FLUX ® 2D Application Scalar command of an induction machine technical paper

8/10/2019 FLUX 2D Application Scalar command of an induction machine technical paper

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CAD Package for Electromagnetic and Thermal

Analysis using Finite Elements

FLUX

2D ApplicationScalar command of an induction

machinetechnical paper

Copyright - September 2004


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FLUX is a registered trademark.

FLUX software : COPYRIGHT CEDRAT/INPG/CNRS/EDF

FLUX2D technical papers : COPYRIGHT CEDRAT

FLUX2D's Quality Assessment

(Electricit de France, registered number AQMIL013)

This technical paper was edited on24 September 2004.

Ref.: K205-R-810-EN-09/04

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FRANCE

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- REMARK -

The files corresponding to different cases studied in this technical paper

are available in the folder:

\DocExamples\Examples2D\DriveMotorWithSimul ink\

FluxFi les\

The corresponding applications are ready to be solved. This allows you

to adapt this technical paper to your needs.


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Introduction

This technical paper shows an example of the simulation of the drive of a rotating machine

through the link FLUX to Simulink Technology.

Precisely we will present the scalar command of an induction machine.

One will first define the FLUX model of the induction machine. After computing a simplified

model of the complete system with Simulink, the computation will be made using FLUX to

Simulink Technology. Results will then be compared.


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FLUX TABLE OF CONTENTS

TABLE OF CONTENTS

PART A: FLUX MODEL 1

1. Geometry .................................................................................................................... 3

1.1 Overview of the geometry ............................................................................................... 3

1.2 Stator geometry............................................................................................................... 4

1.2.1 Geometrical parameters....................................................................................................4

1.2.2 Coordinate systems...........................................................................................................4

1.2.3 Points and lines for the upper half of the stator slot ..........................................................5

1.2.4 Geometric transformations ................................................................................................6

1.2.5 Completing the stator geometry ........................................................................................6

1.3 Rotor geometry ............................................................................................................... 9

1.3.1 Geometrical parameters....................................................................................................9

1.3.2 Coordinate systems...........................................................................................................9

1.3.3 Points and lines for the rotor bar .....................................................................................10

1.3.4 Geometric transformations ..............................................................................................10

1.3.5 Completing the rotor geometry........................................................................................11

1.3.6 Closing the air-gap...........................................................................................................13

1.4 Add and assign regions for the faces............................................................................ 14

1.5 Mesh ............................................................................................................................. 16

1.5.1 Change to the Mesh context............................................................................................161.5.2 Applying the mesh points to the geometry ......................................................................16

1.5.3 Generate, verify and save the mesh................................................................................17

2. Materials.................................................................................................................... 19

3. Definition of the electrical circuit................................................................................ 21

4. Physical properties.................................................................................................... 23

4.1 General information....................................................................................................... 23

4.2 Assign materials to the regions..................................................................................... 23

4.3 Electrical circuit ............................................................................................................. 24

SCALAR COMMAND OF AN INDUCTION MACHINE PAGE A


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TABLE OF CONTENTS FLUX

PART B: SIMULINK MODEL 25

5. Scalar Control............................................................................................................ 27

5.1 Principle of the scalar control.........................................................................................27

5.1.1 Introduction ......................................................................................................................27

5.1.2 Scalar control modeling ...................................................................................................275.1.3 Relations used for the scalar control ...............................................................................29

5.2 Structure of the Scalar control .......................................................................................30

6. Simulink model .......................................................................................................... 33

6.1 Definition of the Simulink model ....................................................................................33

6.2 Definition of the blocks...................................................................................................34

6.2.1 The command..................................................................................................................34

6.2.2 Subsystem of the Instruction block..................................................................................34

6.2.3 Speed controller...............................................................................................................35

6.2.4 Subsystem of the Scalar control......................................................................................36

6.2.5 The induction machines block:........................................................................................37

6.2.6 Outputs ............................................................................................................................41

6.2.7 Other blocks.....................................................................................................................41

7. Solve ......................................................................................................................... 43

PAGE B SCALAR COMMAND OF AN INDUCTION MACHINE


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FLUX TABLE OF CONTENTS

PART C: FLUX TO SIMULINK MODEL 45

8. Definition of the Simulink model ................................................................................ 47

8.1 Description of the Simulink model................................................................................. 47

8.2 Definition of the coupling block ..................................................................................... 48

9. Solve ......................................................................................................................... 49

PART D: RESULTS 51

10. Simulink Results........................................................................................................ 53

10.1 No load torque............................................................................................................... 5310.1.1 150 rad per second..........................................................................................................53

10.1.2 100 rad per second..........................................................................................................55

10.1.3 30 rad per second............................................................................................................56

10.2 With load torque............................................................................................................ 58

10.2.1 Mechanical quantities......................................................................................................58

10.2.2 Comments .......................................................................................................................58

10.2.3 Electrical quantities..........................................................................................................59

11. FLUX Results ............................................................................................................ 61

11.1 No load torque............................................................................................................... 6111.1.1 150 rad per second..........................................................................................................61

11.1.2 100 Rad per second ........................................................................................................63

11.1.3 30 rad per second............................................................................................................65

11.2 With load torque............................................................................................................ 68

