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High-Speed Permanent Magnet Motor Generator for
Flywheel Energy Storageby
Tracey Chui Ping HoSubmitted to the Department of Electrical Engineering and Computer
Sciencein partial fulfillment of the requirements for the degrees of
Bachelor of Science in Electrical Engineeringand
Master of Engineering in Electrical Engineering and Computer Scienceat the
MASSACHUSETTS INSTITUTE OF TECHNOLOGYMay 1999 L> 1~~'
@ Tracey Chui Ping Ho, MCMXCIX. All rights reserved.
The author hereby grants to MIT permission to reproduce and distributepublicly paper and electronic copies of this thesis document in whole or in
part, and to grant others the right to do so. MASSACHUSETTS INcOF TECHNOLOG
Author.....D~partment of lectrical Engineering and
May 20, 1999
Certified by
Professor of Electrical7 -- -
Certified by
Accepted by
Jeffrey H. LangEngineering and Computer Science
Thesis Supervisor
- - --- - -- --c
James L. Kirtley Jr.ProfessorfEfectrical Engineering and Computer Science
is S uer vi s.... .... ....
Arthur C. SmithChairman, Department Committee on Graduate Theses
High-Speed Permanent Magnet Motor Generator for Flywheel Energy
Storage
by
Tracey Chui Ping Ho
Submitted to the Department of Electrical Engineering and Computer Scienceon May 20, 1999, in partial fulfillment of the
requirements for the degrees ofBachelor of Science in Electrical Engineering
andMaster of Engineering in Electrical Engineering and Computer Science
Abstract
This thesis is part of a joint project between MIT and SatCon Technology Corporation todevelop a high-speed motor-generator for a flywheel energy storage system. Such systemsoffer environmental and performance advantages over chemical batteries, with potentialapplications in hybrid electric vehicles and uninterruptible power supplies. The develop-ment of high-energy Neodymium Iron Boron magnets, as well as advances in composites,electric drives and magnetic bearings, has contributed towards making flywheel systemsmore commercially viable.
A 30 kW high-speed permanent magnet synchronous motor-generator was designed,built and tested. The basic electromagnetic design was developed by Professor James Kirt-ley, while much of the mechanical design was done by engineers at SatCon. This thesisfocused primarily on: the development of theoretical models for various loss mechanisms,with particular interest in the modelling of eddy currents in azimuthally segmented rotormagnets; the development of theoretical models for thermal performance; the design of acooling system; and construction details. Finally, several quantities predicted by the elec-tromagnetic analysis and loss models were experimentally measured, to evaluate the valid-ity of the theory. On the basis of this work it is believed that compact permanent magnetsynchronous motor-generators for flywheel energy storage systems can exhibit efficienciesnear 95%, and can operate with idle losses as low as 12 W.
Thesis Supervisor: Jeffrey H. LangTitle: Professor of Electrical Engineering and Computer Science
Thesis Supervisor: James L. Kirtley Jr.Title: Professor of Electrical Engineering and Computer Science
Acknowledgments
I am extremely grateful to my thesis supervisors, Professor Jeffrey Lang and Professor
James Kirtley, who guided me through the project, patiently answered my queries, and
taught me a great deal. I also owe many, many thanks to Wayne Ryan at MIT, who was an
incredible help in all the practical aspects of the project, from ordering parts to constructing
and assembling the machine.
This work was supported by a research grant from the SatCon Technology Corporation
of Cambridge, MA. In this context I wish to thank Ed Godere of SatCon for making the
grant run smoothly. I would also like to thank many people at SatCon: Frank Nimblett for
overseeing the project; John Swenbeck for his invaluable guidance and help in constructing
the machine, without whom the task would have been incredibly difficult; Mike Amaral for
drawing all the manufacturing prints; Jerome Kiley and Ed Ognibene for their help on the
thermal and mechanical aspects; Al Ardolino for machining and altering parts; John Young
for setting up the instrumentation for the spin-down tests; Peter Jones for helping me scan
photos and make slides; Dave Lewis and Ray Roderick and many others at SatCon who
helped me in countless ways. To all these people I am very grateful.
Finally, I would like to express deep gratitude to my friends Philip Tan and Ben Leong
for their selfless computer help in the preparation of my thesis document.
4
Contents
1 Introduction 7
2 Machine Design 92.1 Existing Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Electromagnetic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Modifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3 Magnet Loss Models 193.1 Stator Current Space Harmonics . . . . . . . . . . . . . . . . . . . . . . . 193.2 Eddy Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.2.1 Known Magnetic Field and Thin Magnets mounted on InfinitelyPermeable Surface . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.2.2 Known Stator Excitation Current and Thin Magnets with InfinitelyPermeable Boundaries . . . . . . . . . . . . . . . . . . . . . . . . 26
3.2.3 Known Stator Excitation Current and Magnets with Significant Thick-ness with Infinitely Permeable Boundaries . . . . . . . . . . . . . . 31
3.2.4 Known Stator Excitation Current and Magnets with Significant Thick-ness Without Infinitely Permeable Boundaries . . . . . .
3.2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . .3.3 Loss Calculation . . . . . . . . . . . . . . . . . . . . . . . . .3.4 Application of Model to Rotor Magnet Loss Problem . . . . . .
4 Stator Loss Models, Cooling System Design and Thermal Analysis4.1 Loss calculations . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Conduction Losses . . . . . . . . . . . . . . . . . . . .4.1.2 Eddy Current Losses . . . . . . . . . . . . . . . . . . .4.1.3 Windage Losses . . . . . . . . . . . . . . . . . . . . .4.1.4 Total Losses . . . . . . . . . . . . . . . . . . . . . . .
4.2 Cooling system . . . . . . . . . . . . . . . . . . . . . . . . . .4.2.1 Channel Geometry and Fluid Flow Considerations . . .4.2.2 Channel Outer Wall Material . . . . . . . . . . . . . . .
4.3 Thermal Analysis . . . . . . . . . . . . . . . . . . . . . . . . .
39484950
53. . . . . 53. . . . . 53. . . . . 54. . . . . 57. . . . . 61. . . . . 62. . . . . 63. . . . . 64. . . . . 65
5
4.3.1 Thermal Conductivity Experiments . . . . . . . . . . . . . . . . . 654.3.2 Film Coefficient for Cooling Channel . . . . . . . . . . . . . . . . 674.3.3 Effective Conductivity of Armature Region . . . . . . . . . . . . . 684.3.4 Temperature Calculation . . . . . . . . . . . . . . . . . . . . . . . 69
5 Fabrication of the Experiment 75
6 Testing 896.1 Resistance and Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . 896.2 Spin-down Tests. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
6.2.1 Loss estim ation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.2.2 Back em f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
6.3 Magnetic Field Measurements . . . . . . . . . . . . . . . . . . . . . . . . 96
7 Summary and Conclusions 99
A Inductance Calculation 101
B Matlab code for Rotor Loss Calculation 105
C Thermal Analysis Spreadsheet and Matlab Calculations 111C. 1 Thermal Analysis Spreadsheet . . . . . . . . . . . . . . . . . . . . . . . . 111C.2 Matlab code for windage calculation . . . . . . . . . . . . . . . . . . . . . 114C.3 Matlab code for plotting graphs of loss vs speed . . . . . . . . . . . . . . . 117C.4 Matlab code for plotting graphs of loss vs power . . . . . . . . . . . . . . . 121
D Thermal Conductivity Experimental Results 123
E Manufacturing Drawings 129
F Experimental Results from Spin-Down Tests 135
6
Chapter 1
Introduction
This thesis is part of a joint project between MIT and SatCon Technology Corporation to
develop a high-speed motor-generator for use in a flywheel energy storage system. A major
motivation for interest in such systems is their potential application in hybrid electric vehi-
cles. They can be used either as the main energy source, or as a secondary source, along
with a conventional internal combustion engine or chemical battery, to provide greater
power when needed. Other applications include uninterruptible power supplies for com-
puters, industrial systems and telecommunications.
As described in [1], flywheel energy storage systems have a shorter recharge time,
longer driving range, greater power density and longer operating life than do chemical bat-
teries. They also avoid the environmental problems posed by materials such as lead or
cadmium present in chemical batteries. At present they are still substantially more expen-
sive than the latter. However, over the last decade, technological advances in areas such
as composites, electric drives and magnetic bearings have contributed towards making fly-
wheel systems more commercially viable. Also, newly-developed magnetic materials such
as Neodymium Iron Boron (NdFeB) have made high energy product permanent magnets
available, allowing for more compact machines.
The flywheel system stores kinetic energy in the momentum of the motor/generator
rotor. For this reason, the machine operates in two modes. As a motor, it draws electrical
7
power to reach a steady state rotational speed. If losses are kept low, only a small amount
of electrical power is needed to maintain rotation at this speed. As a generator, the machine
draws on its stored kinetic energy to supply electrical power.
Several factors must be considered in choosing the most suitable type of electric ma-
chine for this application. Major requirements are high two-way efficiency and low "idling"
losses [2]. Magnetic bearings are important in reducing bearing friction losses. Windage
losses can be reduced by having the rotor operate in a vacuum. However, a vacuum im-
pedes heat transfer, so it becomes doubly important to minimize losses in the rotor. Both
induction machines and conventional synchronous machines have rotor windings through
which currents flow, resulting in unacceptably large rotor losses. Permanent-magnet syn-
chronous machines, on the other hand, avoid losses from rotor winding conduction, since
there are permanent magnets rather than windings on the rotor. For magnets with a non-
zero electrical conductivity, losses from eddy currents occur nevertheless.
In order to investigate these losses, and to demonstrate that a practical low-loss machine
of this type can be built, a 30 kW permanent-magnet synchronous machine was designed
and constructed. Theoretical models were developed to predict various loss mechanisms
and other machine quantities, including back emf, efficiency and inductances. Experimen-
tal verification of these predictions is currently proceeding, with the aim of evaluating the
accuracy of the models.
The remainder of this thesis is organized as follows. Chapter 2 presents the design and
an electromagnetic analysis of the machine. The problem of modeling eddy current loss
in the azimuthally segmented rotor magnets is examined in Chapter 3. Models for stator
losses are presented in Chapter 4, along with the design of the stator cooling system and
an analysis of the thermal performance of the machine. The construction of the machine
is described in Chapter 5, and Chapter 6 covers the testing. Chapter 7 concludes the thesis
with a summary and suggestions for future work.
8
Chapter 2
Machine Design
2.1 Existing Design
The motor-generator is based primarily on an existing electromagnetic design completed
by Professor Kirtley, which has been modified slightly through the course of this thesis.
It is an 8-pole permanent-magnet synchronous machine rated at 30 kW and designed for
rotational speeds in the range 15,000 to 30,000 rpm. The permanent magnets are attached
to the rotor, which is on the outside. The stator on the inside contains three-phase windings.
