produces less variation in the torque in the zero to peak sense, but it introduces high frequency ripple in the torque wave form instead. The difference between the triac and the PWM is most significant when the machine is controlled to rotate at low speed. At low speed, the PWM reduces the torque peak with approximately 5 % compared to the triac. Compared to the triac, the PWM controller reduces the THD in the machine current by approximately 24 % at low speed and high torque.

I

II

Acknowledgements This master thesis was written at the Royal Institute of Technology, School of Electrical Engineering, division of Electrical Machines and Power Electronics. The work was carried out in the period of September 2005 - March 2006. First of all, I would like to thank Ankarsrum Industries for initiating the subject of this thesis. A special thanks goes to Mr. Werner Fritz and Mr. Mats Pettersson who provided drawings and measurement data. I would like to thank my supervisor, Dr Juliette Soulard for outstanding assistance and support during the work of this thesis. All staff at the division have been very helpful in many ways, therefore I would like to thank you all. At last I would like to thank Mr. Jan Timmerman for very good times, especially at Roebel’s bar on Friday evenings.

III

IV

Table of contents Abstract ......................................................................................................................................I Acknowledgements................................................................................................................ III Table of contents...................................................................................................................... V 1.

Introduction ...................................................................................................................... 1 1.1. Purpose of this work................................................................................................... 1 1.2. Outline of the thesis.................................................................................................... 1

2.

Universal motors .............................................................................................................. 2

3.

The investigated machine ................................................................................................ 3 3.1. Design......................................................................................................................... 3 3.1.1. Stator .................................................................................................................. 3 3.1.2. Rotor................................................................................................................... 4 3.1.3. Commutator........................................................................................................ 4 3.2. Windings .................................................................................................................... 5 3.2.1. Stator winding .................................................................................................... 5 3.2.2. Rotor winding..................................................................................................... 5 3.3. Measured performance............................................................................................... 8

4.

Linear model..................................................................................................................... 9 4.1. Electrical system ...................................................................................................... 10 4.1.1. Back emf .......................................................................................................... 10 4.2. Mechanical system ................................................................................................... 12 4.2.1. Electromagnetic torque produced by the rotor................................................. 13 4.3. Electromechanical system ........................................................................................ 14 4.4. Implementation in Simulink..................................................................................... 14 4.5. Simulation results using the linear model ................................................................ 15

5.

Finite element model ...................................................................................................... 17 5.1. Geometrical simplifications for the Finite Element Method (FEM) modelling....... 17 5.1.1. Stator ................................................................................................................ 17 5.1.2. Rotor................................................................................................................. 17 5.1.3. Rotor shaft ........................................................................................................ 17 5.2. Flux 2D..................................................................................................................... 18 5.3. Full pitch motor model............................................................................................. 18 5.3.1. Geometry.......................................................................................................... 18 5.3.2. Electric circuit .................................................................................................. 18 5.4. Short-pitch motor model .......................................................................................... 19 V

5.4.1. Geometry.......................................................................................................... 19 5.4.2. Mesh ................................................................................................................. 20 5.4.3. Electric circuit .................................................................................................. 21 5.4.4. End winding reactance ..................................................................................... 24 5.5. Results from full-pitch motor................................................................................... 27 5.5.1. Verification of commutation ............................................................................ 27 5.6. Results from short-pitch motor ................................................................................ 28 5.6.1. Machine resistance ........................................................................................... 28 5.6.2. Machine current and torque at 8000 rpm ......................................................... 29 5.6.3. Simulation time ................................................................................................ 31 6.

Non-linear analytical model .......................................................................................... 32 6.1. Air gap flux for the induced voltage ........................................................................ 32 6.2. Inductance of the field windings .............................................................................. 34 6.3. Inductance of the rotor winding ............................................................................... 36 6.4. Results from non-linear analytical model ................................................................ 37 6.4.1. Machine current................................................................................................ 37 6.4.2. Torque-speed characteristics ............................................................................ 38 6.4.3. Current wave forms.......................................................................................... 38 6.4.4. Torque wave forms........................................................................................... 40 6.5. Conclusions .............................................................................................................. 41

7.

Speed control .................................................................................................................. 42 7.1. Triac ......................................................................................................................... 42 7.2. Sinus PWM .............................................................................................................. 49

8.

Conclusions and future work ........................................................................................ 52

References ............................................................................................................................... 53 List of symbols ........................................................................................................................ 54 Appendix ................................................................................................................................. 56

VI

VII

VIII

Chapter 1 - Introduction

1. Introduction

1.1. Purpose of this work The purpose of this work is to investigate the possibility to model a small universal motor in a non-linear way. The model should be used to compare two different power electronic converters used to control the speed of the machine.

1.2. Outline of the thesis Chapter 2 describes universal motors very shortly. In chapter 3, the machine studied in this thesis is presented thoroughly. The different parts in the machine are described. Drawings of the rotor and the stator are provided. The winding type is described and a drawing of the winding in developed form is included. In chapter 4, the electrical and mechanical equations which describe the machine are derived and a linear model of the machine is presented. Chapter 5 describes the FEM model of the machine. The simplifications that are made to the geometry of the machine are described. A simple type of winding is used at first and then the real machine winding is described. Only the model with the real machine winding is thoroughly described. In chapter 6, a non-linear model of the machine is presented. This model is implemented in Matlab/Simulink. Chapter 7 presents the control of the machine. A triac controller is compared to a sinus PWM controller. In Chapter 8, conclusions are drawn and a discussion of future work is presented.

1

Chapter 2 - Universal motors

2. Universal motors Universal, or series motors, are widely used in fractional horsepower ratings in many domestic appliances such as drilling machines, vacuum cleaners and food mixers. Large universal motors, in the range of 500 hp are used for traction applications. This type of machine is preferable where a high power to weight ratio is required. It can be fed with either DC current or single phase AC current. The single phase AC current is most common to use as supply in most applications [1]. Due to the high starting torque and the possibility to be fed with direct current, this machine is as well used as starter motor for engines. The machine is called a series motor because the excitation winding is connected in series with the rotor winding. Further on, it is called universal motor because it can be operated with both AC and DC supplies. The speed of a universal motor can be very high, typically in the range of 5000-20000 rpm [2]. The speed is entirely dependent on the load and the supply voltage. Without load, a universal motor may reach dangerously high speeds. In fact, the no-load speed is only limited by friction and windage losses. Therefore, a universal motor should never be operated without a load. Since an unloaded universal motor can destroy itself, the load should always be connected directly to the shaft and should not be belt connected [2].

2

Chapter 3 - The investigated machine

3. The investigated machine The different parts of the machine and their purpose are discussed here. The design of the modelled machine is introduced.

3.1. Design Variations of the flux in magnetic materials induce eddy currents. These currents cause unwanted losses in the material. To minimize these losses, the stator and the rotor are made of laminations [2].

3.1.1. Stator The stator lamination is made of an electrical steel of quality M800-65A. This is a cold rolled non grain-oriented electrical steel, which is also called DK70. This material is alloyed with aluminium and silicone which increases the electrical resistance and thereby decreases the eddy currents in the material. Small eddy currents imply small losses in the laminations. The eddy current losses also depend on the thickness of the laminations. Thinner laminations give smaller eddy current losses [3]. A drawing of the cross-section of the stator lamination is found in figure 3.1. The stator consists of two excitation poles and magnetic paths between the poles. The holes in the lamination are used for fixating the sheets together with screws when the stator is assembled. The small rectangular shapes which look like holes are actually flanges, used for guidance when stacking the sheets during assembly. The inner space, to the left and to the right of the field poles, contains the field winding which produces the magnetising flux in the magnetic circuit.

Figure 3.1 – Cross-section of a stator lamination sheet.

3

Chapter 3 - The investigated machine

3.1.2. Rotor The rotor, which supports the armature winding, is laminated in the same way as the stator. The lamination is made of the same steel quality as the stator lamination. The laminations are punched and stacked together on the rotor shaft of the machine. Twelve armature slots hold the coils which conduct the armature current. Twelve rotor teeth complete the magnetic circuit and guide the flux towards the air gap. A drawing of the rotor is presented in figure 3.2.

Figure 3.2 – Drawing of a rotor lamination sheet and a 3D-view of the rotor without the armature coil.

3.1.3. Commutator To feed the rotating armature winding with current, there must be some physical electrical connection between the outer static part of the machine and the inner rotating part. This connection is achieved by using a set of brushes and a commutator. The brushes are fixed and the commutator is attached on the rotor shaft. Moreover, the brush-commutator construction has another purpose which is to reverse the current in the rotor coils when the coil leaves one excitation pole and enters the region of the next pole. The current reversal in each coil must be done in a specific rotor position in order to maintain the direction of the produced electromagnetic torque and thereby the direction of rotation. In order to do this, the commutator is made of several segments, depending on the number of coils and their arrangement in the armature. The segments are isolated from each other with some electrical insulating material. The brushes slide on the commutator surface during rotation and thereby create a connection to the armature coils. The connection of the coils is always made in such a way to make the current in the coils underneath the north pole of the excitation field have an opposite direction compared to the current in the coils underneath the south pole. In other words, with respect to the direction of the magnetizing field, the switching of the commutator ensures a constant current pattern in the rotor [4]. Thus the rotor produces a unidirectional torque which results in a rotating movement. The commutator in the test machine has 24 segments. A drawing showing the principle of the commutator assembled to the rotor shaft is presented in figure 3.3.

4

Chapter 3 - The investigated machine

Figure 3.3 – The commutator assembled on the rotor shaft.