11.2.1 Mechanical quantities at 150 rad per second..................................................................68

11.2.2 Electrical quantities at 1500 rpm .....................................................................................69

12. Conclusion ................................................................................................................ 71

SCALAR COMMAND OF AN INDUCTION MACHINE PAGE C


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TABLE OF CONTENTS FLUX

PAGE D SCALAR COMMAND OF AN INDUCTION MACHINE


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FLUX PART A: FLUX MODEL

PART A: FLUX MODEL

SCALAR COMMAND OF AN INDUCTION MACHINE PAGE 1


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PART A: FLUX MODEL FLUX

PAGE 2 SCALAR COMMAND OF AN INDUCTION MACHINE


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FLUX PART A: FLUX MODELGeometry

1. Geometry

1.1 Overview of the geometry

Our sample problem consists of a 4-pole, 3-phase, 36-slot, 28-bar induction motor. Because of

the motors periodicity, we will model only of it (1 pole). Our model consists of 9 stator slots

and 7 rotor bars. The air-gap is set to 0.25 mm. The figure below is a diagram of our model.

Figure 1.1: Diagram of of the motor



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PART A: FLUX MODEL FLUXGeometry

1.2 Stator geometry

1.2.1 Geometrical parameters

The geometrical parameters used in the geometry part are presented in Table 1.1.

Stator Parameter Name Comment Value (mm)

AIRGAP Air-gap width 0.25

SOD stator outer diameter 170

SID stator inner diameter 117

SSHEIGHT Stator slot height 13

SSOPEN Stator slot opening 3.8

SSBR stator slot bottom radius 3.6

Table 1.1: Data to define Stator Parameters.

Note:

To create parameters, coordinate systems, points, lines and geometrical transformations:double click on their name in the data tree.

1.2.2 Coordinate systems

The geometry of the motor is described using the coordinate system presented in Table 1.2.

Name Comment

Type

of system

coordinat

e

Coordinate

system

of definition

Type of

coordinates

X

coor

d

Y

coor

d

Z rot.

STATMAIN main stator

coordinate system

2D GLOBAL CARTESIAN 2D 0 0 0

STATWORK working system LOCAL STATMAIN CARTESIAN 2D 0 0 0

STATLOC local system LOCAL STATWORK CARTESIAN 2D sid/2 0 0

Table 1.2: Stator coordinate systems.



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1.2.3 Points and lines for the upper half of the stator slot

The description of the stator starts from the geometry of half of a stator slot whose points defined

in STATLOC coordinate system are presented in Table 1.3.

Point X coordinate Y coordinate

P1 0 0

P2 SSHEIGHT 0

P3 0 SSOPEN/2

P4 SSHEIGHT-SSBR SSBR

Table 1.3: Coordinates of points for upper half of the stator slot.

After the creation of these points we will create lines by connecting them as presented in Table

1.4:

Line Type of line Starting point End point Arc radius

L1 Segment defined by 2

points

P3 P4

L2 Arc defined by its radius,

starting and ending point

P2 P4 SSBR

Table 1.4: Lines of stator base geometry.

You should see the screen below:

Figure 1.2: The upper half of the stator slot



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1.2.4 Geometric transformations

The geometrical transformations in Table 1.5and Table 1.6are needed to complete the geometry

of the stator.

Geometrictransformation

Comment Type of geometrictransformation

Firstpoint

Secondpoint

Scaling factor

SMIRROR Mirror image ofhalf stator slot

Affine transformation

with respect to a line

defined by 2 points

P1 P2 -1

(line symmetry)

Table 1.5: Mirror transformation.

Geometric

transformation

Comment Type of

geometric

transformation

Coordinate

system

R

comp.

Theta

comp.

Rot.

Z

SODUPLI Slot duplication Rotation defined

by angles and

pivot point

coordinates

STATWORK 0 0 90

SDUPLI Stator sideduplication

Rotation defined

by angles and

pivot point

coordinates

STATWORK 0 0 10

Table 1.6: Rotational transformations.

1.2.5 Completing the stator geometry

We will use the first transformation, SMIRROR, to duplicate half the stator slot, thusproducing the first stator slot. Using the menu: Choose Actions, Propagate, Propagate Lines.

We will propagate lines L1and L2using SMIRROR.

After that, we will create a line to close the outline of the stator slot by connecting points P5and P3. This line is a small arc based on the inner radius of the stator as it is indicated in Table

1.7.

Choose: Data, Add, Line from the menu:

Line Type of line Starting

point

End

point

Arc

radius

L5 Arc defined by its

radius, starting and

ending point

P5 P3 SID/2

Table 1.7: Line to close the outline.

Now that the slot is closed, that is, the points have been properly connected, the face of the slotcan be generated.



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Note:

Remember that faces are automatically generated in PREFLUX:

Click the Build Faces icon in the toolbar.

You should see the face of the first stator slot as shown below:

Figure 1.3: The face base geometry

Next, we need to modify the local coordinate system to make sure the stator slots will beproperly aligned. Using the data tree menu to modify STATWORKcoordinate System. Then

we choose a Cartesian coordinate system whose origin is [0, 0] and a rotation angle of 5

degrees.

Next we will apply the SDUPLI transformation 8times to duplicate the first stator slot andplace them in the proper positions along the inner outline of the stator.