It is iron-free, which minimizes eddy current losses, and also eliminates side loads from
small displacements of the rotor. Iron tends to attract the magnets, destabilizing the rotor
position. This effect is counteracted by bearings with large positive spring constants in most
machines, but is an issue for a machine with magnetic bearings. A summary of dimensions
and parameters from the original design, along with those of the modified design, is given
in Table 2.1.
The basic layout of the armature winding and magnets is shown in Figure 2-1. In Pro-
fessor Kirtley's original design, the armature windings occupy an annulus of inner radius
Rai = 6.99 cm (2.75 in) and thickness ta = 9.53 mm (0.375 in), with an active length 1
= 10.16 cm (4 in). The windings are constructed using litz wire, which consists of many
separately insulated strands twisted together. This greatly reduces the possibility of eddy
9
Table 2.1: Machine Dimensions and Parameters
Quantity Symbol Original Design Modified DimensionsNo. of pole pairs p 4 4No. of phases q 3 3Wire diameter dw 0.254 mm 0.254 mmActive length 1 10.16 cm 10.01 cmArmature inner radius Rai 6.99 cm 6.73 cmArmature thickness ta 9.53 mm 12.0 mmArmature outer radius Rao 7.94 cm 7.93 cmRotational gap width g 0.508 mm 1.32 mmMagnet inner radius Rmi 7.99 cm 8.06 cmMagnet thickness tm 0.95 cm 0.95 cmMagnet outer radius Rmo 8.94 cm 9.01 cmElectrical angle Owe -r/3 = 1.047 0.856
currents, as compared to having a single thick conductor of equivalent dc resistance. In
this design, the diameter of a single strand of wire is d. = 0.254 mm (0.01 in). Each of
the three phases is wound according to the pattern shown in Figure 2-2. The end turns are
bent outwards at one end and inwards at the other, so as to reduce the axial length of the
machine and make it more compact.
The rotor lies outside of the stator, across a rotational gap width g = 0.508 mm (0.02
in). The rotor has high-energy-product permanent magnets attached to the inside of a fly-
wheel structure. The magnets are 0.375 in (9.53 mm) thick, and segmented azimuthally to
reduce eddy current losses. They are made of bonded Neodymium Iron Boron (NdFeB) and
have a remanent flux density Brem = 0.68 T. There are a total of eight magnets, making
up four pole pairs.
2.2 Electromagnetic Analysis
This section presents an electromagnetic analysis of Professor Kirtley's original design.
Most of the formulae quoted in this section are found in [3]. The results are summarized in
10
v I phas e loelt
spa~cer
-- -- -- -- - -
1--,11-1magnet
Figure 2-1: Cross section showing layout of armature winding and magnets
Table 2.2.
For an iron-free machine with a Halbach array, the magnetic field of the azimuthally
magnetized set is the same as that of the radially magnetized set. Thus the total magnetic
field can be obtained by calculating the field from one set of magnets and multiplying the
result by two. Consider first one set, consisting of p pairs of oppositely polarized magnets,
each subtending an angle of 0m as shown in figure 2-3. Assuming that the magnets occupy
the whole periphery with no spaces in between, 0m = 7r/(2p) for a Halbach array. The
fundamental component of radial magnetic flux at the magnets has the magnitude
4 .(pon s-Brem swr
11
terminaL
Figure 2-2: Winding pattern for one phase. The end turns which are bent outward are atthe connector end, and those bent inwards are at the other end.
This is multiplied by the coefficient km to obtain the radial flux density at the outer radius
of the armature:4 ./p~m'-Bremkm smn Pr7r 2
wherejln ()
(R P - RI,;;P) Rg;O
Multiplying by 2 to obtain the combined field from both sets of magnets, and dividing by
12
km =: ifp=1
: otherwise(2.1)
Table 2.2: Machine Quantities at 15,000 rpm
v to obtain the rms value, we have
Bla = Vf2Bremkm sin A = 0.1624 T7r 2/
To account for the variation in magnetic flux density across the thickness of the arma-
ture, B1a is multiplied by the flux linkage thickness coefficient kt to obtain the effective
fundamental magnetic flux density B 1. The actual flux linked by the thick winding is then
equal to the flux that would have been linked if all its turns were concentrated at outer ra-
dius Rao, and the flux density there were B 1 . To obtain kt, note that the turns density of the
winding is constant in azimuthal angle, while flux linked per turn is proportional to radius
r. Since the flux density is proportional to rP-', we have
Biao - 1 jRao
Rao - Rai RaiBia
R aordr
1 - XP+1
(1l-x)(p +)
where x = Rai/Rao. Thus B 1 = Biakt 0.1278 T.
(2.2)
The internal voltage induced across one turn of the armature winding is given by the
13
Quantity Symbol Original With Modified As-built,Design Dimensions Without Halbach
Rated power P 30 kW 30kW 30 kWRms magnetic field at Rao Bia 0.1624 T 0.1564 T 0.0958 TEffective rms magnetic field B1 0.1278 T 0.1157 T 0.0709 TInternal voltage per turn Ean 3.09 V 2.80 V 1.71 VAmpere-turns NIa 3234 A 3573 A 5835 ASynchronous inductance/N 2 Ld/N 2 2.36 x 10-8 H 2.30 x 10-8 H 2.30 x 10-8 HNormalized reactance Xa 0.155 0.184 0.491Terminal voltage per turn Vn 3.13 V 2.85 V 1.91 V
B
0 O_ T
P
Figure 2-3: Magnetization pattern of one set of magnets
rate of change of flux linked. It has an rms value of
Ean = 2RaoLBikw, = 3.092 V (2.3)P
where the winding breadth factor km is
sin 0km = 2
2
and Owe = r/q is the electrical angle of an armature phase belt. The total induced voltage
across the terminals of one phase winding is Ea = Ean x N, where N is the number of
turns.
For a 3-phase machine, the rated power P is equal to 3 EaIa, assuming that the rms
current Ia can be applied in phase with the internal voltage Ea for maximum power. So for
a 30kW machine, the armature ampere-turns has an rms value of
PN PNIa = -- = 3234 A (2.4)
3Ea En
The synchronous inductance Ld of the 3-phase armature winding is the apparent in-
14
ductance of one phase when balanced currents are used. Since, under these conditions,
the phase currents sum to zero and the mutual phase-phase inductances are equal, Ld =
La - Lab, where La is the self inductance of one armature phase winding and Lab is the
mutual inductance between different phase windings. The self and mutual inductances of
an air gap armature winding with uniform current density in each phase belt have been
calculated previously in [4]. In this machine, however, the number of conductors does not
increase with radius, so current density decreases with radius. In Appendix A we use a sim-
ilar approach to find inductances for this configuration. This calculation underestimates the
inductances, since it takes into account only the straight sections of the winding but not the
end turns, which also have significant inductance. However, an analytical calculation of
the contribution of the end turns is beyond the scope of this thesis. Thus the synchronous
inductance of the machine is somewhat higher than the value calculated from this analysis,
which is
(31Npo\ sin 21-p+x2 p_2xLd = (sNn 2 (1_P+ X2 (1±p)-2xP+l = N 2 x 2.36 x 10-8 H
7r (1 - x)2(i _ p2)p
The internally normalized per-unit synchronous reactance is obtained by dividing the
voltage drop across the armature winding by the internal voltage:
WLdIa _w (Ld/N 2) NIaxa === 0.155
Ea Ean
Maximum power output per unit armature current 'a is obtained when the current is
applied exactly in phase with the internal voltage Ea. Ignoring the voltage drop from
resistance of the windings, the terminal voltage V is given by Ea + jXdIa, SO
V2 = EZ +X3I E2 (1 + x)
15
Thus the terminal voltage per turn is
Vin= En / + X2 = 3.13 VVt nEan a+~.3
2.3 Modifications
A number of parameters in the existing design were altered slightly because of manufactur-
ing constraints. First, the air gap was increased from 0.508 mm (0.02 in) to 1.32 mm (0.052
in) to make manufacturing easier, since the exact gap width is not critical to performance.
Second, it was decided that in practice 0.35 would be an achievable value for the ar-
mature space factor Aa, which is the volume fraction occupied by copper. To achieve this
space factor, rectangular compacted litz wire was chosen, since it has a higher fill factor
than other types of litz wire. Rectangular compacted litz consists of wires twisted and
compressed into a rectangular cross section. The machine was originally planned to have
72 turns, and the number of parallel strands Npa was chosen from commercially available
constructions such that Aa would be close to 0.35. Choosing N to be 77,
A,, 6 x 72 x 77 x (0.0254/2)2/ (7.942 - 6.992) 0.348
The machine eventually ended up being built with 9 litz bundles connected in parallel by
mistake, so the number of turns became 72/9 = 8 and the number of parallel strands
9 x 77 = 693.
The cross section of the 77-strand rectangular litz has dimensions 5.16 mm (0.203 in)
by 1.60 mm (0.063 in). When arranged in two concentric layers as shown in Figure 2-1,
the wires have a radial thickness of 5.16 x 2 = 10.32 mm, which is slightly larger than the
original value ta = 9.53 mm. With insulating tape added between the two layers, as well
around them, the thickness becomes 11.27 mm. The winding had to be put into a mold
to be potted in epoxy, and a bit of extra space was allocated in the mold design, so that
the winding could be inserted without having to force it in. Thus the armature thickness
16
was increased to 12.0 mm (0.47 in), with the additional space gained by reducing the inner
radius Rai of the winding to 6.73 cm (2.65 in).
Since there are only 2 turns in each phase belt, a simpler approach was taken in cal-
culating inductance here than for the general N-turn case. A discussion of this is given in
Appendix A, along with the corresponding calculations. The synchronous inductance Ld
was found to be Ld = N 2 x 2.30 x 10-8 = 1.49 x 10-6 H.
The rotor magnet arrangement was changed from a Halbach array to one consisting of
just radially magnetized permanent magnets, each subtending an angle 0m = -r/6. The rms
fundamental magnetic flux at the armature outer radius becomes
Bi - Bremkm sin " 0.0958 T
and the effective field over the armature is B1 = Biakt = 0.0709 T, where km and kt are
calculated from equations 2.1 and 2.2 using the modified dimensions.
The substantial decrease in magnetic field as a result of the change in magnet arrange-
ment resulted in significant changes in other machine quantities. This can clearly be seen
in Table 2.2, which summarizes, alongside those of the original design, quantities for a ma-
chine with the modified dimensions but the original Halbach array, as well as the machine
as-built. Most notably, the internal voltage per turn decreases from 3.09 V to 1.71 V, and
the ampere turns increases from 3234 A to 5835 A.
17
18
Chapter 3
Magnet Loss Models
Heating caused by rotor magnet dissipation is a significant concern. Eddy current losses can
be substantial, since the high energy product NdFeB permanent magnets have a moderate
electrical conductivity and a relatively low Curie temperature. Furthermore, although the
machine built for this investigation has an air gap, a machine for actual use would have the
rotor in a vacuum, which would limit heat transfer [2].