3.2. Windings 3.2.1. Stator winding The stator winding, wound around the field poles produces the magnetizing flux in the machine [3]. Since the field winding is connected in series with the rotor winding, it carries the same current as the rotor winding. Therefore it must be wound with larger conductors than the rotor winding as the current is split in two parallel circuits in the rotor. The number of turns per coil in the field winding is 110 (there are two field coils connected in series). The diameter of the copper wire is 0.90 mm.

3.2.2. Rotor winding The rotor winding carries the rotor current. It produces a flux which creates the electromagnetic torque in the air gap when interacting with the magnetizing flux. The current in the armature winding is fed by a brush-commutator system. The coil ends are connected to the different segments in the commutator. The coil sides are placed in slots in the rotor. For DC-machines, the most common way of placing these coil sides is by letting the distance between the forward coil side and the backward coil side to be one pole pitch. This is called a full pitch winding. The pole pitch is defined as

One pole pitch = 180 o electrical = where p is the total number of stator poles. 5

o

360 mechanical p

Chapter 3 - The investigated machine There are several ways of inserting the coils in the rotor. The two most common ones are the lap winding and the wave winding (see figure 3.4). In the lap winding, the two ends of one coil are connected to segments in the commutator sitting next to each other. This makes the coil lap back on itself. [4] In the wave winding, the coil ends are connected to two commutator segments spread apart. This way of connecting the coil ends makes the coils look like a wave pattern as they are put into the machine [1].

Figure 3.4 – Left image – Coil configuration in a lap winding. Right image – Coil configuration in a wave winding [6].

The rotor winding in the studied universal motor is a short-pitch lap winding. This means the machine is wound in two layers with two coils side by side in each slot, i.e. four different coils in a slot. Each of the four coils in one slot of the rotor has seventeen conductors. The diameter of the conductor wire is 0.50 mm. For a full pole pitch winding in this machine, each coil would span six rotor slots. In the studied machine, the coil span is equal to five slots. The coil pitch can be calculated in electrical degrees [4] by:

coil pitch=180° ×

y 5 =180° × =150° electrical τp 6

where y - is the coil span, in number of slots, and τ p - is the pole pitch, in number of slots The mechanical coil pitch angle is the same as the electrical since the machine has two poles. The winding principle with end connections to the commutator is shown in figure 3.5. Note that only two coils are shown in the first layer in one slot in this figure. The rest of the coils are wound in a similar manner.

6

Chapter 3 - The investigated machine

Figure 3.5 – The principle of winding the coils and connecting the coil ends to the commutator. The figure is seen from the commutator side of the machine [13].

By studying figure 3.5, a drawing of the winding in developed form has been constructed [4]. The first coils which are illustrated in figure 3.5 are first drawn in respective rotor slots in figure 3.6. The ends of the coils are then connected to the correct commutator segment.

Figure 3.6 – The armature winding in developed form with the coils in figure 3.5 displayed.

7

Chapter 3 - The investigated machine The remaining coils are then drawn in a similar way by doing a one slot translation in a counter-clockwise direction until the fourth coil in slot 1 and slot 6 are drawn. These are the last coils wound in the rotor. The complete rotor winding in developed form is viewed in figure 3.7. This figure is very important for the definition of the finite element model of the machine (chapter 5). The winding type of this machine makes the geometry in the finite element model complex.

Figure 3.7 – The configuration of the coils in the armature winding with connections to the commutator segments.

3.3. Measured performance The performance of the machine has been tested at Ankarsrum and the test results are presented in figure A1 of the appendix.

8

Chapter 4 - Linear model

4. Linear model The system of differential equations that describes the machine is derived from the equivalent circuit of the universal motor. The universal motor is basically a DC-machine with the field winding connected in series with the rotor winding. The equivalent circuit is presented in figure 4.1 [2].

Figure 4.1 – Equivalent circuit of the universal motor

In figure 4.1, the parameters are R a - rotor winding resistance R f - field winding resistance La - rotor winding inductance Lf - field winding inductance u(t) - terminal voltage e(t) - back emf i(t) - current in the machine J - moment of inertia of machine and load D - viscous damping constant T(t) - electromagnetic torque TL - load torque ωm (t) - angular velocity of machine

9

Chapter 4 - Linear model

4.1. Electrical system By applying Kirchoff’s voltage law on the circuit in figure 4.1, the following equation can be derived. u (t ) − ( La + L f

) dtd i(t ) − e(t ) = ( R

a

+ R f ) i (t )

(eq. 4.1)

This differential equation can be solved if an expression for the induced back emf can be determined. This expression will be derived in next section.

4.1.1. Back emf As the rotor rotates in a magnetic field, an electromotive force is induced in the turns of the rotor winding. The emf induced in one turn of a coil is given by equation 4.2. This is the general form of Faraday’s law for a moving conductor in a time varying magnetic field [8]. econductor = ∫ Ea • dl = − ∫ C

S

∂B • ds + ∫ ( v × B ) • dl ∂t C

(eq. 4.2)

where v is the velocity vector of the conductor and B is the magnetic flux density vector in which the conductor is moving. C is the path along the conductor in the magnetic field and S is the surface bounded by C . The first term on the right side of equation 4.2 is the transformer emf due to time variations of the magnetic flux density. If the armature reaction is not considered, the coil short circuited by the brush, i.e the coil undergoing commutation, is the only contribution to the transformer action of equation 4.2. To simplify the analytical model of the machine, this transformer action of the coil undergoing commutation is not considered. The second term to the right of equation 4.2 is the flux cutting emf. A homogenous magnetic field beneath the magnetising poles of the machine is assumed. The flux cutting back emf produced in one turn of a coil according to figure 4.2 together with the simplifications made, is given by equation 4.3 [5]. econductor = ∫ Ea • dl ≈ ∫ ( v × B ) • dl C

C

10

(eq. 4.3)

Chapter 4 - Linear model

Figure 4.2 – Back emf induced in a conductor moving in a homogenous magnetic field [5].

The magnitude of the velocity of the conductors in figure 4.2 is given by: v = ωm r

(eq. 4.4)

where ωm is the angular velocity of the rotor and r is the radius of the rotor. Further on, the magnitude of the electric field along the conductor is Ea = ωm rBn

(eq. 4.5)

Bn is the normal-component of the magnetic flux density, i.e. the component perpendicular to the velocity of the conductor. The electric field Ea is in the opposite direction in conductor 2 compared to conductor 1.

By integrating the electric field along the length of the conductor moving in the magnetic field, the total emf voltage produced in the conductor is given. Assuming a constant electric field along the conductor, the emf induced in one conductor is: econductor = E ⋅ lconductor = ωm rBn ⋅ lconductor

(eq. 4.6)

lconductor is the length of the conductor moving in the magnetic field, i.e. the active length of the machine. Since the electric fields in the two conductors are of opposite directions, the total induced emf voltage in one turn is: eturn = 2 E ⋅ lconductor = 2ωm rBn ⋅ lconductor

(eq. 4.7)

The flux density seen by the conductors beneath each pole in the machine is calculated according to:

11

Chapter 4 - Linear model

Bn =

ψn Apole

=

ψn

2π rpolel pole p

= ⋅c

ψn p

2π rpolel pole c

(eq. 4.8)

Where p is the number of poles in the machine and c is the pole coverage factor. rpole is the radius of the pole which approximately is the same as the radius of the rotor i.e: rpole ≈ r

(eq. 4.9)

Further on, l pole is the same length as lconductor , i.e. l pole = lconductor

(eq. 4.10)

Again, index n stands for the normal component, and ψ n is the normal component of the flux beneath a pole. Substituting equation 4.8-4.10 in 4.7 gives the expression of the induced emf in one turn of a coil. eturn = ωm

p ψ πc n

(eq. 4.11)

This is the induced emf in one turn of a coil due to a conductor moving in a magnetic field. The total emf induced in the rotor winding consisting of N turns is the single turn voltage multiplied by the total number of turns. If the N turns are connected in parallel paths, the total emf induced is divided by the number of parallel paths a . Therefore, the total induced emf in the rotor winding is: e=

N p ωmψ n = K aωmψ n [V] a πc

(eq. 4.12)

N p is called the armature constant (or the rotor constant). This constant contains a πc information on the configuration of the rotor winding. Ka =

Assuming non-saturated operating conditions, the flux ψ n is directly proportional to the current in the field winding. Since the field winding is connected in series with the rotor winding, the flux is proportional to the rotor current according to. e = K a Kψ ωm I a [V]

(eq. 4.13)

Where Kψ is the flux constant.

4.2. Mechanical system The differential equation, describing the mechanical behaviour of the motor is given by:

12

Chapter 4 - Linear model

Te (t ) = TL + Dωm (t ) + J

Where D is the friction constant.

d ωm (t ) dt

(eq. 4.14)

The machine speed and current 1 are connecting the electrical and mechanical differential equations, which leads to an electromechanical system. In next section, an expression for the developed electromagnetic torque is derived.

4.2.1. Electromagnetic torque produced by the rotor Assuming the coils are placed in the air gap, an expression for the developed electromagnetic torque in a universal motor is achieved by summing the forces acting on each conductor in the rotor winding. The force acting on one conductor in the rotor is given by [5]. Fe ,conductor = ∫ (I × B ) dl

(eq. 4.15)

C

Where C is the path along the current carrying conductor located in the magnetic field B . Assuming a constant magnetic field along C , the magnitude of the force acting on one conductor is equal to: Fe ,conductor = IBn l conductor

(eq. 4.16)

Once again, Bn is the normal component of the flux density, i.e. the component perpendicular to the current and the rotor periphery as shown in figure 4.3. lconductor is the length of the conductor located in the magnetic field which is the same as the active length of the machine l .