The proper icon to propagate the faces is .

The building option Add faces and associated linked mesh generator should be selected.

Then, the stator geometry must be closed. A line must connect the upper leftpoint of thefirstslot and the bottom leftpoint of thesecondslot (for the type and the arc radius of the line see

Table 1.7).



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This line will be propagated 7times by SDUPLI transformation, and we obtain the following

display:

Figure 1.4: The slots after duplication

Then the outer stator edges must be created.We create first the bottom edge as the line connecting the two points (created in the coordinate

system STATMAIN) of Table 1.8.


P47 Sid/2 0

P48 Sod/2 0

Table 1.8: Extremities of the bottom edge.

This line will be propagated by SODUPLI. Then, the inner arc of the stators outline must be

completed connecting the bottom and top slots to the straight outer edges we have just created.

So we will connect points P47and P5 (arc radius = sid/2)at the bottom and points

P42and P49 (arc radius = sid/2)at the top. The menu to create the arcs is: Data,Add,Line.

Note:

Remember that arcs must be created in an anticlockwise direction, so be careful to choose

the points in the order shown below (P47and then P5; P42and P49).

Finally to create the outer arc of the stator, we will connect points P48and P50 by an arc

(arc radius =sod/2).



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1.3 Rotor geometry

In the same way as for the stator we present here the data needed for the rotor geometry.

1.3.1 Geometrical parameters

Rotor parameter

name

Comment Value (in millimeters)

RBHEIGHT Rotor bar height 18

RBTOPR Rotor bar top radius 2.75

RBBOTR Rotor bar bottom radius 1.15

ROD Rotor outer diameter 116.5

RBTOP Rotor bar top location 110.26

RID Rotor inner diameter 38

Table 1.9: Rotor parameters.

1.3.2 Coordinate systems

Name CommentType

of system

Coordinate

system of

definition

Type of

coordinates

X

coord

Y

coor

d

Z

rot.

ROTMAIN Main rotorcoordinate

system

2DGLOBAL

CARTESIAN 2D 0 0 0

ROTWORK Working system LOCAL ROTMAIN CARTESIAN 2D 0 0 0

ROTLOC Local system LOCAL ROTWORK CARTESIAN 2D RBTOP/2 0 0

Table 1.10: Rotor coordinate systems



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1.3.3 Points and lines for the rotor bar

The points are defined in the coordinates system ROTLOC.


P51 RBTOPR 0

P52 0 RBTOPR

P53 RBTOPR+RBBOTR-RBHEIGHT RBBOTR

P54 RBTOPR-RBHEIGHT 0

Table 1.11: Rotor points.

Line Type of line Starting point End point Arc radius

L59 Segment defined by 2points P53 P52

L60 Arc defined by its radius,

starting and ending point

P51 P52 RBTOPR

L61 Arc defined by its radius,starting and ending point

P53 P54 RBBOTR

Table 1.12: Lines of stator base geometry.

1.3.4 Geometric transformations

Geometric

transformation

Comment Type of geometric

transformation

First

point

Second

point

Ratio

RMIRROR Mirrortransformationfor rotor bar

Affine transformation with respect

to a line defined by 2 points

P51 P54 -1

Table 1.13: Mirror transformation.

Geometric

transformation

Comment Type of

geometrictransformation

Coordinate

system

R

comp.

Theta

comp.

Rot.

Z

ROTSIDE Rotor sideduplication

Rotation defined

by angles and

pivot point

coordinates

ROTMAIN 0 0 90

RDUPLI Bar duplication Rotation defined

by angles and

pivot point

coordinates

ROTWORK 0 0 90/7

Table 1.14: Rotational transformations.



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1.3.5 Completing the rotor geometry

After the creation of parameters, coordinate systems, points, lines and transformations we

complete the geometry as follows:

Use the first transformation, RMIRROR, to duplicate the half stator slot, thus producing thefirst stator slot. We will propagate lines L60, L59 and L61only once.

Generate the face of the rotor bar.

Then, your screen should display the rotor bar face and the stator as shown below:

Figure 1.5: The entire stator and a rotor bar

Modify the orientation of the rotor bar. Using the data tree menu to modify ROTWORKcoordinate System. Then we choose a Cartesian coordinate system which the origin is [0, 0]

and a rotation angle of 90/(7*2)] degrees.

Apply the RDUPLI transformation 6times to create duplicates of the first rotor bar.

The building option Add faces and associated linked mesh generator should be selected.



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Then, you should see the seven rotor bars in place as shown below:

Figure 1.6: Bars after duplication

Define points for rotor outlines. Define first points P93and P94which coordinates in thecoordinate system ROTMAINare (rod/2, 0) and (rid/2, 0). These two points will be

connected by a segment (L101), which is going to be transformed by ROTSIDE.

Define the inner and outer outlines of the rotor, which will be two arcs connecting points P94

and P96, and points P93and P95. The arc radius is respectively rid/2and rod/2.



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1.3.6 Closing the air-gap

To close the air-gap we will create two connecting segments, one between points P93and P47

and the second between points P49and P95.

