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OTI, STEPHEN EJIOFOR
REG. NO: PG/Ph.D/07/42465
THERMAL MODELLING OF INDUCTION MACHINE USING THE LUMPED PARAMETER
MODEL
FACULTY OF ENGINEERING
DEPARTMENT OF ELECTRICAL ENGINEERING
Ebere Omeje Digitally Signed by: Content manager’s Name DN : CN = Webmaster’s name O= University of Nigeria, Nsukka OU = Innovation Centre
ii
THERMAL MODELLING OF INDUCTION THERMAL MODELLING OF INDUCTION THERMAL MODELLING OF INDUCTION THERMAL MODELLING OF INDUCTION MMMMACHINE ACHINE ACHINE ACHINE
USING THE LUMPED PARAMETER MODELUSING THE LUMPED PARAMETER MODELUSING THE LUMPED PARAMETER MODELUSING THE LUMPED PARAMETER MODEL....
BYBYBYBY
OTI, STEPHEN EJIOFOROTI, STEPHEN EJIOFOROTI, STEPHEN EJIOFOROTI, STEPHEN EJIOFOR
REG. NO: PG/PH.D/07/42465REG. NO: PG/PH.D/07/42465REG. NO: PG/PH.D/07/42465REG. NO: PG/PH.D/07/42465
DEPARTMENT OF ELECTRICAL ENGINEERINGDEPARTMENT OF ELECTRICAL ENGINEERINGDEPARTMENT OF ELECTRICAL ENGINEERINGDEPARTMENT OF ELECTRICAL ENGINEERING
UNIVERSITY OF NIGERIA, NSUKKAUNIVERSITY OF NIGERIA, NSUKKAUNIVERSITY OF NIGERIA, NSUKKAUNIVERSITY OF NIGERIA, NSUKKA
DECDECDECDECEEEEMBERMBERMBERMBER, 20, 20, 20, 2011114444....
SUPERVISORS: PROF. M. U. AGU & PROF. E. C. EJIOGU
iii
THERMAL MODELLING OF INDUCTION MACHINE USING THE LUMPED PARAMETER MODEL
A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE
REQUIREMENT FOR THE AWARD OF DOCTOR OF PHILOSOPHY
(Ph.D) DEGREE IN ELECTRICAL ENGINEERING DEPARTMENT,
UNIVERSITY OF NIGERIA, NSUKKA
BY
OTI, STEPHEN EJIOFOR
REG. NO: PG/Ph.D/07/42465
UNDER THE SUPERVISION
OF
ENGR. PROF. M. U. AGU & ENGR. PROF. E. C. EJIOGU
DEPARTMENT OF ELECTRICAL ENGINEERING
UNIVERSITY OF NIGERIA, NSUKKA
DECEMBER, 2014.
iv
TITLE PAGE
THERMAL MODELLING OF INDUCTION MACHINE USING THERMAL MODELLING OF INDUCTION MACHINE USING THERMAL MODELLING OF INDUCTION MACHINE USING THERMAL MODELLING OF INDUCTION MACHINE USING
THE LUMPED PARAMETER MODELTHE LUMPED PARAMETER MODELTHE LUMPED PARAMETER MODELTHE LUMPED PARAMETER MODEL
v
APPROVAL PAGE
THERMAL MODELLING OF INDUCTION MACHINE USING THE LUMPED PARAMETER MODEL
By
Oti, Stephen Ejiofor. Reg. No: PG/Ph.D/07/42465
DECEMBER, 2014
A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE AWARD OF DOCTOR OF PHILOSOPHY
(Ph.D) DEGREE IN ELECTRICAL ENGINEERING DEPARTMENT, UNIVERSITY OF NIGERIA, NSUKKA
Oti, Stephen Ejiofor: Signature……………. Date………… (Student) Certified by: Engr. Prof. M.U. Agu Signature……………..Date…………. (Supervisor I) Engr. Prof. E. C. Ejiogu Signature………………Date………... (Supervisor II) Accepted by: Engr. Prof. E. C. Ejiogu Signature……………..Date………... (Head of Department) Engr. Prof. E.S. Obe Signature……………..Date………... (PG Faculty Rep.) Engr. Prof. O. I. Okoro Signature………………Date……….. (External Examiner)
vi
CERTIFICATION PAGE
I hereby certify that the work which is being presented in this thesis entitled,
“Thermal Modelling of Induction Machine Using the Lumped Parameter Model”, in
partial fulfillment of the requirements for the award of Doctor of Philosophy (Ph.D)
Degree (Electric Machines & Drives) in the Department of Electrical Engineering,
University of Nigeria, Nsukka is an authentic record of the research carried out under
the supervision of Engr. Prof. M.U. Agu and Engr. Prof. E. C. Ejiogu except where
due reference has been made in the work. Therefore, opinions and assertions
contained herein are those of the authors as they are indicated on the reference pages.
The work embodied in this thesis has not been submitted for the award of any degree
of any other University.
Oti , Stephen Ejiofor: Signature……………. Date………… (Student) This is to certify that the above statement made by the candidate is correct and true to the best of my knowledge. Engr. Prof. M.U. Agu Signature……………..Date…………. (Supervisor I) Engr. Prof. E. C. Ejiogu Signature………………Date………... (Supervisor II) Accepted by: Engr. Prof. E. C. Ejiogu Signature……………..Date………... (Head of Department) Engr. Prof. E.S. Obe Signature……………..Date………... (PG Faculty Rep.) Engr. Prof. O. I. Okoro Signature………………Date……….. (External Examiner)
vii
DEDICATION
To godfathers that have the fear of God in them.
viii
ACKNOWLEDGEMENT
I am heartily thankful to my supervisors, Engr. Prof. M.U. Agu and Engr. Prof. E.C. Ejiogu whose encouragement, guidance and support enabled me to develop an understanding of the subject.
I would like to express my profound gratitude to Ven. Prof. T.C. Madueme, Prof. L.U. Anih and Dr. B.O. Anyaka for their warm advice and useful contributions, all towards making this work a success.
At the early stage of this work, and all the way from Germany, Dr. E.S. Obe (now Professor) bombarded me with journal materials that I had more than I needed. This similar feat was repeated of recent by Engr. Chukwuemeka Awah who travelled out for his doctoral programme. May God reward you abundantly.
I owe my deepest gratitude to Professor O.I. Okoro, who has been with me physically and spiritually since the inception of this work, if it gives a farmer joy as the planted seeds sprout, how much is expected of men builder in the person of Prof. Okoro?
I am indebted to many of my colleagues: Engrs. Nwosu, Nnadi, Odeh, Ogbuka, Mbunwe and Ani who have shared with me or supported me in one way or the other to make or mar me. May God bless all of them.
It is an honour for me to thank the men at the laboratory unit- Mr. Okafors, Okoro, Abula, Azu , Eze and Chi for their usual cooperation. Emeka Omeje is also remembered for his prompt response when his attention is needed by me. Many thanks to my friends: Hacco, Chika, Okpoko, Chibuzo, Steve Agada, Alex, Simon, Ejor, Moses, Emma Obollor, Amoke and Engr. Agbo of Mechanical Engineering.
At this juncture, I have to thank my people; brother Mike, sister Uche, Uncle Emma, Amara and Princess for enduring with us until now that God has chosen, and to Him be the Glory.
Lastly, I offer my regards and blessings to all of those who supported me in any respect during the completion of the project.
ix
Abstract
Temperature rise is of much concern in the short and long term
operations of induction machine, the most useful industrial work icon.
This work examines induction machines mean temperatures at the
different core parts of the machine. The system’s thermal network is
developed, the algebraic and differential equations for the proposed
models are solved so as to ascertain the thermal performances of the
machine under steady and transient conditions. The lumped parameter
thermal method is used to estimate the temperature rise in induction
machine. This method is achieved using thermal resistances, thermal
capacitances and power losses. To analyze the thermal process, the
7.5kW machine is divided geometrically into a number of lumped
components, each component having a bulk thermal storage and heat
generation and interconnections to adjacent components through a
linear mesh of thermal impedances. The lumped parameters are derived
entirely from dimensional information, the thermal properties of the
materials used in the design, and constant heat transfer coefficients.
The thermal circuit in steady-state condition consists of thermal
resistances and heat sources connected between the components nodes
while for transient analysis, the thermal capacitances were used
additionally to take into account the change in internal energy of the
body with time. In the course of the simulation using MATLAB, the
response curves showing the predicted temperature rise for the
induction machine core parts were obtained. To find out the effect of
the decretization level on the symmetry, the two different thermal
models, the SIM and the LIM models having eleven and thirteen nodes
respectively were considered and the results from the two models were
compared. The resulting predicted temperature values together with
other results obtained in this work provide useful information to
designers and industries on the thermal characteristics of the induction
machine.
x
TABLE OF CONTENTS
Title page ………………………………………………………....…………….....….iii
Approval page ………………………………………………………………....….…..iv
Certification page………………………………………………………….…..….…...v
Dedication page…………………………………………………………….…..….…..vi
Acknowledgement………………………………………………………….…..….….vii
Abstract……………………………………………………………………..…..….….viii
Table of contents…………………………………………………….….……..….…...ix
List of figures…………………………………………………………….……..……..xii
List of tables………………………………………………………….…….…..….….xiv
List of symbols…………..……………………………………….……………..……..xv
Chapter One: INTRODUCTION ………………………………………………..….…..1
1.1 Background of study…………….……………….…………………………....…1
1.2 Statement of Problem …………………..……….…………….……….….….…3
1.3 Purpose of Study ………………………………..……………..........................3
1.4 Significance of Study …………………………...…………….………………....4
1.5 Scope of Study..…………………………………...….…..................................5
1.6 Arrangement of Chapters ……..………………………………….…................5
Chapter Two: LITERATURE REVIEW …………………………….……………….....6
Chapter Three: HEAT TRANSFER MECHANISMS IN ELECTRICAL MACHINES
3.1 Heat Transfer in Electrical Machines…………….……………….….…......12
3.2 Modes of Heat Transfer …………………..……………..…….………….…13
3.2.1 Conduction ………………………………………………….………...............14
3.2.2 Convection ……………………………………………...……………………..16
3.2.3 Radiation …………………..…………………………………….…................18
3.3. Heat Flow in Electrical Machines ………………….…………..……..…..…20
3.3.1 Heat Transfer Flow Types …………………………………………………..20
xi
3.3.2 Heat Transfer Flow System …………………………………..……….……..21
3.3.3 The Boundary Layers……………………………………...……….…………22
3.4 Determination of Thermal Conductance…………………….......….……....23
3.5 Thermal-Electrical Analogous Quantities ………………………….….……25
3.5.1 Thermal and Electrical Resistance Relationship …………….….…..…….26
Chapter Four: THERMAL MODEL DEVELOPMENT AND PARAMETER COMPUTATION
4.1 Cylindrical Component and Heat Transfer Analysis…………….………......28
4.2 Conductive Heat Transfer Analysis in Induction Motor ………….…….…...28
4.3 Convective Heat Transfer Analysis in Induction Motor………….…….…....34
4.4 Description of Model Components and Assumptions …………….…….….35
4.5 Calculation of Thermal Resistances…………………………….………...….45
4.6 Calculation of Thermal Capacitances ………………………..…………....…56
Chapter Five: LOSSES IN INDUCTION MACHINE
5.1 Determination of Losses in Induction Motors .…………………….…........69
5.1.1 Stator and Rotor Copper Losses ……………………..…………….…….. 69
5.1.2 Core Losses …………………………………………….…….……..….……70
5.1.3 Friction and Windage Losses ………………………….………..………….70
5.1.4 Differential Flux Densities and Eddy-Currents in the Rotor Bars ………..71
5.1.5 Stray-Load Losses …………………………………………………………....72
5.1.6 Rotor Copper Losses ……………………………………………...…….…...72
5.1.7 No Load Losses …………………………………………….…………….…..73
5.1.8 Pulsation Losses ……………………………………………………………...74
5.2 Calculation of Losses from IM Equivalent Circuit…………………………..74
5.3 Loss Estimation of the 7.5 kW Induction machine ….….………………....79
5.4 Segregation and Analysis of the IM Losses……… …………………........82
5.5 Performance Characteristics of the 10 HP Induction machine…..…….....83
5.6.1 Motor Efficiency /Losses ……………………….……………………….......86
xii
5.6.2 Determination of Motor Efficiency ……………………..……...….…….......86
5.6.3 Improving Efficiency by Minimizing Watts Losses ……………………......87
5.7 The Effects of Temperature ……………………………..….…….…...........88
Chapter Six: THERMAL MODELLING AND COMPUTER SIMULATION
6.1 The Heat Balance Equations …………………………………................…...90
6.2 Thermal Models and Network Theory ……………...……………….….....…90
6.3 The Transient State Analysis ……………………………….………....……...98
6.4 The Steady State Analysis …………………………………………………..104
6.5 Transient State Analysis results.………………….….……...………..……..108
6.6 Discussion of Results …………………………….…………...………..…….116 Chapter Seven: CONCLUSION AND RECOMMENDATION
7.1 Conclusion…………………….……………….….…………...………..…….118 7.2 Recommendation …………….………………….…………...………..…….119 REFERENCES …………………………………………………………………..…..….…..120 APPENDIX……...………………………………………..……………….……..……….…..131
xiii
LIST OF FIGURES
Figure 3.1: Illustration of Fourier’s Conduction Law 15
Figure 3.2: Illustration of Newton’s law of cooling 16
Figure 3.3: Simplified diagram for the illustration of thermal and
electrical resistance relationship 26
Figure 3.4: Simplified diagram for further illustration of thermal and
electrical equivalent resistance 27
Figure 4.1: Heat transfer mechanism in squirrel cage IM 28
Figure 4.2: General cylindrical component 28
Figure 4.3: Conductive Thermal circuit- An annulus ring 29
Figure 4.4: Three terminal networks of the axial and radial networks 30
Figure 4.5: The combination of axial and radial networks for a symme-
trically distributed temp about the central radial plane. 32
Figure 4.6: Squirrel Cage Induction Machine Construction 36
Figure 4.7: The geometry of High Speed Induction Machine 36
Figure 4.8: The geometry of Induction Machine rotor teeth 38
Figure 4.9: Squirrel Cage Rotor 41
Figure 4.10: Thermal network model for the Induction machine 43
Figure 4.11: Thermal resistance of air-gap between insulation and iron 45
Figure 4.12: Thermal resistance between the stator iron and the yoke 47
Figure 4.13: Thermal resistance between stator iron and end-winding 49
Figure 4.14: Thermal resistance between Rotor Bar and end ring 50
Figure 4.15: Thermal capacitance for Stator Lamination 56
Figure 4.16: Thermal capacitance for stator iron 57
Figure 4.17: Thermal capacitances for end winding 59
Figure 4.18: Thermal capacitances for rotor iron 61
Figure 4.19: Thermal capacitances for the Rotor bar 63
Figure 4.20: Thermal Capacitance for the Various Rotor-Bar Sections 64
xiv
Figure 4.21: Thermal capacitances for the End rings 65
Figure 5.1. Equivalent Circuit of the AC induction Machine 75
Figure 5.2. Simplified Equivalent Circuit of the AC induction Machine 75
Figure 5.3. IEEE Equivalent Circuit of the AC induction Machine 76
Figure 5.4. Bar chart for loss segregation of 10HP induction machine 82
Figure 5.5. Graph of Torque-Speed characteristics for 10HP IM 83
Figure 5.6. Power against speed for 10HP induction machine 83
Figure 5.7. Stator current against Speed for 10HP IM 84
Figure 5.8. Graph of Torque-Slip characteristics for 10HP IM 84
Figure 5.9. Power factor against speed for 10HP IM 85
Figure 6.1:Transient Thermal model of SCIM with lumped parameter 91
Figure 6.2:Steady State Thermal model of SCIM with lumped parameter 91
Figure 6.3:Thermal network model for the SCIM (SIM Half Model) 95
Figure 6.4:Thermal network model for the SCIM (LIM Full Model) 97
Figure 6.5: Percentage difference in component steady state temperature for the half and full SIM model 107 Figure 6.6: Percentage difference in component steady state temperature for the half and full LIM model 107
Figure 6.7:Response curve for the predicted temp-(SIM Half Model) 108
Figure 6.8: Response curve for the predicted temp-(SIM Half Model contd.)109
Figure 6.9: Graph for predicted temp and symmetry-(SIM Full Model) 110
Figure 6.10:Response curve for predicted steady state temp for LIM 111
Figure 6.11:Response curve for predicted temp - (LIM Model contd.) 112
Figure 6.12: Graph for predicted steady state temp rise for LIM contd. 113
Figure 6.13: Curves to show symmetry in end-ring of LIM model 114
Figure 6.14: Graph for predicted temp and symmetry-(LIM Full Model) 115
xv
LIST OF TABLES
Table 3.1: Thermal conductivities of some materials at room conditions 15
Table 3.2: Emissivity of some materials at 300K 19
Table 3.3: Thermal-Electrical Analogous Quantities 26
Table 4.1: Machine geometric / Dimensional data 44
Table 4.2: Thermal capacitances and thermal resistances from circuit 68
Table 5.1: Induction machine ratings and parameters 79
Table 5.2: Loss Segregation Obtained from Calculation 82
Table 5.3: Efficiency improvement schemes 88
Table 6.1 Steady State predicted temp for different models 106
xvi
LIST OF SYMBOLS
A area [m2]
bA cross-sectional area of rotor bar [m2]
CuA copper area in a stator slot [m2]
rA cross-sectional area of a rotor end ring [m2]
maxB maximum value of the flux density [T]
b thickness or width [m]
rbb width of rotor bar [m]
dsb stator tooth width [m]
drb rotor tooth width [m]
C heat capacity [J/kg.K]
cuC heat capacity of copper [J/kg.K]
d diameter or thickness [m]
ad air pocket thickness [m]
id slot insulation thickness [m]
f frequency [Hz]
FRIwin friction and windage loss
rG Grashof number
g acceleration due to gravity [m/s2]
h height [m]
ch convective heat transfer coefficient [W/m2K]
yh height of yoke
rbh height of rotor bar [m]
0I no load current [A]
sI stator current [A]
rI rotor current [A]
Fk eddy current loss factor
Hyk hysteresis loss factor
mL magnetizing inductance [H]
avl average conductor length of half a turn [m]
slotL entire slot length [m]
barL length of rotor bar [m]
L stator core length [m]
LIM large induction machine
rl length of a rotor end ring segment [m]
xvii
sL leakage inductance of stator [H]
m phase number of motor phases
sN speed of rotating magnetic flux [rad/s]
rL leakage inductance of rotor [H]
mL magnetizing inductance [H]
M mass [kg]
Nu Nusselt number
P power [W]
cusP resistive losses in the stator winding [W]
curP resistive losses in the rotor winding [W]
FesP stator core losses [W]
fwP losses due to friction and windage [W]
strP stray losses in the rotor [W]
outP output power [W]
inP input power [W]
rP Prandtl number
P number of poles per phase
p number of pole pairs
rN rotor slot number
sN stator slot number
q heat flux [W/m2]
cR core loss resistance [Ω]
thR thermal resistance [K/W]
Re Reynolds number
ROTcuL rotor copper loss [W]
ROTaL rotational loss [W]
mR iron (core) loss resistance [Ω]
sR stator resistance [Ω]
rR rotor resistance [Ω]
inr inner radius of tooth [m]
outr outer radius of tooth [m]
r radius [m]
δr average radius of the air gap [m]
s slip
SIM small induction machine
STAcore stator core loss [W]
STAcuL stator copper loss [W]
xviii
SCIM squirrel cage induction machine
T temperature [oC, K]
qrT torque [Nm]
maxT maximum temperature [oC]
shT shaft torque [Nm]
∞T reference temperature [oC],
pitchT tooth pitch [m]
T∆ temperature drop over the air gap [K]
t time [s] TNM thermal network model
2ν kinematic viscosity [m2/s]
sV Voltage [V]
phV phase voltage [V]
ThV Thevenin voltage [V]
tC thermal capacitance matrix
cuρ density of cooper [Kg/m3]
tG thermal conductance matrix
tP loss vector [W]
tθ temperature vector [K]
θ angle between phase voltage and current [degrees]
α heat transfer coefficient [W/m2K]
β volume coefficient of expansion [1/K]
∆ sheet thickness [m]
slotδ air gap of slot [m]
0δ air gap [m]
sagλ stationary air-gap film coefficient
ragλ rotating air-gap film coefficient
ε emissivity
fT temperature rise of the frame [K]
Feλ thermal expansion coefficient of iron [1/K]
ck thermal conductivity [W/m.K]
airk thermal conductivity of air [W/m.K]
insk thermal conductivity of the slot insulation [W/m.K]
sk thermal conductivity of the slot material [W/m.K]
xk
thermal conductivity in x direction [W/m.K]
µ dynamic viscosity [kg/m.s]
xix
0µ permeability of free space [Vs/Am]
υ kinematic viscosity [m2/s]
ρ density [kg/m3]
eρ resistivity [Ωm]
σ Stefan-Boltzmann’s constant [W/m2k
4]
ω angular speed [rad/s]
sR stator resistance [Ω]
IM induction machine
sI stator current [A]
rI rotor current [A]
sV per phase supply voltage of stator [V]
olV volume
rE opposition emf of the rotor [V]
mI magnetizing current [A]
mX magnetizing reactance [Ω]
LsX leakage reactance of the stator [Ω]
rLX leakage reactance of the rotor [Ω]
sX stator leakage reactance [Ω]
mX magnetizing reactance [Ω]
rX rotor reactance [Ω]
sZ stator impedance [Ω]
rZ rotor impedance [Ω]
xx
CHAPTER ONE
INTRODUCTION
1.1 Background of Study
This thesis is concerned with the thermal modelling of the
induction machine. With the increasing quest for miniaturization,
energy conservation and efficiency, cost reduction, as well as the
imperative to exploit easier and available topologies and materials,
it becomes necessary to analyze the induction machine thermal
circuit to the same tone as its electromagnetic design. This would
help in achieving an early diagnosis of thermo-electrical faults in
induction machines, leading to an extensively investigated task
which pays back in cost and maintenance savings. Since failures
in induction machines occur as a result of aging of the machine
itself or from severe operating conditions then, monitoring the
machine’s thermal condition becomes crucial so as to detect any
fault at an early stage thereby eliminating catastrophic machine
faults and avoidance of expensive maintenance costs. Faults in
induction machines can be broadly classified into thermal faults,
electrical faults and mechanical faults. Currently, stator electrical
faults are mitigated by recent improvements in the design and
manufacture of stator windings. However, in case of machine
driven by switching power converters the machine is stressed by
voltages including high harmonic contents. The latter option is
becoming the standard for electric drives. A solution is the
development of vastly improved thermal system cum insulation
material. On the other side, cage rotor design is receiving slight
modifications, apart from that, rotor bars breakage can be caused
by thermal stress, electromagnetic forces, electromagnetic noise
and vibration, centrifugal forces, environmental stress, for example
xxi
abrasion of rotor, mechanical stress due to loose laminations,
fatigue parts, bearing failure, e.t.c.
