66
Thermal modelling of the Synchronous Reluctance Machine MOHAMAD MAHMOUDI Masters’ Degree Project Stockholm, Sweden April 2012 XR-EE-E2C 2012:004
Thermal modelling of theSynchronous Reluctance Machine
MOHAMAD MAHMOUDI
Masters’ Degree ProjectStockholm, Sweden April 2012
XR-EE-E2C 2012:004
Thermal Modelling of the Synchronous Reluctance Machine
MOHAMAD MAHMODUI
Supervisors
Khashayar Javidi, ABB LV Motors
Dr. Freddy Magnussen, ABB LV Motors
Examiner
Prof. Chandur Sadarangani, EEC, KTH
Master ThesisElectrical Machines and Power Electronics
School of Electrical EngineeringRoyal Institute of Technology (KTH)
Stockholm, Sweden 2012
XR-EE-E2C 2012:004
KTH School of Electrical Engineering
SE-100 44 Stockholm
SWEDEN
Akademisk avhandling som med tillstand av Kungl Tekniska hogskolan framlagges till of-
fentlig granskning for avlaggande av Civilinjengorsexamen i Elektroteknik dagen den 28 maj
2012 klockan 13.15 i sal H2, EEC, Kungl Tekniska hogskolan, Teknikringen 39, Stockholm.
c© MOHAMAD MAHMODUI, April 2012
Tryck: Universitetsservice US AB
Abstract
In this paper two methods for thermally modelling the Synchronous Reluctan-
ce Machine (SynRM) have been developed and compared to measured data.
The first is an analytical method for estimating the end winding temperature
based on Brostroms formula. Curve fitting to the measured data was done by
Gauss-Newton estimation. The second method uses a lumped thermal network
for steady state temperature estimation based on machine geometry. Thermal
modelling of the machine parts are portrayed. The frame to ambient thermal
resistance is derived from measured data. Friction and windage losses are calcu-
lated through measurement, the core losses are separated based on finite element
calculations. The results for both methods show good agreement. A comparison
between the different methods is made.
iii
Sammanfattning
I denna rapport har tva metoder for att termisk modellera Synkron Reluktans
Motorn tagits fram och jamforts med matdata. Forst en analytisk metod kal-
lad Brostroms formula, baserad pa kurv-anpassning till matdata enligt Gauss-
Newton optimering, estimerar temperaturokningen i harvandorna. Sedan har ett
termisk natverk utvecklats for stationart tillstand, baserad pa maskingeometrin.
Termisk modellering av olika maskindelar portratteras. Den termiska resistansen
till omgivningen harleds fran matdata. Friktion och lindningsforluster beraknas
fran matdata, jarnforlusterna separeras och fordelas enligt FEM resultat. Resul-
tatet for de olika metoderna visar god overensstammelse. En jamforelse mellan
de olika metoderna har gjorts.
v
Acknowledgment
I would like to thank my family for always being there for me. My supervisors
Khashvad Javidi and Dr. Freddy Magnussen for their guidance and motiva-
tional speeches when I needed them. My examiner Chandur Sadaragani for his
interesting inputs. Dr. Dan Fors for all the long discussions and encouraging
talks, as well as being a wall to bounce ideas off. Per-Ake Salin, Tech. Lic.
Rathna Chitroju, Gustav Borg and Raimondo for all their help at the begin-
ning as well as during the thesis. I would also like to express my thanks for
anyone not mentioned who has supported me. Thank you.
vii
Contents
Page
Abstract iii
Sammanfattning v
Acknowledgments vii
Contents x
1 Introduction 1
1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Problem description . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Goal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Thermal Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4.1 Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4.2 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4.3 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 Brostroms Formula 9
2.1 Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2.1 Data fitting . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2.2 Machine types . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3.1 Sensitivity analysis . . . . . . . . . . . . . . . . . . . . . . 16
2.3.2 Comparing FS160 and FS250 . . . . . . . . . . . . . . . . 18
3 Lumped thermal model 19
3.1 The finished network . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.2 Thermal resistance of the Machine parts . . . . . . . . . . . . . . 21
3.2.1 Internal air . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2.2 Stator yoke . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.2.3 Teeth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.2.4 Winding . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.2.5 End Winding . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2.6 Air gap . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2.7 Rotor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2.8 Shaft and Bearing . . . . . . . . . . . . . . . . . . . . . . 31
3.2.9 Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
ix
3.3 Setting up the nodal network . . . . . . . . . . . . . . . . . . . . 34
3.4 Power Loss calculations . . . . . . . . . . . . . . . . . . . . . . . 35
3.4.1 Distributed copper losses . . . . . . . . . . . . . . . . . . 35
3.4.2 Friction losses . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.4.3 Iron losses . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.5 Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4 Measurements setup 41
5 Conclusion and future work 45
5.1 Comparing the two models . . . . . . . . . . . . . . . . . . . . . 45
5.2 Error sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
5.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
5.4 Future work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
List of Figures 50
A FS160 measurement data 53
B Stator teeth 54
C Thermal network 59
C.1 The final thermal network . . . . . . . . . . . . . . . . . . . . . . 60
C.2 Resisitances and Losses for the FS160 @ 3000 rpm and . . . . . . 62
D Machine data for FS160 63
x
Chapter 1
Introduction
1.1 Background
The population of the world has now passed the seven billion mark and the en-
vironmental footprint we leave on earth is becoming more and more strenuous.
The expansion of the electrical infrastructure in developing countries as well as
the rapid growth of the world population has increased the demand for electrical
energy. As the resources of the earth are not unlimited, the rising demand has
pushed up the electricity prices, see figure 1.1.
2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 201010
15
20
25
30
35
40
45
50
55
60
Year
Price
[EU
R/M
Wh
]
Figure 1.1: The electricity price in Sweden over
the last decade [1]. The fluctuation is due to the
energy consumption being lower during warmer
years.
1
Figure 1.2: The different types of machines used in Sweden [6].
Figure 1.3: The energy consumption in Sweden [6].
The higher energy price in combination with the climate-change awareness has
led to energy conservation laws being formed [2]. The majority of the electrical
energy in the world is consumed by electrical machines, see figures 1.2 and 1.3.
Due to this fact, there exist today many government enforced restrictions, stan-
dards and guidelines on manufacturers, regarding the efficiency of the machines
produced i.e. International Electrotechnical Commission (IEC), National Elec-
trical Manufacturers Association (NEMA) etc. [3–5].
The most common machine in the world by far is the Induction Machine (IM).
Since it was invented in 1883 it has been the front runner for its simplicity and
ability to be connected directly on the grid line (DOL). The efficiency of the IM
has continuously been improved over its lifetime and the possibilities to further
improve the design are reaching saturation. The stagnated efficiency of the IM
and the lower prices of inverter drives have promoted research of other machine
2
designs. The Synchronous Reluctance Machine (SynRM) has been researched
since 1923 and is used in certain applications today [7]. However unlike the IM
the SynRM cannot be run Directly On Line (DOL) but only through Variable
Speed Drive (VSD). Only as recently as 1980 did the mass production of the
frequency controllers take off enabling affordable VSD solutions [8].