Figure 4.3 – A conductor carrying a current in a homogenous magnetic field

The electromagnetic torque produced by one conductor is then given by multiplying the force acting on the conductor, by the radius of the rotor. Te ,conductor = IBn lr

(eq. 4.17)

The total number of conductors in the rotor winding is 2 N , where N is the total number of turns. As stated in section 4.1.1, the number of parallel circuits in the rotor is a , and the I current in each conductor is a This means that the total electromagnetic torque acting on a the rotor is given by: 1

See section 4.3

13

Chapter 4 - Linear model

Te =

2 NI a Bn lr a

(eq. 4.18)

With Bn replaced by equation 4.8, the torque is obtained by: Te =

Np I aψ n = K a I aψ n [ Nm ] πac

(eq. 4.19)

Again, with the flux ψ n directly proportional to the machine current, equation 4.19 can be written as [5].

Te = K a Kψ I a2

(eq. 4.20)

According to equation 4.20, the torque pulsates at twice the supply frequency.

4.3. Electromechanical system The machine is now described by the following set of differential equations forming an electromechanical system according to equation 4.21. d ⎧ ⎪⎪u (t ) − ( La + L f ) dt i (t ) − K a Kψ i (t )ωm (t ) = ( Ra + R f ) i(t ) ⎨ ⎪ K K i 2 (t ) = T + Dω (t ) + J d ω (t ) L m m ⎪⎩ a ψ dt

(eq. 4.21)

The back emf and the developed torque have been substituted with the expressions derived in section 4.1.1 and 4.2.1 respectively.

4.4. Implementation in Simulink The system of differential equations given by equation 4.21 is implemented in Simulink as a block diagram. The input voltage u (t ) and the load torque TL are input signals. The current and rotor speed are outputs. The linear simulink model of the machine is presented in figure 4.4.

Figure 4.4 – Block diagram modelling the universal motor.

14

Chapter 4 - Linear model

Further on, the block denoted Universal Motor in figure 4.4 consists of a block representing the electrical system in equation 4.21 and another block representing the mechanical system as shown in figure 4.5. 1 Machi ne current

2

Load Torque

Load T orque

i

i

w

2

w

Rotor speed Flux

1

Flux

Voltage Ka

Input voltage

Ka

Mechani cal system

El ectrical system

Figure 4.5 – The electrical and mechanical systems of the universal motor.

The contents of the electrical and the mechanical system are shown in figure A2 and figure A3 in appendix.

4.5. Simulation results using the linear model Even though this model does not take saturation into account, simulations are run in order to verify the behaviour of the model. The machine constant K a is calculated according to equation 4.12. The unknown parameters included in the model are at this stage only guessed. The parameters used in the simulations are: Known parameters Ra = 1.651 Ω

Guessed parameters La = 10 mH

R f = 1.164 Ω

L f = 10 mH

u = 120 V / 60 Hz

K a Kψ = 0.21 J = 0.0001 kg ⋅ m2 D = 0 Nm ⋅ s

Figure 4.6 shows the rotor speed when a constant load torque TL = 0.5 Nm is applied. The maximum speed reached is approximately 5300 rpm with a load torque of 0.5 Nm.

15

Chapter 4 - Linear model

Figure 4.6 – Simulated rotor speed vs time with TL =0.5 Nm .

The speed given by the simulations is far lower than the speed of the machine measured at this load torque which would be approximately 8000 rpm at this load torque. This is due to the fact that the saturation is not taken into account in this linear model. The current and torque wave form are of perfect sine shape which are shown in figure 4.7 and figure 4.8. The torque pulsates with twice the frequency of the current as expected.

Figure 4.7 – Machine current as function of time

16

Figure 4.8 – Produced electromagnetic torque as function of time.

Chapter 5 - Finite element model

5. Finite element model A finite element model of a full pitch machine is introduced in this chapter and a FEM model of the short pitch machine described in chapter 3 is thoroughly described.

5.1. Geometrical simplifications for the Finite Element Method (FEM) modelling The simplifications of the geometry are presented in this section. These simplifications are made to • •

Obtain a better mesh of the geometry Simplify the drawing of the geometry

The results from FEM simulations are more trustworthy if the triangles of the mesh are equilateral. If sharp edges and small radii’s are present in the geometry, the number of elements in those regions has to be very high in order to get a satisfactory mesh. Therefore it is desirable to have as few small radii’s and sharp corners as possible.

5.1.1. Stator In the finite element model, some holes in the stator geometry were not included. This simplification of the geometry can be made because the positions of the holes are not critical. This is the case for the holes situated in regions where the area of the magnetic path is large enough to ensure that the iron in these regions is not saturated. Further on, small smoothening radii’s in the stator were not taken into account in the finite element model. Besides simplifying the drawing of the geometry in the finite element program, this also makes the meshing of the geometry simpler to implement.

5.1.2. Rotor Some simplifications are also made to the smoothening radii’s of the rotor in the finite element model.

5.1.3. Rotor shaft The rotor shaft is the part of the machine that mechanically transfers the electromagnetic torque produced by the rotor to a connected load. The material properties of the shaft are critical in a mechanical point of view, but since the shaft is made of a solid non-magnetic steel material, it is not necessary to include the real properties of the shaft material in an electromagnetic model. The only crucial thing that must be correct is the diameter of the shaft. 17

Chapter 5 - Finite element model In the finite element model, the shaft is treated as a hole filled with vacuum in the centre of the rotor.

5.2. Flux 2D The software used in the FEM modelling in this project is the 2D 8.1 version of Flux which is a dedicated tool for making FEM models of electrical machines. This is due to the possibility to assign rotating properties to the air gap, and external connections with electrical circuits [7]. These are two features which are used in the simulation of this machine.

5.3. Full pitch motor model A simplified finite element model is obtained for a machine with two full-pitch rotor coils in each slot. This model is used to verify that the commutation works properly. This simplified motor is also used to test how to implement the non-linear model in Simulink. The model is only briefly presented here because the procedure used to create it is very similar to the one for the short-pitch motor.

5.3.1. Geometry The geometry of the full-pitch model is almost the same as for the test motor model. The only difference is that instead of having four regions in each rotor slot, the full-pitch machine only has two. The geometry of the full-pitch machine is presented in figure A9 in appendix. The rotor slots are divided in two regions which contain the rotor coils. The region coils are then linked to the electric circuit.

5.3.2. Electric circuit The rotor coils in the electric circuit in figure A10 in appendix have twice as many turns as for the short pitch machine presented in section 5.4. This is because the total number of coils in the full pitch winding is half the number of coils in the short pitch winding and the number of ampere turns in the rotor must be kept constant in order to have similar properties as the real machine.

18

Chapter 5 - Finite element model

5.4. Short-pitch motor model

5.4.1. Geometry Making the geometrical simplifications stated in the previous sections, the geometry is drawn in Flux2D. The final drawing of the universal machine is presented in figure 5.1. The rotor, the stator and the air are sectioned into regions which are assigned the respective material properties. Each rotor slot is separated into four regions so that the rotor coils in the electric circuit can be associated with the right region. The black slot corresponds to rotor slot 1a at rotor position zero, i.e.

θ mec = 0 The winding description given by Ankarsrum is shown in figure 3.5. It is drawn to give a clockwise rotation at the commutator side of the machine. In Flux, a counter clockwise rotation is defined positive. Instead of redrawing the drawing of the winding and in this way give the machine a counter clockwise rotation, a negative speed of the rotor is chosen in Flux. The coils in the rotor are at first inserted in the bottom of the slots, then in the outer region closer to the air gap. This removes the symmetry in the machine and the entire geometry of the machine must therefore be drawn in Flux. This implies that totally 48 rotor coil regions must be described. The large number of coils and regions give rise to very time- consuming simulations. The simulation time is especially high when the machine is fed with a current source of high amplitude. This is because the iron becomes highly saturated and the solver has difficulties to find a solution of the problem.

Figure 5.1 – The finite element model of the short pitch machine seen from the commutator side. The black slot corresponds to rotor slot 1a at rotor position zero degrees.

19

Chapter 5 - Finite element model

5.4.2. Mesh The geometry has to be meshed into a number of triangular surface elements. The most sensitive regions need a very fine mesh in order to get good results. The air gap is divided into three separate regions, one closest to the rotor, one in the middle and one closest to the stator. This is usually done when the variations in the torque are to be studied. The mesh can be seen in figure 5.2 which also includes a view of the fine mesh in the air gap region. The middle gap is remeshed at each timestep, the rotor air gap moving as the rotor and the stator air gap being fixed. To eliminate so called mesh noise, the mesh has to be implemented in such a way that the mesh in the air gap appears to be constant in time [9]. When the simulation time is increased by one time step the mesh in the air gap should appear to be exactly the same as in the previous time step.

Figure 5.2 – The mesh of the short pitch model. The right image shows an enlargement of the air gap region with its three-layer mesh.

20

Chapter 5 - Finite element model

5.4.3. Electric circuit Figure 5.3 shows a sketch of the position of the commutator segments relative to the rotor slots and relative to the brushes at position θ mec = 0 . The figure also contains information about how the machine is wound. This can be seen under Winding Arrangement in the right side of the figure. The brushes are on a fictional horizontal axis.

Figure 5.3 – The rotor slots with its coils and their relative position to the commutator at rotor angle zero degrees.