In the design of the induction machine, the manufacturers take
many factors into consideration to ensure that it works efficiently.
One of the most important factors in the design of an induction
motor is its thermal limits for different operating conditions because
if a machine works beyond its thermal limit for a prolonged time,
the life span of the machine is reduced.
The lumped-parameter thermal method is the most popular
method used to estimate the temperature rise in electrical
systems. The thermal model is based on thermal resistances,
thermal capacitances and power losses. To analyze the thermal
process, the electrical system is divided geometrically into a
number of lumped components, each component having a bulk
thermal storage and heat generation and interconnections to
flanking components through a linear mesh of thermal
impedances. It may be a simple network as demonstrated in [1] or
may have many tens of nodes. For any given configuration, the
designer looks for a matching design tool for the analysis. Motor-
Cad is a design tool used by some authors in [2-3] for thermal
analysis of electrical motors. This design tool gives a detailed
model, based on the geometry and the type of the motor. It was
predominantly used to analyze the parameter sensitivity of the
thermal models. In [4], D. A. Staton et al also used Motor-Cad to
determine the optical thermal parameters for electrical motors.
Here in, the thermal circuit is solved in matlab as is the case in [5]
through a system of linear equations.
The lumped parameters are derived from entirely dimensional
information, the thermal properties of the materials used in the
xxii
design, and constant heat transfer coefficients. The thermal circuit
in steady-state condition consists of thermal resistances and heat
sources connected between the components nodes while for
transient analysis, the heat thermal capacitances are used
additionally to take into account the change in internal energy of
the body with time. The associated equivalent thermal network,
would have the heat generation in the component concentrated in
its midpoint. This point represents the mean temperature of the
component.
1.2 Statement of Problem
The main limiting factor for how much an electric machine can
continuously be loaded is usually the temperature. When a
machine exceeds its thermal limit there are various outcomes: The
oxidation process in insulation materials is accelerated, which
eventually leads to loss of dielectric property. Bearing lubricants
may deteriorate or the viscosity may become too high, resulting in
reduced oil film thickness. Other problems are mechanical stress
and changes in geometry caused by thermal expansion of the
machine elements. Statistics show that despite the reliability of the
induction machine, there is a little annual failure rate in the
industries and from research it has been shown that most of the
failures are caused by extensive heating of different motor parts
involved in the machine operation.
1.3 Purpose of Study
The objectives of this research work include:
To study the various parts or components of the induction machine;
xxiii
To study the thermal behaviour or temperature limits of the
induction machine and its components under various operating
conditions;
To review the losses and methods of heat transfer in the induction
machine;
To develop an accurate thermal model for an induction machine;
To predict the temperature in different parts of the induction
machine using the thermal model and software program and lastly,
To investigate how the machine symmetry is affected by the nodal
configuration.
1.4 Significance of Study
The essence of this research work is to develop a thermal
model for an induction machine that will enable the prediction of
temperature in different parts of the machine. This is very
important first to the manufacturer or designer of an induction
machine because with these predictions one can decide on the
insulation class limits the machine belongs to. Also modern trends
in the construction of machines is moving in the direction of
making machines with reduced weights, costs and with increased
efficiency. In order to achieve this, the thermal analysis becomes
very crucial in deciding on what types of insulators and other
materials that would be used to make these machines.
In industries, the knowledge of the thermal limits of machines
increases the life span of their machines and reduces downtime;
thereby increasing production and profit. Finally, it is hoped that
this work would be an important tool for other researchers who
may desire to carry out further work in this topic or similar topics.
xxiv
1.5 Scope of Study
This research work reviews the thermal characteristics of the
induction machine in general and focuses on the thermal modelling
of totally enclosed natural ventilated induction machine.
1.6 Chapter Arrangement
Chapter one introduced the work by presenting the background of
the study and the statement of problem. The purpose, significance
and scope of the work were also presented in this chapter.
Chapter two exclusively took care of the literature review while in
Chapter three, the heat transfer mechanisms in electrical
machines were discussed. The thermal model development and
parameter computation were treated in Chapter four. It involved
the conductive and convective heat transfer analyses and details
of the calculation of the thermal resistances and capacitances.
In Chapter five, the losses in induction machine were discussed
while in Chapter six, the thermal modelling and computer
simulation were carried out, the simulation results were also
presented in this chapter. Lastly, Chapter seven was presented in
the form of conclusion and recommendations.
xxv
CHAPTER TWO
LITERATURE REVIEW
An electrical machine is said to be well designed when it exhibits
the required performance at high efficiency with operation within the
range of the maximum allowed temperature. Several motors used
in industrial applications rely on electromechanical or thermal
devices for protection in the overload range [6] but thermal
overheating and cycling degrade the winding insulation which
results in the acceleration of thermal ageing. The consequence is
insulation failure which eventually leads to motor failure. Presently,
there is high reliability on thermal motor protection schemes using
the thermal devices or the microprocessor embedded thermal
models, all of which are based on the thermal heat transfer model
of the induction machine.
The analysis of the heat transfer process is usually achieved
by choosing an idealized machine geometry. It is then carefully
divided into the fundamental elements and characterized by a node,
thermal resistance, thermal capacitance and a heat source. In
describing the fundamental elements, much about the machine
construction cum the thermal properties of the materials used have
to be known. A careful division of the machine into several parts
gives a better result but poses a great deal of complexity in the
computation task; this may have informed the suggestion of [7] that
a compromise between a detailed model and an oversimplified one
must be reached as the former can be very cumbersome to use
both in computer simulation and software development.
In the market today, there exist many general purpose advanced
computational fluid dynamic (CFD) packages. The CFD codes are
designed using sophisticated and modern solution technology to
xxvi
enhance the handling of high demanding cases of thermal
modelling
of flow system whether external or internal. The electrical machine
manufacturers have depended on this to a large extent especially
in the cooling and ventilation modelling [8] and in the thermal
management of alternating current electrical motors [9].
The thermal network models, (TNM) [10, 9] popularly called
the lumped parameter model is one of the schemes adopted in
studying thermal models for the determination of rise in
temperature in electrical machines.
The finite-element method (FEM) is another scheme used in
the determination of the temperature rise in electrical machines,
and also in analyzing the thermal behavior of electrical machines.
Many researchers [12, 13] have adopted this rather later method in
one way or the other.
A number of thermal circuits of induction motors [14, 15], radial flux
[16], stationary axial flux generators [17] and many others that
have been proposed in the past were all studied using the lumped
parameter model (LMP) approach and the results so obtained
suggest a good agreement with the experimental data.
Here in, the thermal network model, that is, the lumped
parameter model approach is adopted. The lumped parameters
are derived entirely from the dimensional information, the thermal
properties of the materials used in the design and the constant
heat transfer coefficients. This translates to high level adaptability
to various frame sizes.
The calculations of the parameter values arising from this lumped
arrangement are comparatively complex and result in sets of
thermal equations which mathematically describe the machine in
xxvii
full and which can be solved and adapted for online temperature
monitoring for many applications including motor protection [11, 14,
18, 19].
The above approach is better in that it saves one the hurdles
involved in the solution of heat conduction by Fourier analysis
approach and that of convective heat transfer by use of Newtonian
equations. The duo adopts the analytical models for the simulation
of the temperature distribution within a generator [19, 21].
The thermal circuit method has been in vogue for the estimation
of temperature rise in electrical machines through the aid of real
resistance circuits but the calculation was enhanced by the
introduction of computers in the early seventy’s. This computer
time enabled the use of numerical methods such as the finite
element and the finite difference analysis in the thermal modelling
of electrical machines [22].
Among the early researchers is Soderberg who in [23] published
work on thermal networks for electrical machines. He derived the
equivalent thermal circuit for steady-state heat flows in stators and
rotors having radial cooling ducts where he obtained good results
for large turbine generators.
The adequacy of lumped parameter thermal network for any
kind of component divided into arbitrary subparts having uniform
heat generation was confirmed by Bates et al in [24]. They
adopted an open circuit in the thermal model so that the heating of
the cooling fluid was included in the calculations. It was reported in
[22] that within the same time, though after Kotnik’s work using
equivalent circuit [25], Hak’s work on the calculation of
temperature rise by thermal networks was published. He did not
stop at that as he also published another work which looked at a
xxviii
model for the air-gap. The next were models for: axial heat transfer
in electrical machines in 1957 and models for stator slot, tooth as
well as yoke in 1960. It was further reported that by 1960-1963,
Kessler has developed a thermal network, where he was able to
extend the work so as to study the transient state calculations of
electrical machines. However, the contents of the work could not
be totally understood because of the difference in language of the
texts and perhaps too, it has not been translated. Later research
reports have been published by Kaltenbacher et al in [26],
Mukosiej in [27, 28], Mellor et al in [11, 14, 18] and Kylander in [29].
One of the most recent works is the one published by O.I. Okoro in
[30] where he studied the dynamic and thermal modelling of
induction machine with non linear effects. He also published so
many other works [31 - 37] in thermal modelling of electrical
machines some of which are duely cited herein.
Of the earliest works that dealt with temperature calculations in
electrical machines by finite element method (FEM) are the ones
published by Armor et al [38, 39] and later by Armor in [40]. They
determined the steady state heat flow and the iron losses in the
stator core of large turbine generators by using three-dimensional
finite elements. Alain et al in [41] also used FEM approach in the
thermal analysis of brushless direct current motor where he
compared the result with that from lumped scheme.
Doi et al also looked at the temperature rise of stator end-cores
by three-dimensional finite elements in [42]. They were able to
investigate the local heat transfer coefficients occurring in the end
winding space and also measured the thermal resistances of the
various materials.
xxix
Roger et al as well reported the steady and transient state thermal
analysis of induction motors with the finite element method in [43].
In 1990, a work on coupled electrical thermal calculation was
published by Garg et al [44] and was later developed in [45] by
Hatziathanassiou et al. Dokopoulos et al were in [22] reported to
have adopted the finite difference method for the thermal analysis
of electrical machines in 1984. Their study was restricted to the
rotor of cage – induction motors. Tindall et al in [46] also adopted
the finite difference approach to model the transient and steady
state temperature distribution of salient pole alternators.
The method of predicting the temperature rise of and the
determination of heat state of normal load for induction machine,
both based on the no- load test were suggested in [47, 48]. This
method has relatively low precision as the work centers on the
analysis of the equivalent thermal circuit of induction motor, the
parameters which were approximately estimated. A simple
empirical thermal model which estimates the stator and rotor
winding temperatures in an inverter-driven induction machine
under both transient and steady-state conditions was proposed in
[49]. The model centers on thermal-torque derating for inverter-
driven induction machine, and features a single frequency
dependent thermal resistance and time constant for each winding.
The demerit of this method is seen from the fact that only one
thermal source and only one thermal resistance are used for the
thermal model which predicts the temperatures rise of the stator
winding, or rotor winding. According to [49], this simple model
gives a temperature error of about 10oC which is of relatively low
accuracy.
xxx
In that work, a method for obtaining a generalized thermal model
of induction machine which gives good accuracy in predicting the
temperature rise in its full load tune was proposed. The method
was based only on a no-load test, though, simple and energy
saving as they sounded, the work was silent on thermal
capacitance effect. The inclusion of actual full load test would also
have produced a better and more detailed result.
The thermal networks are more often used than the numerical
method owing to their simplicity, accuracy and speed. For design
purposes the thermal networks give the global temperature
distribution of the machine particularly well. However, the
numerical calculation method is preferred when a transient state
analysis or a local temperature distribution is required. In this work,
the temperature rise of the machine parts is computed under
steady and transient conditions from the state equation using the
Runge-Kutta numerical method [51] by incorporating the ambient
temperature and that of the various core parts computed.
xxxi
CHAPTER THREE
HEAT TRANSFER MECHANISMS IN ELECTRICAL
MACHINES
3.1 HEAT TRANSFER IN ELECTRICAL MACHINES
Heat is popularly defined as the form of energy that is transferred
between two systems, usually a system and its surroundings by
virtue of temperature difference [52, 53]. This gives thermal energy
a clearer meaning in thermodynamics when we refer to adiabatic
processes. Since from the first law of thermodynamics or the
conservation of energy principle, energy cannot be created or
destroyed [52], we have therefore, that the amount of heat
transferred during a process between two states, say 1 and 2 is
denoted by 12Q or simply Q . Hence, heat transfer per unit mass, m
of a system is denoted by q which is obtained from
q = m
Q KJKg-1...................................................................................
(3.1)
The amount of heat transferred per unit time to be simply called
the rate of heat transfer is denoted by Q•
where the over dot
stands for the time derivative of Q . If Q•
varies with time, the
amount of heat transfer during a process is obtained by integrating
Q•
over the time interval of the process as follows.
Q = dtQt
t
•
∫2
1
KJ……………………….…………………………(3.2)
If Q remains constant during a process the relation above reduces
to Q = Q•
∆t where ∆t = t2 – t1 is the time interval during which the
process occurs.
xxxii
In electrical machines as is represented in figure (4.1), page 28,
heat is transferred from various parts to another. The transfer from
the stator to the outside surrounding and that of the rotor to the
stator plus many other transfers are not of the same mode. Hence
we look at the various modes of heat transfer.
3.2 MODES OF HEAT TRANSFER
A major aspect of thermal modelling involves the determination of
the thermal resistances of the thermal network. To achieve the
calculation of this, one has to be grounded in the areas of heat
transfer. Hence, there is need to study briefly the various modes of
heat transfer. It is good to remember once more that all modes of
heat transfer require the existence of a temperature difference,
and all modes of heat transfer are from the high-temperature
medium to a lower temperature one.
It’s good to quickly remind us about a common issue that
insulation reduces heat transfer and saves energy and money. The
decisions as regards the amount of insulation are based on heat
transfer analysis. The financial implication gets to us after the
economic analysis of the energy loss involved.
Adding insulation to a cylindrical pipe or spherical shell decreases
the rate of heat transfer Q•
; also, the outer radius of the insulation
is less than the critical radius of insulation defined in [54] as:
........................................................................................................,c
inscylindercr h
kr = ……
….(3.3)
.........................................................................................................2
,c
insspherecr h
kr = ……
…..(3.4)
xxxiii
Where insk is the insulator’s thermal conductivity )./( TmW and ch is
the convective heat transfer coefficient )./( 2 TmW . Materials or
aggregates of materials used primarily for the provision of
resistance to heat flow are referred to as thermal insulators.
Thermal insulations are useful in some areas for varying reasons
like in energy conservation, regulation of process temperature and
even in personnel protection to mention but a few. Insulation
materials are classified as fibrous, cellular, granular and reflective.
The degree or effectiveness of an insulation is often given in terms
of its ,valueR − the thermal resistance of the material per unit
surface area, expressed as
......k
LvalueR =− …………………………………………..………………………….
… (3.5)
Where L is the thickness and k is the thermal conductivity of the
material. To enhance heat transfer, the use of finned surfaces are
commonly adopted. Fins enhance heat transfer from a surface by
exposing a larger surface area to convection.
The basic modes of heat transfer are conduction, convection
and radiation [52 - 54]. However, [55] recognized convection and
radiation as thermal radiation and so has just two modes of heat
transfer. No matter the classification, all of them are associated
with the induction machine operations in one way or the other.
3.2.1 CONDUCTION: Energy transfer by conduction can take
place in solids, liquids and gases. This can be thought of as the
transfer of energy from the more energetic particles of a substance
to the adjacent particles that are less energetic due to interactions
between particles.
xxxiv
The time rate of energy transfer by conduction is quantified
macroscopically by Fourier’s law as illustrated in figure (3.1), T(x)
is the temperature distribution. The time rate at which energy
enters the system by conduction through the plane area A
perpendicular to the coordinate x is given by dx
dTkAQ x −=
•
(W) …………………..……………. (3.6)
The proportionality factor k , which may vary with position, is a
property of the material called the thermal conductivity.
Substances, like copper and silver with large values of thermal
conductivities are good conductors. Table 3.1 shows the thermal
conductivities of some materials at room conditions [52] together
with the thermal conductivity values as used by [22, 33].
Table 3.1: Thermal conductivities of some materials at room conditions [22,
33, 52]
Substance W/(m.K) Substance W/(m.K)
Diamond 2300 Al-Si 20 for frame 161
Silver 429 Steel(0.5%C) for shaft 54
Air at 50 o C 0.0280 Stator core (radial) 29
Human skin 0.3700 Aluminium for rotor cage 235-240
Gold 317 Copper for stator winding 370-401
Steel (0.1%C) 52 slot insulation (casted) 0.2-0.3
Stator core (axial ) 1- 4 Unsaturated polyester 0.2000
•
xQ
System boundary
Plane surface
T(x)
Figure 3.1: Illustration of Fourier’s Conduction Law
xxxv
Iron 80.2000 Air at 300K for air-gap/ ambient air 0.02624
Water (l) 0.6130 Stator core (axial) 2.5000
Stator core (radial) 18-40 Stainless steel 15-25
Iron ( casted) 58 Enamel coating(conductors) 0.2
3.2.2 CONVECTION: Here we refer to energy transfer between a
solid surface at one temperature and an adjacent moving gas or
liquid at another temperature. The energy conducted from the
system to the adjacent moving fluid is carried away by the
combined effects of conduction within the fluid and the bulk motion
of the fluid.
The rate of energy transfer from the system to the fluid can be
quantified by the empirical expression
•Q = hA )( fb TT − ………………………………………………………………
….. (3.7)
which is known as the Newton’s Law of cooling or Newtonian’s
equation. In equation (3.7) A is the surface area, bT is the
temperature on the surface and fT is the fluid temperature away
from the surface. For bT > fT energy is transferred in the direction
indicated by the arrow on figure (3.2). The proportionality factor ch
is called the heat transfer coefficient. ch is not a thermo dynamic
property, it is higher for forced convective operations relative to
free or natural ones as seen when fans and pumps are used.
F A
Velocity variation System boundary
Solid
xxxvi
Figure 3.2: Illustration of Newton’s law of cooling
The natural convection heat transfer coefficient ch in a cylindrical
isotherm surface is dependent on the Grashof’s number rG and
the Prandtl’s number rP , according to the expressions in [33, 56,
57];
.........................................................................................................1−= dkNh uc …….
.. (3.8)
............................................................................................)(59.0 25.0rru PGN = …...…
... (3.9)
.........................................................................................)( 23 −∞−= νβ dTTgG wr ……
..(3.10)
......................................................................................................1−= tpr kCP µ …… ..
(3.11)
..............................................................................................................1−= fTβ …….