Then we build the faces for the rotor and the air-gap:

If you click the Build Faces icon in the toolbar, you should see the complete geometry with

the 19 faces constructed, as we show below:

Figure 1.7: Complete geometry



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1.4 Add and assign regions for the faces

First, regions are created by entering names, comments (reflecting the material or source

properties, in this case) and colors. Using the menu : Choose Data, Add, Region Face.

Table 1.15 below indicates the name, comment and color to be entered for each region face of

our model.

Region Face Region Face

NameComment Color

1 RB_1 Aluminum Turquoise





6 RB_6 Aluminum Turquoise7 RB_7 Aluminum Turquoise

8 Rotor Iron Cyan

9 Airgap Moving airgap Yellow

10 SSA Plus A Red

11 SSB Plus B Magenta

12 SSC Minus C Turquoise

13 Stator Iron Cyan

Table 1.16: Regions of faces.

Then, regions must be assigned for the faces. The figure below shows which features of the

geometry are assigned to each named region face.

Figure 1.8: correspondence between faces and regions



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To assign faces to the regions we have created, from the Geometrymenu:

Open the Assign Regions dialog with the icon in the toolbar.

Finally, you should see the solid colored surfaces as shown in our figure below.

Figure 1.9: complete colored geometry



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1.5 Mesh

1.5.1 Change to the Mesh context

The Mesh commands are available only in the Mesh context.

Add the Mesh points

In our example we will create additional mesh points to default mesh points that exist in

PR LUX

. This will give us better control over the mesh density across the geometry.

The following Table contains information to define the six mesh points:

Name of mesh point Comment Value (millimeters)

MSSBOT Stator slot bottom 2

MRBBOT Rotor bar bottom 2.5

MRBTOP Rotor bar top 0.8

MAIRGAP Moving airgap 0.5

MSOD Stator outer diameter 9.5

MRID Rotor inner diameter 8

Table 1.17: Mesh points.

Using the menu: choose Data, Add, Mesh point.

1.5.2 Applying the mesh points to the geometry

The mesh points are assigned as presented in Table 1.17:

Name of

mesh point

Points

MSSBOT P4, P2, P6

MRBBOT P53, P54, P55

MRBTOP P52, P56MAIRGAP P51 and

all the points in RELATIONwith the Surfacic regionAIRGAP

MSOD P50, P48

MRID P94, P96

Table 1.18: Assignation of mesh points.



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1.5.3 Generate, verify and save the mesh

Mesh all lines of the geometry. Using the icon in the toolbar or choose Actions, Mesh Linesand Faces, Mesh the Lines

Generate the surface elements of the mesh. By using the same command choose Mesh Faces.

Then save your project with a name of representing your FLUX2D problem; for example weenter INDUCTION_PHYSIQUE.

The window seen will be as shown below:

Figure 1.10: complete colored mesh geometry



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FLUX PART A: FLUX MODELMaterials

2. Materials

The B(H) dependence of stator and rotor magnetic cores is tabulated in Table 2.1.

B (T) 0.50 1.10 1.60 1.70 1.85 2.00 2.10

H (A/m) 129.50 243.25 1850.00 3700.00 9900.00 22100.00 43000.00

Table 2.1: Points for B(H) curve

The FLUX2D scalar-spline model is represented in Figure 2.1. Scalar spline model allows us to

define B(H) curve starting from experimental values of B and H. This curve represents the

interpolation of the values presented in Table 2.1for the saturation value Js = 2.07 T.

Figure 2.1: B(H) curve of magnetic cores

Note:

The curve will be approximate by 2 values representing the saturation value Js = 2 T and

the relative slope a = 1100. This takes less time to calculate than the original curve.



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PART A: FLUX MODEL FLUXMaterials

The properties of the materials used in this paper are summarized in Table 2.2.

Material

name

Comment Property Model Value

Iron Nonlinear steel Isotropic

Permeability Iso_MU

B_scalar_a_sat Js = 2.0 T

a=1100

Copper Linear copper Isotropic resistivityIso_RHO

Scalar_cst 0.172.10-7m

Aluminu

m

Linear

aluminum

Isotropic resistivityIso_RHO

Scalar_cst 0.278.10-7m

Table 2.2: Materials properties



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FLUX PART A: FLUX MODELDefinition of the electrical circuit

3. Definition of the electrical circuit

The machine in our example is delta connected; its external circuit is shown below (seeFigure

3.1), with the data corresponding to different components (seeTable 3.1).The voltage sources

are defined as constant as they will be fully controlled from Simulink. Kirchoffs law will

deduce immediately the voltage of Phase C. Thus, there is no need to connect phase C with anexternal source.

Figure 3.1: Electrical circuit



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PART A: FLUX MODEL FLUXDefinition of the electrical circuit

Component

name

Type Model Data

VAC Voltage

source

Constant 134.35 V rms

VBA Voltage

source

Constant 134.35 V rms

PA Coil Total value Number of turns: 132

Resistance: 0.46557Ohm

PB Coil Total value Number of turns: 132


MC Coil Total value Number of turns: 132


Resis4 Resistor Constant 0.5575 Ohm



Induc7 Inductance Constant 0.0021 H



Squirrel_cage Squirrel cage Constant 7rotor bars

Ring resistance: 2.5.10-6

Ring inductance: 4.0.10-9H

Looping type of thedisplayed part: -1(anticyclic)

Table 3.1: Electrical components

Note:

The sources are defined as constant. Indeed, as it will be explained in chapter 2, the

machine will be controlled by the magnitude of the input voltages. There is no need then to

define the frequency or the phase of the sources



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FLUX PART A: FLUX MODELPhysical properties

4. Physical properties

4.1 General information

The case has a constant cross section (plane problem) with a depth of 145 mm. It is solved with

the magneto-transient application.