Consumers are also becoming more aware of the machine Life Cycle Cost (LCC)
as the initial and maintenance costs are only 10% of the total cost while the
remaining 90% is energy cost. This awareness further promotes high efficiency
machines [8, 9]. The higher initial investment of the machine is thus often paid
off within a couple of years, through savings made by lower electricity bills [8].
The SynRM is proven to not only be colder but also more efficient than the
IM [7]. The drawback of the machine is that, due to physical properties has
a lower power factor. This thesis will not cover the physical properties of the
machine design; instead the reader is referred to [7] for further reading.
1.2 Problem description
The major limitations of contemporary machines are due to thermal restraints.
Continuous operation for a machine under rated load, is usually limited by the
maximum winding temperature rise the machine can handle. Surpassing the
thermal boundaries can reduce the lifetime of the machine dramatically [8]. A
rule of thumb is that the lifetime of the machine is halved for every 10 degree
K that is over the insulation class [8].
Potential problems that can occur if the temperature is too high are oxidation of
the isolation material on the copper wires, mechanic expansion of machine parts
exceeding their tolerances. Over 40% of the failures can be derived to bearing
faults; causes being the lubrication structure changing at the molecular level at
higher temperatures. Between the stator winding failure and the bearing failure
they make up just less than 80% of all failures [9].
Being able to predict the temperature rise under different operating points is
crucial when designing machines and to do this accurately, thermal models of
the machine are needed. The models need to be a correct representation of how
the actual physical machine behaves and for this a certain complexity to the
model is needed.
1.3 Goal
The goal of this thesis is to thermally model a Synchronous Reluctance Ma-
chine. There are several methods to do this of varying complexities, among
many which are computer aided [10–13]. This thesis will look into and compare
an analytical method with a thermal lumped modell. The results will also be
compared to measured data.
3
The analytical model is derived from Brostroms formula, see equation 2.2 [8].
It is a simple and elegant empirical formula based on data fitting. This formula
was formulated by ASEA engineer Brostrom. The second model is based on
building up a thermal flow network using the lumped model approach. The
complexity of this method is high and a great deal needs to be known about the
design and geometrical dimensions of the machine.
1.4 Thermal Basics
A grasp of the thermal basics is needed in order to fully understand thermal
modelling. Where ever there is a temperature gradient, heat transfer will occur
from the warmer to the colder side [14, 15]. The transfer can occur in different
ways, this section will try to give a brief overview of the subjects addressed in
this thesis. For more information the reader is referred to [14,15].
1.4.1 Conduction
Conduction describes heat transfer in any material through molecular collisions,
lattice vibrations and unbound electron flow. The heat transfer rate in a mate-
rial is given by Fouriers Law
P = −λAdT
dx(1.1)
where A (m2) is the cross sectional area, λ (W/m·K) is the thermal conductivity
of the material, which describes how well a material will transport heat and
dT/dx is the temperature gradient in the material. The negative sign is due
to that the heat transfer occurs opposite to the temperature gradient see figure
1.4.
Figure 1.4: Heat transfer in one dimension, T1 > T2.
4
From figure 1.4 it can be seen that
dT
dx=
T2 − T1
L(1.2)
which gives
P = −λAT2 − T1
L= λA
T1 − T2
L(1.3)
By comparing equation (1.3) to Ohms law for electrical circuits
P =λA
L· (T1 − T2) ≡ I =
A
ρL· (U1 − U2) (1.4)
the thermal and electrical analogy can be seen, giving the expression for thermal
resistance as
Rth =L
λA. (1.5)
In the cylindrical case it is assumed for simplicity that there is only heat transfer
in the radial direction and the thermal resistance is derived in [15] for figure 1.5
as
Rth =ln (ro/ri)
λLϕ. (1.6)
Figure 1.5: Heat transfer in the radial direction for a cylinder, T1 > T2.
where ϕ (rad) is the angle of the sector, ro (m) and ri (m) are the outer and
inner radius of the cylinder. In this thesis radial heat transfer is assumed unless
specified otherwise.
5
1.4.2 Radiation
Thermal radiation is the heat dissapation to the environment through electro-
magnetic waves, see figure 1.6. Radiation can occur regardless of solid or fluids.
The amount of heat dissipated through radiation is given by
P = Aεσ(T 4s − T 4
amb) (1.7)
where σ is Stefan-Boltzmann constant (σ = 5.67 · 10−8W/m2 ·K4) and ε is the
surface emissivity (0 < ε < 1) relative to a blackbody. The surface emissivity
depends on the coating of the frame and can be obtained from manufacturers.
Ts and Tamb are the surface and ambient temperatures respectively given in
absolute temperature (K).
Figure 1.6: An example of radiation [16].
1.4.3 Convection
Convection is the heat transfer that occurs between a surface and a fluid when
there is a temperature gradient. Convection can occur as free or forced. Free
convection is when the fluid around a surface is not being moved in any way
by an external force. Forced convection on the other hand is when there is an
external force being applied to the fluid, which in the case of Totally Enclosed
Fan Cooled (TEFC) machines is by the end-mounted fan. The heat transfer by
convection is given by equation 1.8. When determining h a clear insight to the
fan characteristic is needed. Since the fins are open, the pressure will decrease
along the axis of the machine, see figure 1.7. The main obstacle when modelling
convection is therefore the determination the fluid thermal conductivity h.
6
P = hA(T1 − T2) (1.8)
where h (W/m2 ·K) is the fluid thermal conductivity, consequently the thermal
resistance for convection is given by
Rth =1
hA. (1.9)
Figure 1.7: A rough estimation of how the air moves around the machine.
7
Chapter 2
Brostroms Formula
2.1 Description
ASEA (now ABB) engineer Brostrom formulated an equation 2.1 in the middle
of the last century to describe the temperature rise of the end winding from
empirical data. It has continuously been used by engineers for its straightfor-
wardness and simplicity to implement [8]
∆T = kcs ·√
Pcu · Ptot (2.1)
where Pcu (W ) are the copper losses, Ptot (W ) the total losses and kcs (K/W )
an empirical tuning constant determined based on measuring data. Note that
the copper losses are accounted twice as they are the biggest contributor to the
temperature rise.
The Brostroms formula was formulated before the introduction of Variable
Speed Drive (VSD) and thus it presumes the machine to be connected Direct
On Line (DOL). A modified version suggested by Moghaddam takes VSD into
consideration [7], see equation 2.2.
∆T = kcs ·√
P xcu · P y
tot · nz (2.2)
where n is the speed in rounds per minute (rpm). Note that there now are four
unknowns, kcs, x, y and z, that need to be found instead of just one. As the
values of x, y and z are arbitrary the dimension of kcs will no longer be the
same, but instead have a dimension to guarantee that the resulting dimension
is in kelvin (K).
2.2 Calculations
The total power losses Ptot was determined by comparing the power that was
fed into the system and the mechanical output of the system
Ptot = Pin − Pm (2.3)
9
where Pin (W ) is the input power and Pm (W ) the mechanical output power.
The total copper losses Pcu for all the phases are calculated by Ohms law as the
nominell phase current and phase resistance are known
Pcu = 3 · I2phase ·Rphase (2.4)
where Iphase (A) and Rphase (Ω) are the phase current and resistance. The
resistance of copper is not fixed but changes with regard to the temperature.