.(3.12) where pC is fluid’s specific heat KKgJ ./( ), uN is the
Nusselt’s number, tk is fluid’s thermal conductivity KmW ./( ), ν is
fluid’s kinematic viscosity sm /( 2 ). g is acceleration due to
gravity 2/( sm ),
µ is fluid’s dynamic viscosity smKg ./( ).
β is volume coefficient of expansion )/1( K ,
∞TTT wf and . are temperature values )(K .
d is diameter of the cylindrical surface m( ).
The coefficient of heat transfer is dependent on flow type - laminar
or turbulent, geometry of the body, the average temperature,
physical characteristics of the fluid and whether the heat transfer is
natural or forced. The fluid motion obtained in the free convective
xxxvii
case is possible due to the buoyancy forces just as those of forced
convection cases are as a result of such external forces from fans,
pumps or rotating parts. The forced convection types prevail in
most activities with electrical machines. The mode of convection
mechanism, according to [10] is determined from the ratio of
Grashof number rG to the Reynold number eR as given below:
............................................................................................................2e
rconv
R
GM = …
… (3.13)
And free convection dominates if 1>>convM .
3.2.3 RADIATION: This is the energy emitted by matter in the form
of electromagnetic waves (or photons) as a result of the changes
in the electronic configurations of the atoms or molecules. Unlike
the other modes, it does not require a medium. Although all bodies
at a temperature above absolute zero emit thermal radiation; the
analysis here will not concentrate much on this mode of transfer.
However, the maximum rate of radiation that can be emitted from
surface at an absolute temperature ST is given by Stefan-
Boltzmann law as:
4sATQ σ=
•
(W) ……… ……………………………..……….………(3.14)
Where A is the surface area and σ = 5.67 x 10-8 w/(m2T4) is the
Stefan-Boltzmann constant. The black body is the idealized
surface.
The energy emitted by black body is greater than that emitted by
all real surfaces and it is also expressed by [52] as
4sATQ εσ=
•(W)…(3.15)
xxxviii
where for two real bodies [33, 57, 58] put the net heat transfer in
the form
)( 44fir TTAQ −=
•εσ ……………………………………...………………….………
.(3.16)
ε is the emissivity of the surface )10( ≤≤ ε . Table 3.2 that follows
shows the emissivity of some materials at 300K
Table 3.2: Emissivity of some materials at 300K [22, 59]
Material Emissivity Material Emissivity
Aluminum foil 0.07 Black body 1.00
Anodized Aluminum 0.82 Cast iron (rough) 0.97
Polished Copper 0.03 Forging iron (oxidized) 0.95
Polished Gold 0.03 Forging iron (polished) 0.29
Polished Silver 0.02 Copper (oxidized) 0.40
Polished Stainless steel 0.17 Copper (polished) 0.17
Black paint 0.98 Aluminium 0.08
White paint 0.90 Water 0.96
Another important radiation property of a surface is the absorptivity,
bα which is the fraction of radiation energy incident on a surface
that is absorbed by the surface. Kirchhoff’s law of radiation states
that the emissivity and absorptivity of a surface are equal at the
same temperature and wavelength. The thermal resistance for
radiation between two surfaces is given by [60] as:
xxxix
............................])273()273)][(273()273[(
111
22
2121
22
2
12111
1
++++++
−++−
=TTTT
AFAARthrad σ
εε
εε
………
…(3.17)
From the above, the radiative thermal resistance thradR , depends on
the difference of the third power of the temperature T , the surface
spectral property ε , and the surface orientation taken into account
by a form factor F ; A is the surface area.
3.3 HEAT FLOWS IN ELECTRICAL MACHINES
3.3.1 Heat Transfer Flow Types
Laminar flow, sometimes known as streamline flow, occurs when a
fluid flows in parallel layers, with no disruption between the layers.
In fluid dynamics, laminar flow is a flow regime characterized by
high momentum diffusion, low momentum convection, pressure
and velocity independent from time. It is the opposite of turbulent
flow. In nonscientific terms laminar flow is "smooth," while turbulent
flow is "rough."
The dimensionless Reynolds number is an important parameter in
the equations that describe whether flow conditions lead to laminar
or turbulent flow. In the case of flow through a straight pipe with a
circular cross-section, Reynolds numbers of less than 2300 are
generally considered to be of a laminar type [61]; however, the
Reynolds number upon which laminar flows become turbulent is
dependent upon the flow geometry. When the Reynolds number is
much less than 1, creeping motion or stokes flow occurs. This is
an extreme case of laminar flow where viscous (friction) effects are
xl
much greater than inertial forces. For example, consider the flow of
air over an airplane wing. The boundary layer is a very thin sheet
of air lying over the surface of the wing (and all other surfaces of
the airplane). Because air has viscosity, this layer of air tends to
adhere to the wing. As the wing moves forward through the air, the
boundary layer at first flows smoothly over the streamlined shape
of the airfoil. Here the flow is called laminar and the boundary layer
is a laminar layer.
3.3.2 Heat Transfer Flow System
One of the important factors controlling heat transfer is the
resistance to heat flow through the various ‘layers’ that form the
barrier between the two fluids. The driving force for heat transfer is
the difference in temperature levels between the hot and cold
fluids; the greater the difference the higher the rate at which the
heat will flow between them and the designer must optimize the
temperature levels at each stage to maximize the total rate of heat
flow. The resistance to the heat flow is formed by five layers as
follows [61]:
i The inside ‘boundary layer’ formed by the fluid flowing in close
contact with the inside surface of the tube.
ii The outside ‘boundary layer’ formed by the fluid flowing in close
contact with the outside surface of the tube.
iii The fouling layer formed by deposition of solids or semi-solids
on the inside surface of the tube (which may or may not be
present).
xli
iv The fouling layer formed by deposition of solids or semi-solids
on the outside surface of the tube (which may or may not be
present).
v The thickness of the tube wall and the material used will govern
the resistance to heat flow through the tube itself.
The values to be used for (iii) and (iv) are usually specified by the
client as the result of experience while the designer will select the
tube size, thickness and materials to suit the application. The
resistance to heat flow resulting from (i) and (ii), (designated the
partial heat transfer coefficients) depend greatly on the nature of
the fluids but also, crucially, on the geometry of the heat transfer
surfaces they are in contact with. Importantly the final values are
heavily influenced by what happens at the level of the boundary
layers.
3.3.3 The Boundary Layers
When a viscous fluid flows in contact with a tube at low velocity it
will do so in a way which does not produce any intermixing of the
fluid, the boundary layer, the fluid in contact with the tube, will have
its velocity reduced slightly by viscous drag and heat will flow
through the fluid out of (or into) the tube wall by conduction and/or
convection. As the velocity of the fluid is increased it will eventually
reach a level which will cause the fluid to form turbulence eddies
where the boundary layer breaks away from the wall and mixes
with the bulk of the fluid further from the tube wall. The velocity at
which this occurs is influenced by many factors, the viscosity of the
fluid, the roughness of the tube wall, the shape of the tube, size of
the tube etc. By experimentation [61], it has been found that
xlii
Reynolds numbers of less than 1200 describe the condition at
which there is no breaking away from the tube wall which is termed
laminar flow. The physical properties of the fluid are the
determining factors for the heat transfer in this area which is
inefficient in heat transfer terms. At Reynolds numbers above 2000
there is substantial breaking away from the tube wall and the
condition is described as turbulent flow with significant mixing of
the boundary layer and the bulk fluid. This is the most efficient
area for heat exchangers to work in. In order to quantify the
turbulence in practical terms heat transfer Engineers use a
dimensionless number called the Reynolds number which is
calculated as follows:
...................................................................................................................µG
DRe = (3.
18)
Where:
D = the hydraulic diameter of the tube (m)
G = the Mass velocity (kg/m²s)
µ = the viscosity of the fluid (kg/ms)
Many techniques have been tried in order to reduce the Reynolds
number value at which turbulent flow is produced but most have
the disadvantage of increasing the resistance to fluid flow, the
pressure loss, at a rate which increases more rapidly than the
decrease in boundary layer resistance. Some are not useable if
there are solids present, others if the fluid is very viscous. One
technique which is universally useful and does not have the
disadvantages of the others is that of deforming the tube with a
continuous shallow spiral indentation or an intermittent spot
xliii
indentation. Research has shown that by choosing the depth,
angle and width of the indentation carefully, the Reynolds number
at which turbulent flow is produced can be reduced significantly
below 2000. At values of Reynolds number above 2000 this form
of deformation also increases significantly the amount of
turbulence and therefore the rate of heat transfer which can, when
balanced correctly with the other factors reduce the surface area
requirement and therefore the cost of the heat exchanger.
3.4 DETERMINATION OF THERMAL CONDUCTANCE
According to [62], the thermal conductivity of a component is the
most important factor when determining the discretisation levels for
a thermal model. They however warned against increasing the
discretisation level unjustifiably as it would complicate the model
analysis without yielding better, more accurate result.
The popular methods of determining thermal conductivity are the
dynamic and static. The dynamic approach can be achieved by
employing highly sensitive instrumentation scheme. Also, the
diffusion solution equation has to be employed so as to determine
the diffusibility of the material through the measurement of the
thermal motion involved. The static approach in the other hand
promises a better accuracy, though takes a reasonable time. It
requires the knowledge of the heat flow density and temperature
gradient along the normal to the isothermal surface [63] leading to
the solution of Fourier’s law of conduction so as to determine the
thermal coefficient. Because of the relatively low temperatures
involved in electrical machine, the static method is often applied.
It is also reported in [50] that the determination of thermal
conductivity involves the synthesis of the induction machine
xliv
thermal model using the experimentally obtained results of
measuring the temperature of different parts and the power losses.
This above method which was adopted in [64] considered it very
necessary to execute the precise measuring of the loss densities
within the motor and to measure the temperature in the various
parts of the machines. To obtain all the necessary data, it furthered,
the number of the required tests is (N+1)/2, where N is the number
of the thermal network nodes. The tests which must be carried out
under full and half load conditions cum all the tasks involved make
this method in [64] very difficult and complicated. From this work, it
is possible to predict the temperature distribution within a machine.
To achieve this, the quantity of heat loss and the location have to
be known as well as the thermal characteristics of the materials.
However, inconsistencies arising from measurement of thermal
conductivities of material abound and therefore, introduce error in
the real or exact values. It is reported in [63] that increased
difficulty also exists in the characterization of composite materials
and the evaluation of conductances in interface regions. He further
suggested an infusion of correct data through the use of more
reliable measurement techniques as a way of eliminating these
uncertainties.
3.5 THERMAL-ELECTRICAL ANALOGOUS QUANTITIES
This section attempts to compare the basic thermal quantities to
that of electrical [21, 59, 65] for ease of understanding. A thermal
equivalent circuit is essentially an analogy of an electrical circuit in
which the rate of the heat analogous to current flowing in each
path of the circuit is given by a temperature difference analogous
xlv
to voltage divided by a thermal resistance analogous to electrical
resistance. The thermal resistance depends on the thermal
conductivity of the material k , the length l , and the cross sectional
area dA , of the heat flow path and may be expressed as:
...................................................................................................kA
lR
dd =
(3.19)
The thermal resistance for convection is expressed as:
...................................................................................................1
ccc hA
R =
(3.20)
Where cA , is the surface area of the convective heat transfer
between two regions and ch is the convective heat transfer
coefficient. The quantities are simplified in the table 3.3 that follows.
Table 3.3: Thermal-Electrical Analogous Quantities [54]
3.5.1 Thermal and Electrical Resistance Relationship
Thermal Electrical
Through variable Heat transfer rate q watts Current (I)amperes
Across variable Temperature θ )(T , C0 Voltage volts
Dissipation element Thermal resistance thR wattC /0 Electrical resistance
V/I =ohms
Storage element Thermal capacitance thC CJ 0/ Electrical capacitance
Q/V =farads
•Q
x
Tc
•Q
R
Th
xlvi
Figure 3.3: Simplified diagram for the illustration of thermal and electrical
resistance relationship
Considering figure (3.3), we observe that the temperature gradient
is
x
TT
x
T
x
T hc −⇒
∆=∂∂ …………………………………….…………………..……(3
.21)
Also, the rate of energy transfer is
dx
dTkAQ =
•………………..…………… ..(3.22)
This is Fourier equation. When steady state has been established
x
TTkAQ ch )( −=
• or
kAx
TT ch − …………………………………….…………..
(3.23)
This is exact analogy to Ohms laws of electrical resistance R
EI =
where, •Q is analogous to I and =∆T ch TT − is analogous to E so
that kA
x becomes thermal resistance thR .
Thus
th
ch
R
TTQ
−=•
………………………………………………………….………(3.24)
•Q
•Q
•Q
Th
T1
T2
Tc
R1 R2 R3
x2 x1 x3
xlvii
Figure 3.4: Simplified diagram for further illustration of thermal and electrical
equivalent resistance
The thermal resistances in series will be equivalent to electrical
resistances in series, hence, total resistance given by
321 RRRR ++= implies that the thermal resistance between two
points 1x and 2 x is as given in [33]: =R
kA
xx )( 12 −……………………………………………….……….(3.25)
CHAPTER FOUR
THERMAL MODEL DEVELOPMENT AND PARAMETER
COMPUTATION
4.1 CYLINDRICAL COMPONENT AND HEAT TRANSFER ANALYSIS
The heat transfer processes is summarized in the simplified
diagram of induction motor shown in figure (4.1) below.
Conduction also occurs in the air-gap, between stator slots and
stator iron and between rotor bars and rotor iron.
Ambient (convection and radiation)
Rotating stator flux
Stator (conduction)
3-Phase Supply
Rotor (conduction)
xlviii
4.2 CONDUCTIVE HEAT TRANSFER ANALYSIS IN INDUCTION
MOTOR
The rotor, stator, shaft and some other parts of the induction motor
are analyzed on the basis of the general cylindrical component as
shown in figure (4.2) below.
Figure 4.2: General cylindrical component T1 ,T2 and T3 , T4 represent the inner and the outer surface
temperatures of the components while r2 and r1 denote the inner
and the outer radius respectively. In the same way, if one end of
the cylinder is cut out, it will give rise to a ring or what is referred to
as annulus ring as shown in figure (4.3).
Figure 4.1: Heat transfer mechanism in squirrel cage induction machine
r1
r2
T3
T1
T2
T4
r1
L
r2
Figure 4.3: Conductive Thermal circuit- An annulus ring
T1
T2
T3
T4
L
xlix
In arriving at the expression for the thermal resistance networks in
line with the conduction of heat across the general component, the
following assumptions are made:
i The heat flows are of axial and radial type and are independent.
ii A unique mean temperature represents the heat flows in both
directions.
iii Circumferential heat flow is not present.
iv The thermal capacity and heat generation are uniformly
distributed.
In [45], the surfaces in the air-gap were further considered to be
smooth so that they can make use of the experimental results of
Ball et al in [66]. According to [11], on adoption of those
assumptions listed above, the solution of the heat conduction
equations in each of the axial and radial directions yields two
separate three-terminal network as shown in figure (4.4) below.
3T
u,Cth 2T
4T
1T
R2r
R1a
R2a
mT
R1r
R3a R3r
l
Figure 4.4: Three terminal networks of the axial and radial networks In the above figure, 1T and 2T , 3T and 4T represent the surface
temperature of components, and the third, the mean or average
temperature mT of the component at which any internal heat
generation u or thermal storage thC is introduced. The central
node of each network is to give the mean temperature of the
component but for the internal heat generation or storage. The
values of the thermal resistance according to [67] and also in [14]
come directly from the independent solutions of the heat
conduction equation in the axial and radial directions. These are
obtained considering the physical and cylindrical dimensions cum
the axial )( ak and radial )( rk thermal conductivities [11, 21]. The
expressions for the thermal resistances obtained from the thermal
networks are as follows:
)(2 22
21
1 rrk
LR
aa −
=π
……………………………………………………………….….(
4.1)
)(2 22
21
2 rrk
LR
aa −
=π
……………………………………………………………….….(
4.2)
)(6 22
21
3 rrk
LR
aa −
−=π
……………………………………………………………….….(
4.3)
li
−
−=2
22
1
2
122
1
2
14
1
rr
r
rr
LkR
n
rr
l
π……………………………………………………
(4.4)
−−
= 1
2
4
12
22
1
2
121
2 rr
r
rr
LkR
n
rr
l
π……………………………………………………
(4.5)
( ) ( )
−
−−−−=
22
21
2
122
21
22
212
22
13
4
8
1
rr
r
rrr
rrLkrr
Rn
rr
l
π…………………………….……
(4.6)
The total thermal capacitance of the cylinder is determined from
the density of the material ρ , the specific heat capacity pC and the
motor dimensions as follows:
( )LrrCC pth2
22
1 −= πρ ………………………….………………………………………
….(4.7)
The variation in the internal energy of the machine components
with time will be accounted for by the transient analysis hence the
introduction of the thermal capacitance.
......................................................................................................PPolth MCCVC == ρ….(4.8)
where
lii
M is mass and olV is volume
The networks of figure (4.4) are in one-dimension and can be
combined by connecting the two points of mean temperature ( aR3
and rR3 ) together. The thermal network can be reduced to a much
simpler one as in figure (4.5) if we assume a symmetrically
distributed temperature in the cylinder about the central radial
plane such that the temperature 3T and 4T on the faces of the
cylinder are equal. This will warrant that the modelling of half of the
cylinder be carried out with half of the heat generation and thermal
capacitance considered. This will appreciably reduce figure (4.4) to
figure (4.5) as shown below.
Figure 4.5: The combination of the axial and radial networks for a symmetrically distributed temperature about the central radial plane. A close observation of figure (4.5) reveals four thermal resistances;
cba RRR ,, and mR lumped together to two internal nodes. The thermal
resistances are now given as:
( )22
21
31 62
rrk
LRRR
aaaa −
=+=π
………………………..………………………..….…
…(4.9)
1θ
2θ
u, C
mθ Ra
Rc
43 θθ =
Rb
Rm
1T
2T
u, Cth
mT Ra
Rc
43 TT =
Rb
Rm
liii
−
−==2
22
1
2
122
1
2
12
12
rr
r
rr
LkRR
n
rrb
l
π………………………………………………..
…(4.10)
−−
== 1
2
2
12 2
22
1
2
121
2 rr
r
rr
LkRR
n
rrc
l
π…………………………………………………..….
(4.11)
( ) ( )
−
−+−−==
22
21
2
122
21
22
212
22
13
4
4
12
rr
r
rrr
rrLkrr
RRn
rrm
l
π…………………………..…
(4.12)
The model of figure (4.5) can be adapted for different thermal
conductivities in both directions which makes for easy
consideration of the thermal effects of the stator and the rotor
laminations.
The general cylinder models the solid rod, say the shaft of
induction motor if the expressions given above as the radius 2r ,
tends to zero and the node corresponding to the central
temperature 2T is removed.
liv
4.3 CONVECTIVE HEAT TRANSFER ANALYSIS IN INDUCTION
MOTOR
Thermal resistance value given by cR , models the convective heat
transfer between open parts of the solid materials and the cooling
air both inside and outside. As stated earlier it has the value 11 −−= ccc AhR ……………………………………………………… ……...…….……
…(4.13)
Where =ch boundary film coefficient (convective heat transfer
coefficient) and =cA surface area in contact with the cooling air.
Film coefficients normally used in the study of convective heat
transfer in induction motor according to [11] are four in number
namely: between
(a) frame and external air
(b) stator or rotor and air-gap
(c) stator iron, rotor, end-windings or end-cap and end-cap air
(d) rotor cooling holes and circulating end-cap air.
It further stated that for a given surface, a film coefficient applies
when the machine is stationary, that is, the external and internal
fans are not functional; a second film coefficient applies when the
machine is rotating. Hence, the film coefficient for the stages (a) –
(d) can be denoted by , ; aras hh , ; brbs hh crcs hh ; and drds hh ; respectively.
The work hinted that coefficients due to case ‘a’ above can be
found directly from test if the motor is run at constant load until
thermal equilibrium is reached, arh is then determined from the
surface ambient temperature gradient and the total machine loss,
the ash being similarly found from a low voltage locked rotor test,
where under thermal equilibrium, the heat dissipated from the
motor surface is equal to the total electric power input. The rest of
lv
film coefficients were obtained through various means as
described in [68-71].
Concerning the air-gap, the two main parts, the rotor and the stator
are in the likes of two concentric cylinders in relative rotatory
motion to each other. Aside from the large induction motor types
any heat emitted from the rotor surface moves unhindered and
across the air-gap to the stator. The axial heat flow, if any, from the
air gap to the adjoining endcap air is very negligible, and is not
given regard. The film coefficients of the air-gap 1h , in terms of a
dimensionless Nusselt number uN , the air-gap width agwL and
thermal conductivity of air cT is related thus:
agw
airu
L
kNh =1 …………………………………….……………………………………(4.