4.2 Assign materials to the regions

There are three materials that should be assigned to regions as follows:

Name of

the region

Material Property Model Data

SSA VACUUM Source External circuit

SSB VACUUM Source External circuit

SSC VACUUM Source External circuit

RB1 ALUMINUM Source External circuit







AIRGAP Y Rotational

air gap

Mechani

c values

Constant J = 0.02 Kg/m

F = 0.001 N.m

Number of pole pairs: 2

ROTOR

and

STATOR

IRON Source No source

Table 4.1: Materials and correspondent region



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PART A: FLUX MODEL FLUXPhysical properties

You will see the boundary conditions that are applied automatically:

Figure 4.1: boundary conditions (automatically assigned)

4.3 Electrical circuit

The different electrical components are described in the second chapter.



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FLUX PART B: SIMULINK MODELPhysical properties

PART B: SIMULINK MODEL



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PART B: SIMULINK MODEL FLUXScalar Control



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FLUX PART B: SIMULINK MODELScalar Control

5. Scalar Control

5.1 Principle of the scalar control

5.1.1 Introduction

The principle of this type of control is to adjust the motor speed Vs by varying the output

frequency fssuch as the magnetic state of the machine is about fixed, through the preservation of

sin a constant value, and such as the torque follows the wished law of variation according to

the speed. This type of command does not allow to control the electric and magnetic transients,

thus it must be operated only for laws of control with slow dynamics for which we can consider

that the machine keeps on electric and magnetic steady state.

The simplicity of this type of control favors its implementation in many industrial variators

conceived for these types of application. Indeed, it is characterized by a simplified structure, but

requiring a speed sensor and a speed controller for position servo-control.

5.1.2 Scalar control modeling

We will use the vector expressions of the asynchronous machine in a coordinate system, which is

bound to stator:

V R Id

dt

V R Id

dtj

S S SS

R R RR

R

= +

= = +

0 (1)

][Imag *RSmpelm IILp=

in an electrical sinusoidal steady state we obtain the following model :

+==

+=

RRSRRR

SSSSS

jjIRV

jIRV

0

(2)



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and with the relations :

RS = ,

+=

+=

RRSmR

RmSSS

ILIL

ILIL (3)

the model can be written also as :

++=

+=

RRSSRS

m

SSR

R

mS

SS

jjL

L

jL

LV

)1

(1

0

)(1

R

(4)

][Imag *RSRS

mpelm

LL

Lp =

to simplify the expressions we will introduce the following arguments:

)( RRArctg = , so2)(1

sin

RR

RR

+= and

2)(1

1

RR

+=cos

)( RRArctg = , so2)(1

sin

RR

RR

+= and

2)(1

1

RR+cos = (5)

and some relations can be deduced from the first model:

=

R

RRR

RjI , S

RR

RR

R

mR I

j

j

L

LI

+=

1or S

j

RR

RR

R

mR Ie

L

LI )2/(

2)(1

+

+= (6)

Sj

RR

mR Ie

L

+=

2)(1

also from the second model we can deduce:

S

j

RRS

mR e

L

L

+

=

2)(1

1 2

2

2

)(1

)(1

S

RR

RR

S

m

R

pelm

L

L

L

p

+

=

(7)

the last relation allows to calculateR

elm

d

d

which is zero for the value

R

R

1

= (8)

so we can deduce the maximal torque: 22max2

1)(

1)( S

S

m

R

pelmL

L

Lp =

(9)



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5.1.3 Relations used for the scalar control

The scalar control may allow to obtain, for a given flux s, a given torque in a given speed .

From the precedent expressions (8) we can establish the equation:

0])(1

[]) 2222 =+elmRRS

S

m

RRelmR L

L

Lp [( (10)

the solution of this equation is :

=

222

22

2

22

])(1

[

])[(411

])[(2

)(1

RS

S

m

R

p

elmR

elmR

RS

S

m

R

p

R

L

L

Lp

L

L

Lp

(11)

which gives the necessary pulsation for this functioning : += pRS (12)

and the current absorbed to obtain the torque and the flux desired :

Sj

RR

RR

S

S eL

I +

+= )(

2

2

)(1

)(11

(13)

so we can deduce the voltage to apply :

SSSSS jIR += V

These relations show that:

We can determine, for a given flux s, the necessary voltage, in order to obtain a wished

torque in a given frequency2

SSf = in the electrical steady state.

Neglecting the voltage drops R IS S we have

SSS j V (14)

so V SSSSS f = 2

and consequently:

constant2 == SS

S

f

V (15)

This explains the name given generally to this type of control.