The resistance at the temperature T is given by
RT = R0 ·T + 235
T0 + 235(2.5)
where R0 is the reference resistance (Ω) of the winding at the reference temper-
ature T0 (C). Combining equation 2.4 and 2.5 yields
Pcu = 3 · I2phase ·R0 ·T + 235
T0 + 235(2.6)
2.2.1 Data fitting
Data fitting through the least square method was used in order to find the best
value for the unknown coefficients from equation 2.2. A brief overview of the
method will be given below. For deeper understanding the reader is refereed
to [17].
The goal was to try to minimize the sum of the square difference, equation
2.8, between the measured and analytically calculated values, equation 2.7.
ri = Tm−i︸ ︷︷ ︸
Measured
− Ta−i︸︷︷︸
Analytical
(2.7)
minS = min∑
k
r2i (2.8)
This was solved with MATLAB1 using fminsearch that is part of the opti-
mization toolbox. fminsearch can be used for unrestrained nonlinear opti-
mization [18].
2.2.2 Machine types
The frame sizes (FS2) of the machines that were studied ranged from FS90 −FS250. At the time of the fitting there was extensive measurement data on
frame size 160. The data that was available was for different active lengths of
1MATLAB is a registered product of MathWorks Inc.2Standardized sizes for electrical machines. Usually depicting the height between the
ground to the center of the machine shaft in mm.
10
Figure 2.1: An example of a curve fitting.
the machine. Until now the active length has been part of kcs in equation 2.2
but in order to get a better fit, the active length dependency was separated from
kcs giving
∆T = kcs︸︷︷︸
new
·
√
P xcu · P y
tot · nz · Lw
act (2.9)
where L (m) is the active length of the machine. Observe that the dimension
of the kcs in equation 2.9 is not the same as in equation 2.2.
Error in measurements
In order to get the best fitting at rated levels, measurement data in the interval
75%−110% of rated torque was studied and temperature points that were off in
an unreasonable manner, i.e. diving 20K between two working point to reappear
at the next point were assumed to be measuring errors and excluded from the
fitting. A weighted function was used that made sure to give less weight to the
points that diverged from the cluster.
11
2.3 Result
The results below is for the frame size FS160 with different active lengths. The
absolute value of the torque was in the interval 116 − 172Nm. Note that the
double data point are machines of different active length and/or Y/∆ connec-
tion at the same p.u torque level.
∆T = 0.1736 ·√
P 0.2004cu · P 1.9809
tot · n−0.6513 · L−1.5129act
85 90 95 100 105 110 11560
70
80
90
100
110
Framesize 160 @ 1000 rpm.
Torque [%] of nominal
Tem
per
atu
reri
se∆
T[K
]
Tmeasured
Tanalytical
Torque [%] of nominal
Calc
ula
tion
erro
r[K
]
∆ : Tm − Ta
Figure 2.2: The result for frame size 160 at 1000 rpm.
85 90 95 100 105 110 115
60
70
80
90
100
Framesize 160 @ 1500 rpm.
Torque [%] of nominal
Tem
per
atu
reri
se∆
T[K
]
Tmeasured
Tanalytical
Torque [%] of nominal
Calc
ula
tion
erro
r[K
]
∆ : Tm − Ta
Figure 2.3: The result for frame size 160 at 1500 rpm.
12
85 90 95 100 105 110 11550
60
70
80
90
100
Framesize 160 @ 2100 rpm.
Torque [%] of nominal
Tem
per
atu
reri
se∆
T[K
]
Tmeasured
Tanalytical
Torque [%] of nominal
Calc
ula
tion
erro
r[K
]
∆ : Tm − Ta
Figure 2.4: The result for frame size 160 at 2100 rpm.
85 90 95 100 105 110 11560
70
80
90
100
110
Framesize 160 @ 3000 rpm.
Torque [%] of nominal
Tem
per
atu
reri
se∆
T[K
]
Tmeasured
Tanalytical
Torque [%] of nominal
Calc
ula
tion
erro
r[K
]
∆ : Tm − Ta
Figure 2.5: The result for frame size 160 at 3000 rpm.
13
2.3.1 Sensitivity analysis
A look into how sensitive the coefficients are to change was made. The error will
stay within acceptable range for rounding up to the second decimal, see figure
2.6. It was also clear that the most sensitive coefficient was that of the copper
losses, see figure 2.7.
85 90 95 100 105 110 11560
70
80
90
100
110
Framesize 160 for all speeds.
Torque [%] of nominal
Tem
per
atu
reri
se∆
T[K
]
85 90 95 100 105 110 115−6
−4
−2
0
2
4
6
Torque [%] of nominal
Calc
ula
tion
erro
r[K
]
Tmeasured
Tanalytical
∆ : Tm − Ta
Figure 2.6: The result when all the coefficients are rounded to the nearest second
decimal.
Figure 2.7: Comparing the sensitivity of each coefficient.
It is very clear that the initial starting values of the coefficients are impor-
tant as there exist local minimums and sweeping for all values is very resource
consuming and not really an option.
14
2.3.2 Comparing FS160 and FS250
A comparison for the maximum absolute divergence for fitted formula between
the frame sizes FS160 and FS250, see figures 2.8 and 2.9.
Figure 2.8: Result for frame size 160.
Figure 2.9: Result for the frame size 250.
The results have a very good agreement regardless of the speed. Because the
measurement data is distributed stochastically making a perfect fit impossible
but the results are within very good approximation, less than 6%.
15
Chapter 3
Lumped thermal model
The electrical machine has a very complex geometry, making it very difficult
to model thermally. One solution is instead to lump big parts of the machine
together with the help of machine geometrical data, making it easier to handle.
By lumping bigger parts of the machine together the average temperature of
different areas can be found. The complexity of the solution will depend greatly
on how finely the network is setup. From experience it is e.g. known that the
critical temperatures are located at the end windings for an induction motor
and at the magnets for a PM-motor, with the danger of demagnetization at
high temperatures. Therefore, a fine division of these parts can be necessary for
an accurate solution.
This chapter takes a look into how to a make simple yet complex enough model
of the SynRM based on studies done by [11,19–21].
3.1 The finished network
The finished network that was developed is given in figure 3.1.
The resistances in figure 3.1 are defined according to
R1 =1
2Rfr1 +Rfr1−amb
R2 =1
2Rfr2 +Rfr2−amb
R3 =1
2Rfr3 +Rfr3−amb
R4 =1
2Rfr2 +
1
2Ry +Rcy
R5 =1
2Ry +
1
2Rt
R7 =R8 =1
2Rw
R6 =R9 =1
2Rew
17
Figure 3.1: The final thermal lumped network.
RInt−1
RInt−2
RInt−3
Y → ∆
R10 = R16
R12 = R14
R11 = R15
R17 =R22 =1
4Rb
R18 =R21 =1
4Rb +
1
2Rsh +
1
2Rr
R19 =R20 =1
4Rr
R23 =R24 = Rfrf
In section 3.2 it is shown thoroughly how each respective thermal resistance
is derived from machine data and which simplifications that are made in the
process.
3.2 Thermal resistance of the Machine parts
This section explains how the thermal resistance in the thermal network, figure
3.1, are derived from the machine geometrical data, which can be found in ap-
pendix D.
