FLUX ® 2D Application Scalar command of an induction machine technical paper
Transcript of 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
CEDRAT
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38246 Meylan Cedex
FRANCE
Tlphone: +33 (0)4 76 90 50 45
Tlcopie: +33 (0)4 56 380830
E-mail: [email protected]
Web: http://www.cedrat.com
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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
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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
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TABLE OF CONTENTS FLUX
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FLUX PART A: FLUX MODEL
PART A: FLUX MODEL
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PART A: FLUX MODEL FLUX
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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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FLUX PART A: FLUX MODELGeometry
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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PART A: FLUX MODEL FLUXGeometry
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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FLUX PART A: FLUX MODELGeometry
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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PART A: FLUX MODEL FLUXGeometry
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.
Point X coordinate Y coordinate
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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FLUX PART A: FLUX MODELGeometry
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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PART A: FLUX MODEL FLUXGeometry
1.3.3 Points and lines for the rotor bar
The points are defined in the coordinates system ROTLOC.
Point X coordinate Y coordinate
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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FLUX PART A: FLUX MODELGeometry
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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PART A: FLUX MODEL FLUXGeometry
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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FLUX PART A: FLUX MODELGeometry
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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PART A: FLUX MODEL FLUXGeometry
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
2 RB_2 Aluminum Turquoise
3 RB_3 Aluminum Turquoise
4 RB_4 Aluminum Turquoise
5 RB_5 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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FLUX PART A: FLUX MODELGeometry
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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PART A: FLUX MODEL FLUXGeometry
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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FLUX PART A: FLUX MODELGeometry
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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PART A: FLUX MODEL FLUXGeometry
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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
Resistance: 0.46557Ohm
MC Coil Total value Number of turns: 132
Resistance: 0.46557Ohm
Resis4 Resistor Constant 0.5575 Ohm
Resis5 Resistor Constant 0.5575 Ohm
Resis6 Resistor Constant 0.5575 Ohm
Induc7 Inductance Constant 0.0021 H
Induc8 Inductance Constant 0.0021 H
Induc9 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
RB2 ALUMINUM Source External circuit
RB3 ALUMINUM Source External circuit
RB4 ALUMINUM Source External circuit
RB5 ALUMINUM Source External circuit
RB6 ALUMINUM Source External circuit
RB7 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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PART B: SIMULINK MODEL FLUXScalar Control
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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FLUX PART B: SIMULINK MODELScalar Control
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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PART B: SIMULINK MODEL FLUXScalar Control
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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FLUX PART B: SIMULINK MODELScalar Control
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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PART B: SIMULINK MODEL FLUXScalar Control
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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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FLUX PART B: SIMULINK MODELSimulink model
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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PART B: SIMULINK MODEL FLUXSimulink model
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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FLUX PART B: SIMULINK MODELSimulink model
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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PART B: SIMULINK MODEL FLUXSimulink model
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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FLUX PART B: SIMULINK MODELSimulink model
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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PART B: SIMULINK MODEL FLUXSimulink model
%=============================================================================
% 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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FLUX PART B: SIMULINK MODELSimulink model
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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PART B: SIMULINK MODEL FLUXSimulink model
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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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FLUX PART D: RESULTSSimulink Results
10.1.2 100 rad per second
10.1.2.1 Mechanical quantities
Figure 10.4: Angular velocity and Electromagnetic torque
10.1.2.2 Electrical quantities
Figure 10.5: Stator voltages
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PART D: RESULTS FLUXSimulink Results
Figure 10.6: Stator currents
10.1.3 30 rad per second
10.1.3.1 Mechanical quantities
Figure 10.7: Angular velocity and Electromagnetic torque
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FLUX PART D: RESULTSSimulink Results
10.1.3.2 Electrical quantities
Figure 10.8: Stator voltages
Figure 10.9: Stator currents
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PART D: RESULTS FLUXSimulink Results
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
Figure 10.10: Angular velocity and Electromagnetic torque
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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FLUX PART D: RESULTSSimulink Results
10.2.3 Electrical quantities
Figure 10.11: Stator voltages
Figure 10.12: Stator currents
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PART D: RESULTS FLUXSimulink Results
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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
11.1.1 150 rad per second
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
11.1.1.2 Mechanical quantities
Figure 11.2: Angular velocity (=1432rpm) & Electromagnetic torque
11.1.1.3 Electrical quantities
Figure 11.3: Stator voltage (VAC)
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FLUX PART D: RESULTSFLUX Results
Figure 11.4: Stator currents
11.1.2 100 Rad per second
11.1.2.1 Mechanical quantities
Figure 11.5: Angular velocity(954rpm) & Electromagnetic torque
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PART D: RESULTS FLUXFLUX Results
Figure 11.6: Position
11.1.2.2 Electrical quantities
Figure 11.7: Stator voltage VAC
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FLUX PART D: RESULTSFLUX Results
Figure 11.8: Stator currents
11.1.3 30 rad per second
11.1.3.1 Magnetic quantities
Figure 11.9: Equiflux lines for time step 0.1sec
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PART D: RESULTS FLUXFLUX Results
11.1.3.2 Mechanical quantities
Figure 11.10: Angular velocity and Electromagnetic torque
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FLUX PART D: RESULTSFLUX Results
11.1.3.3 Electrical quantities
Figure 11.11: Stator voltage VAC
Figure 11.12: Stator currents
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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
Figure 11.13: Angular velocity and Electromagnetic torque
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FLUX PART D: RESULTSFLUX Results
Figure 11.14: Friction torque
11.2.2 Electrical quantities at 1500 rpm
Figure 11.15: Stator voltage VAC
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PART D: RESULTS FLUXFLUX Results
Figure 11.16: Stator currents
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