14)
The value of the Nusselt number for the convective heat transfer
between two smooth cylinders in rotatory motion is given in [71].
However, there is greater heat transfer across the air-gap than
achieved by the smooth cylinder equations. This is due to the
effect of additional fluid disturbances carried by the winding slots.
According to Gazley [69], his experimental results show that about
ten percent (10%) increase in heat transfer is as a result of the slot
effects.
4.4 DESCRIPTION OF MODEL COMPONENTS AND ASSUMPTIONS The construction of the induction machine under study is as
presented in figures (4.20) and (4.21) below with the parts labeled
lvi
as indicated. A better understanding of the modeling follows from
these few descriptions given below on some of the parts.
1. Ambient Air 6. Fan 11. Rotor iron
2. Rotor winding 7. End winding 12. Cooling rib
3. Stator iron 8. Bearing 13. Stator teeth
4. Air gap 9. Endring 14. Frame
5. Stator winding 10. Shaft
Figure 4.6: Squirrel Cage Induction Machine Construction
1
2
3
4
5
6
7
8
9
10
11
13
12
1
2
8
13
4
6 7
12
14
11
. .
5
8 9 10
3
lvii
FRAME: This is an embodiment of the entire ribbed cooling
structure and the endcaps. The frame absorbs heat from the stator
across the frame-core contact resistance, it also absorbs heat from
the endcap air by convection. The modelling elements of the frame
are different because the frame is thicker at the ends. The entire
frame is considered to be at uniform temperature and can
dissipate heat externally via single frame to ambient convective
thermal resistance. The thermal resistance between two frame
elements is thus: 1)2( −+= ArbLR cc λπλ where A is cross sectional area of the cooling
fins, b is the thickness of frame, r is the mean radius of the frame,
L is the length of frame and cλ is the conductivity.
STATOR IRON: This is made up of the stator lamination pack.
The teeth are not included here. This is modeled using the general
form
which is modified to take care of anisotropy due to the laminations.
This is handled by the introduction of a stacking factor in the radial
direction and by the use of a value lower than that of mild steel for
the axial conductivity obtained from [68]. The stator yoke elements
are considered as hollow cylinders with thermal resistance in the
radial direction given by: 121 )2)( ( −−= LkrnrnR cπll where 1 r and 2 r are
the outer and inner radii of the cylinder.
STATOR TEETH: The stator teeth are modeled as collection of
cylindrical segments connected thermally in parallel as the
1. End-winding cooling duct 6. Stator teeth 11. Rotor end
2. Frame 7. Stator winding 12. Shaft
3. Radial cooling duct 8. Air-gap 13. End-winding space(lower)
4. End-winding space(upper) 9. Rotor core
5. Stator yoke 10. Rotor teeth
Figure 4.7: The geometry of High Speed Induction Machine
lviii
expanded version of the general cylindrical component is
employed.
The heat flow from the slot windings is modeled by an additional
resistance between the slot faces to the point of mean temperature
at the tooth centre. The heat flow coming from the stator teeth is
much more than the heat generated internally.
ROTOR TEETH: The rotor teeth are modeled as being trapezoidal
as presented below. The axial thermal network of the rotor is
analyzed using the equations given in the general cylindrical
component where the letters L and h represent the axial length
and the height and r and R the base and top dimensions.
Figure 4.8: The geometry of Induction Machine rotor tooth
SLOT WINDING: The portions of the winding lying in the slots are
modeled as solid cylindrical rods comprising of array of conductors
and insulations. To obtain the axial and radial conductivities, it is
taken that only the copper conductors transfer heat axially along
the slot. On radial transfer basis the winding acts as a
homogenous solid with conductivity, about two and half times that
of the insulation alone [68], the slot insulation and the air pockets
are modeled by considering a layer that surrounds the slot material.
ENDWINDING: This is modeled as a uniform torroidal material
depicting the circumferential mesh of conductors and insulation,
b1
L b2
h
lix
the legs are considered as short cylindrical extensions of the stator
slot windings. An axial heat transfer is assumed to occur from the
mean temperature point in the torroid to the stator slot winding
along the copper conductors of the legs. The heat transfer from the
end windings is usually due to convection with little trace of
radiation.
AIRGAP: The air gap forms a connection between the stator teeth,
the part of the stator winding exposed in the slot openings and the
rotor surface. The corresponding thermal resistances are found
from the contact areas of these solids and the air-gap film
coefficient. The heat flow in the air-gap is mainly by conduction
and convection. Some researchers [68, 69, 70] have investigated
the heat flow in the air-gap between concentric cylinders with the
thermal effect considered in [71]. Laminar flow was associated with
small motors at low speeds due to absence of axial flow, there was
however a drift to turbulent flow with reasonable increase in speed.
The turbulent mode is defined better using the Taylor number [69]
which can also be presented as
F
aN g
Ta 2
32
νδω
= …………………………………………………………………………
…(4.15)
where ω = angular speed, ga = average air gap radius, δ = air gap
=2ν kinematic viscosity, 1≅F is a factor of geometry, TaN = Taylor number
For the case of small machine at low speed the heat transfer
coefficient becomes ch and is related to Nusselt number uN , as
lx
δ2cu
c
kNh = ……………………………………………………………………………....
(4.16)
From the above equation and with L as the axial length of the
element, thermal resistance between a rotor element and a stator
teeth element is determined as:
.................................................................2
1
LhaR
cgπ= ……….(4.17)
ENDCAP AIR: The circulating air in the endcap is considered as
having a uniform temperature. A single film coefficient is preferred
for the description of its convective heat transfer.
ROTOR IRON: The rotor is made up of several thin steel
laminations
with evenly spaced bars, which are made up of aluminum or
copper, along the periphery. In the most popular type of rotor
(squirrel cage rotor), these bars are connected at ends
mechanically and electrically by the use of rings. According to [72],
more or less 90% of induction motors have squirrel cage rotors.
This is because the squirrel cage rotor has a simple and rugged
construction. The rotor consists of a cylindrical laminated core with
axially placed parallel slots for carrying the conductors. Each slot
carries a copper, aluminum or alloy bar. These rotor bars are
permanently short-circuited at both ends by means of the end rings,
as shown in Figure (4.21). This total assembly resembles the look
of a squirrel cage, which gives the rotor its name. The rotor slots
are not exactly parallel to the shaft. Instead, they are given a skew
for two main reasons. The first reason is to make the motor run
lxi
quietly by reducing magnetic hum and to decrease slot harmonics.
The second reason is to help reduce the locking tendency of the
rotor. The rotor teeth tend to remain locked under the stator teeth
due to direct magnetic attraction between the two. This happens
when the number of stator teeth is equal to the number of rotor
teeth. The rotor is mounted on the shaft using bearings on each
end; one end of the shaft is normally kept longer than the other for
driving the load. Some motors may have an accessory shaft on the
non-driving end for mounting speed or position sensing devices.
Between the stator and the rotor, there exists an air gap, through
which due to induction, the energy is transferred from the stator to
the rotor. The generated torque forces the rotor and then the load
to rotate. Regardless of the type of rotor used, the principle
employed for rotation remains the same.
As was considered for the stator iron, the laminations in the rotor
iron are handled in the same manner. The rotor elements are
taken as having thermal contact with the stator teeth. The thermal
resistance between two rotor elements is given as:
................................................................................................ANk
LR
bc
= ……..……
…(4.18)
where A = cross sectional area of a bar, L = distance between adjacent rotor
elements
ck = conductivity of the bar material, bN = number of bars
1
5
lxii
SHAFT: The shaft is modeled as a cylindrical rod with no internal
heat generation. The axial heat conduction is modeled as three
sections. A good thermal contact is assumed to exist between the
shaft and the frame across the bearings. Any shaft external to the
bearing is therefore considered to act as part of the frame.
Thermal resistance between a rotor element and a shaft element
includes that due to the thermal resistance via the rotor core which
is given by:
Lk
r
rn
Rc
a
b
π2
=l
..................................................................................... …..………………..(4.19)
where ar = radius of the shaft , ck = conductivity of the core material
L = distance between adjacent elements and br = radius of the bottom of rotor slots.
In light of the descriptions given so far on the machine, an
equivalent circuit representing some core parts is given in the
figure below. Table 4.1 showing some geometric values and
dimensions for the machine parts is also presented.
1. End Ring 3. Conductors 5. Shaft
2. Bearing 4. Skewed Slots
Figure 4.9: Squirrel Cage Rotor
2
3
4
lxiii
11
C4
C2
C1
R35 P5
R810
R511
End-ring
P8
R10a
Өc
P12 R312
R1213
C12 Stator teeth
P13
Rotor teeth
C13
R713
R79
Rotor Iron
Rotor bar (winding)
End-ring
End-winding End-winding
Frame
Stator lamination
Stator winding
Ambient
P1
R12
P2
Өa
R11c
Өb
R1b
R23
P3
P4 10
C8
C5
C3
P6
R67
P7 R78
R911
P9
C6
R26
R34 R410
C7 C9
lxiv
Table 4.1 MACHINE GEOMETRIC / DIMENSIONAL DATA [2 9, 30, 33]
Machine elements Values Height of slot 16.9 mm Width of slot 7.76 mm Length of air -gap between slot t eeth and insulation 0.1 mm Thickness of insulation 0.2 mm Area of conductor at the end -winding 40.38 mm2 Length of end-winding connection 216.79 mm Height of stator iron teeth 17.5 mm Number of rotor slots 28 Outer radius of stator 100 mm Inner radi us of stator 62.5 mm Base of rotor slot 4.06 mm Slot -die ratio 1:12 Thickness of slot insulation 0.3 mm Inner radius of rotor 15 mm Height of end-ring 13.2 mm Width of end-ring 4.4 mm Copper winding cross section in slots 40.38 mm2 Iron core length 170 mm Total slot length 239 mm Length of rotor bar for sectioning 12.144 mm Mean roughness of air-gap 3e-7 m Air- gap length between stator core and lamination 0.7 mm Width of bar 3.86 mm Area of insulation 2570.4 cm2 Thickness of air 0.001mm Radius of end -ring 2.03 mm Height of rotor bar 13.7 mm Length of frame 250 mm Radius of frame 135 mm Number of end-caps 40 Number of rotor slots 28 Coil pitch 12 Diameter of wire 0.71mm
Figure 4.10: Thermal network model for the Induction machine
lxv
4.5 CALCULATION OF THERMAL RESISTANCES
4.5.1 Thermal resistance of the air-gap between insulatio n and
stator iron
Perimeter of the air-layer similar to that of the insulation
Pair ≅ 2 (16.9) + 7.76 ≅ 42mm per slot
Area of air-layer is also similar to area of insulation
Aair = Ains = Pair.L ; where L = stator core length = 170mm
Aair = 42 x 170 = 7140mm2
Total area = Aair –T = Aair x Ns . where Ns = number of stator slots =
36
Aair –T = 7140 x 36 = 257040mm2
Aair –T = 2570.4cm2 = Ains-T
R23a = Tairair
airsslot
Axk −
−
δ where sslot is stator slot
Height of end-ring 13.2mm Width of end-ring 4.4mm Length of half-turn of stator winding 39.667 mm Equivalent stacking factor for rotor and stator 0.9 5 Permeability of free space -710 x 4π H/m
Temperature coefficient of copper at 20 0 C 0.0039 /K Number of turns in the stator w inding 174 Specific heat capacity KkgJCcu ./385= , KkgJC fe ./460= , KkgJCC frameendR ./960==
Thermal conductivity KCmWkcu ./8.3= , KCmWk fe ./5.0= , KCmWxkins ./102 3−=
Density 3/8900 mKgcu =ρ ,
3/7800 mKgfe =ρ ,3/2650 mKgframeendR == ρρ
7.76
Insulation, mmins 2.0=δ
16.9
Air layer thickness,
mmairsslot 1.0≈−δ
Figure 4.11: Thermal resistance of the air-gap betw een insulation and iron
P2
R23a
P3
Fig 4.11a R23
lxvi
airk = 0.28 x 10-3 W/cm.K = thermal conductivity of air
R23a = 4.2570 1028.0
01.03 xx − hence R23a = 13.9x10-3 K/W
For half of the machine, we have,
R23a.half = 27.8 x10-3 K/W
4.5.2 The thermal resistance of the insulation slot , R23b
R23b = Tinsins
ins
Ak −x
δ , insk = 2x10-3 W/cm.K
R23b = 4.2570x 102
02.03−x
R23b = 3.89 x 10-3 K/W
For half of the machine
R23b.half = 7.78 x 10-3 K/W
Therefore, the thermal resistance between the stator winding and
stator iron becomes,
R23 = R23a + R23b
R23= 13.9 x 10-3 + 3.89 x 10-3
R23 = 17.79 x 10-3K/W
For half of the machine,
R23 = 35.58 x 10-3K/W
P2
R23b
P3
Fig 4.11b R23
lxvii
4.5.3 Thermal resistance between the stator iron an d the yoke,
R12
Stator outside radius, ro = 100mm
Stator inside radius, ri = 62.5mm
Stator core length, L = 170mm
fek = 0.5 W/cm.K = thermal conductivity of iron
hy = ro – (17.5 + ri) = height of yoke
hy = 100 – 62.5 – 17.5, hy = 20mm
Radius of yoke = Ry
Ry = ro -2
yh = 100-10
5mm
ro
Ry
Figure 4.12: Thermal resistance between the stator iron and the yoke, R 12
R12a
P2
Fig 4.12a R12
hy
ri
17.5
lxviii
Ry = 90mm
∴R12a = yfe
y
Ak
h
Ay = Area of the yoke
Ay = 2π RyL Kfe (where Kfe is iron stacking factor)
Ay = 2π x 90 x 170 x 0.95 = 91326.1mm2
Ay = 913.26cm2
R12a = 26.913 x 5.0
2
R12a = 4.38 x 10-3 K/W
For half of the machine
R12a.half = 8.76 x 10-3 K/W
4.5.4 Thermal resistance of the air-layer between i nsulation and
yoke, R 12b
R12b = yokeairair
airyoke
Ak −
−δ
δ yoke-air = 0.01mm
airk = 0.28 x 10-3 W/cm.K
ro = outside stator radius = 100mm
Ls = stator core length = 170mm
Aair – yoke = 2π ro Ls
= 2π x 100 x 170
= 1068.14cm2
R12b = 14.1068 x 10x28.0001.0
3−
R12b = 3.34x10-3 K/W
For half of the machine, R12b.half = 6.68 x10-3K/W
R12 = R12b + R12a = 3.34x10-3 + 4.38 x 10-3
R12b
P2
Fig 4.12b R12
lxix
R12 = 7.72 x 10-3 K/W
For half of the machine,
R12 = 0.01544 K/W = 15.44 x 10-3K/W
4.5.5 Thermal resistance between stator iron and en d- winding,
R34
Considering a slot and half of the machine, we have,
R34 – slot = ccu
ew
Ak
LL
x 44
+
Conductor area = Ac = 40.38 mm2 = 0.4038 cm2
Length of end winding = ewL
ewL = 216.79 mm
cuk ≅ 3.8 W/cm.K = the thermal conductivity of copper
L = 170 mm
L/2 Endwinding
Figure 4.13: Thermal resistance between the stator iron and the end-winding, R 34
L/4
L
L/4
L/4 Endwinding
P3
P4 R34
Fig. 4.13a R34 =R35
lxx
R35-slot = 4038.0 x 8.34
7.214
17 + = 6.305 K/W
Considering the entire slots, we have
R35 = s
slot
N
R −35 = 36
305.6
R35 = 0.1751 K/W
For the whole machine, we have
R35 = 2
1751.0 = 0.08755 K/W
4.5.6 Thermal resistance between Rotor Bar and the end ring,
R67
Nr = 28 is Number of rotor slots
Lendr
L 2
LLbar −
Figure 4.14: Thermal resistance between Ro tor Bar and the end ring, R 67
∆ L1
∆ L
∆ L2
lxxi
L = 170 mm is the stator iron core length
slotL = 239 mm is the entire slot length
cuk = 3.8 W/cm.K
∆ L 1 = 4
170
4=L
= 42.5 mm
∆ L 2 =
−2
170240x
2
1 = 17.5 mm
∆ L = ∆ L 1 + ∆ L 2 = 60mm
Arbar = hrbar x brbar
hrbar = 13.17mm
brbar = 4.06 mm
Arbar = 13.17 x 4.06 = 53.47 mm2
For one rotor bar, R67’ becomes,
R67’ = rbarcu Ak
L∆
R67’ = 5347.0 x 8.3
6
R67’ = 2.953 K/W
∴The thermal resistance for all the rotor bars with the half of the
machine considered, gives
R67 = R67’xrN
1
R67 = 2.953 x 28
1
R67 = 0.1055 K/W
For the whole machine, we have
R67 = 2
1055.0 = 0.05275 K/W
R67 = 0.05275 K/W
P7
R67
P6
Fig 4.14a R67
lxxii
4.5.7 Thermal resistances of the rotor bar
Acu = 2
L brslot =
2
17x 4.06 x 10-1 = 3.451 cm2
with a base of 35mm
R45rb = 451.3x 8.3
350.0= 26.76 x 10-3 K/W (since cuk =3.8
W/cm.K)
For all the rotor slots, Nr = 28, hence R45= 28
1 x R45rb
R45 = 28
1 x 26.76 x 10-3 K/W
R45 = 0.956 x 10-3 K/W
4.5.8 Thermal resistances from the Rotor slot to en d ring, R 78
brslot = 4.06 mm
cuk =3.8 W/cm.K
∆ L = 6 cm (as in full slot calculated in page 51)
With area of calculated as Acu =3.451 cm2
R78 = 451.3x 8.3
6= 0.45753
WKR /46.078 =
However for half of the machine
WKxR half /46.0278 =
WKR half /92.078 =
lxxiii
4.5.9 Thermal resistance between the rotor bar and rotor-iron,
R56
airk = 0.28 x 10-3 W/Cm.K = Thermal conductivity or air
rN = 28 = Number of rotor slots
δ air = 0.01mm thickness of air
For the whole machine R56 = airendr khL 1
air
2
∆δ
endrL , total endring to endring lenght= 239 mm
∆h1, the width of the sectioned rotor bar = 0.827 mm
R56,1 = 31028.0 x 0827.0 x 9.23 x 2001.0
−x = 0.9035 K/W
R56,2 = 31028.0 x 1686.0 x 9.23 x 2001.0
−x = 0.4432 K/W
R56,3 = 31028.0 x 344.0 x 9.23 x 2
001.0−x = 0.2172 K/W
R56,4 = 31028.0 x 6191.0 x 9.23 x 2001.0
−x = 0.1207 K/W
R56,5 = R56,1 = 0.9035 K/W
∴ 5,564,563,562,561,5656
111111
RRRRRR++++=
9035.01
1207.01
2172.01
4432.01
9035.011
56
++++=R
=
( )9035.0x 1207.0x 2172.0x 4432.0x 9035.0
010497.00785807.00436680.00214005.00104977.0 ++++
lxxiv
356 10484690.9
1646446.01−=
xR ; R56 = 1646446.0
10484690.9 3−x = 0.05761
K/W
For half of the machine we have,
R56,1.half = 1.807 K/W , R56,2.half = 0.8864 K/W
R56,3.half = 0.4344 K/W , R56,4.half = 0.2414 K/W
R56,5.half = 1.807 K/W , R56.half = 0.11522 K/W
The number of rotor slots 28r =N , therefore
For all the slots, and for the whole machine
R56,1 = rN
R56,1
= 28
0.9035 = 0.03227 K/W
R56,2 = 0.01583 K/W , R56,3 = 7.757 x 10-3 K/W
R56,4 = 4.3107 x 10-3 , R56,5 = 0.03227 K/W
∴ R56 = 2.0575 x 10-3 K/W
For half of the machine, we have
R56,1.half = 0.0645 K/W , R56,2.half = 0.031657 K/W
R56,3.half = 0.01551 K/W , R56,4.half = 8.6214 x 10-3 K/W
R56,5.half = 0.0645 K/W . R56.half = 4.115 x 10-3 K/W
4.5.10 Air-Gap Thermal Resistance, R 25 (the thermal resistance
between the stator iron and the rotor iron).
R25 = nLD
e2
log214.118.0
+ δ
where,
δ = air-gap width (mm) = 0.3mm
e = the mean roughness of the air-gap wall [mm] = 0.0003mm
lxxv
L = iron length [m] = 0.170m
n = machine rated speed (rpm) = 1440rpm
D = stator bore diameter (m) = 0.2m
R25 = 170.0 1440 2.0
0003.03.0
log214.1 18.0
2 xx
+ = K/W0.131
4.5.11 Thermal Resistance of stator teeth
pitchT = 0.0106m is tooth pitch ,
sagλ = 65 W/m2.K is stationary air-gap film coefficient
dsb = 0.053m is stator tooth width,
inr = 0.1075m is inner radius of tooth
L = 0.170m is stator length.