Figure 5.4 shows one commutator segment and one brush. The commutator segment is approaching the brush. The connection is established when the edge of the commutator segment reaches the edge of the brush. The brush starts to conduct current through the commutator segment. The connection is maintained until the other edge of the segment leaves the brush at the other edge of the brush.

Figure 5.4 – The commutation process.

A virtual set of brushes and a commutator are attached to the rotor in the FEM model as shown in figure 5.3. It is important to notice that no physical commutator or brush is actually included in the FEM model. These parts are solely modelled by variable resistances called 21

Chapter 5 - Finite element model brush-segment components in the electrical circuit. The instantaneous resistance of the brushsegment is governed by the rotor angle θ mec . The virtual commutator–brush approach of the problem simplifies the determination of the angular parameters of the switches. The brush-segment component is characterized by five parameters [9]: • •

The on state resistance Ron in ohms. The off state resistance Roff in ohms.

•

The angular position, θ pos in degrees, of the commutator segment with respect to the brush position. The brush opening angle, θbru in degrees. The opening angle of the commutator segment, θbar in degrees.

• •

The brush-segment component is in full conduction whenever the mechanical angle of the rotor θ mec is opposite to the angular position of the commutator segment with respect to the brush θ pos [9].

Figure 5.5 – Brush-segment conduction characteristics [9]

22

Chapter 5 - Finite element model The drive circuit presented in figure A5 in appendix is fed by a 120 V / 60Hz voltage source in series with a small inductance representing the end windings of the machine. The 48 square shaped brush-commutator segments in the upper and lower section of the drawing represent the brush-commutator connections. Figure 5.6 shows θ pos for commutator segment number 22 with respect to the right brush and to the left brush at θ mec = 0 . In figure A5 in appendix, this gliding contact to the right brush is handled by brush-segment number G22 and brushsegment G46 handles the contact to the left brush. A table showing the numbering of the brush-segments and their corresponding segment and brush is found in appendix. The angular parameters of the brush-segment components for the modelled machine at position zero, i.e. θ mec = 0 , are presented in figure 5.7.

Figure 5.6 – The commutator and its segments with positions relative to the right and left brush. These angles are used to set the switch parameter θ pos .

23

Chapter 5 - Finite element model

Brush-segment number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Brush-segment number 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

Pos. r.r.b.

-7.5 7.5 22.5 37.5 52.5 67.5 82.5 97.5 112.5 127.5 142.5 157.5 172.5 -172.5 -157.5 -142.5 -127.5 -112.5 -97.5 -82.5 -67.5 -52.5 -37.5 -22.5

Figure 5.7 – The position of the segments relative the brushes,

Pos. r.l.b.

172.5 -172.5 -157.5 -142.5 -127.5 -112.5 -97.5 -82.5 -67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5 82.5 97.5 112.5 127.5 142.5 157.5

θ pos at θ mec = 0 . Pos r.r.b: Position relative

right brush, Pos r.l.b: Position relative left brush.

5.4.4. End winding reactance The effect of the end windings is not included in a two dimensional FE problem such as the one that is used. Therefore the end-winding reactance should be added in the circuit. In [10], an empirical method to determine the end-winding reactance is proposed. A simplified drawing of the end winding is viewed in figure 5.8.

Figure 5.8 – The simplified shape of a coil end [10].

The angle γ is the angle between the end winding and the radial rotor plane. ly is the length of the end winding. α is calculated approximately by measuring the length of the end winding and the nearest length between the forward and the backward coil, which is the distance τ. 24

Chapter 5 - Finite element model

cos γ =

τ

(eq. 5.1)

ly

[7] states that inductance of a coil end Ly is equal to:

⎛ b ' sin γ Ly = 0.2nl2l y ln ⎜ kz ⎜ 2g 1,1 ⎝

⎞ −6 ⎟⎟ ⋅10 H ⎠

(eq. 5.2)

where nl - Total number of conductors in a coil end bkz' - Ideal reduced commutation zone, the length along the anchor where the current in the winding is reversed g1,1 - A coil ends average geometrical distance from itself

g1,1 = 0.2235 ( b1 + h1 )

(eq. 5.3)

b1 is the width of the coil and h1 is the height of the coil showed in figure 5.9.

Figure 5.9 - Cross section of a coil end [10].

For simplicity, the reduced commutation zone length bkz' , is substituted with the non-reduced length bkz . This gives a somewhat larger inductance.

⎛ β − 1 + fλ ⎞ bkz = τ s ⎜1 + ⎟ uc ⎝ ⎠ where uc is the number of coils, side by side in a slot .

25

(eq. 5.4)

Chapter 5 - Finite element model

β=

bc − bi

λτ

(eq. 5.5)

with bc - the width of the brush bi - the thickness of the insulation between two adjacent commutator segments

λτ - the distance between two adjacent commutator segments λτ =

π dk nk

(eq. 5.6)

Where d k is the diameter of the commutator and nk is the number of segments in the commutator.

τ s is the distance between two adjacent rotor slots τs =

2π r Q

(eq. 5.7)

D is the outer diameter of the rotor and Q is the total number of rotor slots. f λ is the reduction from full pole pitch expressed in number of commutator segments. fλ =

nk − ucτ p p

(eq. 5.8)

τ p is the pole pitch in number of rotor slots. With expressions 5.1-5.8 and some geometrical data from drawings and approximations, a first value of the end winding inductance can be calculated. It is found to be approximately. Ly ≈ 5μ H

(eq. 5.9)

At 60 Hz this inductance corresponds to a reactance equal to

X Ly = ω Ly ≈ 2mΩ

(eq. 5.10)

There are one end winding per coil on both sides of the machine and there are a = 2 number of parallel paths in the rotor. With 4 coils per rotor slot, the total end winding reactance can be written. 1 X Ltoty = 2 4Qω Ly = 48 ⋅ 2mΩ = 96mΩ a

(eq. 5.11)

At maximum current in the machine, this reactance results in a total reactive voltage drop of less than 1V. Therefore the effect of the end windings could be neglected. 26

Chapter 5 - Finite element model

5.5. Results from full-pitch motor The FEM model of the full-pitch motor is primarily used to verify that commutation works properly i.e. that the brush-segments in the electric circuit is described properly.

5.5.1. Verification of commutation The method used to describe the brush-commutator connection was explained in detail in section 5.4 which deals with the FEM model of the short-pitch machine. The model of the full-pitch machine is fed with a constant current and the commutation process is studied. To speed up the simulation time, no stator current is present. To further decrease the simulation time, the coils in the rotor have less turns than they should have (reduced level of magnetization). If the commutation is described properly, the coils situated in the region under one stator pole should always have opposite current direction compared to the coils situated in the region under the other stator pole. The coils in rotor slots 1-6 should have the same current direction at θ mec = 0 . Every time the rotor has rotated to a position where a commutator segment enters the brush, the associated coils should start to commutate. The first time step corresponds to θ mec = 10o , which means that the coils in slot 1 and slot 7 are fully commutated. When the rotor has rotated to position θ mec = 180o , the coils in slot 1 and slot 7 should have reversed their current direction. The commutation order of the coils in slot 1-3 and their corresponding coils on the opposite side of the rotor, i.e. coil 7-9, are shown in figure 5.10 and figure 5.11 respectively. All the curves are as expected, so the commutator was well described.

Figure 5.11 – Current in conductors in slot 7-9

Figure 5.10 – Current in conductors in slot 1-3

27

Chapter 5 - Finite element model

5.6. Results from short-pitch motor 5.6.1. Machine resistance The winding resistance is represented by the resistors connected in series with the coils. Their values depend on the loading of the machine since losses increase the temperature of the winding and therefore its resistance. This heating results in a linearly increasing resistance for the coils. It was assumed that the rotor and the stator resistances provided by Ankarsrum have been measured at room temperature. In the electric circuit associated to the FEM model, the resistance at the unknown working temperature should be used. To identify the value of the machine resistance at a load corresponding to 8000 rpm, the FEM model is simulated. The current and the torque are compared to the measured values. This was done for different values of the machine resistance. Using the results for the current, a linear approximation is done and an approximate value of the resistance is determined. Finally, the machine is once again simulated with the identified resistance and the simulated current is compared to the measured current. It was intended to do this procedure for several speeds, i.e. different loads, but the simulation time turned out to be long. Therefore this simulation was only performed at 8000 rpm. Figure 5.12 shows the relationship between the total machine resistance and current together with the linear approximation at 8000 rpm. It was found that a total machine resistance of 5.7 Ω corresponds to a winding temperature of approximately 280 οC which is non-realistic. The high resistance can however be explained by the contact between the brush and the commutator that is not so well characterized.

Figure 5.12 – Relationship between the current and the total machine resistance at 8000 rpm.

28

Chapter 5 - Finite element model

5.6.2. Machine current and torque at 8000 rpm The most relevant results from the FEM simulations are presented here. The machine speed is fixed at 8000 rpm and the applied voltage is 120V / 60 Hz. The value of the sum of the stator and rotor resistances is 5.7 Ω.

Figure 5.13 – The voltage applied to the machine in the simulation.

Figure 5.14 shows the current drawn by the machine. The steady-state current seems to be reached almost instantaneously. Therefore the simulation time can be set to two periods. This amount of time ensures that the steady-state is obtained in the second period of the current wave form. The RMS value of the current is 6.21 A which can be compared to the measured value 6.29 A. It can be seen that the current is distorted due to saturation of the magnetic circuit. This implies that a non-linear model is necessary to predict the performance of this machine. Since the current is adjusted with the total resistance of the machine, the produced electromagnetic torque should also be analyzed to verify the model.