18
3.2.1 Internal air
Internal air to frame
According to [11] the heat transfer from the internal air to the internal frame
area is given by
RInt−1 =1
α1A1(3.1)
where A (m2) is the area exposed to the internal air and is sum of the internal
shield and frame areas, see figure 3.2
A1 = π(r2s−o + rs−o(lsh − lfe)) (3.2)
where α1 (W/m ·K) is given as
α1 = 15 + (6.75 · 2π · rps︸ ︷︷ ︸
ω
·rr−i)0.65 (3.3)
Figure 3.2: Cross-section of the machine.
It is debated whether or not the frame length should be the whole lsh, in equation
3.1, in this thesis the active length area has been subtracted as it will be part
of the yoke to frame resistance.
Internal air to rotor
According to [11] the convection from the rotor wings to the internal air is given
by the empirically determined equation
RInt−2 =1
α2A2(3.4)
19
where
α2 = (16.4 · 2π · rps · rr−i)0.65 (3.5)
with the rps being the machine revolution per second and rr−i (m) being the
outer radius of the rotor. For the induction machine the area A2 is the sum of
the area of the wings and the cross-section area of the rotor see figure 3.3.
A2 = 2 · bfin · hfin · nfin + πr2δ (3.6)
where hfin (m) and bfin (m) are the height and width respectively of the rotor
fins, nfin is the number of fins and rδ (m) is the average air gap length
rδ =rs−i + rr−o
2(3.7)
Figure 3.3: Calculating the area of the rotor wings.
Unlike the induction machine the synchronous reluctance machine does not have
rotor wings to move the internal air around. However, it has end plates that
keep the lamination sheets together, which give an equivalent area for the first
term of A2.
End winding to internal air
The thermal resistance between the end windings and the internal air is calcu-
lated according to the same principal as above
RInt−3 =1
α3A3(3.8)
with
α3 = 6.5 + (5.25 · 2π · rps · rr−i)0.6 (3.9)
20
and
A3 = Aα +Aβ +Aγ +Aδ +Aε (3.10)
where the different areas are given according to figure 3.4
Figure 3.4: Surface area of the end winding.
Aα =π · (r2h−o − r2h−i)
Aβ =2π · rh−o · lh
Aγ =2π · rh−i · lh
Aδ =Qs · 2π · rlk · lh−s
Aε =Aα −Qs · π · r2lk
The radius rlk (m) is calculated from the average copper area. In ref. Kylander
describes other ways to approximate the A3 when the dimensions are not fully
available [11]. However it should be noted that this is a simplification as well.
Once all the thermal resistances to the internal air have been calculated, the
internal air node is eliminated through Y − ∆ transformation. This simplifies
the solution as the internal air temperature is not desired.
3.2.2 Stator yoke
The thermal resistance of the stator frame and the stator yoke are calculated
according to equation 1.6. However an approximation can be made for the frame
21
Figure 3.5: Cross-section of the stator.
as the height is much smaller than the diameter
Rfr2 =hfr
λfr2π(rs−o − hfr)(3.11)
Ryoke =ln (rs−o)− ln (rs−i + ht)
2πλ · Lact(3.12)
where the rs−o is the outer diameter of the stator and rs−i the inner diameter.
As it is the yoke thermal resistance that is sought, the height of the tooth is
added to the inner stator diameter, see figure 3.13.
When it comes to the contact resistance between the stator body the contact
area is random and jagged, see figure 3.6a. Lindstrom [22] mentions studies
made by Kotrba where an equivalent air gap is added to compensate for this.
Kotrba suggested adding an equivalent thermal resistance, given by equation
Rcy =ge
λgeAge(3.13)
where ge (µm) is the equivalent air gap length, see table 3.1, Age (m2) the area
and λge the heat conductivity of the equivalent area, being air in this case.
Machine type ge(µm)
Cast iron frame or larger than 100 kW 50 · · · 75
Aluminum frame or larger than 100 kW 30 · · · 40
Table 3.1: Equivalent air gap dimensions.
The final result will then look as
Rcy =ge
λair2π(ge + rs−o)(3.14)
However the value of Rcy is not very precise. Compared to the other resistances
22
(a) Real contact (b) Equivalent contact
Figure 3.6: The contact area between the stator and frame.
in the network it is very high thus raising the average temperature. The way
that this can be dealt with is by neglecting Rcy and compensating for it by
adding the yoke thermal resistance twice [23]. Based on that the same error
arose in this thesis together with more information being unavailable, the same
procedure was utilized here.
3.2.3 Teeth
In order to thermally model the teeth a clear understanding of the tooth ge-
ometry is required. Figure 3.7 shows a basic setup of a stator tooth. As the
area of the tooth is changing with the height the thermal resistance needs to be
integrated, see figure 3.7. We have
Rt =
∫ y
0
1
Qs · λ · Lact · x(y)dy (3.15)
where Qs is the number of stator slots. This is then divided into smaller and
more manageable integrals. To simplify the integration can further be made
over half the tooth and divide the final result by a factor two. The derivation
can be found in Appendix B. The final result is
Rt =1
2QsLactλ
[2y1x1
+2y2
x2 − x1ln(
x2
x1) +
2y3x3 − x2
ln(x3
x2) +Arc
]
(3.16)
where
Arc =2y4
x4 − x3
[
−π
2+
k√k2 − 1
tan−1(1
√k2 − 1
)
]
. (3.17)
23
Figure 3.7: Teeth geometry.
3.2.4 Winding
In order to find the thermal resistance of the slot in the transverse direction it
is transformed into a rectangular shape, see figure 3.8,
(a) Real slot area. (b) Equivalent slot area.
Figure 3.8: The modelling of the stator slot.
where
24
d =dI + dA
b =xsl−3 + xsl−2
2− 2 · d
h =2 ·AQ
xsl−3 + xsl−2− 2 · d
dA is the equivalent length of the air between the insulation and teeth and given
to be around 0.3 mm [11]. dI is the length of the insulation which is given by
the manufacturer or measured. It is set to 1 mm here.
The heat dissipation in X- and Y-direction inside the equivalent winding area
is given by
Rx0 =b
hλs
Ry0 =h
bλs
where λs is the equivalent slot thermal conductivity and depends on the fill
factor of the machine according to figure 3.9 [11].
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.81
2
3
4
5
6
7
8Equivalent thermal conductivity
Fillfactor
λ s /
λ imp
Fitted curveSampled data
Figure 3.9: Higher fill factor gives better conductivity.
The contribution from the insulation material and the air film are given by
Rix =dIhλI
+dAhλA
Riy =dIbλI
+dAbλA
25
giving together
Ry =1
2·
(
Riy +Ry0
6
)
Rx =1
2·
(
Rix +Rx0
6
)
.
The winding thermal resistance Rw is then given by
Rw =RxRy
QsL(Rx +Ry)
(
1−Rx0Ru0
720RxRy
)
. (3.18)
3.2.5 End Winding
Rew is given according to [11] as
Rew =lav
λcu ·Acu ·Qs · 6(3.19)
where lav (m) is the average coil length of half a turn, see figure 3.10.