4.5.12 Thermal Resistance of rotor teeth
pitchT = 0.0106m is tooth pitch ,
ragλ = 96.89 W/m2.K is rotating air-gap film coefficient
dsb = 0.067m is rotor tooth width,
outr = 0.1351m is outer radius of tooth
L = 0.170m is rotor bar height.
saginds
pitchthst Lrxb
TR
λπ=
ragoutdr
pitchthrt Lrxb
TR
λπ=
lxxvi
4.6 CALCULATION OF THERMAL CAPACITANCES
4.6.1 Thermal Capacitance for Stator Lamination
L = Iron core length = 170 mm
ri = Inside stator radius = 62.5 mm
ro = Outside stator radius = 100 mm
Cfe = Iron specific heat capacity = 460 J/kg.K
Kfe = Lamination stacking factor = 0.95
feρ = Lamination iron density = 7800kg/m3
CTotal = feρ Cfe V
V = π )r - (r 2i
2o Kfe L
V = (0.12 – 0.06252) π x 0.95 x 0.170
V = 3.092 x 10-3 m3
∴ CTotal = 7800 x 460 x 3.092 x 10-3
CTotal = 11094 J/K (for the whole lamination)
CTotal.half = 5547 J/K (for half of the lamination)
4.6.2 Thermal Capacitance for Stator Iron
7.76
14.2 16.9
ro
L
ri
Figure 4.15: Thermal capacitance for Stator Lamination
lxxvii
Area of the stator slot = slotA
slotA = 2
5.42) (7.76+ x 14.2 + 2
x2.712 π
= 93.578 + 11.536
slotA = 105.114 mm2
Total volume of slot, slotTV
slotTV = Ns x slotA x Ls
Ns = total number of stator slots = 36
L = iron core length = 170 mm
slotTV = 36 x 105.114 x 10-6 x 0.170
= 6.433 x10-4 m3
∴ Stator slot thermal capacitance, thslotC is
thslotC = Kfe slotTV CL Lρ
thslotC = 0.95 x 6.433 x 10-4 x 7800 x 460
thslotC = 2192.75 J/K
∴ The thermal capacitance of the stator lamination, CthsLam is
CthsLam = CT - thslotC
17.5
ri = 62.5
Figure 4.16: Thermal capacitance for stator iron
C2
P2
Fig. 4.16a Cslam
lxxviii
CthsLam = 11094 – 2192.75
CthsLam = 8901.25 J/K = C2
If half of the machine is considered, Cthslam becomes
Cthslam.half = 4450.625 J/K
4.6.3 Thermal Capacitance for Stator Windings, C 3
C3 = cuC cuρ cuA x Ls x Ns
cuC = Copper specific heat (385 J/kg.k)
cuρ = Copper density (8900 kg/m3)
cuA = Copper winding cross section in slots (40.38 mm2)
Ns = Number of stator slots (36)
L = Stator length (170 mm)
D = Wire diameter (0.71 mm)
C3 = 385 x 8900 x 40.38 x 10-6 x 0.170 x 36
C3 = 846.776 J/K
For half of the machine, C3.half = 423.388 J/K
4.6.4 Thermal Capacitance for End Windings, C 4
L = 170 mm
Stator winding
P3
C3
Fig4.16b C3
lxxix
L = Stator core length
Lm = Mean length of the end winding
Slot die = 1: 12
Number of stator slots, Ns = 36
Lm = 2π avslotr x36
11
avslotr = stator inside radius ( ir ) + the 2
height slot
= +5.62 2
5.17
Lm = 2π x 71.25 x 36
11
Lm = 136.79 mm; then the total length = Lst
L = 170 mm
Lm
1 12
40 mm
Lm
Figure 4.17 a,b,c: Thermal capacitances for end winding
(a)
(b)
(c)
lxxx
Lst = Lm + 2 x 40
Lst = 136.79 + 80
Lst = 216.79 mm
∴Average conductor length = Lst + L
= 216.79 + 170
= 386.79 mm
Total winding length (Lmt) = 2 (Lst + L)
Lmt = 773.58 mm
C4 = cuC cuρ cuA x Lst x Ns
C4 = 385 x 8900 x 40.38 x 10-6 x 216.79 x10-3 x 36
C4 = 1079.84 J/K
Half of the machine, C4.half = 539.92 J/K
4.6.5. Rotor Iron Thermal Capacitance, C 6
End ring
239
170 RR
34.5
C4
P4
Fig.4.17d C4
lxxxi
(a) Volume of Rotor Lamination + Shaft (solid cylin der)
rRrlam LKRV π2=
L = Stator Core Length (170 mm)
rK = Equivalent Rotor stacking factor (0.95 ≈1)
RR = Radius of rotor lamination = δ−ir
δ = air – gap (0.7 mm)
ir = Inside stator radius (62.5 mm)
RR = 62.5 – 0.7
RR = 61.8 mm
rlamV = (61.8 x 10-3)2 xπ x 0.170 x 1.0
rlamV = 2.04 x 10-3 m3
(b) Volume of the Rotor bar, rbV
Number of rotor slots, 2N = 28
Width of rotor bar, rbb = 4.06 mm
Height of rotor bar, rbh = 13.17 mm
Figure 4.18: Thermal capacitance for rotor iron
Rotor Iron
P6
C6
Fig.4.18a C6
lxxxii
Equivalent Rotor stacking factor = (0.95≈1)
rbV = 2N rbb rbh L rK
rbV = 28 x 4.06 x 10-3 x 13.17 x 10-3 x 0.170 x 1
rbV = 2.545 x10-4 m3
(c) Total Volume, TV
TV = rlamV – rbV
TV = 2.04 x10-3 – 2.545 x 10-4
TV = 1.786 x 10-3 m3
(d) Rotor thermal capacitance, C 6
C6 = TV FeFeCρ
Feρ = Iron density [7800 kg/m3]
FeC = Iron specific heat [460 J/kg.K]
C6 = 1.786 x 10-3 x 7800 x 460
C6 = 6408.17 J/K
For half of the machine,
C6.hallf = 3204.08 J/K
4.6.6. Rotor Bar Thermal Capacitance, C 7
bSt
hrb
hSt
bL
hL
a
a
lxxxiii
Lb = abrb 2− ; Lh = ahrb 2−
insδ = insulation thickness (0.1 mm)
rbb = 4.06 mm ; rbh = 13.17 mm
Lb = 3.86 mm ; Lh = 12.97 mm
Volume of the active part of the rotor bar, rbV
rbV = Lb Lh 2N L
= 3.86 x10-3 x 12.97 x 10-3 x 28 x1.0 x 0.17
rbV = 2.383 x 10-4 m3
rbarC = C7 = rbV CuCuCρ
C7 = 2.382 x 10-4 x 8900 x 385
C7 = 816.535 J/kg
Half of the machine, C7.half = 408.267 J/kg
4.6.7 Thermal Capacitance for Various Rotor- Bar Se ctions
Crb1
Crb2
Crb3
Crb4
0.827
1.686
3.44
6.191
0.827
brb
12.97 mm
Figure 4.19: Thermal capacitance for the Rotor bar
Figure 4.20: Thermal Capacitance for the Various Rotor-Bar Secti ons
Crb5
C7
P7
Fig. 4.19a C7
lxxxiv
C7 = Crb1 + Crb2 +Crb3 + Crb4 + Crb5
∴ Crb1 + Crb2 +Crb3 + Crb4 + Crb5 ≡ 816.535
Crb1 = 97.12
827.0 x 816.535 = 52.06 J/kg
Crb2 = 97.12
686.1 x 816.535 = 106.14 J/kg
Crb3 = 97.12
44.3 x 816.535 = 216.57 J/kg
Crb4 = 97.12
191.6 x 816.535 = 389.76 J/kg
Crb5 = Crb1 = 52.06 J/kg
When half of the machine is considered we have,
Crb1.half = kgJ / 03.26 , Crb2.half = kgJ / 3.075 , Crb3.half = kgJ / 08.291 ,
Crb4.half = kgJ / 99.881 , Crb5.half = kgJ / 6.032 .
4.6.8 End Rings Thermal Capacitance, C 8
(i) Part of the slot outside the active part
4.06
RRing
13.17
hs
170
239
RRotor
dRing
Figure 4.21: Thermal capacitances for the End rings
lxxxv
Total slot length = 239 mm = LsL
Slot length outside the active part, La = LsL – L
La = 239 – 170
La = 69 mm
VsL = bL hL La . N2
= 3.86 x 12.97 x 69 x 28
VsL = 96724.03 mm3
= 9.672 x 10-5 m3
CsL = VsL cuρ cuC
= 9.672 x10-5 x 8900 x385
CsL = 331.41 J/K
(ii) The end- ring part
Area of the end-ring, Ar = rbb rbhx
Ar = 4.06 x13.17 mm
Ar = 53.47 x 10-6 m2
RR = Radius of rotor lamination = δ−ir
RRing = RR – hs – hrb - 2Ringd
= 61.8 – 0.5 – 13.17 – 2
06.4
RRing = 46.10 mm
VRing = 2π RRing. Ar
= 2π x 46.10 x 10-3 x 53.47 x 10-6
VRing = 1.549 x 10-5 m3
CRing = VRing cuρ cuC
1.549 x 10-5 x 8900 x 385
CRing = 53.08 J/K (for one ring)
CRingtotal = 106.16 J/K
End-ring
C8
P8
Fig.4.21a C8
lxxxvi
Therefore the total thermal capacitance of the end-rings with the
slot part outside the active part included is
C8 = 331.41 + 106.16
C8 = 437.57 J/K
For half of the machine, C8.half = 218.785 J/K
4.6.9 Frame Thermal Capacitance, C 1
C1 = Ce δe Vec + Cf δf Vf-e
δe = δf = 2650 kg/m3
Ce = Cf = 960 J/kg.K
(i) Vf-e = volume of frame without endcap
Vf-e = π 2fr Kfe Lf
rf = ro + da
rf = 100 + 35 = 135 mm ; Lf = 250 mm
Vf-e=π x (0.135)2 x 0.25 x 0.95 m3 where,
Vf-e=13.6 x 10-3
ro = outside radius of the stator
Lf = length of the frame
Kfe = lamination stacking factor
da = distance between the stator winding and the frame
(ii) Vec = volume of end cap
Vec = ha wa La x ne x Kfe
ne = Number of end cap = 40
Vec = 23.66 x 10 -3 x 4.44 x 10-3 x 0.226 x 40 x 0.95
lxxxvii
Vec = 9.02 x 10-4 m3
C1 = Cf δf Vf-e + Ce δe Vec
= 960 x 2650 x 13.6 x 10-3 + 960 x 2650 x 9.02 x 10-4
= 34598.4 + 2294.69C1 = 36893.09 J/K
For half of the machine, C1.half = 18446.55 J/K.
The calculated values of the thermal resistances and the thermal
capacitances used for the simulation are as shown in the table (4.2)
below, other values marked (*) are not calculated herein but are as
given in [25, 28 and 31]:
TABLE 4.2: Calculated thermal capacitance and thermal
resistance values obtained from the thermal circuit.
Thermal Capacitances
Description of component location in the thermal circuit
SIM (J/kg)
LIM (J/kg)
C1 Frame thermal capacitance 18446.55 18446.55 C2 Thermal capacitance of stator lamination 4450.625 4450.625 C3 Thermal capacitance of stator winding 423.388 423.388 C4 End-windingR thermal capacitance 539.92 539.92 C5 Thermal capacitance of rotor iron 3204.08 3204.08 C6 Rotor bar thermal capacitance 408.267 408.267 C7 Thermal capacitance of end-ringR 218.785 218.785 *C8 Thermal capacitance of ambient air 1006 1006 C9 Thermal capacitance of end-ringL 218.785 C10 Thermal capacitance of ambient air 1006 C11 Thermal capacitance of end-windingL 539.92 *C12 Thermal capacitance of the stator teeth 341.33 *C13 Rotor teeth thermal capacitance 871.566 Thermal Resistances
(K/W)
(K/W)
*R1b between ambient and frame 0.0416 0.0416 R12 between frame and stator lamination 15.44e-3 15.44e-3 R23 between stator lamination and stator winding 35.58e-3 35.58e-3 R25 between stator lamination and rotor iron 0.131 0.131 R34 between stator winding and end-winding 0.1751 0.1751 *R48 of the end-winding 1.886 1.886 R56 between rotor bar (winding) and rotor iron 4.115e-3 4.115e-3 R67 between rotor bar and end-ring 0.1055 0.1055 R78 of the end-ring 0.932 0.932
lxxxviii
R8c for ambient air 0.015 0.015 * R713 rotor bar and rotor teeth 0.002703 *R312 between stator teeth and stator winding 0.02245 *R1213 between stator teeth and rotor teeth 0.12576
CHAPTER FIVE
LOSSES IN INDUCTION MACHINE
5.1 DETERMINATION OF LOSSES IN INDUCTION
MOTORS
Power losses that occur during the transfer of power from the
electrical supply to mechanical load give rise to the heating of the
induction machines. Some of the loss components were described
in [72] under iron losses, copper losses, harmonic losses, stray
load losses and mechanical losses.
There are five main losses that occur in an induction machine and
these are identified as follows:
1. Stator copper losses that occur as a result of the current flowing
in the stator.
2. Core losses linked to the magnetic flux in the machine, which is
independent of the load.
3. Stray load losses that vary with the driven load.
4. Rotor copper losses.
5. Friction and windage (rotational) losses that occur in the
bearings and ventilation ducts.
lxxxix
5.1.1 Stator and Rotor I2R Losses
These losses are major losses and typically account for 55% to
60% of the total losses. I2R losses are heating losses resulting
from current passing through stator and rotor conductors. I2R
losses are the function of a conductor resistance, the square of
current. This is one of the major harmonic losses, a resistive loss
of the rotor expressed as:
.......................................................................3 2rrr RIP = ……………………………….…
..(5.1)
where rI and rR are the current and resistance per phase
respectively. Resistance of conductor is a function of conductor
material, length, temperature and cross sectional area. The suitable
selection of copper conductor size will reduce the resistance.
Reducing the motor current can be accomplished by decreasing the
magnetizing component of current. This involves lowering the
operating flux density and possible shortening of air gap. Rotor I2R
losses are a function of the rotor conductors (usually aluminum)
and the rotor slip. Utilization of copper conductors will reduce the
winding resistance. Motor operation closer to synchronous speed
will also reduce rotor I2R losses.
5.1.2 Core Losses
Core losses are those found in the stator-rotor magnetic steel and
are due to hysteresis effect and eddy current effect during 50 Hz
magnetization of the core material. These losses are independent
of load and account for 20 – 25 % of the total losses [73]. The
hysteresis losses which are a function of the flux density are
reduced by utilizing low loss grade of silicon steel laminations. The
xc
reduction of flux density is achieved by suitable increase in the
core length of stator and rotor. Eddy current losses are generated
by circulating current within the core steel laminations. These are
reduced by using thinner laminations.
5.1.3 Friction and Windage Losses
Friction and windage losses result from bearing friction, windage
and circulating air through the motor [74-76] and account for 8 –
12 % of total losses. These losses are independent of load.
5.1.4 Differential flux densities and Eddy-currents in the rotor
bars
The rotor copper losses arise from the flux pulsations in the rotor
teeth. The differential flux densities of two adjacent rotor teeth will
be an indication of flux pulsation seen by a rotor bar. This occurs
under no-load which means that currents will flow in each bar.
The flux pulsations at no-load means eddy-currents and to prove
this, the rotor copper losses are in [77] calculated in a separate
solution where the rotor short circuit rings are neglected. The only
loss that occurred was that of the eddy currents. This shows that
even under no-load the rotor copper loss is significant and in this
case the cause for overheating. In [78], the eddy current losses of
stator esP and rotor erP are calculated using these formulae:
( ).................................................................................*5.1
22
s
dsqses R
VVP
+= …………….
…. (5.5)
( ).....................................................................................*5.1
22
r
drqrer R
VVP
+= …………
….. (5.6)
xci
while the copper losses at stator side cusP and at rotor side curP are
computed using the conventional formulae below.
( ) ..................................................................................*5.1 12
12
ssscus diqiRP += ……
… (5.7)
( ) .................................................................................*5.1 12
12
rrrcur diqiRP += ……….
(5.8)
where qsV , qsI are q-axis voltage and current, dsV , dsI are d-axis
voltage and current while sR is the resistance at the stator side.
Other symbols are the equivalent at the rotor side.
Measuring the no-load copper losses is very difficult. However, it
has been shown that the numerical calculation of iron and
pulsations losses can lead to design improvements.
5.1.5 Stray Load-Losses
These losses vary according to the square of the load current and
are caused by leakage flux induced by load currents in the
laminations and account for 4 to 5 % of the total losses. These
losses are reduced by careful selection of slot numbers, tooth/slot
geometry and air gap. The stray-load loss is that portion of losses
in a machine not accounted for by the sum of friction and windage,
stator RI 2 loss, rotor RI 2 loss and core loss. This statement gives
no special hints to uncover the origin of the losses but theory of
stray load losses enjoys some levels of documentation according
to [80], who further listed several ways of determining the stray–
load losses to include: No Load Test, Differential method, Input-
Output method, AC/DC Short Circuit method and Reverse Rotation
method [82]. Expression for the calculation of no load loss is
documented in [83].
xcii
There are two different classes belonging to eddy current losses
and to hysteretic losses which are in fact often summarized under
the idea of additional iron losses. Most of the theory tackles the
eddy current losses and states that the hysteretic losses (heat loss
caused by the magnetic properties of the armature) are difficult to
grasp [41].
5.1.6 Rotor copper losses
The eddy-current in the rotor arises from flux pulsations in the rotor
teeth. These flux pulsations can be calculated by defining some
model parameters in the rotor teeth so as to simplify the
calculation of the average flux densities. The average flux density
in each of the rotor teeth at each time step of the transient analysis
can then be calculated as:
............................................... 1
, dABA
BA
avgtooth ∫−
= ………………………..
…...(5.2)
Where =B magnetic flux, =A bar cross sectional area
Once the flux density in each tooth as a time function is known, a
Fourier analysis is used to determine the DC-flux component as
well as the higher order harmonics under no-load. The differential
flux densities between two adjacent rotor teeth will be an indication
of the flux pulsation seen by the rotor bar between the teeth. Using
a 2D finite element model the rotor currents only have a
component in the z-direction. Similar to the average flux density in
a tooth, the loss of each rotor bar is calculated by means of a
program after each time step as given in [77]:
xciii
∑=n
nCu IRP1
222 ............................................................................................ ……….
……(5.3)
........................................................................... 1
2
22 ∑∫=n
A
zCu dAJRP ………..
…..(5.4)
where 2R is the resistance of a rotor bar; n , the total number of
bars and A , the cross-sectional area of a bar. =zJ Current density
5.1.7 No-load losses
The no-load test on an induction machine gives information with
respect to the exciting current and no-load losses. At no-load only
a very small value of rotor current is needed to produce sufficient
torque to overcome friction and windage. The rotor copper losses
are therefore usually assumed to be negligibly small while the
stator copper losses may be appreciable because of the larger
exciting current. The core losses are usually confined largely to the
stator iron.
5.1.8 Pulsation losses
Generally there are discontinuities in magnetic field components
as rotor teeth and slots sweep past the stator, hence, the rotating
stator fields produce losses in both the stator and rotor laminations
that aren’t accounted for by the hysteresis and dynamic losses in
the steel [81]. Flux pulsations in the rotor teeth for example will
cause eddy-currents in the rotor bars, even at no-load. This
additional eddy-current loss is what is referred to as pulsation loss.
xciv
5.2 CALCULATION OF LOSSES FROM IM EQUIVALENT
CIRCUIT
Different schemes exist in an attempt to evaluate the
electromagnetic losses in electrical machines, this is most
probably because they contribute substantially to the temperature
distribution in the machine, and more so, when there is need for
estimating the efficiency [84].
Here, a classical approach based on the equivalent circuit
methodology as shown in figure (5.1) and simplified to figure (5.2)
is adopted. The induction machine equivalent circuit model, shown
in figure (5.2) is constructed by using the following set of induction
machine parameters: ( sR , sX ), ( mR , mX ), and ( rR , rX ). Each pair
represents resistance and leakage reactance, respectively. The
first pair deals with the stator parameter, the next pair refers to
magnetizing parameters while the third one deals with the rotor.
The second pair of parameters takes care of magnetizing effects
and models the generation of the air gap flux within the induction
motor.