Figure 5.14 - The machine current wave form at 8000 rpm, 120V / 60Hz and total machine resistance 5.7Ω.

29

Chapter 5 - Finite element model

Figure 5.15 – Produced electromagnetic torque

In figure 5.15, the electromagnetic torque is shown. It pulsates with twice the supply frequency and also presents some high frequency ripple at the torque peaks. The average torque from the simulations is 0.496 Nm. This value corresponds very well to the measured value which is 0.498 Nm. The simulated current is approximately 0.1 A smaller than the measured value. Increasing the simulated current by reducing the total resistance would lead to an increased average torque, but the results for the torque would still be very close to measurements.

Figure 5.16 – The input power to the machine

Another parameter that can be used to test the validity of the model is the input power, which is shown in figure 5.16. The average power consumed by the motor in the simulation is 697.8 W. This is 9.5 % higher than the measured value which is 637 W.

30

Chapter 5 - Finite element model

5.6.3. Simulation time The simulation time for this machine with so many coil regions in the geometry is very high. It should be mentioned that the results from one working point of the short-pitch motor model presented in this chapter took more than a weekend to produce. The simulations were carried out on a Pentium 4, 2.40 GHz with 512 Mb of RAM. Due to the high simulation time, it was decided to use the full pitch motor model in the following FEM simulations.

31

Chapter 6 - Non-linear analytical model

6. Non-linear analytical model As the short-pitch machine turned out to be very time consuming to simulate, the full-pitch machine will instead be investigated. Unfortunately, the full pitch machine also requires long simulation time. Therefore only four loads/speeds have been simulated. To further decrease the simulation time, the time stepping procedure in FEM is terminated after one period of the electromagnetic torque, i.e. half the period of the current wave form. This is believed to give results accurate enough according to the results obtained for the 8000 rpm simulation of the short-pitch machine. This simulation showed no transient behaviour at the start. It is therefore assumed that the full-pitch machine does not show transients either. To begin with, two ways of making a non-linear analytical model in Simulink have been considered. These are:

• •

Combined Simulink-Flux model Pure Simulink model

In the combined Simulink-Flux model, a block representing the FE model of the machine would be implemented in Simulink. This method would probably yield more accurate result compared to the pure Simulink model. The risk of serious difficulties which may occur when the programs are connected is considered high. Further on, the simulation time is expected to be even higher than with FEM only. These are the reasons why the pure Simulink model, was chosen. As the iron in the machine becomes saturated, the air gap flux will no longer show a linear behaviour with the machine current. Therefore, a method to calculate the flux in the air gap for different currents is needed in order to determine the induced back emf in equation 4.13 and the electromagnetic torque in equation 4.20. The air gap flux calculation procedure is presented in section 6.1. Further on, the inductance of the field winding will also become saturated. The calculation of the non-linear field winding inductance is presented in section 6.2. No method has been investigated to calculate the rotor winding inductance but it is believed to be small compared to the inductance of the field winding.

6.1. Air gap flux for the induced voltage The normal component of the flux density under one pole is calculated using Flux. The flux density is not homogenous in the air gap beneath a field pole due to armature reaction. Therefore an average value of the normal component of the flux density is calculated according to

Bn,avg =

1 Bn dl lC C∫

Where C is the integration path shown in figure 6.1. lC is the length of C .

32

(eq. 6.1)

Chapter 6 - Non-linear analytical model

Figure 6.1 – The path where the flux density is calculated

An increasing current is applied to the full-pitch machine in the FEM simulation and equation 6.1 is applied for the different current levels. This gives the average value of the flux density which generates the back emf in the rotor winding. The average flux density in the air gap beneath a pole as function of the machine current is shown in figure 6.2. With Bn available, it is straight forward to calculate the average flux cut by the conductors in the rotor winding using the width of the excitation pole. A lookup table is added to the linear Simulink model in section 4. This table has the machine current as input and the flux, which is used in equation 4.12 and 4.19 as output. With the air gap flux known for any current, the back-emf can be calculated.

Figure 6.2 – The average flux density in the air gap calculated by FEM.

33

Chapter 6 - Non-linear analytical model

6.2. Inductance of the field windings In the linear analytical model, it is assumed that the flux produced by the field windings is directly proportional to the machine current. Unfortunately, this is not true when the machine operates with high currents at high loads, i.e. the iron in the machine is highly saturated. To deal with the non-linearity in the inductance of the field windings, the following procedure is used. The objective is to calculate the field winding inductance of the machine for different rates of saturation. This is achieved by simulating the flux linkage seen by the stator- and the rotor winding for different current levels with FEM. The method to obtain the flux linkage from magneto-static FEM simulations is thoroughly described in [11] and is shortly presented here. The method is based on the magnetic vector potential. This quantity reflects how much flux per unit length is circulating around a certain point. Starting with one of Maxwell’s equations

∇• B = 0

(eq. 6.2)

Equation 6.2 implies that the curl of another vector field A can be used to express B such as: B = ∇× A

(eq. 6.3)

where A is the magnetic vector potential. The total flux seen by a coil is given by integrating the flux density over the area of the coil.

φ=

∫∫ B • ds = ∫∫ (∇ × A ) • ds

Acoil

(eq. 6.4)

Acoil

By applying Stoke’s theorem, equation 6.4 becomes:

φ=

∫∫ B • ds = ∫ A • dl

Acoil

(eq. 6.5)

C

where C is the countour bounding the area Acoil . In a 2D model A , has only one component in the axial direction (A z ) . Equation 6.5 can therefore be written as:

φ = ∫ A • dl = Leff ⋅ ( A1 − A2 ) + Lend ⋅ ( Aend 1 − Aend 2 )

(eq. 6.6)

C

where Leff is the length of the coil, i.e the effective length of the machine. A1 is the averaged vector potential on one side of the winding and A 2 is the averaged vector potential on the other side of the winding. Neglecting the end windings, equation 6.6 becomes: 34

Chapter 6 - Non-linear analytical model

φ = ∫ A • dl ≈ Leff ⋅ ( A1 − A2 )

(eq. 6.7)

C

This method is applied to the field winding of the full-pitch motor model, with A1 the averaged vector magnetic potential in the right slot holding the stator winding and A 2 the averaged vector magnetic potential in the left stator winding slot. The total flux seen by the field winding is:

ψ field = N field Leff ⋅

(∑

p

Ar ,i − ∑ i =1 Al ,i p

i =1

)

(eq. 6.8)

N field is the number of turns in the field winding, Ar ,i and Al ,i are the averaged vector potential in the right and left region holding field coil i .

For a non-linear magnetic circuit, two inductances can be defined. The absolute inductance is defined as the proportionality factor between the flux and the current i.e. the usual way in a linear case. L f ,abs =

φ Ia

(eq. 6.9)

An incremental inductance can also be defined as: L f ,inc =

dφ dI a

(eq. 6.10)

Figure 6.3 illustrates the values of the absolute and incremental inductance for a level of current where the circuit is saturated. As can be seen, the value of the incremental inductance is much lower than the absolute inductance at saturation. They are equal in the linear region. For the Simulink model of the universal motor, it is the incremental inductance that should be used as the inductive voltage drop in the field winding can be expressed as: ef =

dφ f dt

=

dφ f dia di ⋅ = L f ,inc (ia ) ⋅ a dia dt dt

(eq. 6.11)

The saturated flux linkage seen by the stator widning is shown in figure 6.3. The two different interpretations of the inductance concept are also shown in the same figure.

35

Chapter 6 - Non-linear analytical model

Flux linkage

0.25 L

= dΨ/di

inc

Flux linkageΨ [Wb]

0.2 Instantaneous operating point 0.15 L

= Ψ/i

abs

0.1

0.05

0

0

5

10

15 I [A]

20

25

30

Figure 6.3 – The flux linkage seen by the stator winding and the two ways of interpreting the inductance.

As for the air gap flux used to model the back emf, the incremental self inductance of the field winding L f , is implemented in the non-linear Simulink model as a lookup table. With the machine current I a as input, the lookup table returns the incremental field winding inductance, saturated or non-saturated depending on the level of I a . The incremental inductance of the field winding as function of the machine current is presented in figure 6.4. The electrical system of the non-linear analytical model is shown in figure A4 in appendix.

Field winding inductance 100 90 80

Lf = dΨ/dI [mH]

70 60 50 40 30 20 10 0

0

5

10

15 I [A]

20

25

30

Figure 6.4 – The field winding inductance calculated based on equation 6.10.

6.3. Inductance of the rotor winding The inductance of the rotor winding is chosen based on trial and error. It is considered constant and independent of the load. The inductance of the rotor winding is set to 6 mH.

36

Chapter 6 - Non-linear analytical model

6.4. Results from non-linear analytical model The interesting results from the non-linear analytical model are presented here. These results are compared to the FEM results for the full-pitch machine.

6.4.1. Machine current The RMS value of the current drawn by the machine is plotted versus the produced electromagnetic torque in figure 6.5. It is found out that the current obtained from the non-linear analytical model compares well to the current in the FEM model. The FEM value of the current is between 5-7 % (depending on the torque) less than with the Simulink model at a given torque.

Electromagnetic torque vs current 0.8

FEM Simulink

0.7 0.6

Te [Nm]

0.5 0.4 0.3 0.2 0.1 0

3

4

5

6 Irms [A]

7

8

9

Figure 6.5 – Electromagnetic torque vs. current.