Observe that unlike Rw the heat transfer through the impregnation material oc-
curs in parallel instead of in series with the copper windings. As the conductivity
of copper is much higher, the heat transfer contribution from the impregnation
material is neglected. This is assumed for all the parallel slots. If no other
information is available, lav can be approximated as [22]
lav = Lact + 1.2τp + l′ (3.20)
where τp is the pole pitch and l′ (m) is empirically determined for small machines
to 0.05 m.
Figure 3.10: The half coil length as shown in the machine.
26
3.2.6 Air gap
Modeling the heat transfer over the air gap is a bit tricky since the transfer is
made by conduction, convection and radiation see section 1.4. Radiation being
the smallest part compared to the other two, is often neglected [24]. Studies
has shown that for a smooth surfaced rotor the heat transfer is given by [25]
h =Nu · λa
2 · δ(3.21)
where Nu is the Nusselt number and is calculated from the Taylor number
Ta =(2π · rps)2 · r · δ3
ν2(3.22)
If the value of the Taylor number is smaller than 1740 the value of Nu is set to
2 and if it is larger (as for higher speeds) Nu is defined as
Nu = 0.409 · T 0.241a − 137 · T−0.75
a (3.23)
where ν (m2/s) is the kinetic viscosity of the fluid and is defined as
ν =µ
ρ(3.24)
where µ (Pa · s) is the dynamic viscosity and is defined for air at the absolute
temperature T as
µ = µ0 ·T0 + C
T + C
(T
T0
)
(3.25)
where C (K) is Sutherlands constant for gases T (K) is the temperature of the
gas and µ0 is the dynamic viscosity of air at the reference temperature T0. The
corresponding values are
C =120K
T0 =291.15K
µ0 =18.27 · 10−6 Pa · s.
Further, ρ is the density of the air in the machine and is defined as
ρ =p0
T ·Rs
where Rs (J/kg · K) is the specific gas constant and p0 (N/m2) the absolute
pressure. For dry air being
Rspec =287.058J/kg ·K
p0 =1.01325 · 105 N/m2
27
This shows that the heat transfer over the air gap depends on the temperature
of the air gap as well as the speed and therefore needs to be solved iteratively
to find the final solution.
3.2.7 Rotor
Rr is the thermal resistance through the rotor lamination down to the shaft.
Rr is defined according to the principles described by [20], see figure 3.11a. Un-
like [20] we have air between the bars. Computational fluid dynamic studies
(CFD) has shown that there is no air passing through the rotor from one side
to the other [26]. This implies that there is only natural conduction, see section
1.4. As the thermal conductivity of the iron parts are much higher than that of
the air, it was decided to neglect the effects of the free convection.
The thermal resistances are calculated according to section 1.4 by using equation
1.5 for R2−R8 and equation 1.6 for R1, R9−R12, with the geometrical given in
figure 3.11b. During the time of this thesis verification through measurements
could not be made. The thermal resistances are therefore summed together
to one thermal resistance that will represent the rotor. As this is only an 8th
of the rotor lamination the result is divided by a factor of 8 to get the final result.
3.2.8 Shaft and Bearing
Shaft
The heat transfer in the shaft is assumed to only be in the axial direction, using
equation 1.5
Rsh =lsh
πr2r−iλsh(3.26)
where lsh is the length of the shaft from bearing to bearing, rr−i the outer radius
of the shaft and λsh the thermal conductivity of the shaft.
Bearing
The thermal resistances of the bearings are based on an empirical equation
based on studies made by Kylander [11], given as
Rb−d = 0.45(0.12− dbd)(33− 2π · rps · dbd) (3.27)
Rb−nd = 0.45(0.12− dbnd)(33− 2π · rps · dbnd) (3.28)
where dbx (mm) (x = d, nd) is the diameter of the bearings. As the diameter
of the bearing on the drive side is larger than that of the non drive side the
thermal resistance will also differ.
28
(a) An 8th of an principle rotor lami-nation.
(b) Heat transfer paths for the prin-ciple lamination.
(c) The equivalent thermal resistanceof the heat paths.
Figure 3.11: The principal rotor lamination and the equivalent circuit resistance.
3.2.9 Frame
The resistances R23 and R24 connect the end shields with the middle frame
part as there will be heat transfer due to the high temperature gradient from
the N-side to the D-side of the machine. The equation for the resistance is given
below. Note that an assumption is made to set the length from the end of the
shield until the beginning of the active part and also that they are symmetrical,
see figure 3.12
29
R23,24 =lsh − lfe
2 · λf r · π · ((rs−o + hfr−2)2 − r2s−o)(3.29)
Figure 3.12: The resistances between the frames and to the ambient.
When it comes to calculating the heat dissipation from the frame to the ambient
surrounding a lot of information needs to be known, i.e. the airflow of the fan
at different speed, the air pressure and so on. Machines with open fins will have
an airflow drop along the machine axis, see figure 1.7. The drop depends on
the spacing between the fins, the height of the fins and so on, making it very
difficult to get reliable results unless measurements are made along the whole
of the machine providing a reliable source to rely on. As such an approach was
out of the scope of this project, an empirical thermal resistance was instead
derived. [11] suggests that temperature measurements should be done along the
frame and then by knowing the total losses dissipated in the machine the ther-
mal resistance can easily be calculated according to 3.30.
∆θ
PLoss−Total= Rfr (3.30)
and by setting
Rfr =1
c1 + c2 · ω0.8(3.31)
Kylander [11] suggest doing locked-rotor test and no-load test to find the coeffi-
cients. However that is not possible in this project and so another method was
used. By setting (3.30) = (3.31) and shifting around a bit we will get
30
1 nk1
1 nk2
1 nk3
1 nk4
[c1c2
]
=
Ploss−1/∆T1
Ploss−2/∆T2
Ploss−3/∆T3
Ploss−4/∆T4
(3.32)
It is important to note that since this is an over-determined system, there ex-
ists no exact solution. This problem is solved by the means of the least square
formula
[c] = (NT ·N)−1 ·NT [Ploss/∆T ] (3.33)
0 500 1000 1500 2000 2500 3000 3500 40000
0.02
0.04
0.06
0.08
0.1
0.12
0.14
The
rmal
res
ista
nce
[K/W
]
Speed [rpm]
Frame to ambient thermal resistance
AnalyticalMeasured
Figure 3.13: The thermal resistance of the frame
Solving this will yield
c1 =11.4939W/K
c2 =0.0734W · s0.8/K.
3.3 Setting up the nodal network
The way to solve the network is to simply set up the equation for the circuit in
accordance with Kirchhoff current laws (KCL), see figure 3.14.
31
Figure 3.14: Heat flow example.
Setting up the equation for the figure 3.14 will give
Pi =θi − θjRij
+θi − θhRih
(3.34)
By doing this for all the nodes in the network and setting up the problem in
matrix for it will give
P = Gθ (3.35)
where G is the matrix containing the thermal resistances, G is also sometimes
called the Y-bus matrix and all the elements are symmetrical around the diag-
onal. The full matrix can be found in appendix C.
3.4 Power Loss calculations
In order to find the correct temperatures in the desired nodes the correct losses
need to be injected into the right nodes. One way to do this would be to calculate
the losses from the equivalent electrical circuit as has been shown in [11,19]. In
this project another method is applied, based on measured data.