Figure 5.1. Equivalent Circuit of the AC induction Machine
Io
Rs jXs Is Ir
Im Ic
Rc jXm
jX r
s
R r Vs
xcv
Figure 5.2. Simplified Equivalent Circuit of the AC induction Machine
The equivalent circuits shown in figures (5.1 and 5.2) are all
convenient to use for predicting the performance of induction
machine, in some other cases, a step by step approach can be
followed to treat the shunt branch, that is cR and mX , particularly
the resistance cR , representing the core loss in the machine. Not
much effort is required to get such cases analyzed according to
[85].
For a machine operating from a constant-voltage and constant-
frequency source, the sum of the core losses and friction and
windage losses remains essentially constant at all operating
speeds. These losses can thus be lumped together and termed
rotational losses of the induction machine. If the core loss is
lumped with the windage and frictional losses, then the resistance
due to core losses can be ignored and the component
representing it, cR can be removed from the circuit of figure (5.2) to
give rise to the IEEE recommended equivalent circuit of figure
(5.3). The circuit of figure (5.1) is analyzed herein and used in the
calculation of the machine losses and the associated machine
performances respectively. The rotor values are those of referred
quantities.
Is
Vs Im
Xm
Rm Ir
Rr Xr Xs
Rs
rRs
s−1
xcvi
Figure 5.3. IEEE Equivalent Circuit of the AC induction Machine
Other parameters of interest presented in the figure above are sV
(per-phase supply of the stator) and sI , mI , rI (the phase currents of
stator, magnetizing and rotor circuit, respectively). These
parameters can vary in the model with different operational
conditions. The required electromagnetic losses are calculated as
follows: 2ss ImRSTAcuL = (stator copper losses),
……………………….…………....(5.9) 2mm ImRSTAcore = (stator core
losses), ………………………….………….…(5.10) 2rr ImRROTcuL = (rotor copper
losses), ………………………….…………..(5.11)
where m is the phase number of the motor (in this case m = 3).
Formulae (5.9)-(5.11) are used with the values for the phase
currents computed as follows
rmrsms
rmss ZZZZZZ
ZZVI
+++= )( ,
…………………………………..…….………….(5.12)
Rr Xr Xs
Rs
Is
Vs Im
Xm
Ir
rRs
s−1
xcvii
rmrsms
msr ZZZZZZ
ZVI
++=
,
…………………………………..….…………….(5.13)
rmrsms
rsm ZZZZZZ
ZVI
++=
. ………………………………………..…………(5.14) Where sZ , mZ and rZ are the phase impedances of stator,
magnetizing and rotor circuit, respectively
mmmrrrsss jXRZjXsRZjXRZ +=+=+= ,/, ……………………………
…(5.15)
Finally, the leakage reactances sX , rX and mX in (5.15) are
computed by using the following formulae
,2,2,2 mmrrss fLXfLXfLX πππ === where sL , rL and mL are leakage
inductance of stator, rotor and the magnetized inductance
respectively. In this work, various losses formula shown below
which were used in [10] are also adopted in calculating the
associated losses.
The iron loss ( FeP ) is principally made up of the hysteresis ( hysP )
and the eddy current ( eddP ) losses.
fMB
P hyshys
2
10
= σ ……………………………………………….………….……(5.
16)
MfB
P Feeddedd
22
1010
∆
= σ ………………………………………….……………(5.
17)
MBf
fPPP FeeddhyseddhysFe
22
10
10
∆+=+= σσ …………………………………..(5.
18)
xcviii
and for squirrel induction machine ,
FerFesTFeyFeTot PPPP ++= …………….(5.19)
where
eddσ is eddy current loss coefficient
hysσ is hysteresis loss coefficient
B is the magnetic flux density,
Fe∆ is the thickness of lamination
f is the frequency and M is the mass.
,FeyP FesTP and FerP are loss components of yoke, stator teeth and
rotor.
To distribute the total iron losses FeTotP between the stator and
rotor a factor ( sK ) is used such that: for stator we have
FeTotsFes PKP = ……. (5.20)
and for the rotor we have ( ) FeTotsFer PKP −= 1 ………………………………..
(5.21)
Stator iron losses are in itself re-distributed between the teeth and
yoke components with another factor ( tK ) such that: for the yoke
we have
FestFey PKP = …………………………………………………….…………………….
(5.22)
and for the rotor we have
( ) FestFesT PKP −= 1 ………………..………….…(5.23)
xcix
5.3 LOSS ESTIMATION OF THE 7.5kW INDUCTION MACHINE
The values of the parameters of this 400V, 50 Hz, 10Hp machine
having 4 poles, whose synchronous and measured rated speeds
are 1500 rpm and 1440 rpm with rated current 13.5A analyzed in
this work are given in Table 5.1 below.
Table 5.1: Induction Machine rating and Parameters [98]
10HP Induction Motor Parameters Value No of poles 4 Rated speed 1440 rpm Rated frequency ( f ) 50 Hz Output power 7.5 KW Rated voltage 400 V Stator Current ( sI ) 13.4699 A Stator Resistance ( sR ) 0.7384 Ω Rotor Resistance ( rR , ) 0.7402 Ω Stator Leakage Inductance ( sL ) 0.003045 H Magnetizing Inductance ( mL ) 0.1241H Excitation Current ( 0I ) 5.5534 A Stator Core Loss Resistance ( cR ) 680.58 Ω Motor inertia ( J ) 0.0343
Kgm2 Wind frictional coefficient ( F ) 0.000503 NmS Calculated rated values Rotor Current ( rI , ) 11.6627 A Stator Core Loss Current ( cI ) 0.3393 A Magnetizing Current ( mI ) 5.5430 A Slip ( s ) 0.04 Rotor Leakage Inductance ( rL, ) 0.003045 H Stator Leakage Inductance ( sL ) 0.003045 H
c
Magnetizing Inductance ( mL ) 0.1241H Angle between sV and sI 28.90o Shaft Load Torque ( shT ) 49.471 N-m Developed electromagnetic torque ( eT ) 48.079 N-m All the loss estimation of this 10HP induction machine has also
been summarized in table (5.2) while formulas for the detailed
calculation are provided as m-files in the appendices, the results
obtained here followed careful usage of some of these formulas:
Input power: θCosIVP ssin 3= ………………………………………………….……………….…. (5.24)
The input power of the induction machine is W 8158.91
Efficiency calculation:
%100*1
−=
in
losses
P
Pη ……………………………………………….…………
(5.25)
Losses calculation:
If power losses(Plosses), stator copper loss ( PSTAcuL), rotor copper
loss (PROTcuL), stator core loss (PSTAcore), Friction and windage
losses (PFRIwin ) and stray load loss( PSTR ) are represented in this
way, then;
Plosses = PSTAcuL + PROTcuL + PSTAcore + PFRIwin + PSTR ……………………..
(5.26)
The Shaft Load Torque:
)( NmP
Tin
outSh ω
= ……………………………………………….…….……
…(5.27)
Air gap power:
ci
s
RIP r
rAG23=
……………………………………………….……….…….....
… (5.28)
Per-Phase Stator Core Loss Resistance (neglecting the stator impedance voltage
drop):
c
sc I
VR =
……………………………………………….………………...……
… (5.29)
Per-Phase Stator Magnetizing Inductance:
m
sm fI
VL
π2=
……………………………………………….…………..…..… .
.(5.30)
Based on IEEE 1112-B standard [81], the PSTR value at 1 kW is
2.5% of the full-load input power, dropping at 10kW to 2%, at
100kW to 1.5%, at 1000kW to 1%, and at 10MW to 0.5% as
reported in [87, 88], the stray load loss and rotational losses can
be calculated .
Since the machine under study here is a 10 hp machine, therefore,
PSTR(IEEE) = W 18.163%0.2*8158.91 = …………………………………………..
(5.31)
However, in the IEC 34-2 standard, these losses were not
measured but were arbitrarily estimated to be equal to 0.5% of the
full-load input power [82, 83], so that PSTR(IEC) =
W 40.795%5.0*8158.91 = ……………………….(5.32)
A suggested solution in Ontario Hydro’s simplified segregated loss
method assumed a value for a combined windage, friction and
core losses [84, 88]. The study recommends that these combined
losses be set to 3.5% of the input rated power which translates to:
WPP inROTaL *%5.3 = ……………………………………………….…………
… (5.33)
cii
Therefore, obtained rotational losses:
8158.91*%5.3=ROTaLP
W 285.562= ……………………………………………… (5.34)
5.4 SEGREGATION AND ANALYSIS OF THE IM LOSSES The estimated losses are summarized in table (5.2) below and
presented in the following bar and pie charts for ease of
understanding.
TABLE 5.2: Loss Segregation Obtained from Calculation
Losses Segregation Calculated Value (W)
Input Power (Pin) 8159.2 Stator Copper Loss 400.8250 Rotor Copper Loss 302.0875 Stator Core loss 235.0474 Friction and Windage Losses 50.5247
Stray Losses (PstrayIEEE-12B Standard) 163.1840 Total Losses (Watts) 1151.7 Output Power (Pout) 6968.5
ciii
1 2 3 4 50
50
100
150
200
250
300
350
400
450
Class of Losses
Loss
es (
[wat
ts])
1- STAcuL2- ROTcuL3- STAcore4- STRieee5- FRIwin
Figure 5.4. Bar chart representing loss segregation of 10HP induction machine
5.5 PERFORMANCE CHARACTERISTIC OF THE 10HP INDUCTIO N MACHINE When the parameters of table 5.1 are further used for the
equivalent circuit of figure 5.1, steady state performance curves
are generated as indicated in figures 5.5 to 5.9.
civ
0 500 1000 15000
20
40
60
80
100
120
140
160
180Torque vs speed curve for IM
Speed in RPM
Tor
que
in N
-m
Figure 5.5. Torque against speed characteristics for the 10HP induction machine
0 500 1000 15000
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2x 10
4Power vs speed curve for IM
Speed in RPM
Pow
er in
wat
ts
Figure 5.6. Power against speed characteristics for the 10HP induction
machine
cv
0 500 1000 15000
10
20
30
40
50
60
70stator current vs speed curve for IM
Speed in RPM
stat
or c
urre
nt in
Am
pere
s
Figure 5.7. Stator current against Speed for 10HP induction machine
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
20
40
60
80
100
120
140
160
180Torque vs slip curve for IM
Tor
que
in N
-m
Slip in p.u. Figure 5.8. Graph showing the Torque-Slip characteristics for 10HP induction
machine
cvi
0 500 1000 15000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1Power factor vs speed curve for IM
Speed in RPM
pow
er fa
ctor
Figure 5.9. Graph of Power factor against Speed characteristics for the 10HP IM
The starting current for an induction motor is several times the
running current and the starting power factor is much lower than
the power factor at rated speed. Both of these features tend to
cause the supply voltage to dip during start-up and can cause
problems for adjacent equipment. The torque-speed/slip
characteristic of this induction motor is shown in figures (5.5 and
5.8) above along with mechanical load torque. The rated torque is
usually slightly smaller than the starting torque so that loads can
be started when rated load is applied. The curve has a definite
maximum value which can only be supplied for a very brief period
since the motor will overheat if it is allowed to stay longer.
In figure (5.7), the response of current to the speed is plotted. The
starting current is several times larger than the rated current since
cvii
the back emf induced by Faraday’s law grows smaller as the rotor
speed decreases. Whenever a squirrel-cage induction motor is
started, the electrical system experiences a current surge while the
mechanical system experiences torque surge. With line voltage
applied to the machine, the current can be anywhere from four to
ten times the machine’s full load current. The magnitude of the
torque (turning force) that the driven equipment sees can be above
200% of the machine’s full load torque [89]. These wastages of
power due to losses account for a reduced internal and thermal
efficiency of the machine [90, 91]. The associated current and
torque surges can be reduced substantially by reducing the
voltage supplied to the machine during starting as one of the most
noticeable effects of full voltage starting is the dimming or flickering
of light during starting.
5.6.1 Motor Efficiency /Losses
The difference - watts loss is due to electrical losses plus those
due to friction and windage. Even though higher horsepower
motors are typically more efficient, their losses are significant and
should not be ignored. In fact, according to [94] higher horsepower
motors offer the greatest savings potential for the least analysis
effort, since just one motor can save more energy than several
smaller motors.
5.6.2 Determination of Motor Efficiency
Every AC motor has five components of watts losses which are the
reasons for its inefficiency. Watts losses are converted into heat
which is dissipated by the motor frame aided by internal or external
fans. Stator and rotor RI 2 losses are caused by current flowing
cviii
through the motor winding and are proportional to the current
squared times the winding resistance ( RI 2 ). Iron losses are mainly
confined to the laminated core of the stator and rotor and can be
reduced by utilizing steels with low core loss characteristics found
in high grade silicon steel. Friction and windage loss is due to all
sources of friction and air movement in the motor and may be
appreciable in large high-speed or totally enclosed fan-cooled
motors. The stray load loss is due mainly to high frequency flux
pulsations caused by design and manufacturing variations.
5.6.3 Improving Efficiency by Minimizing Watts Loss es
Improvements in motor efficiency can be achieved without
compromising motor performance at higher cost within the limits of
existing design and manufacturing technology. The formula for
efficiency in equation (5.47) shows that any improvement in motor
efficiency must be the result of reducing watts losses. In terms of
the existing state of electric motor technology, a reduction in watts
losses can be achieved in various ways. All of these changes to
reduce motor losses are possible with existing motor design and
manufacturing technology. They would, however, require
additional materials and/or the use of higher quality materials and
improved manufacturing processes resulting in increased motor
cost. In summary, we can say that reduced losses imply improved
efficiency.
cix
Table 5.3: Efficiency improvement schemes [94]
Watts Loss Area
Efficiency Improvement
1 Iron Use of thinner gauge, lower loss core steel reduces eddy current losses. Longer core adds more steel to the design, which reduces losses due to lower operating flux densities.
2 Stator RI 2 Use of more copper and larger conductors increases cross sectional area of stator windings. This lowers resistance ( R ) of the windings and reduces losses due to current flow ( I ).
3 Rotor RI 2 Use of larger rotor conductor bars increases size of cross section, lowering conductor resistance ( R ) and losses due to current flow ( I ).
4 Friction/ Windage
Use of low loss fan design reduces losses due to air movement.
5 Stray Load Loss Use of optimized design and strict quality control procedures minimizes stray load losses.
5.7 THE EFFECTS OF TEMPERATURE
Temperature effect in induction machine has a very important
influence in the assessment of the machines performance. Many
works could not consider the effects due to the difficulty
encountered in the measurements. This difficulty according to [95]
is due to the strong coupling between the electrical and thermal
phenomena inherent in the machine. Attempts at modelling it by
the variation of the stator and rotor equivalent resistances as a
function of their average temperatures which were measured
directly using a microprocessor-based data acquisition apparatus
was carried out in [77]. The measured resistance mR at the test
cx
temperature tT is corrected to a specified temperature sT as
follows;
..........................................................................................................KT
KTRR
t
stm +
+= …
…(5.48)
where mR is the corrected resistance at sT , and 5.234=K and 225
for copper and aluminum respectively [32, 77].
In induction motor thermal monitoring by [90], the rotor
temperature was monitored from its resistance identification and
then its temperature dependence given by:
..............................................................................................].........1[0 TRR ∆+= α …..
(5.49)
where 0R is resistance at reference temperature 25 C0 , however
20 C0 is used herein, α is resistance temperature coefficient and
T∆ is temperature increase. The resistance method allows for the
measurement of stator winding temperatures. However the main
source of error in the use of the resistance method is from
impurities associated with copper.
cxi
CHAPTER SIX
THERMAL MODELLING AND COMPUTER
SIMULATION
6.1 THE HEAT BALANCE EQUATIONS
In the lumped parameter thermal circuit analysis, it is often
assumed that the temperature gradient with certain parts of the
machine is negligible. According to [60] this assumption can only
be made if the internal resistance to the heat transfer is small
compared with the external resistance. The Biot number iB , is
usually used for determining the validity of this assumption. In the
case where internal conduction resistance is compared with
external convective resistance, iB is defined as:
...............................................................................................................s
ci k
LhB = ……
…(6.1)
where sk is the thermal conductivity of the solid material
L is the characteristic length of the solid body
ch is the convective heat transfer coefficient.
The criterion 1.0∠iB ensures that the internal temperature will not
differ and in the words of [96], the assumption of uniform
temperature is acceptable except for the early times of the step
change in temperature and for such, the time for the change is
localized in a thin ‘skin’ near the fluid or solid surface.
6.2 THERMAL MODELS AND NETWORK THEORY
cxii
In modelling a thermal network, the material is discretized giving
rise to aggregates of thermal elements that join at a given node
through thermal resistances. Inadequate discretisation has been
considered in [97] as one source of discrepancies between
experimental and simulated results. If well considered, the thermal
network so formed can be likened to electrical network as
explained in section 3.5.
The simplified diagrams of figures (6.1) and (6.2) below depict a
generalized thermal model as proposed in this work.
Figure 6.1: Transient Thermal model of SCIM with lumped parameter
RT
Heat Source
Tts
C
Ambient air Ta
Ps
Conductor
Tss
RT Heat Source
Ta
Ps
Figure 6.2: Steady State Thermal model of SCIM with lumped parameter
cxiii
If we consider the conductor temperature rise T∆ as the rise in
relation to ambient temperature caused by the presence of heating
loss, then the temperature rise is generally given by aTTT −=∆
hence in figure 6.1 we have that atsts TTT −=∆ while in figure 6.2 we
have that Tsss RPTT =∆=∆ . This will thus give us
0=−∆+∆S
T
PR
T
dt
TdC ……………………………………….(6.2)
The ambient air temperature aT , serves as the thermal reference
while a deviation from the reference, that is, a rise in temperature
denotes the machine elements. Assuming that we have ‘ N ’
number of loads singly linked to other nodes via thermal
resistances baR , in which ba and are the number of the nodes, with
baR , as the thermal resistance between the reference and node ‘b ’
then the steady-state rise in temperature at the node ‘a ’ can be
derived from the relation below:
............................................. 1 a 01 ,,∑
=
=
+−=
N
b ba
b
ba
aa R
T
R
TP ……………………
….(6.3)
Where
b. anda nodes adjoining twobetween resistance thermal
a node of re temperatuthe
a nodeat generationheat the
, ===
ba
a
a
R
T
P
For multinode consideration, 1T to NT represent the temperature
rises of each node while 1P to NP represent the losses at the
various nodes. The matrix defined by ‘G ’ in equation (6.4) is a
conductance matrix which when joined with the column vectors
represented by TP and TT as given below give rise to equation
cxiv
( 6.5 ) which finally leads to a stationary solution using equation
(6.6).
−−−−
−−−−
−−−−
−−−−
−−−−
=
∑
∑
∑
∑
∑
=
=
=
=
=
N
a aNNNNN
N
N
a a
N
N
a a
N
N
a a
N
N
a a
RRRRR
RRRRR
RRRRR
RRRRR
RRRRR
G
1 ,4,3,2,1,
,41 ,43,42,41,4
,34,31 ,32,31,3
,24,23,21 ,21,2
,14,13,12,11 ,1
1...
1111..................
1...
1111
1...
1111
1...
1111
1...
1111
,…………………………….
.. (6.4)
=
N
T
P
P
P
P
P
P
...4
3
2
1
and
=
N
T
T
T
T
T
T
T
...4
3
2
1
TT GTP = ………………………….……………………………………….. (6.5)
Hence
TT PGT 1−= ………………………………………………………..………… (6.6)
The SIM thermal network in full form as shown in figure (6.3) has a
total of twelve nodes and fifteen thermal resistances, while that of
LIM as shown in figure (6.4) has fourteen nodes and eighteen
thermal resistances. It was assumed in [15] that the heat
transferred from the rotor winding through the air-gap goes directly
to the stator winding with negligible impact on the stator teeth,
cxv
however this assumption did not go down well with the LIM model
here as the teeth is fully considered and the effects studied
alongside others. Hence, the rotor part of the machine is divided
into the rotor iron, rotor windings, rotor teeth and end rings while
the stator of the machine has networks for the stator iron, stator
winding, and end winding together with the stator teeth. The
connection of the above mentioned networks for rotor, stator and
frame gives rise to the thermal network models of figures (6.3 and
6.4) as shown below. The separate temperatures of the nodes are
evaluated using this set of heat balance equation as given below.
( ) 1 b a, 1 =
−−= ba
abaa TT
RP
dt
dTC ………………..……………………….