37

Chapter 6 - Non-linear analytical model

6.4.2. Torque-speed characteristics The torque available at different speeds is presented in figure 6.6. The torque from the non-linear analytical model is overestimated by approximately 7 % for 10000 rpm. This might be explained by the inductance of the rotor winding which may show non-linear behaviour in reality. However, no method of modelling this inductance in a non-linear way has been investigated. Further on, the commutation is not modelled in Simulink. This factor also influences the torque/speed characteristics of the machine. In FEM time-stepping simulations with a voltage source and the rotor forced to rotate at a certain speed, the time steps are coarse in order to keep the simulations reasonably long. The large time steps generate solutions which are somewhat noisy. This is a source of error when the rms value of the current and the average produced electromagnetic torque are calculated from FEM. The FEM simulations for the two highest loads are only simulated for half the current period. If the entire period were investigated, the current would increase. This was seen by comparing calculations from the third highest torque (8000 rpm in figure 6.6). The current increase led to better results for this point. Electromagnetic torque vs speed 0.8

FEM Simulink

0.7 0.6

Te [Nm]

0.5 0.4 0.3 0.2 0.1 0 5000

6000

7000

8000 9000 speed [Rpm]

10000

11000

Figure 6.6 – Electromagnetic torque vs. speed.

6.4.3. Current wave forms Some current wave forms generated by the non-linear analytical model at different loads and speeds are presented in this section. It is interesting to note that the wave form at 8000 rpm for the simulated full-pitch machine is of similar shape as for the short-pitch machine at 8000 rpm which is presented in figure 5.14, in section 5.6.2. The current wave forms for 6500, 7000, 8000 and 10000 rpm and their related load torque are presented in figure 6.7. For comparison, the current from the 8000 rpm FEM simulation is provided in figure 6.8.

38

Chapter 6 - Non-linear analytical model

Current vs time at 6500 rpm, 0.69 Nm from non-linear analytical model

20

Current vs time at 7000 rpm, 0.631 Nm from non-linear analytical model

15

Machine current Ia

Machine current Ia

15 10 10 5 Ia [A]

Ia [A]

5 0

0

-5 -5 -10 -10

-15 -20

0

0.005

0.01

0.015 0.02 Time [s]

0.025

-15

0.03

0

Current vs time at 8000 rpm, 0.524 Nm from non-linear analytical model

10

5

5

-5

-10

-10

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.025

0.03

-15

0.03

Machine current Ia

0

-5

-15

0.015 0.02 Time [s]

Current vs time at 10000 rpm, 0.321 Nm from non-linear analytical model

10

0

0.01

15

Machine current Ia

Ia [A]

Ia [A]

15

0.005

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.03

Figure 6.7 – The current wave forms simulated at different loads and speeds by the non-linear analytical Simulink model.

Figure 6.8 – The current wave form simulated with FEM at 8000 rpm, 0.52 Nm load torque.

39

Chapter 6 - Non-linear analytical model

6.4.4. Torque wave forms The electromagnetic torque corresponding to the currents presented in figure 6.7 in section 6.4.3 is plotted in figure 6.9. The torque simulated by FEM is plotted in figure 6.10 as a reference. Electromagnetic torque vs time at 6500 rpm, 0.69 Nm from non-linear analytical model

Electromagnetic torque vs time at 7000 rpm, 0.631 Nm from non-linear analytical model

2

Electromagnetic torque

1.8

1.8

1.6

1.6

1.4

1.4 Torque [Nm]

Torque [Nm]

2

1.2 1 0.8

1 0.8 0.6

0.4

0.4

0

0.2 0

0.005

0.01

0.015 0.02 Time [s]

0.025

2

0

0.03

Electromagnetic torque vs time at 8000 rpm, 0.524 Nm from non-linear analytical model

1.8

1.6

1.6

1.4

1.4

1 0.8

0.4

0.2

0.2 0.01

0.015 0.02 Time [s]

0.025

0.025

0.03

Electromagnetic torque

1

0.6

0.005

0.015 0.02 Time [s]

0.8

0.4

0

0.01

1.2

0.6

0

0.005

2

1.8

1.2

0

Electromagnetic torque vs time at 10000 rpm, 0.321 Nm from non-linear analytical model

Electromagnetic torque

Torque [Nm]

Torque [Nm]

1.2

0.6

0.2

Electromagnetic torque

0

0.03

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.03

Figure 6.9 – The torque wave forms simulated at different loads and speeds with the non-linear analytical Simulink model.

Figure 6.10 – The torque wave form simulated with FEM at 8000 rpm, 0.52 Nm load torque.

40

Chapter 6 - Non-linear analytical model

6.5. Conclusions The current waveforms clearly show that a non-linear model is required to predict the performance of such universal motors (full-pitch and/or short-pitch). A way to obtain the required non-linear functions from FEM has been presented. Good correlation has been obtained for the current and torque (wave forms, RMS value and mean value respectively). At high speed, the commutation might be added in the model to help obtaining better correlation. Further on, the non-linearity of the rotor winding inductance should be investigated. The good correlation between FEM and the non-linear analytical model offers to use the latter for investigating control strategies.

41

Chapter 7 - Speed control

7. Speed control In this chapter, two speed control strategies are implemented in Matlab/Simulink together with the non-linear analytical model of the full-pitch motor. These two strategies are implemented with triacs or controlled switches. The two circuits are shown in figure 7.1 and figure 7.2. The influence of the torque and current waveforms are compared.

Figure 7.1 – Triac motor controller block.

Figure 7.2 – PWM motor controller block.

7.1. Triac To control the speed of the machine, an open loop triac controller that controls the effective applied machine terminal voltage, is implemented in Simulink. It is believed that the switching of the machine voltage caused by the triac adds harmonics to the supply current and affects the torque wave form of the machine in a negative way.

42

Chapter 7 - Speed control The triac block is shown in figure 7.1. The control angle block is high whenever the control angle has passed in the electrical period of the voltage. If the triac is feeding a pure resistive load, the zero current detection block can be removed. The machine is an inductive load. Therefore, for each half period, the triac can only turn off the applied terminal voltage once the current has reached zero. The sr-latch in figure 7.1 holds the on-state of the switch until the zero current detection block detects a zero crossing in the current wave form so that the behaviour of a triac is described. Some signals in the triac block are plotted in figure. This plot corresponds to a control angle of 45ο and a current that lags the voltage by 30ο. For simplicity, the amplitude of the load current and the supply voltage are both set to unity. The triac is turned on as the control angle has elapsed in each half period. For a resistive load, the triac is always turned off as the control angle signal turns low. However, if the load is inductive (as in figure 7.3), it is seen that the current is not zero as the control angle turns low. This means that the triac can not turn off. As the current crosses zero, the current state (Zero current detection block in figure 7.1) turns high for a very short time. This spike is sensed by the reset input in the flip-flop and the output voltage is set to zero by the switch. The triac is then turned on once the control angle turns high in the next half cycle.

Signals in the triac block

1

Supply voltage

0 -1 10

100

200

300

400

500

600

700

Load current

0 Amplitude

-1 1 0.5 0 1 0.5 0

0

100

200

300

400

500

600

700

Control angle

0

100

200

300

400

500

600

700

Current state

10

100

200

300

400

500

600

700

Output voltage

0 -1 0

100

200 300 400 500 Electrical angle [degrees]

600

700

Figure 7.3 – The signals in the triac controller block in figure 7.1.

For control angles α higher than the current lag ϕ , the operation of the triac is perfectly stable. However, with control angles lower than the current lag, the triac is unstable. This is due to the fact that the current need to cross zero during the period when the control angle signal is low. If this is not the case, the triac is constantly on. The phenomena is presented in figure 7.5 where the current lag is ϕ = 40 o and the control angle is α = 30 o . However, it is assumed that the torque and current wave forms are mostly affected at high control angles. Therefore this problem is not investigated further as only values of alpha higher than the current lag will be chosen from now on.

43

Chapter 7 - Speed control To verify that the triac controller works properly, the output voltage at different control angles and pure resistive load are plotted in figure 7.4. The output voltages with inductive load at different control angles are plotted in figure 7.5. Note that the upper left plot for the inductive load shows unstable operation.