Rotor magnet losses result from stator current harmonics that appear non-stationary
with respect to the rotor. These harmonics produce asynchronous magnetic fields that cause
eddy currents in the rotating magnets. Since the stator winding is made of discrete phase
belts, there will be space harmonics; there may also be time harmonics in the terminal
currents. These harmonics cause eddy currents to flow in the rotor magnets. Total eddy
current loss would be obtained by estimating the loss from each harmonic component of
stator current separately, and adding these up.
3.1 Stator Current Space Harmonics
The armature winding is made up of 2pq phase belts, where p is the number of pole pairs
and q the number of phases. Each phase belt subtends an angle of Owe/p. If each phase belt
of one phase has a current density of J, the overall current density of this phase, expressed
19
as a Fourier series, is, from [2],
4 nwE - sin J cos (npo)
n odd ?l f
For a balanced q-phase source of amplitude J and frequency w, total current density is
) J cos(wt -F npO)for n = 2kq ± 1, integer k
The armature windings are constructed such that current density is inversely proportional
to radius r for this machine. As shown in Appendix A, the current density in one phase belt
is J = J0 /r, whereNI
Owe (Rao - Rai)(3.1)
So the overall current density at radius r is
Z " cos(wt - npO)n T
for n = 2kq ± 1, integer k
q 4 .nweJn - -sin Jo
2 n7 ( 2)
We model the stator current as a current sheet at r = Rao, choosing the magnitudes of
its components, Ka, such that the magnetic fields produced by the thick armature and the
current sheet are the same for r > Rao. Now the stator current sheet
K = K, cos (wt - npO)2n
gives rise to magnetic fields that can be expressed as the negative gradient of scalar poten-
tials
'is for r < Rao
20
q 4_.- -
n 2mrsin ( nOwe
2
where
(3.2)
= Aln sin (Lot - npO)n Ra
= fo A2n Rao)psin (wt - npO)n r
Substituting His = -VTi, and Hos = -V To, into the boundary conditions Hos, - Hi,, =
K, and Ho,, = Hjs, at r = Rao, we have
KnRao2np
KnRaoA =
2np
Thus, for r > Rao,
Ho E Kn (Rao np+1
n 2 r
Hr = Kn (Rao np+1
n 2 r
cos (wt - np6)
sin (wt - npO)
(3.3)
(3.4)
Equating fields from the armature and the equivalent current sheet, we obtain
Kn (Rao np+1
2 r ) JRaoR ai
JndR2R
(R) np+1
r
Jn (R np+1 _ Ra+
2(np + 1)rnp+l
=> Kn, -Rai \np+1)
Rao(3.5)
Since the windings and magnets have a finite axial length 1, we introduce another
Fourier summation in m:
004K = Kn COS(Wtt - np) --- sinn m=1 MT
modd
(7WZ)
for 0 < z < 1. This is an approximation since it implies that current exists for z < 0
and z > 1, alternating in direction every length 1, which is not the case in reality, but it is
21
T O, for r > Rao
Jn
np+1
adequate over most of the range of interest, 0 < z < 1.
If the rotor has mechanical speed w/p, the angle 0' in the rotor frame is equal to the
angle 0 in the stator frame minus wt/p [2]. Then wt T- npO = wt ~F nwt -F npO'. So the rotor
sees the current distribution
004 m7rzK = Kncos ((1 T n)wt -F np') - sin
n M=1 MITmodd
for n = 2kq i 1, integer k (3.6)
3.2 Eddy Currents
This section presents an electromagnetic analysis of eddy currents in azimuthally seg-
mented magnets. The first three subsections describe simplifications of the problem of
interest, namely that of estimating the three-dimensional eddy current distribution caused
by a given stator excitation. The first subsection assumes a given magnetic field at the
magnets, the first two assume that the magnets are thin, and the first three assume that the
magnets and stator windings are fastened to infinitely permeable boundaries.
In the following analyses, the magnets have finite length 1 in the axial direction and
subtend an angle Om in the azimuthal direction. The geometry of the problem is illustrated
in Figure 3-1. Since the thickness of the air gap and magnets is small relative to the radius of
the motor, we use a rectilinear approximation to simplify the geometry of the problem. We
have x as the azimuthal coordinate, y as the radial coordinate and z as the axial coordinate.
The magnet width then becomes d = 0nR, where R is the average radius of the magnet.
3.2.1 Known Magnetic Field and Thin Magnets mounted on Infinitely
Permeable Surface
As an initial simplification of the problem, we assume that the thickness of the magnets,
Am, is small compared to the skin depth, and that normal magnetic field at the magnet layer
22
air gap
Kz,
magnets
Figure 3-1: Geometry of magnet loss problem
(y = 0) is known to be
B ~ ~00 M1
Y ( Bn, sin (wnt - nkx) sin
mcdd
This problem is diagrammed in Figure 3-2.
The time-varying B field induces an electric field E according to Faraday's Law
x = ORat
the y-component of which yields the relation
aE Ez B--. - - - Y
az ax at
23
(3.7)
y
Bmagnets
0 d2 3x
Figure 3-2: Diagram for problem with known magnetic field and thin magnets
This electric field gives rise to eddy currents in the magnets, but eddy currents result from
only those components of electric field that match the boundary conditions imposed by
the magnet dimensions. Since the current K is constrained to circulate in thin magnets
of length 1 and width d, the x-component Kx must be 0 at x = 0 and x d, and the
z-component Kz = 0 at z = 0 and z = 1. These conditions are equivalent to K being given
by V x (CQ), where C is of the form
C = E EChm(t) sin hdxsin (' Zh m
Then
O9Z h~r m lk d IkK = - Z - 7 Cm sin Cosm
Kz = Chncos sin (3.8)
Thus
Amo : for modes satisfying Ex = 0 at x = 0, d; Ez = 0 at z =0, 1
0 : otherwise
24
where o is the conductivity of the magnets.
To extract the components that induce eddy currents, we express OBy/Ot as a Fourier
series over the width d of one magnet according to
00/m'r
aBy t= ( E BnmWn cos (writ - nkx) sin zn mB1
modd
= BnmWn (COSwUtcosnkx ± sin wt sinrtkx) sin(m )n m=
modd
The functions cos nkx and sin nkx can in turn be expressed as
E z ainuU
= (a 3 uU
COS (27rx) + a2 u sin 2u7rx)
COS (2u7x) + a4 n. sin 2u7rx)
a1 ""
a2n""
a3n""
a4n""
=sin nkd I + Is nkd + 2uir nkd - 2u7r)
(1 - cos nkd)(nkd + 2uw + nkd - 2u7)
=(1 -- cos nk d) 1 + 1(nkd +2uxr nkd - 2nx1 1
= sinrnkd - +nkdd - 2uir nkd + 2ur
This Fourier series is valid over the interval 0 < x < d when nk > i, and nkd is
not an even multiple of 7r. In this case the expression is approximate, since it assumes
discontinuities at x = 0 and x = d, which imply the existence of artificial current sheets
at those boundaries. If nkd = 2m1 7r, where mi is an integer, then the Fourier series for
cos nkx and sin nkx each reduce to a single term, cos (2nyx) or sin (2nyr2) respectively,
25
cos nkx
sin nkx
where
(3.9)
(3.10)
: foru=mi
: otherwise
: foru=mi
: otherwise
Since only the terms in sin (2ur) give rise to eddy currents, Equation 3.7 becomes
+ sin wnt (E a4 . sin (2wr ) sin (m7rz
1'AmoU
OK
=1 2- Am EEm
OKz)
d )Chm sin (h7rx) sin m7rz
(
Comparing the expressions termwise, we have
En BnmLn (a2nh/2 cOS wat+a4nh/2 sin Lot): when h even, m odd
d I
: otherwise
This is substituted into Equations 3.8 to solve for the eddy currents.
3.2.2 Known Stator Excitation Current and Thin Magnets with In-
finitely Permeable Boundaries
As in the previous section, we assume that the magnets have thickness Am, which is small
compared to the skin depth. In this case however, the source of excitation is the stator
26
ie.
alau
a2nu
a3nu
a4nu
1
0
-0
-0
1
0
-En
(3.11)
SBnmn
modd
(cos wn
Ch0Amo O
a2,, Sin 2ndr ))
+ M72
current, represented by a current sheet at y = A whose distribution is
K= ( Knm COS (wnt - nkx) sin Z
modd
We assume that the magnets and the stator windings are fastened to infinitely permeable
surfaces, so H = 0 for y > A and y < 0. A diagram of the geometry of the problem is
given in Figure 3-3.
inFinitely permeable material
stator current sheet K
air gap
magnets
2 d
infinitely permeable
3d x
material
Figure 3-3: Diagram for problem with known stator current and thin magnets
The current sheet sets up a magnetic field in the air gap, which gives rise to eddy
currents in the rotor magnets. As before, the eddy currents are constrained to be of the
form
acBC
mir-- 55 Chm sin
h m I
aC h7rKz E E Chm COS
ax h m
h7rx
d
h7rx)
( z )zco
(my) (3.12)
27
Y
A~
Am0
Since there is no current in the air gap, the magnetic field is irrotational and can be
obtained as the negative gradient of a magnetic scalar potential T. We find 4' as the su-
perposition of two solutions 4 1 and 4'2, each of which satisfies the boundary condition at
one boundary and is zero at the other. The boundary conditions are obtained by applying
Ampere's Continuity Condition at the boundaries. At y = A, -Q x H = K, which implies
Hx Kz = Knm cos (wnt - nkx) sin(m7zfl m=1 M7
modd
Hz =-K = 0
Now
|_a= - Hxdx = E Knm sin (wnt - nkx) sin (m7z)nk n m=1
modd
satisfies Hz -&89/Oz = 0 at y = A for 0 < z < 1, since the summation in m yields a
constant in z for 0 < z < 1. Thus the scalar potential
001 mn_ _ zE Knm k sinh 1 in (nt - nkx) sin sinh (#nmy)
n m=1 (k3sinh(#1ammodd
which is zero at y 0, matches K at y A. Since V 2 91 = 0,
012m = (nk)2 + (rl)2
At y = 0, Q x H=K,ie.
h7F hwx M71zH = -Kz =-ZZ d Chm COS(d sin
h m
rn h'rx m7zHz =K - EE Ch, sin ) cosh m
28
Since
IF|,l =- Hdx = Z Chm sin
satisfies H2 = -849/Dz at y = 0, the scalar potential
h m sinh(-#2hmA)sin h7rx)
(d )
(hwrxd )
sin (mirz1'mw
sin (mlrz)
sinh ( 3 2hm (Y - A))
matches K at y = 0, and is zero at y = A.
(h r 2#2m =d )
since V 24' 2 = 0.