3.4.1 Distributed copper losses
The copper losses are calculated according to Ohms law. As the length of the
copper winding is known, the losses are calculated according to equation (2.6).
However since the machine coil has been divided into 4 parts the losses need to
reflect this. The winding resistance is therefore multiplied with a ratio factor
defined as
kw =Lact
0.5 · lav
kew =1
2− kw
The losses are then calculated
Pew−n = 3 · I2n ·R0 · kewTew−n + 235
T0 + 235
32
Pew−d = 3 · I2n ·R0 · kewTew−d + 235
T0 + 235
Pw−d = 3 · I2n ·R0 · kwTw−d + 235
T0 + 235
Pw−n = 3 · I2n ·R0 · kwTw−n + 235
T0 + 235
These losses will change iteratively with the change of the air gap thermal
resistance, making the equilibrium shift.
3.4.2 Friction losses
The friction losses were measured under a no load test. All parasitic losses were
neglected. The measured results then fitted to the equation 3.36
Pfri = kc + kf · ω3 (3.36)
The values were found to be
kc =0.0045W
kf =1.04 · 10−8 W · s3
0 500 1000 1500 2000 2500 3000 3500 40000
100
200
300
400
500
600
700
Speed [rpm]
Loss
es [W
]
Friction losses
Figure 3.15: The friction losses as function of the speed.
33
3.4.3 Iron losses
The total losses are known from the measurements. Since the copper losses and
friction losses are known as well, the losses that are left would befall the iron
losses if other stray losses are neglected.
3.5 Result
Even though the total frame resistance is known, the loss dissipation ratio be-
tween the different frame areas is not. As the results were too far off, different
combinations of heat dissipation ratios were tested to find the best result, based
on minimizing the least mean square error. In order to get a better fit another
level of freedom was added to the magnitude of the total resistance as well as
the magnitude of the resistance connecting the end shields with the frame, see
figure 3.12. The final result gives a very good accuracy after the tweaking. One
source of error could be that the temperature rise of the frame is based on one
measuring point and does not need to represent the average rise of the whole
shield, see figure 4.2, Kylander suggests measuring along the whole axis to get
a good average. Also the length used in equation 3.29 may not be the optimal
value and need further looking into.
Figure 3.16: The result for the lumped model at 3000 rpm.
34
Figure 3.17: The result for the lumped model at 2100 rpm.
The results show a good estimation but we can see that when the speed is
decreased the end winding error is increased; this can be due to several things.
One reason that the heat dissipation to the ambient is not as high because is
the flow of air reaching the end side is not sufficient to cool it. Another reason
being that the internal air thermal resistances are based on empirical formulas
derived for the induction machine. At higher speeds the thermal resistances will
converge for the two machine designs but not for lower speeds. This will need
to be further investigated.
35
Chapter 4
Measurements setup
In this chapter a brief description of the setup for the measurements is given. In
the setup two machines were connected together back-to-back, one acting as a
motor, the other as a generator. Each machine was connected to a VSD of the
model ABBACS850. The generated power was feed back to the motor through
a DC-link between the VSDs. This means that the only consumption from the
grid are the total system losses. In order not to include the efficiency of the
VSD drives, the power consumption Pin of the machine is measured between
the drive and the machine. The measurement is made by a digital power meter
by Yokogawa Electric Corp. The mechanical power output and the speed of the
machine is measured through a torque meter by HBM , see figure 4.1.
Figure 4.1: The setup for the measurements.
By measuring the torque and speed, Pm which is the same as Pout can be cal-
culated by
37
P = T · ω (4.1)
where P (W ) is the power in, T (Nm) the torque and ω (rad/s) the angular
speed. The currents were measured with the Yokogawa as well. With the line-
to-line current and the phase resistance know the copper losses can easily be
calculated. In order to warrant steady state conditions for the measurements
the machine was run at the same work load for 4 − 5h. All the measurement
data used are steady state values. The measurement points are given in figure
4.2.
Figure 4.2: The measurement points for the machine.
Figure 4.3: The setup for the measurements.
38
Chapter 5
Conclusion and future work
5.1 Comparing the two models
Even though both models solve the problem in their own way, there are pro and
cons for each method. Brostrom formula is the quick and easy way to solve the
problem, if you have access to measurement data. It will estimate the hot spot
temperature of the end windings. Even though it will give a good approximation
the model is looked to a certain machine frame size. The formula needs to be
fitted for all other frame sizes as well. Another source of error being the guess
of the initial values when fitting, the fit might converge to a local minimum.
The lumped model unlike the Brostroms formula is based on the geometri-
cal data of the machine. It gives a higher flexibility and freedom for future
change in the model design. Change of the frame material i.e. from cast iron to
aluminum would for Brestoms formula require new tests to be done while the
thermal lumped model only requires the change of the thermal heat conductiv-
ity of the frame, assuming all other parameters are constant. The model can
also estimate the temperature in other parts of the machine such as the bearings
which are the second most important part concerning temperature rise.
A clear recommendation of which method is better than the other cannot be
made; it all comes down to the application, time and money. The complexity of
the thermal lumped model is higher. The initial work time is high but it gives
higher overview. The lumped thermal model is the clear choice if the resources
are available.
5.2 Error sources
Even though the thermal lumped model is more finely tuned it is not without
fault as many simplifications are made when defining lumped resistances, i.e.
heat transfer only in the radial direction. The assumption is made that the losses
are distributed evenly along the whole area. The empirical relations derived and
used in earlier works based on the Induction Machine and not the SynRM.
39
5.3 Conclusion
The basis of this thesis was to thermally model the Synchronous Reluctance
Machine in order to predict the temperature rise of the machine during op-
eration. The goal has been fulfilled in two different ways. One simple and
analytical equation based on empirical data and a more detailed model based
on the lumped model. The lumped model is based on works done for the IM and
since the only difference between the IM and the SynRM is the rotor, the same
models can be used for all parts except for the rotor. The rotor was modeled
with the free convection neglected as the thermal conductivity is of the air is
much lower than iron. The frame to ambient thermal resistance was calculated
based on thermal measurement data rather than geometrical data as the con-
vection to the ambient is difficult to model. This resistance was divided into
the three parts: drive shield, non-drive shield and the middle frame. Thermal
resistances connecting these parts were added as well to account for heat trans-
fer in the axial direction. The ratios and magnitude were found by comparing
the result of different combinations with the measured values and finding which
set minimized the total least square error.
5.4 Future work
The thermal lumped model that has been developed gives a good estimation of
the machine temperature compared to the values measured values, but there is
more work that can be done to further expand 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 based on the measured
data. Setting up a separate electrical circuit for loss calculation based on
geometrical data will give the freedom of estimation the temperature on
theoretical machine design without need for testing.
• Accounting for the Fan characteristics. The frame to ambient thermal
resistance has been decided based on measured data, giving an empirical
relation as the fan characteristics were not available, future work needs to
take the fan characteristic into consideration making the model functional
for any kind of fan.
• Include free convection inside the rotor bars. Account for the heat transfer
from the rotor bars to the air inside by free convection.
• Calculate the thermal losses in a FEM simulation program. Validate the
model through finite element method (FEM) calculations.
• Recalculate the empirical internal air coefficients. Looking into and recal-
culating the empirical values governing the heat transfer to the internal
air.