44

Chapter 7 - Speed control

Figure 7.4 – Triac output voltage at different control angles with pure resistive load

Figure 7.5 - Triac output voltage at different control angles with inductive load

45

ϕ = 40o

Chapter 7 - Speed control It is assumed that the torque wave form is mostly affected at high torque and high control angles α . Therefore, the motor and its controller are simulated at 0.69 Nm which is the highest load simulated in chapter 6. The terminal voltage, machine current and the produced electromagnetic torque are shown in figure 7.6-7.8. The speed of the machine varies between 5000 rpm down to 2000 rpm in steps of 1000 rpm. The control angles corresponding to the simulated speeds are α = 56 o − 78.5 o . Applied terminal voltage, α =56o

Applied terminal voltage, α =65o

200

Terminal voltage

150

150

100

100

50

50

Voltage [V]

Voltage [V]

200

0 -50

0 -50

-100

-100

-150

-150

-200

0

0.005

0.01

0.015 0.02 Time [s]

0.025

-200

0.03

Terminal voltage

0

0.005

Applied terminal voltage, α =72o

150

100

100

50

50

0 -50

-150

-150

0.01

0.015 0.02 Time [s]

0.025

-200

0.03

Terminal voltage

-50 -100

0.005

0.03

0

-100

0

0.025

Applied terminal voltage, α =78.5o

150

-200

0.015 0.02 Time [s]

200

Terminal voltage

Voltage [V]

Voltage [V]

200

0.01

0

0.005

0.01

0.015 0.02 Time [s]

Figure 7.6 – The terminal voltage at different control angles

46

0.025

0.03

Chapter 7 - Speed control

Current vs time at 5000 rpm, 0.69 Nm, α =56o

Current vs time at 4000 rpm, 0.69 Nm, α =65o

20

Machine current Ia

15

15

10

10

5

5 Ia [A]

Ia [A]

20

0

0

-5

-5

-10

-10

-15

-15

-20

0

0.005

0.01

0.015 0.02 Time [s]

0.025

-20

0.03

Machine current Ia

0

Current vs time at 3000 rpm, 0.69 Nm, α =72o

15

10

10

5

5

0

-5

-10

-10

-15

-15

0.005

0.01

0.015 0.02 Time [s]

0.025

0.025

0.03

-20

0.03

Machine current Ia

0

-5

0

0.015 0.02 Time [s]

Current vs time at 2000 rpm, 0.69 Nm, α =78.5o

15

-20

0.01

20

Machine current Ia

Ia [A]

Ia [A]

20

0.005

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.03

Figure 7.7 – The machine current at different control angles

Electromagnetic torque vs time at 5000 rpm, 0.69 Nm, α =56o

Electromagnetic torque vs time at 4000 rpm, 0.69 Nm, α =65o

2

Electromagnetic torque

1.8

1.8

1.6

1.6

1.4

1.4 Torque [Nm]

Torque [Nm]

2

1.2 1 0.8

1.2 1 0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0

0.03

Electromagnetic torque

0

Electromagnetic torque vs time at 3000 rpm, 0.69 Nm, α =72o

1.8

1.6

1.6

1.4

1.4

1.2 1 0.8

0.4

0.2

0.2 0.015 0.02 Time [s]

0.025

0

0.03

Electromagnetic torque

1

0.6

0.01

0.03

0.8

0.4

0.005

0.025

1.2

0.6

0

0.015 0.02 Time [s]

Electromagnetic torque vs time at 2000 rpm, 0.69 Nm, α =78.5o

1.8

0

0.01

2

Electromagnetic torque

Torque [Nm]

Torque [Nm]

2

0.005

0

0.005

0.01

0.015 0.02 Time [s]

0.025

Figure 7.8 – The produced torque at different control angles

47

0.03

Chapter 7 - Speed control The total harmonic distortion (THD) are calculated for the current wave forms according to [12].

I THD

⎛ I (n) = 100 ⋅ ∑n = 2 ⎜ ⎜I ⎝ (1) ∞

⎞ ⎟ ⎟ ⎠

2

(eq. 7.1)

[%]

where I (1) is the fundamental component and I (n ) are the harmonics. This calculation is performed in Simulink. The triac controller definitely adds harmonics to the machine current as shown in table 7.1 where the THD for different control angles are shown. Table 7.1 – The total harmonic distortion of the current at different speeds at 0.69 Nm load torque with triac control.

Control angle (degrees)

Speed (rpm)

THD (%)

56

5000

38.13

65

4000

40.79

72

3000

42.76

78.5

2000

44.46

As a direct consequence to the current distortion, the torque wave form is also affected due to the triac controller. This is seen by comparing figure 7.8, with figure 6.9 which shows the torque wave form without speed control. The peak of the produced electromagnetic torque when the machine is speed controlled with a triac is higher while the rise time of the torque seems to be shorter. The machine without triac speed control has a smoother variation of torque with time. With triac speed control, the current drops to zero under a certain period. Therefore the machine does not produce any torque during that phase. At a certain time, the triac starts to conduct the current and the torque increases rapidly. This is most likely to produce noise as well as affect the wearing of gears and transmissions connected to the machine.

48

Chapter 7 - Speed control

7.2. Sinus PWM Instead of having a triac controller controlling the speed of the machine, a PWM controller is tested in the model. The PWM block is shown in figure 7.2. A triangle wave of high frequency is compared to a reference voltage. Depending on which signal is the highest, the switch in figure 7.2 is high, + Vd or low, − Vd . The fundamental of the output voltage can be calculated as [12] ∧

V 1 = m a ⋅ Vd

(eq. 7.2)

Where m a is the modulation index, which is defined as [12] ∧

ma =

V ref

(eq 7.3)

∧

V tri The switching frequency used is f s = 3 kHz . The time step that is used for the simulations is 10 μs i.e, the sample frequency is 100 kHz. The current and torque wave forms are shown in figure 7.9 and figure 7.10 respectively.

Current vs time at 5000 rpm, 0.69 Nm, ma=0.93

Current vs time at 4000 rpm, 0.69 Nm, ma=0.88

20

Machine current Ia

15

15

10

10

5

5 Ia [A]

Ia [A]

20

0

0

-5

-5

-10

-10

-15

-15

-20

0

0.005

0.01

0.015 0.02 Time [s]

0.025

-20

0.03

Machine current Ia

0

10

10

5

5 Ia [A]

Ia [A]

15

0

-5

-10

-10

-15

-15

0.005

0.01

0.015 0.02 Time [s]

0.025

0.025

0.03

0.03

Machine current Ia

0

-5

0

0.015 0.02 Time [s]

20

Machine current Ia

15

-20

0.01

Current vs time at 2000 rpm, 0.69 Nm, ma=0.78

Current vs time at 3000 rpm, 0.69 Nm, ma=0.825

20

0.005

-20

0

0.005

0.01

0.015 0.02 Time [s]

0.025

Figure 7.9 – The current wave form when sinus PWM-control is used.

49

0.03

Chapter 7 - Speed control

Electromagnetic torque vs time at 5000 rpm, 0.69 Nm, ma=0.93

Electromagnetic torque vs time at 4000 rpm, 0.69 Nm, ma=0.88

2

Electromagnetic torque

1.8

1.8

1.6

1.6

1.4

1.4 Torque [Nm]

Torque [Nm]

2

1.2 1 0.8

1 0.8 0.6

0.4

0.4

0.2

0.2 0

0.005

0.01

0.015 0.02 Time [s]

0.025

0

0.03

0

1.8

1.6

1.6

1.4

1.4 Torque [Nm]

1.8

1 0.8

0.4

0.2

0.2 0 0.01

0.015 0.02 Time [s]

0.025

0.03

0.03

Electromagnetic torque

1

0.4

0.005

0.025

0.8 0.6

0

0.015 0.02 Time [s]

1.2

0.6

0

0.01

2

Electromagnetic torque

1.2

0.005

Electromagnetic torque vs time at 2000 rpm, 0.69 Nm, ma=0.78

Electromagnetic torque vs time at 3000 rpm, 0.69 Nm, ma=0.825

2

Torque [Nm]

1.2

0.6

0

Electromagnetic torque

0

0.005

0.01

0.015 0.02 Time [s]

0.025

0.03

Figure 7.10 – The torque wave form when sinus PWM-control is used.

The distortion in the current wave form is calculated in the same way as for the triac controller. The results are gathered in table 7.2. Table 7.2 – The total harmonic distortion of the current at different speeds at 0.705 Nm load torque with PWM-control.

Modulation index

Speed (rpm)

THD (%)

0.93

5000

33.68

0.88

4000

33.66

0.825

3000

33.70

0.78

2000

33.80

The current wave form is now less distorted compared to the triac control even though it contains high frequency ripple due to the switching. The reduction of the current distortion is most significant at low speeds. At 2000 rpm, the THD is approximately 24 % lower with PWM control. Figure 7.11 shows the harmonic content in the current at 2000 rpm, 0.69 Nm, for the two different control strategies. The third harmonic is significantly reduced when PWM control is used. On the other hand, a switching frequency at 3000 Hz is introduced. 50

Chapter 7 - Speed control

Current harmonics, 2000 rpm 0.69 Nm, Sinus-PWM controlled

Current harmonics, 2000 rpm 0.69 Nm, Triac controlled 15

15

Ia(n)

Ia(n)

10 Amplitude [A]

Amplitude [A]

10

5

5

0

0

100

200

300

400

500

600

700

800

900

0

1000

Frequency [Hz]

0

500

1000

1500 2000 2500 Frequency [Hz]

3000

3500

4000

Figure 7.11 – The harmonic content in the current at 2000 rpm, 0.69 Nm, for the two different control strategies. Triac control to the left and sinus-PWM to the right.

The produced electromagnetic torque for the machine with a PWM controller consists of high frequency ripple. The long period where the machine produces no torque introduced by the triac is not present for the PWM. The high frequency ripple in the current can most likely be reduced by the use of filters. This would lead to less ripple in the torque wave form as well. The peak torque is approximately 5 % lower with PWM-control compared to triac at 2000 rpm, 0.69 Nm. This would probably reduce the 120 Hz noise produced by the machine. Further on, with PWM, the torque wave forms are similar independent of the machine speed. This is not the case for the triac where the zero current period prevents the machine from producing a torque during a certain period of time. This dead time has to be compensated with a high peak in the torque in order to keep the speed. It would be interesting to simulate at least one point with FEM together with a triac controller, in order to look at the variations of the normal forces acting on the rotor teeth and excitation poles.