By superposition, the magnetic scalar potential T in the air gap (0 < y < A) is 'I1'+ 2.
The normal magnetic field is then given by
_8DI91 D42H + aay ay
=- _E Km ." sin (wt
modd
- nkx) sinmJrz
I)cosh (#Anmy)
-- Y E Chm- m -sinh m sinh (-#2hmA)
hwx
d )sin cosh (02hm (Y - A))
As in the previous section, the electric field induced by the time-varying magnetic field
satisfies
z Dx aB
Now
00 !nm-Po 1Z Z Knmk. AWn cos
m1 nk smh#A(wnt - nkx) sin
dChm #hm .+ po E E sin
h m dt sinh(#aA)
h7rx)
d )sin (7). mIZ cosh (2hmA)
29
OB"Dt (mwz
+ M72
1
-0 Il Z/n n 3 nm - (opo Knm * w (cos w\Wt cos nkx + Sin wnt sin nkx) sinn menk sinh1mmodd
dChm #3hm
h m dt sinh (32hm)
( h7rx)s( - ) s( "z) cosh (/3 2hmA)
can be expressed as a Fourier series over the magnet width d, by using Equations 3.9- 3.11
for cos nkx and sin nkx. Noting that only the terms in sin ( 2 7 ) couple to eddy currents,
we have
n =modd
Knm nk snnm onrik sinh /3inm A (coszt a 2 nu
+ sinont (
-Po E dChm #2hm . h-ixh m dt sinh(/ 2hmA) d
1
Amo-
(KxOz
sin ( MFZ )
sin (2ndrx
a4,, sin (2udrx)
cosh (#2hmA)
1
m3 h m
OK
x )
h7 2 rn) 2] h7rx (m7rzCam Sin si1
Comparing the expressions termwise, we have, for h = 2u and m odd,
dChk1 d
k 2 Ch= 1 (k 3n cos wnt + k4n sin wnt)n2
where
ki = 32hmcoth (3 2hmA)
k2 Amct1 h7 2 + (m7r)2
AMUo- ( d )
k3 = Knm#1inmWn a2nk sinh (/31nm) 2/2
k4 = Knm/3 ino ank sinh (#1/ma) 4n /2
30
sin (m7rz)
The solution to the differential equation is of the form
Chm = E (C1, cos wt + C2, sin wt)n
Substituting this into the differential equation, we obtain
C1"
C2,"
k2k3, - k 1 k 4 ,wn
(kiWn) 2 + k2k 2 k 4 n + k 1 k3swn
(kin)2 + k
Thus
(k2k3, -kik4w.) cos wt+(k2k4n+kIk3,W.)sin wt :
(kW" )2 +k2
0
when h even and m odd
otherwise
Expressions for the eddy currents can be obtained by substituting this result into Equa-
tions 3.12.
3.2.3 Known Stator Excitation Current and Magnets with Significant
Thickness with Infinitely Permeable Boundaries
Here we assume that the magnets have thickness T, and that the source of excitation is the
stator current sheet
00
1: E K,,,, cosfl m=1
modd
(wt - nkx) sin
at y = A. We analyze the magnetic field in two regions, as shown in Figure 3-4. Region 1
consists of the magnets (-T < y < 0) and region 2 is the air gap (0 < y < A).
First we consider the region inside the magnets. We assume that there are no radial
components of the eddy currents, ie. Jy is 0. In this case, the current density J = Joi + JZz
31
Chm{ =
m7rz
I )
inFinitely permeable material
stator current sheet K
air gap
magnets
materialinfinitely permeable
Figure 3-4: Diagram for problem with known stator current and magnets with significantthickness
is given by V x (Ce), where C is of the form
C = Z E Cm(y, t) Sinh m
hyrx
d )sin (7)
so as to match the magnet dimensions and the assumption of no radial current. Thus
aC m7r siJx - Bz lCm(Y, t)inh m
OC h~cJz - =CdZChm(y,t)
Xh m
(cos ( d J
hirx
d )sin (7)
From Ampere's Law, the magnetic field induced by the eddy currents satisfies V x H = J,
ie.
aHz aHy
ay aziJx
32
Y
0
-T
Region 2
Region 1 t 3 d x
(mrzCos 1)
!) k,-n) - ". I 7j 7-) 77 V- -, 77 7- '
Thus Hz, Hy and Hz are of the form
H - Ahm (, t) Cosh m
Hy = Z Ayh(Y, t)h m
Hz =h
S:m
Azhm (y, t) sin
(h7rx)sin (hdx)
(hwx)
sin 7Z
sin (M7FZ)
Cos (1)
Now H satisfies the diffusion equation
pLoa O = V2)7
within region 1. Therefore,
cos
2- Axh (7) 2 ± a2xh
Similarly,
- OA - Axh h w 2
o y - A Y m hat - hm \d)
IL _ a z _ - A h (hwN )2
[U at dh
+ 2 Axhm- Axhm Tr2
- AYhm
-Azhm
+ a 2 AYm
±g2 Ah
Since the excitation is a sum of sinusoids with frequencies w, we can write Ajhm (Y)
33
OH_ OHx
Ox Oy0HOz
- 0OHzOx
OAxm
atILOUh m
=5Ih
sin ( 7IZ
Em (-Axh (d) Cos ( hdx) sin )
in the form E Re AZhmn (y)ewnt}I for i = x, y, z. Substituting this into the differential
equation, we obtain
I-10Or)WnA Ahmn
02 A ihmn
Oy2
E A ihmn( (h7dT
+2 ( w2)
+(rnlr)2- Azh
A ihmn ai+ hmn -hmnY + a -hmn-
where
7hmn Chmn+ dhmnJ =)
(Wn0 )2
2
+ (Wn0IoC)2)
h m n
Hy zz55h m n
h mE
n
Re { (ax+eO" + ax e-7) eiwnt} cos
Re (ay+ e
Re {(az+ey
+ aye-Y) eiW sin
(h7xd J
(+ az-e YT) eint sin (
h7rx)
d'rx
sin TT
sin
Cos (1
where the subscripts hmn have been omitted from ai±hmn and 7hmn for notational com-
pactness.
34
En
+±En
2hmn
Oy 2
± jn/l0)
d)2
(3.13)
Chmn
dhmn
± 4WnP;OU
h7r2
d)
2
Thus
( )+
2)2
(3.14)
+ MT2
+M72
T 2
(h)2 M,)2 2
Applying the boundary condition that H, = 0, Hz = 0 at y = -T, we have
ax+e-T + axe? yT 0
az+e~-T + az-e T 0
Thus
HX = 1:1h
h m n
Re { ax+ (eY - e
Re {az+ (ey - e
-2-yTe 19Y) ej"t} Cos (-2-Te -YY) ejUn} sin (
hurx)sin (m1rZ)
Cos mIrz
Next we consider the air gap, region 2. The boundary condition at y = A is the same
as in the previous case, where we have from our earlier analysis that
00
Hxdx = 1n m=1
modd
Knk sin (wnt - nkx)
As before, the magnetic scalar potential T is found as the superposition of T1 and W2, each
of which satisfies boundary conditions at one boundary and is zero at the other.
Knm I sin (otnkn
modd
- nkx) sin mrz sinh (#1nmY)J sinh (AinmA)
where
i,2m = (nk) 2 + (7)2
matches K at y = A, and is zero at y = 0.
AF2Z= EDhm(t)h m
n h7rx)si d sin
sinh (#2hm (Y - A))sinh (U 3 2hmA)
matches the form of HM at y = 0, and equals zero at y = A. The magnetic field in
this region is then found by taking the negative gradient of the total scalar potential T' =
35
(3.15)
(3.16)
F Iy = - fxJ W t ksin (n7z
S ax- _ -
ax+az- -e -2-yr
q1 + X2.
At y 0, tangential H is continuous since the current density is finite, and normal H is
also continuous since we assume that the permeability y of the magnet is close to yo. Now
h7rx)d )- ZDhyh7r COS
h m dsin( Z)
=- EE Knm- (sin wat cos nkx - cos wat sin nkx) sinm = 1 n k oddmodd
+ EDhm sin
h mwm7r
=- Z Dhm sh m
h7rx)
#1""sinh (i1nmA)
si(") 32,m coth (/ 3 2hmA)
h7rx) m7rzCOS 1)
However, only some components of H' 2 ) couple to the eddy currents. Using the Fourier
expansions from Equations 3.9- 3.11 for cos nkx and sin nkx, we have
h7rDhm d COS (h7rx
d )= EEE Re ax+ (I
h m n
sin (7)
- e-2yT) ejwnt cos (h7rx
00 1 {EKnmk (sinwnt
n m_= rikmodd
/3 inm
sinh (#1inmA)
( a 2 nusin 2u7rx) -cos wnt (E a 4 nu
2urxsin d I
sin (n1Fz)
+ E E Dhm sinh m
= E E Eh m n
h7rx
dsin #Z) /2hm coth (#2hmA)
Re ((ay+ + ay_) ejwnt}sin h7rx)
si d )sin (7)
mr h7rx- E EDhm 17 sin hdx
h m
-EE Re jaz+ (Ih m n
- e-2yT) ejwt} sin h~x
Writing Dhm = En Re {fhmneWnt}I, a termwise comparison of the expressions yields, for
36
HX(2 , lyzz
H| 2 , " " --o
H(2 ) jY=
-Eh m
sin (mrz
(m7rz
in (
m~rzCOS (
h = 2u and m odd,
ax+ (I - e-2yT)
ay+ + ay-
a2+ (I - e 2-T)
= -- and bhmn
-Knm! 3 inm (j a2nh,2nk sinh #1nm A
= - Dmn
+ a4h/2 ) + Dhmn 3 2hm coth ( 3 2hmA)
which gives the following expressions for ax+ and az+ in terms of Dhmn:
hir bhmnax+ = - e2 r
m = -hmnI I 1- e-2 y
(3.17)
(3.18)
We can obtain corresponding expressions for ay+ and ay_, by noting that the magnetic
continuity condition V -H = 0 holds for all x, y and z in region 1:
OHxOH(1)
+09Y
- - E E E hi Re {ax+ (eY - e2T e~Y7) ejw"h m n
+ E E E Re (7 (ay+ey - ay-e~7Y) ej'nt} sinh m n
- E S E n7 Re {az+ (eYY - e 2,Te-YY) ejwnt}h m n I
-0
- Re ey ( dL ax+ + -yay+ -
+e-7y9( ax+e 2- - -yay_
- 0
M7r \i az+)
+ -az+e 2r
11 sin (h7rx)
d )
hux)
m7rzsin
sin h7rx
h7r mir
- -dax++ -yay+ - az+ -0
-ax+e-r - -yay- + -az+e-r= 0
37
sin (7)
sin 7
eiwt
Oz
Substituting in the expressions for ax+ and az+ in terms of Dhmn obtained earlier, we have
- ax(a++ az+
((h7 2
d /)
- ax+ +
d ()
+ (Mnr)2
rraz+ e27T
± (TnT )2)
We can substitute these expressions into the expression for a,+ + ay to solve for hmn:
Knm/'3 nm
nk sinh (#1nA)(
ay+ + ay_
h7 2
d )
d )
::: Dhmn -
ja2.h/2 + a4h/ 2 ) + bhmn2hm coth (/32hm/A)
S(7r)2 bmn (1 + e--Y)\ l } (1- -,) -y
(7r)2 Dhmn coth (yT)
Knkinm n a/ +aflk sifh(O 1innA) kj
2nh/2 nh/2,
#2hm coth (2hmA) +
Knm 3 1nm7 (ja2 h/2 + a4fh/2 )
nk sinh (#1nmA) ) + (m)) coth (7T) + '72hm coth (#32hmA) ]when h even and m odd
The coefficients a,+, a,-, ay+, ay_, az+ and az_ can then be obtained from Equations 3.15,
3.16, 3.17, 3.18, 3.19 and 3.20. Substituting these coefficients in Equation 3.14 yields the
magnet field components that couple to the eddy currents.