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5.2 Structure of the Scalar control

The scalar control assures, in permanent speed, the flux module control. It is combine with an

automatic drive where:

ws = wr + wm (w is the angular electric speed in rad/sec)

ws is the stator pulsation

wr is the rotor pulsation

wm is the mechanical speed

The basic structure is represented on the following figure:

Bridge Filter Inverter

PWM

VS

s

m

ref

MAS

P

Figure 5.1: Structure of control with variable frequency in open-loop

One of the main difficulties is the determination of the values of the V s(fs) table for lowfrequencies in order to take into account correctly the influence of the resistive term because this

one varies rapidly with:

Is , that is with the desired torque in low speeds, and especially during start-up.

Rs ,that is with the thermal state of the machine.

With some approximations we can have: 1)( 2 +SS

SSSS

L

RV

(16)

This simple structure is satisfying only with slow dynamics because the law of variation of Vs isestablished in the steady state. To overcome this problem we can complete the precedent

structure with a speed loop (see Figure 5.2).



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We complete the precedent structure, with a speed loop.

The structure is represented on the following figure:

Bridge Filter Inverter

PWM

VS

S

Speed

controller

r

Ref

Pm

MAS

Figure 5.2: Structure of control in close-loop

Note:

The scalar control allows, in steady state, to minimize the input-current for a constant

torque.



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FLUX PART B: SIMULINK MODELSimulink model

6. Simulink model

6.1 Definition of the Simulink model

In the following is presented the whole Simulink model. The coupling with FLUX is detailed in

chapter 3.

Figure 6.1: Whole Simulink model

The model includes:

The command, which is a scalar control. (medium grey blocks) The induction machine. For this block we can use a model deduced from the equations of the

machine, use a model with S-functions or use the machine block of Simulink library. In our

example we will use the model based on S-functions. (dark grey block) in the Concordia

frame.

A filter to soften the reference values. A PI regulator for machines servo-control.

The outputs to be displayed. Some accessories to have specific measures (light grey blocks).



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PART B: SIMULINK MODEL FLUXSimulink model

6.2 Definition of the blocks

6.2.1 The command

This part controls the machine. It will control the value of the 3-phase voltage source of the

machine (i.e. the amplitude and the pulsation).

Figure 6.2: Command part of Simulink model

6.2.2 Subsystem of the Instruction block

Figure 6.3: Subsystem : Instruction

6.2.2.1 Generation of the pulsation

The blocks parameters of the pulsation is defined as follows:

The final value ws of the step block must be the value

of the reference pulsation. For example, ws=50rad/sec

so Ws=477rpm.

Figure 6.4: step block



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6.2.2.2 Filter

A filter is required to soften the reference value of the speed with the transfer function equal to:

110400

1

++=

ssF ;

Figure 6.5: Filter block

6.2.3 Speed controller

In this part, the use of PI controller is necessary for the asynchronous machine servo-control. In a

empirical way and with some computation tries, we can determine some PI regulators

coefficients. In the structure shown above, we can see the detail of speed controllers that ensure

the servo-control from the reference control variables.

With: P=25 and I=0.02

Figure 6.6: PI Regulator block



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6.2.4 Subsystem of the Scalar control

Inside the subsystem of scalar control, we have the following model, which use some defined

functions:

Figure 6.7: Subsystem scalar control

The stator pulsation of the induction machine is the result of the addition of the referencepulsation (or the rotor pulsation) and the mechanic pulsation.

In consequence, this command enslaves the stator pulsation to the motor angular velocity by

controlling the rotor pulsation wr. By the way of a gain representing the statoric flux we have a

proportional voltage amplitude (see Paragraph 1.1.3).

The value of the statoric flux is given by the program containing the parameters of the machine.

The defined functions used in the scalar control are in the following path in the Simulink library

browser:

Figure 6.8:the Simulink function Fcn



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6.2.5 The induction machines block:

It is a subsystem who has voltage sources as

inputs, and currents, flux, speed and

electromagnetic torque as outputs. The machine is

also connected to a load torque input.

Figure 6.9: The induction machines subsystem

In our example we use the model based on S-functions. Inside the inductions machine

subsystem we find the following blocks:

Figure 6.10: The interior of the induction machines subsystem



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We can see the mechanical system represented by the transfer function, and the S-function,

which use a Matlab program. The name of the S-function program and the machines parameters

used in this program must be mentioned in the S-functions block parameters as follows:

Figure 6.11: The S-function block parameters



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The Matlab program used masyn.m is the following is the same as in chapter B

% /////////////////////////////////////////////////////////////////////////////////

% /// Asynchronous machine in the Concordia frame///

% ////////////////////////////////////////////////////////////////////////////////

%

% Inputs :% u(1)=Vsalpha

% u(2)=Vsbeta

% u(3)=w=pOmega

%

% States :

% x(1)=Fsalpha

% x(2)=Fsbeta

% x(3)=Fralpha

% x(4)=Frbeta

%

% Outputs :% y=[Fsalpha,Fsbeta,Fralpha,Frbeta,Isalpha,Isbeta,Iralpha,Irbeta,Couple]

%

% Parameters :

% Rs,Ls,Rr,Lr,Lm,p

function[sys,x0,str,ts] = masyn(t,x,u,flag,Rs,Ls,Rr,Lr,Lm,p)

% some coefficients used afterward

a2=Rs*Lm/(Ls*Lr*sig);

a1=-Rs/Ls-Lm/Ls*a2;

b1=-Rr/Lr/sig;

b2=-b1*Lm/Ls;

switchflag,

% Initialization %

case0,

[sys,x0,str,ts]=mdlInitializeSizes;

% Derivatives %case1,

sys=mdlDerivatives(t,x,u,a1,a2,b1,b2);

% Outputs %

case3,

sys=mdlOutputs(t,x,u,Ls,Lr,Lm,sig,p);

case{ 2, 4, 5, 9 }

sys = [];



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%=============================================================================

% mdlInitializeSizes

% Return the sizes, initial conditions, and sample times for the S-function.