……(6.7)
Where
a nodeat generationheat the
b. anda nodes adjoining twobetween resistance thermal
a node of re temperatuthe
a node of ecapacitanc thermal
==
==
a
ab
a
a
P
R
T
C
The power losses ( 1P - 11P ) associated with the model of figure (6.3)
are outlined in equations (6.8 – 6.20). However, in the simulation
for the half model of the induction machine, equations (6.8 – 6.15)
representing ( 1P ) to ( 8P ) are used. This is equivalent to losses
equations ( 1P - 8P ) and are shown at the right hand side of figure
(6.3) with shaded resistors.
cxvi
8
Figure 6.3: Thermal network model for the squirrel cage induction machine
SIM Half Model --Considered
Rotor Iron
Rotor bar (winding)
End-ring
End-winding End-winding
Frame
Stator lamination
Stator winding
Ambient
R12
P2
T1a
R8c
T1b
R1b
R23
P3 P11
10
C9
C4 C3
P5
R56
P6
R69 R78
P7
C5
R25
R311 R1011
C6 C7
C11
C2
C1
R34
P4
R910
R48
End-ring
P9
R10a
R67
T1c
cxvii
In the case of the complete (LIM) model, equations (6.16 and 6.17)
for ( 3P ) and ( 6P ) are respectively modified as ( 3'P ) and ( 6
'P ) while
equations (6.18 – 6.20) for ( 9P ), ( 10P ) and ( 11P ) as derived from the
complete model are added so as to obtain the following set of
equations.
( ) ( ) ........................................................11
11
2112
111 b
b
TTR
TTRdt
dTCP −+−+= ………
… (6.8)
( ) ( ) ( ).....11152
2532
2312
12
222 TT
RTT
RTT
Rdt
dTCP −+−+−+= ………………. ………
… (6.9)
( ) ( ) ..........................................11
4334
2332
333 TT
RTT
Rdt
dTCP −+−+= …………………….
..(6.10)
( ) ( ) ............11
3434
8448
444 TT
RTT
Rdt
dTCP −+−+= ……………………………………
..(6.11)
( ) ( ) ...........................................11
6556
2552
555 TT
RTT
Rdt
dTCP −+−+= …………………
..(6.12)
( ) ( ) .........................11
7667
5665
666 TT
RTT
Rdt
dTCP −+−+= ……………………………..
..(6.13)
( ) ( ) ...................................11
8778
6776
777 TT
RTT
Rdt
dTCP −+−+= ………………………..
…(6.14)
( ) ( ) ( ) ...........111
88
4884
7887
888 c
c
TTR
TTR
TTRdt
dTCP −+−+−+= ……………………..
… (6.15)
( ) ( ) ( ) ................................111
113311
4334
2332
333
' TTR
TTR
TTRdt
dTCP −+−+−+= ………...
(6.16)
cxviii
( ) ( ) ( ) ......................111
9669
7667
5665
666
' TTR
TTR
TTRdt
dTCP −+−+−+= … …………..
.(6.17)
( ) ( ) ........................................11
109910
6969
999 TT
RTT
Rdt
dTCP −+−+= ………………….
(6.18)
( ) ( ) ( )........11110
10910
9101110
1011
101010 a
a
TTR
TTR
TTRdt
dTCP −+−+−+= …………………
(6.19)
( ) ( ) .................................11
10111011
311311
111111 TT
RTT
Rdt
dTCP −+−+= …….……………
.(6.20)
11
C4
C2
C1
R35 P5
R511
End-ring
R10a
Tc
P12 R312
R1213
C12 Stator teeth
P13
Rotor teeth
C13
R713
R
Rotor Iron
Rotor bar
End-winding End-winding
Frame
Stator lamination
Stator winding
Ambient
P1
R12
P2
Ta
R11c
Tb
R1b
R23
P3
P4 10
C8
C5
C3
P6
R67
P
P
C6
R26
R34 R410
C7 C9
cxix
6.3 THE TRANSIENT STATE ANALYSIS
The general transient equation for thermal network system of ‘ N ’
nodes linking others through thermal resistances baR , is
represented as follows:
[ ] [ ]dt
TdCa = 1 a
1 ,,∑
=
=
+−
N
b ba
b
ba
aa R
T
R
TP ……………………………………….
…(6.21)
where
a nodeat generationheat the
b. anda nodes adjoining twobetween resistance thermal
a node of re temperatuthe
a node of ecapacitanc thermal
,
==
==
a
ba
a
a
P
R
T
C
The existence of thermal capacitance in the network demands that
a thermal capacitance matrix as given below will be incorporated.
Figure 6.4: Thermal network model for the squirrel cage induction machine
LIM Full Model
cxx
.....................................................................
...0000
..................
0...000
0...000
0...000
0...000
4
3
2
1
=
NC
C
C
C
C
C …….
(6.22)
Hence we have,
[ ] [ ] [ ] [ ][ ]TGPdt
TdC −= ……………………………………………………………...(
6.23)
Or
[ ] [ ] [ ][ ] ................................................................ ][][ 11 TGCPCdt
Td −− −= ……………...(
6.24)
where
[ ][ ][ ] generators thermalofmatrix column a
esconductanc internodal ofmatrix square a
escapacitanc thermalofmatrix column
===
P
G
C
The power associated with each thermal node is expressed as
shown in this system of algebraic and differential equations which
sum up the thermal behaviour of the developed thermal model of
figure 6.4.
( ) ( ) ........................................................11
11
2112
111 b
b
TTR
TTRdt
dTCP −+−+= …………..
(6.25)
( ) ( ) ( ) ....................................111
6226
3223
1212
222 TT
RTT
RTT
Rdt
dTCP −+−+−+= ……..
(6.26)
( ) ( ) ( ) ( ) ..............1111
123312
4334
5335
2332
333 TT
RTT
RTT
RTT
Rdt
dTCP −+−+−+−+= ……….
(6.27)
cxxi
( ) ( ) ...........................................................11
3434
104410
444 TT
RTT
Rdt
dTCP −+−+= ……..
(6.28)
( ) ( ) ..........................................................11
115511
3553
555 TT
RTT
Rdt
dTCP −+−+= …….
(6.29)
( ) ( ) ..................................................11
2662
7667
666 TT
RTT
Rdt
dTCP −+−+= ……………
(6.30)
( ) ( ) ( ) ( ) .............1111
137713
9779
8778
6767
777 TT
RTT
RTT
RTT
Rdt
dTCP −+−+−+−+= ………
(6.31)
( ) ( ) .......................................................11
108810
7878
888 TT
RTT
Rdt
dTCP −+−+= ………..
(6.32)
( ) ( ) ....................................................11
119911
7979
999 TT
RTT
Rdt
dTCP −+−+= …………
(6.33)
( ) ( ) ( ) .....................111
1010
810810
410410
101010 a
a
TTR
TTR
TTRdt
dTCP −+−+−+= …………
(6.34)
( ) ( ) ( )..........111911
11911
11511
511
111111 TT
RTT
RTT
Rdt
dTCP c
c
−+−+−+= …………… ……..(
6.35)
( ) ( ) ............................................11
13121213
312312
121212 TT
RTT
Rdt
dTCP −+−+= …………
(6.36)
( ) ( ) .........................................11
12131213
713713
131313 TT
RTT
Rdt
dTCP −+−+= ……………(
6.37)
cxxii
The constants ,aT bT and cT are the ambient temperature values
and are equal, the equations are further rearranged to make the
differential the subject as shown below. Matlab programs [98 - 101]
are developed to solve the steady state and transient state
mathematical models of the machine.
( ) ( ) .............................................111
11
2112
11
1
−−−−= b
b
TTR
TTR
PCdt
dT…………………
…(6.38)
( ) ( ) ( ) .............................1111
6226
3223
1212
22
2
−−−−−−= TT
RTT
RTT
RP
Cdt
dT………..
…(6.39)
( ) ( ) ( ) ( ) ........11111
123312
4334
5335
2332
33
3
−−−−−−−−= TT
RTT
RTT
RTT
RP
Cdt
dT………….
..(6.40)
( ) ( ) ............................................111
3434
104410
44
4
−−−−= TT
RTT
RP
Cdt
dT……………
…(6.41)
( ) ( ) .................................................111
115511
3553
55
5
−−−−= TT
RTT
RP
Cdt
dT…………
…(6.42)
( ) ( ) ...............................................111
2662
7667
66
6
−−−−= TT
RTT
RP
Cdt
dT…………..
…(6.43)
( ) ( ) ( ) ( ) ..........11111
137713
9779
8778
6767
77
7
−−−−−−−−= TT
RTT
RTT
RTT
RP
Cdt
dT………
…(6.44)
( ) ( ) ...............................................111
108810
7878
88
8
−−−−= θT
RTT
RP
Cdt
dT…………...
.(6.45)
cxxiii
( ) ( ) .............................................111
119911
7979
99
1
−−−−= TT
RTT
RP
Cdt
dT……………...
(6.46)
( ) ( ) ( ) .........................1111
1010
810810
410410
1010
10
−−−−−−= a
a
TTR
TTR
TTR
PCdt
dT……..
(6.47)
( ) ( ) ( ) .......1111
911119
1111
511511
1111
11
−−−−−−= TT
RTT
RTT
RP
Cdt
dTc
c
…………………
…(6.48)
( ) ( ) ..............................................111
13121213
312312
1212
12
−−−−= TT
RTT
RP
Cdt
dT …….(
6.49)
( ) ( ) .............................................111
12131213
713713
1313
13
−−−−= TT
RTT
RP
Cdt
dT. ……..
(6.50)
Having arranged them in that form, the next thing is to put them in
the matrix form and according to this expression:
[ ] [ ] [ ] [ ][ ] ........................................................ *11 TGCPCT tttt−−
•−=
…………...........(
6.51) where
;
T
T
T
T
T
T
T
T
T
T
T
T
13
12
11
10
9
8
7
6
5
4
3
2
1
=
•
•
•
•
•
•
•
•
•
•
•
•
•
•
T
T
cxxiv
[ ]
1
13
12
11
10
9
8
7
6
5
4
3
2
1
1
000000000000
000000000000
000000000000
000000000000
000000000000
000000000000
000000000000
000000000000
000000000000
000000000000
000000000000
000000000000
000000000000
−
−
=
C
C
C
C
C
C
C
C
C
C
C
C
C
Cand t
[ ]
+
=
13
12
11
10
9
8
7
6
5
4
3
2
11
*
*
*
P
P
GT
GT
P
P
P
P
P
P
P
P
GTP
P
cc
aa
bb
t , [ ]
=
13
12
11
10
9
8
7
6
5
4
3
2
1
T
T
T
T
T
T
T
T
T
T
T
T
T
T and
cxxv
[ ] (6.52)
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
00000000
0000000000
0000000000
0000000000
00000000
000000000
00000000000
13131312137
12131212123
1111119115
1010108104
9119997
8108887
71379787776
676662
5115553
4104443
31235343332
26232221
1211
−−−−
−−−−
−−−−
−−−−−−
−−−−
−−−−−−−
−
=
GGG
GGG
GGG
GGG
GGG
GGG
GGGCG
GGG
GGG
GGG
GGGGG
GGGG
GG
Gt
Some of the entries of the tG -matrix are given as follows:
..........................................................................................1211 GGG b += ……………...
(6.53) ...........................................................................26232122 GGGG ++= …………………
(6.54) ..........................................................................31234353233 GGGGG +++= …………..
(6.55) ...........................................................................................3441044 GGG += ……………
(6.56) ............................................................................................5115355 GGG += …………...
(6.57) ..............................................................................................626766 GGG += …………..
(6.58) .............................................................................71379787677 GGGGG +++= ………...
(6.59) ..........................................................................................8108788 GGG += ……………..
(6.60) ..............................................................................................9119799 GGG += …………..
(6.61) .................................................................................101081041010 aGGGG ++= ………....
(6.62) ........................................................................................111191151111 cGGGG ++= …..…
(6.63)
cxxvi
......................................................................................12312131212 GGG += ………….....
(6.64) ........................................................................................13121371313 GGG += …………...
(6.65)
6.4 THE STEADY STATE ANALYSIS
Equation (6.23) holds firm for the induction motor when it is
rotating. However, at stand still, a different conductance matrix
[ ]ssG is used because of the attendant change in the value of the
convective elements of the branch thermal impedances. The stand
cxxvii
still equation when there is no supply (no heat generation), is given
as:
[ ] [ ] [ ] [ ][ ]TGPdt
TdC ss −= ………………………………………………………………..
(6.66)
During the steady state, the thermal capacitance is at maximum so
that the derivative [ ]0=
dt
Td hence loses its contribution just as it
renders equation(6.21)
as ............. 1.......N a 1 ,∑
=
=
−=N
b ba
baa R
TTP ………………..(6.67)
Hence the algebraic steady-state temperature rise in the proposed
thermal network model in matrix form can be written as follows:
[ ] [ ][ ] ................................................................................... ttt TGP = ……………. …….
(6.68)
So that we have on arranging
[ ] [ ] [ ] ........................................................................................... 1ttt PGT −= …………...(
6.69)
where all the three variables are also defined thus;
cxxviii
=tP
+
13
12
11
10
9
8
7
6
5
4
3
2
11
*
*
*
P
P
GT
GT
P
P
P
P
P
P
P
P
GTP
cc
aa
bb
and
=
13
12
11
10
9
8
7
6
5
4
3
2
1
T
T
T
T
T
T
T
T
T
T
T
T
T
T t
……………………………………………………..(6.70)
=tG
−−−−
−−−−
−−−−
−−−−−−
−−−−
−−−−−−−
−
13131312137
12131212123
1111119115
1010108104
9119997
8108887
71379787776
676662
5115553
4104443
31235343332
26232221
1211
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
00000000
0000000000
0000000000
0000000000
00000000
000000000
00000000000
GGG
GGG
GGG
GGG
GGG
GGG
GGGCG
GGG
GGG
GGG
GGGGG
GGGG
GG
…...(
6.71)
cxxix
With NN2211 G.......... G ,G taking their usual values, the results of the
simulated work is presented in tabular and in graphically forms. In
the simulation, the temperature vector TT which is given by
t4321 ] .... T [ NT TTTTT = is used instead of the temperature rise
vector T . The first node is taken as ambient temperature and is
updated during the simulation so as to get the transient solution
from
[ ] [ ]dt
TdC a
a = .................................... 1.......N a 1 ,,∑
=
=
+−
N
b ba
b
aa
aa R
T
R
TP ………
(6.72)
which in matrix form appears as:
.........................................................................................................GTPdt
dTC −= …
…(6.73)
The summarized equation (6.73) is simulated and the results are
presented in table (6.1) which also shows the percentage
difference in the steady state values for the component parts of the
SIM and Lim models.
Table 6.1: Steady State predicted temperatures: (a) SIM half model; (b) SIM
full model; (c) LIM half model; (d) LIM full model;
(a) (b)
SIM Model Component (full)
Steady State Predicted Temperature (oC)
Percentage Difference [x100]
1 Frame 61.5100 1.0042 2 Stator lamination 76.9287 1.3683 3 Stator winding 78.9422 0.9659 4 End-windingR 80.8451 0.8903 5 Rotor iron 68.5502 6.2275 6 Rotor winding 68.2484 6.3325 7 End-ringR 63.8564 5.6907 *8 Ambient 20.0000 0.0000 9 End-ringL 63.8564 -
cxxx
(d)
(d)
The bar charts of figures 6.5 and 6.6 represent the percentage difference in predicted steady state temperature rise of the component parts. It is very clear that the components of the LIM model have higher percentages of the steady state temperature rise than the SIM model.
1 2 3 4 5 6 70
1
2
3
4
5
6
7
Model component [SIM]
Per
cent
age
diffe
renc
e X
10
Frame 1Stator lamination 2Stator winding 3End windingR 4Rotor iron 5Rotor winding 6End ringR 7
*10 Ambient 20.0000 0.0000 11 End-windingL 80.8451 - SIM Model
Component (half) Steady State Predicted Temperature (oC)
1 Frame 62.5042 2 Stator lamination 78.2970 3 Stator winding 79.9081 4 End-windingR 81.7354 5 Rotor iron 74.7295 6 Rotor winding 74.5809 7 End-ringR 69.5471 *8 Ambient 20.0000
LIM Model Component
(half)
Steady State Predicted Temperature (oC)
1 Frame 58.7013
2 Stator lamination 73.0789
3 Stator winding 69.1442
4 End-windingR 78.6627
5 Rotor iron 78.1632
6 Rotor winding 75.6776
7 End-ringR 70.0144
*8 Ambient 20.0000
9 Stator teeth 71.7736
10 Rotor teeth 78.0752
* Not shown on the graph
LIM Model Component
(full)
Steady State Predicted Temperature (oC)
Percentage Difference [x100]
1 Frame 61.1247 2.4234 2 Stator lamination 76.4032 3.3243 3 Stator winding 74.1461 5.0019 4 End-windingR 83.2488 4.5861 5 Rotor iron 83.0035 4.8403 6 Rotor winding 83.6902 8.0126 7 End-ringR 78.0733 8.0589 *8 Ambient 20.0000 0.0000 9 End-ringL 56.0439 - *10 Ambient 20.0000 0.0000 11 End-windingL 82.2513 - 12 Stator teeth 76.7514 4.9778 13 Rotor teeth 82.9178 4.8426
cxxxi
1 2 3 4 5 6 7 8 90
1
2
3
4
5
6
7
8
9
Model component [LIM]
Per
cent
age
diffe
renc
e x
10
Frame 1Stator lamination 2Stator winding 3End windingR 4Rotor iron 5Rotor winding 6End ringR 7Stator teeth 8Rotor teeth 9
Figure 6.6: Percentage difference in component steady state
temperature for the half and full LIM model
6 .5 TRANSIENT STATE ANALYSIS RESULTS
Here, the graphs of the transient state analysis are presented to
show the rise in temperature of the component parts with time.
Figure 6.5: Percentage difference in component steady state
temperature for the half and full SIM model
cxxxii
0 20 40 60 80 100 120 14020
30
40
50
60
70
80
90
Time[Mins]
Tem
pera
ture
ris
e[°C
]
Graph of temperature rise against time at rated Load
Stator lamination T2Stator winding T3End winding T4Rotor winding T6
This graph above is obtained using power losses equations (P2, P3,
P4 and P6) as outlined in equations (6.9, 6.10, 6.11 and 6.13).
Figure 6.7: Response curve for the predicted temperatures- half SIM model
cxxxiii
0 20 40 60 80 100 120 14020
30
40
50
60
70
80
Time[Mins]
Tem
pera
ture
ris
e[°C
]
Graph of temperature rise against time at rated load
Rotor iron T5End Ring T7Frame T1
This graph above is obtained using power losses equations (P5,
P7, and P1) as outlined in equations (6.12, 6.14, and 6.8).
Figure 6.8: Response curve for the predicted temperatures-
half SIM model continued
cxxxiv
0 50 100 15020
40
60
80
100
Time[Mins]
Tem
pera
ture
ris
e[°C
]
End-WindingR T5
0 50 100 15020
30
40
50
60
70
Time[Mins]
Tem
pera
ture
ris
e[°C
]
End RingR T9
0 50 100 15020
40
60
80
100
Time[Mins]
Tem
pera
ture
ris
e[°C
]
End-WindingL T4
0 50 100 15020
30
40
50
60
70
Time[Mins]
Tem
pera
ture
ris
e[°C
]
End RingL T8
The above graph is obtained using power losses equations (P4, P5,
P8 and P9) as outlined in equations (6.11, 6.12, 6.15 and 6.18).
Figure 6.9: Response curve for predicted temperature and
symmetry for full SIM models
cxxxv
0 20 40 60 80 100 120 14020
30
40
50
60
70
80
90
Time[Mins]
Tem
pera
ture
ris
e[°C
]
Graph of temperature rise against time at rated Load
Stator lamination T2 Stator winding T3End windingL T4 Rotor iron T6
This graph above is obtained using power losses equations (P2, P3,
P4 and P6) as outlined in equations (6.26, 6.27, 6.28 and 6.30).
Figure 6.10: Response curve for the predicted transient state
temperatures for LIM
cxxxvi
0 20 40 60 80 100 120 14020
30
40
50
60
70
80
90
Time[Mins]
Tem
pera
ture
ris
e[°C
]
Graph of temperature rise against time at rated load
End windingR T5Rotor winding T7Frame T1End-ringL T8
The above graph is obtained using power losses equations (P1, P7,
P5 and P8) as outlined in equations (6.25, 6.31, 6.29 and 6.32).