38
Dhmn
-e-2-yr) -Y(3.19)
Dhmne2r(3.20)
2 ()2) coth(yT)
3.2.4 Known Stator Excitation Current and Magnets with Significant
Thickness Without Infinitely Permeable Boundaries
Without the simplifying assumption of infinitely permeable boundaries, obtaining the mag-
net boundary conditions requires the analysis of magnetic fields in the regions interior to
the armature (r < Rao) and exterior to the magnets (r > Rmo). Magnetic fields arise from
the stator current sheet
00 4 m,,rzK ( K cos (wut - npo) - sin (
modd
at r = Rao, and the eddy currents in the magnets.
We first consider the field due to the stator current sheet. Since K does not vary with z in
the region of interest, 0 < z < 1, we use the two dimensional solutions from Equations 3.3
and 3.4. This solution is valid only over the finite axial length of the stator, and only
approximately so near its ends, where the end turns of the stator winding have not been
included in our model. We introduce a Fourier series in z so as to facilitate comparison
with the modes of the magnetic fields from the magnet eddy currents.
Kn Rao np+1 oo 4 mrzHod8 = ( cos (wnt - npO) ( --- sin
n m 1modd
Kn Rao np+1 o 4 m~szHoSr = sin (wnt - np6) ( sin
modd
Within the thin magnet layer, we can use a rectilinear approximation to the geometry,
as we have done in the previous sections. From previous analysis (Equations 3.13- 3.14),
the magnetic field in the magnets that couples to the eddy currents is of the form
Hx = E E Re (ax+e" + axe~-Y) ent} cos (hx sin (m7zNh m n d
Hy= - § Re ((ay+eYY + aye-Y7) eiwn'} sin hdx sin (7r z)
39
Hz EEh m n
Re Iaz+e?" + az-e --YY) e j f sin rxe~( d )
'Yhmn = Chmn + dhmnj = \t() + (7)2 n/1o
The fields induced outside the magnets by the eddy currents have a similar form, and
can be expressed as the negative gradient of scalar potentials
= EEEA 3hm.h m n
= EEE A 4 mh m n
F3hm (r) sin
F4h (r) sin
(hTO)
hvO)hi0
sin (1UZ
sin m7Z)
for r < Rmi
for r > Ro
which satisfy Laplace's equation
I a (rO) 1 &2 T
r2 002 + =0
in cylindrical coordinates. Substituting Tir and Wo, into Laplace's equation, we obtain
82F 1 FOr2 r Or
h7r 2 m (' 2~
+ r
The form of this differential equation matches a version of Bessel's Equation, which has
as solutions the hyperbolic Bessel Functions I (TjZMr), which grows with radius, and
Kh (r), which decreases with radius and is singular at the origin. Thus
F 3h.(r)
F 4hm(r)
mT r= Ih r
om i< r
In the air gap between the stator current and the magnets, the total field is the superpo-
40
where
Cos ( ) (3.21)
Tor
or
F = 0 for i = 1, 2
V2T = 0
sition of H,, = -V4' 0 and Hir = -VWTr. At the inner surface of the magnets, r = Rmi,
0K (Rao nP+1 4 m7z
Ho = f ( ao cos (wnt - npO) sinn m=1 2 Rmi mr 7
modd
1mr h7r hr mirzR ZEA3 h R Cos Ihir sinRmi h m n 1 t m \ m) 1
oo K /Ra np+1 4 mrHr = ( ( - -- sin (wnt - npO) - sin("~z
E 2 Rmi mrmodd
- E E Y A I/ R sin /sin z
h m n1 j-(1(0) (1
m7r h7rO m7 m7r zHz =-EE( ( As Ih Rm sin 6 - Cos
h m n hirO m 0 ( 1
Assuming a magnet permeability close to po, the normal and tangential components of
magnetic field are continuous across the magnet boundary. However, only some compo-
nents of magnetic field couple to eddy currents within the magnet boundaries. These can
be identified by expressing Ho, as a Fourier series over the angle subtended by one magnet,
Om. Now
cos(npO) (a1n. cos + a2 . Sin (3.22)
sin(npO) = a3nu cos + a4 ,. sin (2ur) (3.23)
where
a1u= sin (npom) + (3.24)np6m + 2u7r up6, 2u7r)
a2n = (1 - cos (npO)) (3.25)(np6m + 2ugr np6, 2u~r)
a =(1 - Cos (npOm)) ( + (3.26)npm + 27 np6 , - 27 )(
a4n = sin (np~m) I - 1(3.27)(npom - 2u-c np6m + 2ugr)
41
unless npOm is an even multiple of 7, of the form np0m2 m17, in which case
-0
-0
1 : foru=mi
0 : otherwise
1 : forun=mi
0 : otherwise(3.28)
For those components of magnetic field that couple to eddy currents, we have, noting
that Hr = -Hy,
0Kn(a np+1
modd
M Sin ( M"Z
(cost1 ainu + sin Wnt (E a 3 u cos (2urO)))
cos ( hrO) sin ( m wzF1 mr h7rREEEAshmnnIh Rn -
Rmi h m n miM )0
= EEERe (a-+h m n
+ ax) ejW't cos (hurO)sin (n7IZ
_0K ( a np+1
modd
4.sin
m7r
(sin wnt (Ea 2nu sin m2 - cos wnt (1a 4 nu sin 2zurO
( Orl))
mr z1
- E E A3h m n
- - E E E Re (ay+h m n
mir m (,h7r miI' W R( I
h76sin (0)(him
+ ay_) en't I sin ( hm )
- EEE5A 3 hmi rh m0
=E E E Re(h m n
/RmiMTr h7
I sin ( m)c(
h7rO)(az+ + az- ) ej''t ) sin (0)
CS(m7rz
os )
(mrz
42
alnu
a2nu
a3nu
a4nu
En
sin (M17Z)
sin (n7z
Cos (2uyr6
Comparing these expressions termwise, with A3hmn = Re {A3hm eiWn}, we have
ax+ + ax_ -
ay+ + ay_ -
az+ + az- =
2Kn Rao np+l1
m7 Rmi
2Kn Rao )np+1 (.a
mm~r
-A 3hm I h RmiT yT
n u jaa,) --
Z3hhir
Rmi -m
mir2n + a4,) + A3hm I' hir
)7
for h = 2u and m odd. A similar calculation at the outer surface of the magnets yields
2Knmr
Rao) np+l
Rmo
RAo A4hn
2Kn Rao) np+1
m7r (Rmo
( mR mo h7rI I m
(ja 2 u + a4n.)
ax+e y + axe
ay+e +yT ay-eT
Z 4 hK' ( '7A lh 1 n OM(Rmo)
- A 4 hKh, (7 Rmo)
for h = 2u and m odd. Finally, from the magnetic continuity condition V -HM = 0 Vx, y
and z we have
h7r
d- ax+hTr
d
mir+±yay+ - 1az+
m7- -yay_ -az_
1
= 0 (3.35)
(3.36)
The eight equations 3.29- 3.36 can be solved simultaneously for the eight unknowns
43
m7rRmi
h7
OM
(T Rmi)
(3.29)
(3.30)
(3.31)
(3.32)
az+e yT+ ae yT
(3.33)
(3.34)
( a1nu - ja3nu )
ax+, ax_, ay+, ay_, a2+, az-, A3hm and A4hm, in terms of K,:
az-
A4hm
- A-ly
where
1 1 0 0 0 0 h^* I ( mrRmi)
0 0 1 1 0 0 "'"I'h7, mRir
0 0 0 0 1 1 " I ("'fni1 om (
e-yT e-T 0 0 0 0
0 0 e-7T eT 0 0
0 0 0 0 e-T eyT
-h 0
0 -^lr
y0 -~
0
0
0
0
0
0
0
0 ME
0
0
0
hr K h, mirRo)Rmo m m )
""~rK' h~r m~rRmo
"Kh,, (mirRmo)
0
0
44
(3.37)
2K, (Rao \np+l,mir \Rmi)2K, (Rao" )np-f-rni7r kRmt)
2K, (Rao ")np+1
m.7r Rmo)
2Kn (Rao )n~
mir \Rmo)
( ah/2 - ja3nh/2 )
(ja2h/2 + a4fh/2 )0
( lnh/2 - 3/2)
a2lh/ 2 + a4fh/ 2)0
0
0
The values of the Bessel functions Ih- (m''m") and Kh - (mRmi ) vary greatly with h,m Tm
so to avoid having very large or very small values in the matrix to be inverted, the following
modified form of Equation 3.37 is used:
a,+
a._
ay+
ay _
az+
az-
4 hK , (miRmiAhm n
(3.38)
45
and
1 1 0 0 0 0 Rr
hir
0 0 1 1 0 0 myram
0 0 0 0 1 1 mit
-T eyT 0 0 0 0 0
0 0 e-?T eT
0
070
0
0
hit
0
0Y
0
0
0
- I
0 0
-00
eyT
0
mt
0
0
0
0
0
0
0hitr
Rmo~m
K hr (mrRmo)
mr
0
0
and y is, as before,
2Knmir
2K.m7r
2Knmit
2Knmitr
SRaonP+1(\Rmi)J(i,, )flP+1
(Rao np+1
Rnp+1
( nh/2 - Ja3nh/2)
a2nh/2 + a4nh/
2 )
0
nh/ 2 3nh/ 2 )
(Ja2nh/ 2 +a4h/2
0
0
0
The coefficients a,+, a_, ay+, ay_, az+ and az_ are then substituted in Equation 3.21 to
find the magnetic field components that couple to the eddy currents.