%=============================================================================

function[sys,x0,str,ts]=mdlInitializeSizes

sizes = simsizes;

sizes.NumContStates = 4;

sizes.NumDiscStates = 0;

sizes.NumOutputs = 9;

sizes.NumInputs = 3;

sizes.DirFeedthrough = 0; %0 because the input u doesn't intervene directly in the outputs computation

sizes.NumSampleTimes = 1; % at least one sample time is needed

sys = simsizes(sizes);

% initialize the initial conditions

x0 = [0 0 0 0];

% str is always an empty matrix

str = [];

% initialize the array of sample times

ts = [0 0]; % continuous time

% end mdlInitializeSizes

%

%=============================================================================

% mdlDerivatives

% Return the derivatives for the continuous states.

%=============================================================================

functionsys=mdlDerivatives(t,x,u,a1,a2,b1,b2)

x5=x(1);

x6=x(2);

x7=x(3);

x8=x(4);

dx5=a1*x5+a2*x7+u(1);

dx6=a1*x6+a2*x8+u(2);dx7=b1*x7+b2*x5-u(3)*x8;

dx8=b1*x8+b2*x6+u(3)*x7;

sys = [dx5 dx6 dx7 dx8];

% end mdlDerivatives

%

%=============================================================================



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6.2.6 Outputs

Figure 6.12: Outputs

We can visualize all outputs that we need by scopes or workspaces. The Scope allows a

simultaneous visualization, whereas the workspace allows the transfer of results from Simulink

to Matlab where they can be easily postprocessed. In Figure 6.12, we have an illustration of theuse of the two methods.

6.2.7 Other blocks

Figure 6.13: Selector block

Selector: Select or re-order the specified elements of an input vector or matrix. It is available in

the Simulink library browser.

Figure 6.14: Concordia subsystem block

Concordias transformation:performs the (a,b,c) to (,) transformation on a set of three-phase signals.



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FLUX PART B: SIMULINK MODELSolve

7. Solve

To initialise the Simulink model, we need a Matlab program which supplies the machines

parameters to the Simulink model. The parameters are computed in the Flux case.

The program is as follows:

disp(

data loaded) ;



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PART B: SIMULINK MODEL FLUXSolve

Before starting the solving, the simulation parameters should be defined:

Figure 7.1: Simulation parameters



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FLUX PART C: FLUX TO SIMULINK MODEL

PART C: FLUX TO SIMULINK MODEL



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PART C: FLUX TO SIMULINK MODEL FLUX



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FLUX PART C: FLUX TO SIMULINK MODELDefinition of the Simulink model

8. Definition of the Simulink model

8.1 Description of the Simulink model

The whole Simulink model looks as follows:

Figure 8.1: Simulink model of the coupling

The model includes:

A coupling with FLUX (2D) (V8) block: it refers to FLUX during the computation.

The command: it is the same command used in the Simulink model in part B. It commandsthe value of the voltage sources needed by the FLUX case. It will supply sinusoidal voltages.

It need only to supply the two first phases aand b because, as it is explained inparagraph

3 of Part A, the electrical circuit of the Flux case needs only two voltage sources, the third

phase will automatically have the correspondent value. Connect the third phase with a

terminator in order to avoid warning messages concerning unconnected output ports.

Moreover, a gain of is added to take into account the fact that only a quarter of the

machine is simulated.

A PI regulatorfor machines servo-control, whose value are P=2 and I=0.142857. The inputsmay also include a load torque of 20 N.m (will be used in the second part of the

results). The outputsto be displayed (see Part B).A gain to convert the speed from rpm to rad/sec is also added (its value is pi/30).



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PART C: FLUX TO SIMULINK MODEL FLUXDefinition of the Simulink model

8.2 Definition of the coupling block

This block enables a direct co-simulation with both FLUX (2D) and Matlab-Simulink. It is

available in the Simulink Library Browser, in the folderflux_link.

It is defined by: the TRA file name: it is the TRA file that will be solved, we must give its name without the

extension .TRA. In our example: Motor (or your problems name).

FLUX (2D) inputs: the voltage sources VBA and VAC should be defined as inputs to FLUX(2D). The syntax to use is described in the users guide.

[VOLTAGE:VBA; VOLTAGE:VAC]

Note:

The components names correspond to the name given to components in Table 3.1 of Part

A. Do not forget to check that it corresponds to your circuit.

FLUX (2D) outputs: the mechanical values are displayed (torque and angular velocity) as wellas the electrical values (current in phase 1, 2 and 3 and in the rotor bars 40 and 41).