Figure 6.11: Response curve for the predicted temperatures for LIM continued
cxxxvii
0 20 40 60 80 100 120 14020
30
40
50
60
70
80
90
Time[Mins]
Tem
pera
ture
ris
e[°C
]
Graph of temperature rise against time at rated load
End ringR T9AmbientStator teeth T12Rotor teeth T13
Figure 6.12: Response curves for the predicted transient state
temperature rise for LIM continued The above graph is obtained using power losses equations (P9, P12,
and P13) as outlined in equations (6.33, 6.36, and 6.37).
cxxxviii
0 20 40 60 80 100 120 14020
30
40
50
60
70
80
Time[Mins]
Tem
pera
ture
ris
e[°C
]
Graph of temperature rise against time at rated load
End ringR T9End ringL T8
The above graph is obtained using power losses equations (P8 and
P9) as outlined in equations (6.32 and 6.33)
Figure 6.13: Comparing the response curves to show extent of
difference in symmetry in end-ring of LIM model
cxxxix
0 50 100 15020
40
60
80
100
Time[Mins]
Tem
pera
ture
ris
e[°C
]
0 50 100 15020
40
60
80
Time[Mins]
Tem
pera
ture
ris
e[°C
]
0 50 100 15020
40
60
80
100
Time[Mins]
Tem
pera
ture
ris
e[°C
]
0 50 100 15020
30
40
50
60
Time[Mins]
Tem
pera
ture
ris
e[°C
]
End-WindingR T4End RingR T7
End-WindingL T11 End RingL T9
Figure 6.14: Response curve for predicted temperature and
symmetry for full LIM models
The above graph is obtained using power losses equations (P4, P5,
P8 and P9) as outlined in equations (6.28, 6.29, 6.32 and 6.33)
6 .6 DISCUSSION OF RESULTS
cxl
It is obvious from table 6.2(a) representing SIM half model that the
predicted steady state temperature values recorded are slightly
less than that obtained from table 6.2(b) representing SIM full
model.
However, in table 6.2(b), the predicted steady state temperature
values recorded for SIM full model shows that thermal symmetry
effect was at play. This is easily noticed when end ring and end
winding steady state temperature values are considered.
From table 6.2(c) representing LIM half model, the predicted
steady state temperature values recorded are also less than that
obtained from table 6.2(d) representing LIM full model with that of
left end ring giving a reasonable difference.
In table 6.2(c), the predicted steady state temperature values
recorded for LIM full model shows that the effect of thermal
symmetry cannot be noticed again. This is easily observed when
end ring, end winding, stator teeth and rotor teeth steady state
temperature values are considered. By extension, the higher the
size of the machine, the more the influence on the symmetry.
In figures 6.7 and 6.8, the response curves showing the predicted
temperature rise for the machine (LIM) core parts are shown.
Figure 6.9 shows the response curve for predicted temperature for
full SIM model showing the symmetry effect. It is observed that the
left and right parts of the machine core parts exhibited the same
graphical characteristics showing good symmetry. This is not the
same with the LIM model as is evident in table 6.2(c).
Figures 6.10–6.13 present the response curve for predicted
temperature for LIM model. While the predicted temperature rise is
relatively small for the left end-ring and the frame part, the end
cxli
winding, the rotor teeth and the rotor iron showed a remarkable
increase with the end-winding showing the highest value. Figures
6.13 and 6.14 are there for the comparison of response curve for
predicted temperature for LIM and SIM models in terms of
symmetry effect. It is just clear that unlike in the case of SIM, there
is no associated symmetry exhibited in the LIM configuration.
CHAPTER SEVEN
CONCLUSION AND RECOMMENDATIONS
7.1 CONCLUSION
In the work presented so far, the need for thermally modeling a
system such as this machine is highlighted. The basics of the
cxlii
thermal modeling are introduced and the general equation for the
implementation obtained. The calculation of thermal capacitances,
thermal resistances and the consideration of losses all led to the
determination of the thermal conditions of the core parts. For the
full nodal configuration, the predicted temperature rise in degree
centigrade for the core parts of the machine are as follows: frame
(61.51), stator lamination (76.93), stator winding (79.94), end-
windingR (80.85), rotor iron (68.55), rotor winding (68.25), end-
ringR (63.86), end-ringL (63.86) and end-windingL (80.85) for SIM
model and frame (61.13), stator lamination (76.40), stator winding
(74.15), end-windingR (83.25), rotor iron (83.00), rotor winding
(83.69), end-ringR (78.07), end-ringL (56.04), end-windingL
(82.25), stator teeth (76.75), rotor teeth (82.92) for LIM model.
It is observed that contrary to the research results of some authors,
the machine does not have a uniform increase in temperature in
some of the core parts. The larger the machine, the more the
difference in temperature meaning reduced asymmetry effect.
The transient and steady state models are analyzed. Tabular and
graphical results from the steady and transient states simulation
are presented leading to a clearer comparison of results obtained.
Some discrepancies as may be noticed in this work are likely
coming from the neglect of radiation effect cum errors due to
assumptions and approximations.
In conclusion, this work can appropriately be employed to predict
the temperature distribution in induction machine especially when
used for wind energy generation. The results obtained here
provide useful information in area of machine design and thermal
characteristics of the induction machine.
cxliii
7.2 RECOMMENDATION
The thermal lumped model that has been developed gives a good
estimation of the machine temperature but there is more work that
can be done to further improve the model, some of which are:
• Setting up an equivalent electrical model for loss calculation. The loss
calculation for the lumped circuit model has been partly based on the
estimated data. Setting up a separate electrical circuit for loss
calculation based on geometrical data will give the free will of estimating
the temperature on theoretical machine design with much ease.
• Accounting for the Cooling characteristics. The frame to ambient
thermal
resistance has been decided based on measured data, giving an
empirical relation as the cooling characteristics were not available,
future work needs to take the cooling characteristic into consideration
so as to make the model functional for a realistic range of temperature
condition.
• Calculation of the thermal losses in a FEM simulation program and
validating the model through finite element method FEM calculations is
likely to give a more sound result.
Generally, temperatures variations should be given considerable
importance in the design and protection of our machines. A data
base should be produced from several generated thermal results
for predictive purposes. This will go a long way in the improvement
of loadability schedules especially in wind energy generation
schemes.
REFERENCES
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cxliv
[2] A. Boglietti, A. Cavagnino and D. A. Staton, “TEFC Induction Motors Thermal Model: A parameter Sensitivity Analysis“, IEEE Transactions on Industrial Applications, Vol. 41, no. 3, pp. 756-763, May/Jun. 2005. [3] A. Boglietti, A. Cavagnino and D. A. Staton, “Thermal Analysis of TEFC Induction Motors”, 2003 IAS Annual Meeting, ,Salt Lake City, USA, Vol. 2, pp. 849-856, 12 – 16 October 2003. [4] D. A. Staton and E. So, “Determination of optimal Thermal Parameters for Brushless Permanent Magnet Motor Design”, IEEE Transactions on Energy Conversion , Vol. 1, pp. 41-49, 1998. [5] X. Ding , M. Bhattacharya and C. Mi , “ Simplified Thermal Model of PM Motors in Hybrid Vehicle Applications Taking into Account Eddy Current Loss in Magnets”, Journal of Asian Electric Vehicles, Volume 8, Number 1, pp.1337, June 2010.
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cxlv
Induction Motors Using Computational Fluid Dynamic Modelling,”
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APPENDIX
Program data
Program-A: Thermal network model for the squirrel cage induction machine
(11n), HALF OF SIM MODEL --C onsidered global R1b R12 R34 R25 C1 C2 C3 C4 P1 P2 P3 x0 t0 tf tspan xb global C5 C6 C7 C8 R8c R48 R67 R78 R56 R23 global P4 P5 P6 P7 P8 xc Thermal Differential equations Theta(1)=(1/C1)*(P1-(x(1)-xb)/R1b-(x(1)-x(2))/R12); Theta(2)=(1/C2)*(P2-(x(2)-x(1))/R12-(x(2)-x(3))/R23-(x(2)-x(5))/R25); Theta(3)=(1/C3)*(P3-(x(3)-x(2))/R23-(x(3)-x(4))/R34); Theta(4)=(1/C4)*(P4-(x(4)-x(8))/R48-(x(4)-x(3))/R34); Theta(5)=(1/C5)*(P5-(x(5)-x(2))/R25-(x(5)-x(6))/R56); Theta(6)=(1/C6)*(P6-(x(6)-x(7))/R67-(x(6)-x(5))/R56); Theta(7)=(1/C7)*(P7-(x(7)-x(8))/R78-(x(7)-x(6))/R67); Theta(8)=(1/C8)*(P8-(x(8)-x(4))/R48-(x(8)-x(7))/R78-(x(8)-xc)/R8c); global R1b R12 R34 R25 C1 C2 C3 C4 P1 P2 P3 x0 t0 tf tspan xb global C5 C6 C7 C8 R8c R48 R56 R78 R67 R23 global P4 P5 P6 P7 P8 xc Initial temperature x0=[20]; Thermal Capacitances C1=18446.55; C2=4450.625; C3=423.388; C4=539.92; C5=3204.08; C6=408.267;
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C7=218.785; C8=1006; Heat Losses PfeT=292.196; Ks=0.975; P2=(Ks*PfeT); P3=384.27824; P4=84.35376; P5=(1-Ks)*PfeT; P6=72.503; P7=111.04; Thermal resistances R1b=0.0416; R12=15.44e-3; R23=35.58e-3; R25=0.131; R34=0.1751; R48=1.886; R56=4.115e-3; R67=0.1055; R78=0.932; R8c=0.015; xb=20; xc=20; t0=0.0; tinterval=0.5; tf=7560; tspan=t0:tinterval:tf; figure(1); plot(t/60,x(:,2),'r'); grid on hold on plot(t/60,x(:,3),'b'); plot(t/60,x(:,4),'g'); plot(t/60,x(:,6),'c'); xlabel('Time[Mins]') ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated Load') legend('Stator lamination','Stator winding','End winding','Rotor winding') figure(2); plot(t/60,x(:,5),'r'); grid on hold on plot(t/60,x(:,7),'g'); plot(t/60,x(:,1),'b'); xlabel('Time[Mins]')
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ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated load') legend('Rotor iron','End Ring','Frame') display('computed steady-state temperatures')
Program-B: Thermal network model for the squirre l cage induction machine
(11n), FULL SIM MODEL –Considered global R1b R12 R34 R25 C1 C2 C3 C4 P1 P2 P3 x0 t0 tf tspan xb xa xc
R56 R23 global C8 C9 C10 C11 C12 R69 R910 R1011 R10a R311 R8c R48 R67
R78 global P4 P5 P6 P7 P8 P9 P10 P11 xc Thermal Differential equations Theta(1)=(1/C1)*(P1-(x(1)-xb)/R1b-(x(1)-x(2))/R12); Theta(2)=(1/C2)*(P2-(x(2)-x(1))/R12-(x(2)-x(3))/R23-(x(2)-x(5))/R25); Theta(3)=(1/C3)*(P3-(x(3)-x(2))/R23-(x(3)-x(4))/R34-(x(3)-x(11))/R311); Theta(4)=(1/C4)*(P4-(x(4)-x(8))/R48-(x(4)-x(3))/R34); Theta(5)=(1/C5)*(P5-(x(5)-x(2))/R25-(x(5)-x(6))/R56); Theta(6)=(1/C6)*(P6-(x(6)-x(7))/R67-(x(6)-x(5))/R56-(x(6)-x(9))/R69); Theta(7)=(1/C7)*(P7-(x(7)-x(8))/R78-(x(7)-x(6))/R67); Theta(8)=(1/C8)*(P8-(x(8)-x(4))/R48-(x(8)-x(7))/R78-(x(8)-xc)/R8c); Theta(9)=(1/C9)*(P9-(x(9)-x(6))/R69-(x(9)-x(10))/R910); Theta(10)=(1/C10)*(P10-(x(10)-x(9))/R910-(x(10)-x(11))/R1011-(x(10)-xa)/R10a); Theta(11)=(1/C11)*(P11-(x(11)-x(3))/R311-(x(11)-x(10))/R1011); global R1b R12 R34 R25 C1 C2 C3 C4 P1 P2 P3 R23 x0 t0 tf tspan xb xa xc global C5 C6 C7 C8 C9 C10 C11 C12 R69 R910 R1011 R10a R311
R8c R48 global P4 P5 P6 P7 P8 P9 P10 P11 xc Initial temperature x0=[20]; Thermal Capacitances C1=18446.55; C2=4450.625; C3=423.388; C4=539.92; C5=3204.08; C6=408.267; C7=218.785; C8=1006;
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C9=C7; C10=C8; C11=C4; Heat Losses PfeT=292.196; Ks=0.975; P2=(Ks*PfeT); P3=384.27824; P4=84.35376; P5=(1-Ks)*PfeT; P6=72.503; P7=111.04; P9=P7; P11=P4 Thermal resistances R1b=0.0416; R12=15.44e-3; R23=35.58e-3; R25=0.131; R34=0.1751; R48=1.886; R56=4.115e-3; R67=0.1055; R78=0.932; R8c=0.015; R69=R67; R910=R78; R1011=R48; R10a=R8c; R311=R34; xb=20; xc=20; xa=20; t0=0.0; tinterval=0.5; tf=7560; tspan=t0:tinterval:tf; figure(1); plot(t/60,x(:,2),'r'); grid on hold on plot(t/60,x(:,3),'b'); plot(t/60,x(:,4),'g'); plot(t/60,x(:,6),'c'); xlabel('Time[Mins]') ylabel('Temperature rise[°C]')
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title('Graph of temperature rise against time at rated Load') legend('Stator lamination','Stator winding','End winding','Rotor winding') figure(2); plot(t/60,x(:,5),'r'); grid on hold on plot(t/60,x(:,7),'g'); plot(t/60,x(:,1),'b'); xlabel('Time[Mins]') ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated load') legend('Rotor iron','End Ring','Frame') figure(3); plot(t/60,x(:,9),'r'); grid on hold on plot(t/60,x(:,11),'b'); xlabel('Time[Mins]') ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated load') legend('End ringL','End windingL') display('computed steady-state temperatures')
Program-C: Thermal network model for the squirrel cage induction IM (13n), Half (LHS) of the LIM model --Cons idered
function Theta=oti3(t,x) global R1b R12 R34 R25 C1 C2 C3 C4 P1 P2 P3 x0 t0 tf tspan xb global C5 C6 C7 C8 R8c R48 R67 R78 R56 R23 global P4 P5 P6 P7 P8 xc Thermal Differential equations Theta =zeros(8,1); Theta(1)=(1/C1)*(P1-(x(1)-xb)/R1b-(x(1)-x(2))/R12); Theta(2)=(1/C2)*(P2-(x(2)-x(1))/R12-(x(2)-x(3))/R23-(x(2)-x(5))/R25); Theta(3)=(1/C3)*(P3-(x(3)-x(2))/R23-(x(3)-x(4))/R34); Theta(4)=(1/C4)*(P4-(x(4)-x(8))/R48-(x(4)-x(3))/R34); Theta(5)=(1/C5)*(P5-(x(5)-x(2))/R25-(x(5)-x(6))/R56); Theta(6)=(1/C6)*(P6-(x(6)-x(7))/R67-(x(6)-x(5))/R56);
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Theta(7)=(1/C7)*(P7-(x(7)-x(8))/R78-(x(7)-x(6))/R67); Theta(8)=(1/C8)*(P8-(x(8)-x(4))/R48-(x(8)-x(7))/R78-(x(8)-xc)/R8c); global R1b R12 R34 R25 C1 C2 C3 C4 P1 P2 P3 x0 t0 tf tspan xb global C5 C6 C7 C8 R8c R48 R56 R78 R67 R23 global P4 P5 P6 P7 P8 xc Initial temperature x0=[20]; Thermal Capacitances C1=18446.55; C2=4450.625; C3=423.388; C4=539.92; C5=3204.08; C6=408.267; C7=218.785; C8=1006; Heat Losses PfeT=292.196; Ks=0.975; P2=(Ks*PfeT); P3=384.27824; P4=84.35376; P5=(1-Ks)*PfeT; P6=72.503; P7=111.04; Thermal resistances R1b=0.0416; R12=15.44e-3; R23=35.58e-3; R25=0.131; R34=0.1751; R48=1.886; R56=4.115e-3; R67=0.1055; R78=0.932; R8c=0.015; xb=20; xc=20; t0=0.0; tinterval=0.5; tf=7560; tspan=t0:tinterval:tf;
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figure(1); plot(t/60,x(:,2),'r'); grid on hold on plot(t/60,x(:,3),'b'); plot(t/60,x(:,4),'g'); plot(t/60,x(:,6),'c'); xlabel('Time[Mins]') ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated Load') legend('Stator lamination','Stator winding','End winding','Rotor winding') figure(2); plot(t/60,x(:,5),'r'); grid on hold on plot(t/60,x(:,7),'g'); plot(t/60,x(:,1),'b'); xlabel('Time[Mins]') ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated load') legend('Rotor iron','End Ring','Frame') display('computed steady-state temperatures')
Program-D: Thermal network model for the squirrel cage induction IM
(13n), Complete LIM model –C onsidered global R1b R12 R34 R25 C1 C2 C3 C4 P1 P2 P3 x0 t0 tf tspan xb xa xc global C8 C9 C10 C11 C12 R69 R910 R1011 R10a R311 R8c R48 R67
R78 global P4 P5 P6 P7 P8 P9 P10 P11 xc R56 R23 Thermal Differential equations Theta(1)=(1/C1)*(P1-(x(1)-xb)/R1b-(x(1)-x(2))/R12); Theta(2)=(1/C2)*(P2-(x(2)-x(1))/R12-(x(2)-x(3))/R23-(x(2)-x(5))/R25); Theta(3)=(1/C3)*(P3-(x(3)-x(2))/R23-(x(3)-x(4))/R34-(x(3)-x(11))/R311-(x(3)- x(12))/R312); Theta(4)=(1/C4)*(P4-(x(4)-x(8))/R48-(x(4)-x(3))/R34); Theta(5)=(1/C5)*(P5-(x(5)-x(2))/R25-(x(5)-x(6))/R56); Theta(6)=(1/C6)*(P6-(x(6)-x(7))/R67-(x(6)-x(5))/R56-(x(6)-x(9))/R69-(x(6)-
x(13))/R613); Theta(7)=(1/C7)*(P7-(x(7)-x(8))/R78-(x(7)-x(6))/R67); Theta(8)=(1/C8)*(P8-(x(8)-x(4))/R48-(x(8)-x(7))/R78-(x(8)-xc)/R8c); Theta(9)=(1/C9)*(P9-(x(9)-x(6))/R69-(x(9)-x(10))/R910);
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Theta(10)=(1/C10)*(P10-(x(10)-x(9))/R910-(x(10)-x(11))/R1011-(x(10)-xa)/R10a); Theta(11)=(1/C11)*(P11-(x(11)-x(3))/R311-(x(11)-x(10))/R1011); Theta(12)=(1/C12)*(P12-(x(12)-x(3))/R312-(x(12)-x(13))/R1213); Theta(13)=(1/C13)*(P13-(x(13)-x(6))/R613-(x(13)-x(12))/R1213); global R1b R12 R34 R25 C1 C2 C3 C4 P1 P2 P3 x0 t0 tf tspan xb xa xc R613
R312 global C5 C6 C7 C8 C9 C10 C11 C12 C13 R69 R910 R1011 R10a
R311 R8c global P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 C13 R613 R312 R1213 P13 R56 Initial temperature x0=[20]; Thermal Capacitances C1=18446.55; C2=4450.625; C3=423.388; C4=539.92; C5=3204.08; C6=408.267; C7=218.785; C8=1006; C9=C7 C10=C8 C11=C4 C12=341.33 C13=871.566 Heat Losses PfeT=292.196; Ks=0.975; P2=(Ks*PfeT); P3=384.27824; P4=84.35376; P5=(1-Ks)*PfeT; P6=72.503; P7=111.04; P9=P7 P11=P4 P12=68.113 P13=93.445 Thermal resistances R1b=0.0416; R12=15.44e-3; R23=35.58e-3; R25=0.131; R34=0.1751; R48=1.886; R56=4.115e-3;
clxiii
R67=0.1055; R78=0.932; R8c=0.015; R69=67 R910=R78 R1011=R48 R10a=R8c R311=R34 R613=0.002703 R312=0.02245 R1213=0.12576 xb=20; xc=20; xa=20; t0=0.0; tinterval=0.5; tf=7560; tspan=t0:tinterval:tf; figure(1); plot(t/60,x(:,2),'r'); grid on hold on plot(t/60,x(:,3),'b'); plot(t/60,x(:,4),'g'); plot(t/60,x(:,6),'c'); xlabel('Time[Mins]') ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated Load') legend('Stator lamination','Stator winding','End winding','Rotor winding') figure(2); plot(t/60,x(:,5),'r'); grid on hold on plot(t/60,x(:,7),'g'); plot(t/60,x(:,1),'b'); xlabel('Time[Mins]') ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated load') legend('Rotor iron','End Ring','Frame') figure(3); plot(t/60,x(:,9),'r'); grid on hold on plot(t/60,x(:,11),'b'); plot(t/60,x(:,12),'g'); plot(t/60,x(:,13),'c'); xlabel('Time[Mins]')
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ylabel('Temperature rise[°C]') title('Graph of temperature rise against time at rated load') legend('End ringR','End windingR','Stator teeth','Rotor teeth') display('computed steady-state temperatures')