• Adding Transient State. Further expanding the model by adding the
transient state to the lumped thermal model.
40
Bibliography
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http://www.nema.org. Accessed 2012.
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School of Electrical Engineering, KTH, Stockholm, Sweden, October 2011.
[7] Reza Rajabi Moghaddam. Sychronous Reluctance Machine (SynRM) inVariable Speed Drives (VSD) Applications. Doctoral thesis, School of Elec-
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[8] Chandur Sadarangani. Electrical Machines - Design and Analysis of In-duction and Permanent Magnet Motors. Division of Electrical Machines
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42
List of Figures
1.1 The electricity price in Sweden. . . . . . . . . . . . . . . . . . . . 2
1.2 The different types of machines used in Sweden [6]. . . . . . . . . 2
1.3 The energy consumption in Sweden [6]. . . . . . . . . . . . . . . 3
1.4 Heat transfer in one dimension, T1 > T2. . . . . . . . . . . . . . . 5
1.5 Heat transfer in the radial direction for a cylinder, T1 > T2. . . . 6
1.6 An example of radiation [16]. . . . . . . . . . . . . . . . . . . . . 7
1.7 A rough estimation of how the air moves around the machine. . . 8
2.1 An example of a curve fitting. . . . . . . . . . . . . . . . . . . . . 11
2.2 The result for frame size 160 at 1000 rpm. . . . . . . . . . . . . . 13
2.3 The result for frame size 160 at 1500 rpm. . . . . . . . . . . . . . 14
2.4 The result for frame size 160 at 2100 rpm. . . . . . . . . . . . . . 14
2.5 The result for frame size 160 at 3000 rpm. . . . . . . . . . . . . . 15
2.6 Result for rounded coefficients. . . . . . . . . . . . . . . . . . . . 16
2.7 Comparing the sensitivity of each coefficient. . . . . . . . . . . . 16
2.8 Result for frame size 160. . . . . . . . . . . . . . . . . . . . . . . 18
2.9 Result for the frame size 250. . . . . . . . . . . . . . . . . . . . . 18
3.1 The final thermal lumped network. . . . . . . . . . . . . . . . . . 20
3.2 Cross-section of the machine. . . . . . . . . . . . . . . . . . . . . 22
3.3 Calculating the area of the rotor wings. . . . . . . . . . . . . . . 23
3.4 Surface area of the end winding. . . . . . . . . . . . . . . . . . . 24
3.5 Cross-section of the stator. . . . . . . . . . . . . . . . . . . . . . 25
3.6 The contact area between the stator and frame. . . . . . . . . . . 26
3.7 Teeth geometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.8 The modelling of the stator slot. . . . . . . . . . . . . . . . . . . 27
3.9 Higher fill factor gives better conductivity. . . . . . . . . . . . . . 28
3.10 The half coil length as shown in the machine. . . . . . . . . . . . 29
3.11 The principal rotor lamination and the equivalent circuit resistance. 32
3.12 The resistances between the frames and to the ambient. . . . . . 33
3.13 The thermal resistance of the frame . . . . . . . . . . . . . . . . 35
3.14 Heat flow example. . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.15 The friction losses as function of the speed. . . . . . . . . . . . . 37
3.16 The result for the lumped model at 3000 rpm. . . . . . . . . . . . 38
3.17 The result for the lumped model at 2100 rpm. . . . . . . . . . . . 39
4.1 The setup for the measurements. . . . . . . . . . . . . . . . . . . 41
4.2 The measurement points for the machine. . . . . . . . . . . . . . 42
4.3 The setup for the measurements. . . . . . . . . . . . . . . . . . . 43
43
B.1 Tooth geometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
B.2 Half the tooth area D2 tilted. . . . . . . . . . . . . . . . . . . . . 55
B.3 Close up of the arc of the slot. . . . . . . . . . . . . . . . . . . . 57
44
Appendix A
FS160 measurement data
∆T Pcu Ptot n Lact ∆T Pcu Ptot n Lact
64 877 1324 1000 0.205 65 848 1346 1000 0.205
61 885 1593 1500 0.205 65 863 1640 1500 0.205
63 984 1676 1800 0.205 64 870 1825 1800 0.205
64 876 2013 2100 0.205 63 883 2294 2500 0.205
107 1577 2105 1000 0.260 95 1490 2294 1500 0.260
86 1433 2550 2100 0.260 85 1439 3113 3000 0.260
108 1395 1868 750 0.260 77 1086 1588 1000 0.205
79 1059 1602 1000 0.205 100 1206 1676 1000 0.165
71 859 1257 1000 0.165 69 870 1225 1000 0.165
101 1364 2001 1000 0.260 106 1363 2130 1000 0.260
73 1097 1884 1500 0.205 77 1074 1937 1500 0.205
93 1167 1821 1500 0.165 66 857 1460 1500 0.165
64 856 1390 1500 0.165 94 1345 2311 1500 0.260
76 1080 2145 1800 0.205 64 859 1579 1800 0.165
67 993 1516 1800 0.165 75 1094 2325 2100 0.205
64 861 1716 2100 0.165 88 1328 2926 2470 0.260
75 1094 2636 2500 0.205 63 900 2076 3000 0.165
108 1837 3681 3000 0.260 101 1396 3896 3000 0.260
104 1440 5388 4500 0.260 82 1008 1429 1000 0.165
81 1031 1411 1000 0.165 75 1001 1638 1500 0.165
73 1012 1583 1500 0.165 73 1000 1757 1800 0.165
79 1194 1742 1800 0.165 72 1000 1908 2100 0.165
71 1072 2298 3000 0.165
Table A.1: The data for frame size 160 with different active lengths with a 4p
fan, used for fitting.
45
Appendix B
Stator teeth
This appendix will cover the way the stator tooth are modelled.
Figure B.1 shows a basic setup of a stator teeth. As the area of the tooth is
changing with the height, the thermal resistance needs to be integrated.
Rt =
∫ y
0
1
Qs · λ · Lact · x(y)dy (B.1)
Figure B.1: Tooth geometry.
The different parts of the machine will be divided into four areas, namedD1−D4
and the resistance calculated as
46
Rt =1
2 ·Qs · λ · Lact
∫ y1
0
dy
x(y)︸ ︷︷ ︸
D1
+
∫ y
0
dy
x(y)︸ ︷︷ ︸
D2
+
∫ y3
0
dy
x(y)︸ ︷︷ ︸
D3
+
∫ y4
0
dy
x(y)︸ ︷︷ ︸
D4
(B.2)
D1
As the width of the teeth does not change in the radial direction the equation
for D1 is straightforward.
∫ y1
0
1
x(y)dy =
[y
x1/2
]y1
0
= 2 ·y1x1
(B.3)
D2 and D3
Here the width changes with respect to the height and so the equation x(y)needs to be found before we can integrate. If a closer look at the tooth is made,
see figure B.1, it can be seen as a linear equation, see figure B.2.
Figure B.2: Half the tooth area D2 tilted.