[CURRENT:B1;CURRENT:B2;CURRENT:B3;TORQUE;OMEGA;CURRENT:BAR40;

CURRENT:BAR41]

the time step: 1e-3

the initial conditions: there is no initial conditions to set

the initialized by a static computation case must be ticked off

Figure 8.2: Coupling with FLUX2D block

Note:

A time step equal to 1ms is sufficient for 20steps per period because the computationfrequency is 50Hz.



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FLUX PART C: FLUX TO SIMULINK MODELSolve

9. Solve

Note:

There is no need to open FLUX (2D) to solve the case. The simulation can be handled

directly in Simulink. But the .TRA file must be in the same folder as the Simulink model anddata programs (.mdl and .m files).

The computation time step for FLUX (2D) must be the same has been defined in the Coupling

with FLUX (2D) block.

Before starting the solving, the computation range should be defined.

0.5

Figure 9.1: Simulation parameters



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PART C: FLUX TO SIMULINK MODEL FLUXSolve



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FLUX PART D: RESULTS

PART D: RESULTS



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PART D: RESULTS FLUX



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FLUX PART D: RESULTSSimulink Results

10. Simulink Results

Several simulations have been computed for various speed instructions; for 150rad/sec,

100rad/sec and 30rad/sec and for a given flux equal to 1.71wb in a given frequency, in order to

obtain constant2 == SS

S

f

V

,in the electrical steady state. The problem was computed with a PI

regulator with the following tuning: P=25; I=0.02

10.1 No load torque

10.1.1 150 rad per second

10.1.1.1 Mechanical quantities

Figure 10.1: Angular velocity and Electromagnetic torque



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PART D: RESULTS FLUXSimulink Results

10.1.1.2 Electrical quantities

Figure 10.2: Stator voltages

Figure 10.3: Stator currents

10.1.1.3 Comments

The first graph (Figure 10.1) shows the machine's speed reaching 150 rad per second and theelectromagnetic torque of the machine.Because the stator is fed directly by the voltage sources, a transient torque is observed.

However, this noise is not visible in the speed because it is filtered out by the machine's inertia,

but it can also be seen in the stator currents, which are observed above (Figure 10.3).

The second graph (Figure 10.2) shows that the range of the voltage sources evolves in thesame way than the speed transient s (t). If we superimpose them on a same scale, we obtain

an identical style, namely the ratio Vs/ s is constant.

Note:

As expressed in chapter B, we can see that the scalar control allows, in steady state,

minimising the input-current for a constant torque.



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10.2 With load torque

A simulation was computed with a load torque T=20N.m at0.7sec. The results have been

simulated only for 100rad per second. The torque value is defined on simulink interface.

10.2.1 Mechanical quantities


10.2.2 Comments

We can see the torque oscillations in reply to the applied load torque at t = 0.7seconds.

However, this oscillations were not very visible in the speed because it is filtered out by the

machine's inertia, but it can also be seen in the stator currents, which are plotted next (Figure

10.12).



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10.2.3 Electrical quantities





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FLUX PART D: RESULTSFLUX Results

11. FLUX Results

With Simulink, only the values defined as outputs will be displayed, whereas with FLUX, the

computation gives far more results. Indeed, all the quantities usually reachable with FLUX can

be displayed and computed, for example; the Spectrum of the flux density, the equiflux lines for

each time step, the position, etc.

Note:

Theses simulations were computed with PI regulators adjustment P=2 & I=0.142857.

11.1 No load torque


11.1.1.1 Magnetic quantities

Figure 11.1: Flux density distribution at time step 49msec



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PART D: RESULTS FLUXFLUX Results


Figure 11.2: Angular velocity (=1432rpm) & Electromagnetic torque


Figure 11.3: Stator voltage (VAC)



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11.1.2 100 Rad per second


Figure 11.5: Angular velocity(954rpm) & Electromagnetic torque



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Figure 11.6: Position


Figure 11.7: Stator voltage VAC



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11.1.3.1 Magnetic quantities

Figure 11.9: Equiflux lines for time step 0.1sec



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11.2 With load torque

A load torque equal to 20N.m is applied at t=1.2second.

11.2.1 Mechanical quantities at 150 rad per second




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Figure 11.14: Friction torque

11.2.2 Electrical quantities at 1500 rpm




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11.2.2.1 Comments

In first time, we notice that the range of curves in Simulink and FLUX are not very the same and

that because of the regulators tuning. Because FLUX take saturation and no-linear phenomena

into account we have made different regulators adjustment which allows a good enslavement.

Some fluctuations of speed are observed in transient speed for low reference.The lower the speed, the higher the torque oscillations. Then, it is more difficult to allows a good

control of the torque for slow dynamic speed.

The coupling with FLUX enables to see that a proper servo-control of the torque is more difficult

for lower speed while a simple SIMULINK model does not account properly for this difficulty.



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FLUX PART D: RESULTSConclusion

12. Conclusion

The scalar control of induction machines which gives some good results in steady state, is less

efficient during transient period accompanying variation speed.

This fact is also noticed at the starting period, when the motor is in low frequency speed and the

drop in voltage in the stators resistors are not negligible any more compared with fem.

Theses phenomena shows the interest of a direct co-simulation which permit to take into account

precisely overall of the magnetic effect (non-linear phenomena, Eddy currents)



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PART D: RESULTS FLUXConclusion

Documents

FLUX ® 2D Application Scalar command of an induction machine technical paper