The magnetic fields induced by the eddy currents have been expressed as the negative
gradient of scalar potentials 4, and For that match the boundary conditions in cylindri-
cal coordinates. These involve Bessel functions which may be troublesome to compute,
especially for higher orders. Since we are only concerned with the fields at the magnet
46
where
e
boundaries, assuming a rectilinear geometry gives a very good approximation.
h mA3hm e-AmY sin (hwrx
d Jsin m7z
sin ( 7Fz)Z E A4hn e Amy sin hwxm n d)
for r < Rmi
for r > Rmo
Matching fields at the boundaries of the magnets then yields, for h = 2u and m odd,
ax+ + a =
ay+ + ay-
2Knm7
2Knm7r
az+ + az- -A3hm 1
Rao np+1
Rmi
Rao np+l
Rmim7r
(ai1n - ja 3 u) -
(ja 2n + a4nu) + A3hm
at the inner surface, and
ax+e-T + a eyT -2Knmrnz
ay+e-T + = eyT 2Knmrn
Ro)np+1
Rao np+ 1
RmoI
(ain - jas.) --
(ja 2n + a4 nu) -
Z4hm e~OTh7Z4h~4d
A4hm I3e- T
az+e-T + a_ eI = -A4hm e~T mT1
at the outer surface. Matrix A in Equation 3.37 then becomes
Ac=
1 1 0 0 0 o hird1 0
0 0 1 1 0 0 -# 0
0 0 0 0 1 1 Te~yT eyT
0 0 e
0
0 0 0 0 0 e-,T
-1T eT 0 0 0 13e-OT
0 0 0 0 e-?T e
hrd7 0
0 -j
0 0 -T0y -I0
?yT 0 Te- OT
0 0 0
T 0 0
47
Wir
orh
hrA
3 h- (3.39)
(3.40)
(3.41)
(3.42)
(3.43)
(3.44)
(3.45)
t-
The cartesian version Ac also becomes poorly conditioned for larger h and m, owing to
the term e-OT. The following equation works better:
where1
0
0
e-yT
0
0
hitd
0
1
0
0
eyT
0
0
0h-id
ax+
ax-
ay+
ay~
az+
az-
A3hmn
0
1
0
0
e--T
0
7
0
0
1
0
0
e-yT
0
0
-7
(3.46)
0
0
1
0
0
e-?yT
-T
0
0
0
1
0
0
eyT
0mt
hit
-#13
0
0
0
0
0
0
0
0it
d
/3
0
0
and y is unchanged.
3.2.5 Summary
In summary, this section has examined three simplified problems in Subsections 3.2.1, 3.2.2
and 3.2.3, leading up to the problem of interest in Section 3.2.4. In Subsections 3.2.1 and
3.2.2, the magnitudes of the eddy currents were solved for directly, while in Subsections
3.2.2 and 3.2.4, the magnetic fields that couple to the eddy currents were found. The
following section uses the field expressions from Section 3.2.4 to calculate the eddy current
48
=- A'- y
dissipation.
3.3 Loss Calculation
Dissipation from eddy currents is given by integrating the power density
J2 j j 2
over the volume of the magnets. Substituting the field expressions from Equations 3.21
into the relation J = V x H, we have
iDy - z
- EEERe [(az+y-ay+7) ey-h m n
Jz =~ax
hm
Sin hx
OHX19y
Re
(az-7 + ay_ eI7 ) eiwnt
cos ( 1Z)
hirE(ay+ -I( d
e" + (ayhir
+ axj) e-] eiwnt
sin nz
The time-average power dissipation in one magnet is thus
jd- J dzdydx
-Tf U odl E e?" - ±az-7+IT (az+7 -
2
ay_)
e" + ay i + axY) e-t 2dy
dl (IC128rh m n
+ |C3 2) (1 - e-(-y+y)T) (C12 + C304) (I+
- e(--y+)T)
49
+
- ax+7 )
cos (hx
ay+ Mi
h7ray+ - - ax+,l
d
(C201 + 403) (i - e(--)T) (IC2|2 +C4 12) (i -y+y)r
where the constants Cihmf, whose subscripts have been omitted from the notation above,
are
m7rClhmn = az+7 - ay+ ,
m7rC2hmn -az--7 - ay _
hir3.n = -y ax±_y
C4hmn y-- + ax~
3.4 Application of Model to Rotor Magnet Loss Problem
When the motor is operating synchronously, we have, from Equations 3.1, 3.2 and 3.5 in
the first section,
nIr e(iRa )(np+1) 1-Rao) ) when n= 2kq ±1, integer k
0 otherwise
(3.47)
The frequency of the nth harmonic as seen from the rotor frame is wn = w(1 F n), where
w is the frequency of the electrical excitation. The values of the other parameters are given
in Table 3.1.
The values of a,+, a,_, ay+, ay_, az+ and az- are found by solving Equation 3.38 in
matlab. The matlab code implementing the loss calculations is given in Appendix B. For
this motor, which has a rated ampere-turns of 5835 A, the estimated loss from eddy currents
in the rotor magnets is about 40.8 W at 15,000 rpm. The corresponding loss estimate from
the cartesian version is 40.4 W.
Equation 3.38 can also be used to predict eddy current dissipation for a locked rotor test,
in which the stator is excited with a known polyphase current of amplitude I and frequency
50
Table 3.1: Machine parameter values for rotor loss calculation
Parameter Symbol ValueNo. of pole pairs p 4No. of phases q 3Armature inner radius Ri 0.0673 mArmature outer radius Rao 0.0793 mMagnet inner radius Rmi 0.0806 mMagnet outer radius Rmo 0.0901 mMagnet conductivity a 7 x 104 W/mOCMagnet length 1 0.1001 mMagnet angle 6m ir/6Electrical angle owe 0.856Ampere-turns NI 5835 A
w. In this case, wn = w, and K is given by Equation 3.47, as before.
Loss in a single magnet was also calculated for a range of magnet angles. The results
are graphed in Figure 3-5.
51
Eddy current bss in a magnet of angle thetarn
41 1 1 I I
0.2 0.4 0.5 0.8 1 1.2 1.4thetam -rad
Figure 3-5: Graph of eddy current loss vs magnet angle
52
La
1.5
Chapter 4
Stator Loss Models, Cooling System
Design and Thermal Analysis
Winding conduction losses, eddy current losses and windage losses produce heating in
the stator, which is removed by a water cooling system. This chapter presents stator loss
models, the cooling system design, and a theoretical prediction of the thermal performance
of the stator and cooling system. The equations in this chapter are implemented in the
spreadsheet shown in Appendix C; only the results are quoted here.
4.1 Loss calculations
4.1.1 Conduction Losses
The resistance of the copper wire results in conduction losses. From Chapter 2, the rated
ampere-turns NI, = 5749 A at the speed of 15,000 rpm. Therefore, the current in a single
strand is I = 1.037 A, where the number of turns N = 8 and the number of parallel
strands is Npa, = 693. The resulting power loss per unit length of wire from winding
resistance is
PiR - 0.544 Wm 1N o7rr2
53
at the rated ampere-turns, where the conductivity o- of copper is 3.9x 107 S/m at 150'C.
This is multiplied by the total length of the windings to obtain total conduction loss. The
active section of the windings has length la = 10.01 cm, and the straight section is longer
than the active length by a safety margin of 1, = 1.42 cm. The end turns are semicircular,
and have an average length of roughly lend = ir RaojRai sin(22.50) = 8.81 cm. Therefore
total length is estimated to be
2qNNpar (la + is + lend) = 6733 m
and total conduction loss is 3664 W.
Conduction loss, being proportional to the square of current density, varies with rated
power and with rotational speed. Current is proportional to rated power, so conduction loss
is proportional to power squared. The speed dependence is determined by how the machine
is operated. Below 15,000 rpm, the machine operates at constant torque. From 15,000 rpm
to 30,000 rpm, the machine operates at constant power, which means that torque is inversely
proportional to speed. Since current is directly proportional to torque, conduction loss is
constant up to 15,000 rpm, and varies inversely as the square of speed between 15,000 rpm
and 30,000 rpm. Graphs showing the variation of conduction loss with speed and with
power are given in Figure 4-1.
4.1.2 Eddy Current Losses
Eddy current losses occur in the active section of the windings, owing to the time-varying
magnetic field of the spinning rotor magnets. Here only the losses due to the fundamental
component of magnetic field are estimated.
For a sinusoidally varying magnetic field B = B, sin wt perpendicular to the axis of
the wires, the induced electric field is calculated by applying Faraday's Law to the contour
54
Variation of Conduction Loss with Rated Power at 15000 rpmn
3500
3000
02500
5D
1000
1000DO
0 0.5 1 1 5Spead. rpm
2 2.5 3
x 10'
Figure 4-1: Graphs showing variation of conduction loss with rated power and speed
C in Figure 4-2. Accordingly,
B .dsc L
2EL
-d
= d (2xLB,,sin wt)dt
E - -xBow cos wt
The power loss density due to E(x) is given by
-E2 = o(xBow cos Wt) 2.
Power loss per unit length of wire is then
2 1 o x 2 ( Bow cos wt)22fr2 _ X2
10= 4cr(Bow cos wt)2 J(rW cos 6)2 rW sin Orw(- sin )dO
4ur(Bowcoswt) 2r - sidn2 4L
55
Rated Power W 10
Variation df Conduction Loss with Speed at 30 kW
, 10
B sir wt0
Figure 4-2: Contour used in eddy current calculation
= o-(Bow cos wt)2rj.
The time average power loss per unit length of wire due to eddy currents is thus
P = -B 2W2 r (4.1)8 0 W
As discussed in the previous chapter, the radial magnetic flux density varies with radius
as rP- 1, and has an rms value of Bia = 0.0958 T, or an amplitude of \/2Bia, at the armature
outer radius. For this machine, the radial and azimuthal magnetic fields are equal, so the
combined amplitude is 2BIa. The square of the magnetic flux perpendicular to the wires
thus has an average value of
1 Rao, r )P-1)2< B 2 > ={ 2B1a d
Rao Rai Rai Rao
2 1 -x(p(2BIa)2 1_X(2p-1) = 0.0237 T 2
(1 - z)(2p - 1)
56
Substituting this into the previous equation, the eddy current loss per unit length of wire is
determined to be P, = 0.00372 Wm- 1 at 15,000 rpm. Since eddy current losses occur
only in the active section of the windings, total eddy current loss is Pe = 2qNaNp,,,.laP =
12.4 W at this speed.
From Equation 4.1, it can be seen that eddy current loss varies with the square of speed.
This relationship is graphed in Figure 4-3.
Variation of Stator Edy Current Loss with Speed at 30 kW
0 0.5 1 1.5 2 2.5 3Speed rprn , 1
Figure 4-3: Plot of eddy current loss in armature winding vs rotor speed
4.1.3 Windage Losses
While a machine intended for actual use would spin in a vacuum, this machine has an air
gap, and hence experiences loss from fluid friction, or windage. Windage losses for this
machine are estimated here; a machine with a vacuum would have a significantly lower
loss.