51

Chapter 8 - Conclusions and future work

8. Conclusions and future work Two FEM models of a full-pitch and a short-pitch universal motor have been developed. Comparisons with measurements showed that these models are giving realistic results. This project has shown how to derive a non-linear analytical model of a universal motor. Although the simulation times in the FEM simulations are really long for the short pitch machine, it is assumed that a model of the short pitch machine can be created in a similar manner as the non linear Simulink model of the full pitch machine presented in this report. A general conclusion drawn is that nonlinearities due to saturation are difficult to model. To decrease the simulation time for the FEM simulations, only half or even one fourth of the geometry could be used for the full-pitch machine. This would lead to fewer elements and a less complex problem to solve. However, this would not be possible for the short-pitch machine because there is no symmetry with respect to the coils in this machine. The non-linearity of the rotor winding inductance should be analyzed. The errors in the torque/speed characteristics of the machine might be reduced by taking it into account. To further improve the model of the machine, the effects of commutation should be described. A transformer induced emf in the rotor was found in a FEM simulation. The influence of this should be investigated. If the machine is speed controlled with a triac, the wave form of the produced electromagnetic torque is higher in amplitude at low speeds compared to the PWM controller. The level of 120 Hz noise emitted by the machine may therefore be reduced if the machine is controlled by PWM. The higher torque peaks caused by the triac controller also most likely increase wearing of gears connected to the machine. More load levels should be analyzed for the triac and the PWM. A closed loop control should be implemented in Simulink for the triac control as well as for the PWM control. It would be interesting to connect Simulink with Flux and perform the control simulations on this model and compare the results to the simulation results provided in chapter 7. In this way, the effect of slotting and armature reaction would be fully taken into account and the forces acting on the different parts of the machine could be investigated. The FEM simulation used to calculate the back emf and the field inductance should be performed at more rotor angles to investigate how these quantities varies as the position of the rotor is changed.

52

References

References [1] P.C. Sen – “Principles of electrical machines and power electronics”, second edition, ISBN 0-471-02295-0, 1997. [2] Peter F. Ryff, David Platnick, Joseph A. Karnas – “Electrical machines and transformers, Principles and applications”, ISBN 0-13-247222-8, New Yersey 1987. [3] http://www.sura.se – Information of stator- and rotor laminations [4] MG Say EO Taylor - “Direct current machines”, ISBN 0-273-01219-3, Bath, Great Britain, 1980. [5] http://www.ece.msstate.edu – Back emf calculation [6] http://www.reliance.com – Information of winding types in DC – machines [7] http://www.cedrat.com – Flux features [8] David K. Cheng – “Field and wave electromagnetics”, Second edition - United States, 1992, ISBN 0-201-12819-5. [9] Flux Help [10] Emil Alm 1944 – ”D.T.V Elektroteknik III, Elektromaskinlara II D”. [11] P.Thelin, J. Soulard, H.-P. Nee and C. Sadarangani, “Comparison between Different Ways to Calculate the Induced No-Load Voltage of PM Synchronous Motors using Finite Element Methods”, Presented at PEDS’01, Bali, Indonesia, Oct. 2001. [12] Hans-Peter Nee – “Kompendium i Eleffektsystem”, KTH, Stockholm 2003. [13] – Drawings Ankarsrum.

53

List of symbols

List of symbols Number of stator poles Coil pitch Pole pitch

p y τp

[− ] [−] [−]

Active length of machine Instantaneous current Instantaneous voltage Instantaneous emf Instantaneous rotor speed Time Supply frequency Field winding resistance Rotor winding resistance

L eff i u e ωm t f Rf Ra

[m] [A ] [V ] [V ]

[s −1 ] [s] [ Hz] [Ω ] [Ω ]

Field winding inductance Rotor winding inductance

Lf La

[H ] [H ]

Moment of inertia Viscous damping constant Instantaneous electromagnetic torque

[kg ⋅ m 2 ] [ Nm ⋅ s] [ Nm ]

Load torque Rotor constant

J D Te TL Ka

Flux constant

Kψ

[ Wb ⋅ A −1 ]

Electromagnetic force Pole coverage factor Radius of rotor Radius of stator pole Number of parallel paths in the rotor Number of turns in the rotor Length of rotor conductor in magnetic field Length of stator pole Flux linkage Flux Flux density Flux density, normal component Position of rotor Brush resistance at on state

Fe c r rpole a N l conductor l pole Ψ φ B Bn θ mec R on

[ N] [−] [m] [m]

54

[ Nm ] [−]

[−] [−] [m] [m] [ Wb] [ Wb] [T ] [T ] [− ] [Ω ]

List of symbols Brush resistance at off state Position of commutator segment

R off θ pos

[Ω ]

Opening angle of a brush Opening angle of a commutator segment In plane length between forward and backward coil in rotor Length of end winding

θ bru θ bar τ ly

[− ]

Inductance of one coil end Average value of simultaneously commutating coils Reactance of one coil end

Ly β X Ly

[H ]

Number of conductors in a coil end Reduced commutation zone Angle between end winding and plane of rotor lamination A coil ends average geometrical distance from itself

nl b 'kz γ g 1,1

[− ]

Height of a coil end Width of a coil end Distance between two adjacent rotor slots Distance between two adjacent commutator segments

[m] [m] [m]

Width of insulation between commutator segments Number of coils side by side in a rotor slot

h1 b1 τs λτ dk Q nk fλ bc bi uc

Vector magnetic potential Vector differential operator

A ∇

Conductance

G

Diameter of commutator Total number of rotor slots Total number of commutator segments Reduction from full pitch expressed in number of segments Width of brush

55

[− ] [− ] [m] [m] [− ] [Ω ] [m] [− ] [m]

[m] [m] [− ] [− ] [− ] [m] [m] [− ]

[ Wb ⋅ m −1 ] [− ] [ Ω −1 ]

Appendix

Appendix

56

Appendix

Performance test I (A*1000) Cos phi

P1 (W*10) Eta

25,000

1.00

22,500

0.90

20,000

0.80

17,500

0.70

15,000

0.60

12,500

0.50

10,000

0.40

7,500

0.30

5,000

0.20

2,500

0.10

0.00

0 0

10

20

30

40

50

60

70

Ncm

Motortype: KS4242/110 94119202 Armature winding: 4x17t/s Ø 0.50mm Field winding: 110t/c Ø 0.90mm Testvoltage: 120V 60Hz Rotation:

File: PT433.xls Author: ASUL/AH,LC Date: 2004-05-28

Remark: TP20040104 KIA Mörkrotorplåt 02746

Figure A1 – Results from the performance test of the short pitch machine.

57

80

Eta, Cos phi

n (rpm*1) P2 (W*10)

Lf

58

Figure A2 – The electrical system for the linear Simulink model.

Rotor winding resistance

Ra

Field winding resistance

Rf

Voltage

2

Rotor constant

k

Rotor winding inductance

La

Field winding inductance

Ka

3

w

1

f lux

-K-

Gain

flux

2

Product1

E

Product

(Rf +Ra)*i

Vc-(Rf +Ra)*i

Vc-(Rf +Ra)*i-E

Product2

(Vc-(Rf +Ra)*i-E)/(Lf +La)

Integrator

1 s

i Machine current

i

1

Appendix

59

i

Ka

4

Flux

3

2

Product3

1 Load Torque

Ka*phi*i=Te

To Workspace4

Te

Viscous damping

D

J

Product7

Dw

Te-TL

Moment of inertia

Te-TL-Dw

Product8

(Te-TL-Dw)/J

Integrator1

1 s

To Workspace1

w

w

1

Appendix

Figure A3 – The mechanical system for the linear and non-linear Simulink models.

60

Figure A4 – The electrical system for the non- linear Simulink model.

Constant2

Ra

Constant1

Rf

Voltage

2

Constant5

k

La

6e-3

k

3

w

1

Flux

2

Product1

Product

E

f lux

Lf (i)

(Rf +Ra)*i

Vc-(Rf +Ra)*i

Look-Up Table1

Lf +La

Product2

|u|

Abs

Look-Up Table

(Vc-(Rf +Ra)*i-E)/(Lf +La)

abs(i)

Integrator

1 s

i To Workspace

i

1

Appendix

Appendix

Figure A5 – The circuit connected to the short pitch machine in the finite element model.

61

Appendix

Figure A6 – The coils corresponding to the rotor coils in the short pitch machine. This is an enlargement of figure A5.

62

Appendix

Component V1 R1, R2 R3- R26 L1

Component descr i pt i on Vol t age sour ce St at or coi l r esi st ances Rot or coi l r esi st ances End wi ndi ng i nduct ance

Figure A7 – Description of the components in the circuit in figure A5 and A6.

63

Appendix

Coi l B1p B1n B1a B1b B1c B1d B2a B2b B2c B2d B3a B3b B3c B3d B4a B4b B4c B4d B5a B5b B5c B5d B6a B6b B6c B6d

Descr i pt i on Exci t at i on coi l Exci t at i on coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l

in in in in in in in in in in in in in in in in in in in in in in in in

r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or r ot or

sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot

Coi l

1a 1b 1c 1d 2a 2b 2c 2d 3a 3b 3c 3d 4a 4b 4c 4d 5a 5b 5c 5d 6a 6b 6c 6d

B7a B7b B7c B7d B8a B8b B8c B8d B9a B9b B9c B9d B10a B10b B10c B10d B11a B11b B11c B11d B12a B12b B12c B12d

Descr i pt i on

Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l Coi l

i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or i n r ot or

sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot sl ot

7a 7b 7c 7d 8a 8b 8c 8d 9a 9b 9c 9d 10a 10b 10c 10d 11a 11b 11c 11d 12a 12b 12c 12d

Figure A8 – Description of the components in the circuit in figure A5 and A6.

64

Appendix

Figure A9 – The FEM geometry of the full pitch machine.

65

Appendix

Figure A10 – The circuit connected to the full pitch machine in the finite element model.

66