Setting up the linear equation and solving it will give
x(y) = k · y +m
k =x2
2− x1
2
y2
= x2−x1
2·y2
→
m = x1
2
x(y) = x2−x1
2·y2
· y + x1
2 (B.4)
47
Placing (B.5) into (B.2) and integrating over the height will give
∫ y2
0
1
x(y)dy =
∫ y2
0
1x2−x1
2y2
y + x1y2
2y2
dy
= 2y2
∫ y2
0
1
(x2 − x1)y + x1y2dy
=
∫dy
ay + b=
1
a
∫dy
y + b/a
=
[
y + b/a = u; dy =dy
dudu = du
]
=
(1/a) ·
∫dy
u=
1
a· ln(y + b/a)
=2y2
x2 − x1
[
ln(y +x1y2
x2 − x1)
]y2
0
⇒2y2
x2 − x1ln
(x2
x1
)
(B.5)
This eqution can also be used forD3 as they are almost identical. The arguments
in the ln function will always be bigger than zero. However the sign will be minus
for some cases where the area is smaller at the top, e.g. D3, but this is taken
care of by the prefactor.
D4
The last part of the teeth is in the form of an ellipse and to find the correct
form we need to set up the equation of an ellips
48
Figure B.3: Close up of the arc of the slot.
1 =x2
a2+
y2
b2→
x =
√
1−y2
b2
b =y4
a =x4 − x3
2
giving
∫ y4
0
1
x4
2 − a√
1− y2
b2
dy
Substituting with polar coordinates gives
[y = b · sin(θ); dy = b · cos(θ)dθ; 0 = 0; y4 = π/2]
∫ π/2
0
b · cos(θ)x4
2 − a√
1− sin2(θ)︸ ︷︷ ︸
cos(θ)
dθ →b
a
∫ π/2
0
cos(θ)x4
2a︸︷︷︸
K
−cos(θ)dθ
the general solution to the integral is
49
b
a
−x−2K tanh−1
((K+1)·tan(x/2)√
1−K2
)
√1−K2
π/2
0
Simplifying the final form with the boundaries inserted gives
b
a
[
−π
2+
2K√1−K2
tan−1
(1
√K − 1
)]
50
Appendix C
Thermal network
Setting up the thermal network according to equation 3.34 and transforming to
matrix form will give
P2 = θ2R2
− θ0R2
+ θ2−θ1R23
+ θ2−θ3R24
+ θ2−θ4R4
P3 = θ3R3
− θ0R3
+ θ3−θ2R24
+ θ3−θ9R16
+ θ3−θ12R15
+ θ3−θ14R22
P4 = θ4−θ2R4
+ θ4−θ7R5
P5 = θ5−θ1R10
+ θ5−θ12R12
+ θ5−θ6R6
P6 = θ6−θ5R6
+ θ6−θ7R7
P7 = θ7−θ6R7
+ θ7−θ4R5
+ θ7−θ8R8
+ θ7−θ12R13
P8 = θ8−θ7R8
+ θ8−θ9R9
P9 = θ9−θ8R9
+ θ9−θ12R14
+ θ9−θ3R16
P10 = θ10−θ1R17
+ θ10−θ11R18
P11 = θ11−θ10R18
+ θ11−θ12R19
P13 = θ12−θ11R19
+ θ12−θ1R11
+ θ12−θ5R12
+ θ12−θ7R13
+ θ12−θ9R14
+ θ12−θ3R15
+ θ12−θ13R20
P13 = θ13−θ12R20
+ θ13−θ14R21
P14 = θ14−θ13R21
+ θ14−θ3R22
51
C.1 The final thermal network
G =
−1
R1−
1
R2−
1
R30 0 0 0 0 0 0 0 0 0 0
G1 −1
R230 0 −
1
R100 0 0 0 −
1
R170 −
1
R110 0
−1
R23G2 −
1
R24−
1
R40 0 0 0 0 0 0 0 0 0
0 −1
R24G3 0 0 0 0 0 −
1
R160 0 −
1
R150 −
1
R22
0 −1
R40 G4 0 0 −
1
R50 0 0 0 0 0 0
−1
R100 0 0 G5 − 1
R6
0 0 0 0 0 −1
R120 0
0 0 0 0 −1
R6G6 −
1
R70 0 0 0 0 0 0
0 0 0 −1
R50 −
1
R7G7 −
1
R80 0 0 −
1
R130 0
0 0 0 0 0 0 −1
R8G8 −
1
R90 0 0 0 0
0 0 −1
R160 0 0 0 −
1
R9G9 0 0 −
1
R140 0
−1
R170 0 0 0 0 0 0 0 G10 −
1
R180 0 0
0 0 0 0 0 0 0 0 0 −1
R18G11 −
1
R190 0
−1
R110 −
1
R150 −
1
R120 −
1
R130 −
1
R140 −
1
R19G12 −
1
R200
0 0 0 0 0 0 0 0 0 0 0 −1
R20G13 −
1
R21
0 0 −1
R220 0 0 0 0 0 0 0 0 −
1
R21G14
52
where
G0 =1
R1+
1
R2+
1
R3
G1 =1
R1+
1
R10+
1
R11+
1
R17+
1
R23
G2 =1
R2+
1
R4+
1
R23+
1
R24
G3 =1
R3+
1
R15+
1
R16+
1
R22+
1
R24
G4 =1
R4+
1
R5
G5 =1
R6+
1
R10+
1
R12
G6 =1
R6+
1
R7
G7 =1
R5+
1
R7+
1
R8+
1
R13
G8 =1
R8+
1
R9
G9 =1
R9+
1
R14+
1
R16
G10 =1
R17+
1
R18
G11 =1
R18+
1
R19
G12 =1
R11+
1
R12+
1
R13+
1
R14+
1
R15+
1
R19+
1
R20
G13 =1
R20+
1
R21
G14 =1
R21+
1
R22
53
C.2 Resisitances and Losses for the FS160 @
3000 rpm and
R1 0.0896 Pyoke 243.77 W
R2 0.1344 PD−EW 241.6 W
R3 0.0179 PD−W 289.93 W
R4 0.0036 PTeeth 494.59 W
R5 0.0052 PN−W 289.93 W
R6 0.0298 PN−EW 241.6 W
R7 0.0193 PFri 5.22 W
R8 0.0193 Protor 146.57 W
R9 0.0298
R10 1.0565
R11 2.1769
R12 11.1603
R13 0.1416
R14 11.1603
R15 2.1769
R16 1.0565
R17 0.1273
R18 1.9923
R19 0.0337
R20 0.0337
R21 1.9923
R22 0.1585
R23 0.0303
R24 0.0303
Table C.1: The results with and without the additional term for different frame
sizes with a 4p fan.
54
Appendix D
Machine data for FS160
Stator core length 260 mm
Stator inner diameter 165 mm
Stator outer diameter 256 mm
Lamination filling factor 0.98
Number of slots 36
Tooth height 21.5 mm
Number of poles 4
Slot filling factor 0.684
Mechanical air gap 0.45 mm
Rotor inner diameter 53 mm
Rotor outer diameter 164.1 mm
Table D.1: Geometrical data for the machine.
Machine part Heat conductivity (W/m ·K)
Lamination 38
Winding 395
Frame 38
Shaft 51
Winding impregnation 0.19
Air gap 0.0243
Slot insulation 0.2
Table D.2: Heat conductivity of the material in the model.
55