57
The areas in which windage losses occur are shown in Figure 4-4. There is windage
in the air gaps between the rotor and the stator (areas a, b and c), and between the rotor
and the housing (areas e and f). Most of the formulae in this section are found in [5]. The
results quoted below were obtained from the matlab code shown in Appendix C.
coolingchannel
plena
armaturewinding
Figure 4-4: Cross section of the machine showing major parts and indicating regions ofwindage loss. The hashed part of the machine is the rotating rotor.
Windage torque r for a rotating cylindrical surface is conventionally expressed in terms
of a friction factor c1, defined asT
Cf = wR 4lpQ2
where R is the radius of the cylinder, 1 its length, p the density of the fluid, and Q the
rotational speed. Empirical formulae for cf under different flow regimes are given in [5].
58
For the cylindrical surface between the rotor and the stator (area a of Figure 4-4), the Taylor
number is
Ta = pQRaogo go) /2 = 1330P (Rao)
where the density of air p = 1.16 kgm -3, the rotational gap width go = 1.32 mm, and the
dynamic viscosity P = 1.85 x 10-5 Nsm- 2 at the temperature 297K. Flow is either vortex
or turbulent for Ta > 63. The recommended approach is to evaluate cf from the formulae
for vortex and turbulent flow, and to use the higher of the two values. By this criterion, the
flow is turbulent and cf is given implicitly by
pQRaogo 1 + go/Rao ( 858P 1.2V2 7 (1 + 0.5go/Rai) 2 (1 - go/Rai)
For go/Rao << 1, which is the case here, this expression approximates to
pG~ango-0.136Cf = 0.00655 pQRaogo
p= 0.001864
The windage torque is then
Ta = cf 7rR4olaPQ2 = 0.0663 Nm
The windage torque in area b of Figure 4-4 can be found similarly. For Ib = 0.044 m and
gb= 0.017 m, Ta = 61, 477, cf = 0.00156 (turbulent flow), and torque Tb = 0.0245 Nm.
The annular ends of the rotor (areas c and d of Figure 4-4) also experience windage
torque from spinning relative to the stator. Torque on an annular surface is estimated here
as the difference between the torque on a disk of radius R1 and that on a disk of radius
R 2 , where R1 and R 2 are the outer and inner radii of the annulus respectively. Torque on a
spinning disk is evaluated in terms of a disk torque coefficient cmi which is defined as
2T
pQ 2 R 5
59
Reference [5] provides a chart for determining flow regime from the Reynolds number
and the ratio of axial gap to disk radius. Empirical formulae for cmi under different flow
regimes are also given.
For area c of Figure 4-4, RcI = 0.077 cm and Rc2 = 0.020 cm. Since Rc2 is small com-
pared to Rc1 , the annulus can be approximated as a disk of radius Re,. The corresponding
Reynolds number is
Re -Q"" - 619,400
For this value of Re and an axial spacing of gc/Rao = 0.15 1, the flow is turbulent, with the
combined thickness of the boundary layers on the rotor and stator being less than the axial
gap gc. The torque coefficient for drag on one side is then
0.051(gc/Rao)0 .1Cmi = Reo2 = 0.00293
and the torque is
ro/2 = 0.0132 Nm
Area d of Figure 4-4 has an outer radius Rd1 = 0.109 m and an inner radius Rd2 =
0.0806 m. For a disk of radius Rd1 , flow is turbulent with separate boundary layers on
the rotor and stator. The corresponding value of cmi is 0.00234, and windage torque is
rda = 0.0516 Nm. For a disk of radius Rd2 , flow is similarly turbulent, cmi = 0.00273, and
windage torque is Td 2 = 0.0133 Nm. Thus the windage torque in area d is Td = Td, - Td2
0.0383 Nm.
The total windage power loss between the rotor and the stator is
Pw = (ra + Tb + Tc + Td) Q = 224.4 W
at 15,000 rpm. A graph showing the variation of P. with speed, taking into account changes
in flow regime as speed varies, is given in Figure 4-5. The calculations are given in the
matlab code of Appendix C
60
Variation of Windage Loss with Speed at 30 kW
1000--
I BDD --0, 800-
6 00-
400-
200
0 _j0 0.5 1 1.5 2 2.5 3
Speedirpm x 10
Figure 4-5: Plot of windage loss between rotor and stator vs rotor speed
Windage losses also occur between the rotor and the housing (areas e and f). Calcu-
lations similar to those above yield Te = 0.309 Nm and r = 0.0516 Nm. The resulting
power of (-re + rf) Q = 567.0 W is assumed to be lost to the surroundings directly through
the outer housing, and is not considered in the stator thermal analysis.
4.1.4 Total Losses
This analysis has not taken bearing losses into account, but these will be small if magnetic
bearings are used. When the machine is not generating or drawing power, there are no
conduction losses, and the dissipation is Pec + P = 230.8 W. A machine for actual use
would have the rotor spinning in a vacuum to eliminate windage loss. The idling loss would
then be Pec 12.4 W.
When the machine is motoring or generating, total power loss in the stator is obtained
61
by adding the losses from conduction, eddy currents and windage. A plot of total power
loss against speed is shown in Figure 4-6, from which it can be seen that the maximum
total power loss occurs at 15,000 rpm. At this speed total power dissipation is P = PR +
Pec + P,, = 3895 W, and the efficiency of the machine is (1 - 3895/30, 000) = 87%. If
a Halbach array were used, the rated ampere-turns would decrease from 5835 A to 3573
A, and conduction loss would fall by 2249 W, with a corresponding efficiency of (1 -
1646/30, 000) = 94.5%. A vacuum would improve efficiency to 95.2%. For this machine,
dissipation can be reduced by lowering the rated power, as shown in Figure 4-7.
Variation of Total Stator Loss with Speed at 30 kW
1.5 2 2.5 3Speedi rpm Xl1
Figure 4-6: Plot of total dissipation in the stator vs rotor speed
4.2 Cooling system
A water cooling system is employed in the stator. The armature is cooled by a constant
flow of water through a channel adjacent to the inside surface of the windings.
62
Variation of Total Stator Loss with Rated Power at 15,000 rpm
1.5 2 2.5 3Rated Power ' W 1 a"
Figure 4-7: Plot of total dissipation in the stator vs rated power
The flow rate of water is chosen such that the difference in bulk temperature AT be-
tween the inlet and outlet is less than 5C. This necessitates a mass flow rate m of at least
= 0.186 kgs. The specific heat capacity of water at constant pressure c, is roughly
constant over the temperature range 50-59'F, with an average value of 4.19 kJ/kg0 C.
4.2.1 Channel Geometry and Fluid Flow Considerations
Several possible designs were analyzed, some of which involved fins and spiral flows. The
design presented here, and illustrated in Figure 4-4, was chosen for its superior thermal
performance, taking into account strength and pressure requirements, and manufacturing
constraints.
In this design, water flows through a narrow annular channel of width T and length
4c = 10.03 cm (3.95 in). Water is supplied to a plenum at one end, and the pressure in
63
this plenum causes the flow to be even all around the annular channel. The flow rate and
dimensions of the channel are chosen such that the flow is laminar, and the pressure of
ordinary tap water is sufficient to produce the required flow rate.
For laminar flow, the Reynolds number Re must be under 2100, where
Re = vDHPP
the velocity v = m/Ap, and the hydraulic diameter DH, four times the flow cross-sectional
area divided by the wetted perimeter, is 2T. The density of water p is 999.2 kg/m 3 and its
viscosity p is 1.31 x10- 3 at 50 0F.
The total pressure required is ensured to be less than the pressure available from a tap:
55 lbs/sq. inch, or 3.82x 105 Nm--2. The pressure drop across the channel is
pv 2
AP= f le i2DH
where the Moody friction factor f = 64/Re for laminar flow. Choosing T = 0.14 mm,
we have DH = 0.28 mm, and for a mass flow rate m = 0.186 kgs 1 , Re = 676 and
Ap = 169, 800 Nm .
4.2.2 Channel Outer Wall Material
The outer wall of the channel serves to prevent water from coming into contact with the
windings. It also provides strength, and withstands the water pressure in the narrow chan-
nel. However, it is an additional thermal barrier between the windings and the water.
Ideally, we would like a thin but sufficiently strong wall with good thermal conductivity.
However, the choice of wall material is limited to electrically insulating materials, so as
to avoid additional eddy current losses from the rotating magnetic field. This constitutes
a substantial limitation to thermal performance, since most materials with good thermal
conductivity also conduct electrically.
64
The possibility of using a ceramic material was explored but decided against, ceramics
being deemed too brittle. The eventual design choice was two spirals of fiberglass wrap,
impregnated with thermally-conductive epoxy, to separate the water from the windings.
4.3 Thermal Analysis
4.3.1 Thermal Conductivity Experiments
Experiments were carried out to determine the thermal conductivities of the epoxy, epoxy-
impregnated fiberglass, and epoxy-impregnated glass cloth tape. The procedure involved
measuring the heating rates of copper cylinders coated with these materials.
Method
Three nearly identical copper cylinders were cut and polished. Two of them were coated
on their cylindrical surfaces, each with a layer of different test material. The other cylinder
was not coated, and served as a control. A thermocouple was placed in contact with the bare
copper surface at the top of each cylinder, and both ends were insulated with styrofoam.
Figure 4-8 is a diagram of the experimental setup.
Each cylinder was placed, in turn, in a large beaker of water, maintained at a reasonably
constant high temperature by a hot plate and stirrer. Thermocouple measurements of the
copper temperature were taken at 4-second intervals over a temperature range of about
300 C to 90 0 C.
Theory
Assuming negligible heat loss from the insulated ends of the blocks, the rate of heat transfer
through the exposed surface should be equal to the rate of change of internal energy Q of
the copper. Thus,dQ TW -- Te _ dTe
____ ___________= mcdt Rfilm + Rmateriai dt
65
hot waterbath
Lagging
styrofoaminsulation
Figure 4-8: Experimental Setup for Thermal Conductivity Measurements
where T and Tc are the temperatures of the water and copper respectively, m is the mass of
copper, and c is the specific heat capacity of copper, 393.6 J/kg0 C. The thermal resistance
of the film at the copper surface is
1Rfilm hA
where h is the film coefficient and A the surface area, while the layer of test material has
thermal resistance
Rmaterial =ln (r,/ri)
2,rlk
66
testmateriaL
coppercyindler
where r, and ri are the outer and inner radii of the material, 1 is the exposed length, and k
is the thermal conductivity. The differential equation has the solution
Te Ce-t/(Rfji.+Rmateria1) mc ±T
The film coefficient h is estimated from the results for the uncoated copper cylinder,
for which Rmateriai = 0. This value of h is then used in calculations to determine k for the
other cylinders.
Results
Four sets of temperature data were taken for each cylinder. The measured quantities and
temperature plots are given in Appendix D. The thermal conductivities were experimentally
found to be 0.66 W/m0 C for epoxy and 0.31 W/m 0 C for epoxy-impregnated glass cloth
tape.
An attempt was made to measure the thermal conductivity of epoxy-impregnated fiber-
glass using this method, but it was difficult to wind the wide fiberglass strip evenly around
the small cylinder. The rough, uneven surface resulted in a larger surface area that was
difficult to estimate. Consequently, the thermal conductivity could not be accurately deter-
mined. However, we expect the thermal conductivity of epoxy-impregnated fiberglass to
be about the same as that of epoxy-impregnated glass cloth tape, since these materials have
a similar composition.
4.3.2 Film Coefficient for Cooling Channel
Consider now heat transfer at the wall of the cooling channel. For fully-developed laminar
flow in an annulus, with an insulated inner surface and uniform heat flux at the outer wall,
the Nusselt number is about 5 [6]. The heat-transfer coefficient he is thus
he = NuDk = 10, 450 W/m 2 o CD H
67
where the thermal conductivity of water k is 0.585 W/m 0 C at 50'F.
4.3.3 Effective Conductivity of Armature Region
The windings are embedded in a thermally conductive epoxy to improve heat transfer.
Bounds on the effective conductivity of the copper and epoxy composite are obtained as in
[7], where Milton's fourth order bounds are used. These bound the effective conductivity
kce of the composite material consisting of infinitely long conducting circular cylinders of
conductivity cr2 randomly distributed through an insulating material of conductivity o1, for
volume fractions up to 0.65. The effective conductivity kce lies between cYL and UU, where
( 12 + 92 )(u1 + < O- >) - #2(02- 1)2
(Ui + ( ± 2 )(Or2+ < & >) - 42 (1(0 - 0r)2
OrL 9 1(071 + 072)(02 2+ < Or >) - #1 2 (o-2 - oTI)2-
(os + 2 )( u1+ < & >) - #1 2(07 - 03)2
and
< o > = 0-141 + U202
< Or > = Or201 + 91#2
#1 and #2 = 1 - #1 are the volume fractions of the cylinders and insulating material
respectively. (1 and c2 model the cylinder-to-cylinder interactions. From [8], they are
found to be
2= -0.05707423
The thermal conductivity 92 of copper wire is 390 W/m C, and that of the epoxy, or, is
1.8 W/m0 C. For a copper volume fraction of 0.373, the effective conductivity is calculated
to lie between o-L = 1.54 W/m0 C and o-u = 14.4 W/m0 C. As a compromise, a conservative
68
estimate of conductivity is taken to be the geometric mean oLoyU = 4.71 W/m0 C.
4.3.4 Temperature Calculation
Consider the worst-case scenario where the only means of heat removal from the windings
is by conduction to the water channel. Relatively less heat is lost from the end turns, since
they are heavily insulated and embedded in epoxy. We assume that heat generated in the
section adjacent to the water channel is conducted radially towards the water, and that no
heat is lost from the outside surface. We also assume that heat produced in the rest of the
windings is conducted along the wires to this section, from where it is removed by the
water. This gives a conservative estimate of thermal performance.
We analyze heat transfer in one phase belt, approximating the curved cylindrical surface
as a planar surface. One phase belt consists of two layers of wire bundles, with a layer of
epoxy-impregnated glass cloth tape between them. The two layers are surrounded by a
wrap of glass cloth tape. On the side facing the cooling channel, there is a window in the
tape, which is occupied by epoxy alone.
The width of the wire bundles in a phase belt is w = 1.44 cm, and the length of the
cooling channel is le = 10.03 cm. We first examine heat transfer in the section directly
adjacent to the cooling channel. The cross section is shown in Figure 4-9.
The thermal resistance of the fiberglass-epoxy layer separating this portion of the wind-
ing from the water is
Rf tf -0.9160CW--kgwle
where tj = 0.41 mm is the thickness of this layer, and k1 = 0.31 W/m 0 C is the ex-
perimentally determined thermal conductivity of the epoxy-impregnated fiberglass. The
temperature drop across this layer is AT = - x Rf = 148.6 C, where the total power P
divided by the number of phase belts, 24, gives the heat flux across this layer.
69
x=0
outer Layer
x=d_
inner Layer
/x=depoxy window
Lass cLoth tapend epoxy
waterchannet
fioerglassand epoxy
Figure 4-9: Cross section of a phase belt with adjacent cooling channel
The water film at the fiberglass-epoxy surface has resistance
1Rfilm = .06630CW-1
h~wl,
resulting in a temperature drop of ATfium = P x Rfilm = 10.8*C across it.
Between the wires and the fiberglass is a ti = 0.34 mm thick layer, about 45% of whose
area is epoxy-impregnated glass cloth tape, and 55% epoxy alone. The thermal resistance
of this layer is
R1 = k ti k(0.45kg + 0.55ke) wie
= 0.4680CW 1
where the thermal conductivity of epoxy ke = 0.66 W/m0 C and that of epoxy-impregnated
glass cloth tape, k9 = 0.31 W/m0 C. The corresponding temperature drop is AT 1 = P x
R1 = 76.0 0C.
Within the region occupied by the wires, the temperature distribution satisfies the dif-
70
wires andepoxy
ferential equationd2 T <}
+ =0dx 2
kce
whereP
48dwle
d = 5.16 mm and w = 1.44 cm are the cross sectional dimensions of the group of wires
which makes up one layer of a phase belt. A general solution to the differential equation is
T - 2 + C 1 x + C22ke
We first apply this equation to the inner layer of wires, taking x = 0 to be its outer
surface. The heat entering the inner layer from the outer layer is P/48, so the temperature
gradient at x = 0 isdT P/48dx kcewl
Thus C1 = - for the inner layer. The temperature at x = d is Tb + ATfum + ATf +
AT, where T is the bulk temperature of the cooling water. From this boundary condition
we can solve for C2, which is equal to the temperature at x = 0:
T(x =0)=C 2 =T(x = d)+ q d2 -C 1 d2ke
Then the temperature drop across the inner winding layer is
ATi = q d2 - Cid = 33.50 C2kee
The epoxy impregnated glass cloth tape layer between the two wire layers has thermal
resistance
R2_ t2 - 0.759 CW-
where thickness t2 = 0.34 mm and conductivity kg = 0.31 W/m0 C. The temperature drop
71
across this layer is AT 2 = P x R 2 - 61.6 0 C .
For the outer layer of wires, the assumption of no outward heat flow requires T to be
0 at x = 0, so C1 = 0. The temperature drop across it is
ATO q d2 - 30.8 0 C
2kce
Examining heat transfer over the portion of the turn which is not cooled directly, and
for simplicity treating the wire as if it were straight along the x-axis, we have
d2T PiR+R
dx 2 Aou0
We solve this equation to find the temperature in the middle of the end turn, Tend, in terms
of the temperature of the section that is cooled directly, T,:
1 P R lend 2Tend To+ 2 +lxj
2 Ao-cu, 2
The extra distance from the base of an end turn to the cooling channel, 1x = 1.2 cm, is
given by the combined width of the plenum and the G1O sidewall that houses the sealing 0
ring. The temperature difference ATend Tend - To is 43.3 0 C.
The maximum temperature in the windings is thus E AT = 463 0 C above the bulk
temperature of the water. Table 4.1 summarizes the temperature differences across the
various layers, and also includes the results obtained when the upper and lower bounds
of the conductivity of the copper/epoxy composite are used. When the geometric mean
of the upper and lower bounds is used, about two-thirds of the temperature rise occurs
across the windings and insulation, and about a third occurs across the fiberglass wall.
Owing to the constraints of strength and electrical non-conductivity on the wall material,
we found it hard to improve on the cooling system much further. Since most of the heat is
generated by conduction losses, reducing the armature current brings the temperature down
significantly. For instance, having a stronger magnetic field from a Halbach array would
72
Table 4.1: Summary of temperature differences from stator thermal analysis, using the up-per and lower bounds for conductivity of the epoxy-copper composite, and their geometricmean.
Temperature drop across Symbol kce = U VU/oL kce = o-U kce = o-L
Water film ATfilm 10.8 0 C 10.8 0C 10.8 0CEpoxy/fiberglass wall ATf 148.6 0 C 148.6 0 C 148.6 0 CLayer with epoxy window AT 1 76.00 C 76.0 0 C 76.00 CInner winding layer AT 92.4 0 C 30.3 0 C 281.8 0 CEpoxy/glass layer AT 2 61.6 0 C 61.6 0 C 61.6 0 COuter winding layer ATo 30.8 0 C 10.1 C 93.9 0 CEnd turn ATend 43.3 0 C 43.3 0 C 43.3 0 CTotal EAT 463.4 0 C 380.6 0 C 715.9 0 C
reduce rated ampere-turns from 5749 A to 3573 A. The maximum temperature rise then
becomes 194'C.
73
74
Chapter 5
Fabrication of the Experiment
The machine consists of four major components: the stator, the stator cooling system, the
rotor and the shaft. In addition to manufacturing the stator, it was first necessary to manu-
facture a mold for potting the stator in epoxy. The rotor, shaft, cooling system and potting
mold were professionally machined according to the detailed drawings drawn by Mike
Amaral at SatCon. Some of these drawings are shown in Appendix E. My involvement
in the manufacturing was primarily in the construction of the stator windings, along with
Wayne Ryan of MIT and John Swenbeck of SatCon.
The stator windings were constructed from rectangular compacted litz wire that con-
sisted of 11 groups of 7 wires each, for a total of 77 parallel strands. The insulation on each
strand was polyurethane with a nylon overcoat.
Three long bundles of wire were made, one for each phase. Each bundle consisted of
9 sub-bundles of rectangular litz stacked neatly against each other. The 9 litz bundles were
taped together at one end and held together every few inches by fasteners. The fasteners
were made by taping together the adhesive sides of two pieces of cellotape, such that the
tape could be fastened tightly around the wires without the adhesive touching the wire. This
held the wires firmly in place, while allowing them to slide relative to each other, which
greatly facilitated the winding process.
The armature was wound over a winding fixture: a G1O cylinder with two sets of 24
75
evenly spaced dowels radially attached around its circumference. The dowels acted as slots,
holding wires of different phases in place. Two practice windings were constructed before
the actual winding was made. The winding pattern is shown in Figure 2-2.
Initially the straight sections were insulated with Nomex paper while the end turns were
wrapped with glass cloth tape. However, the Nomex paper was later replaced with glass
cloth tape. A spiral wrap of glass cloth tape around the straight section, aided by a wrap of
polyimide (Kapton) tape at either end, was able to hold the two bundles in each phase belt
together more firmly. This was essential for maintaining the form of the straight sections
when the end turns were being bent to fit into the potting mold. Within each phase belt, the
two groups of wire were insulated from each other by a layer of glass cloth tape. One layer
of tape was deemed adequate, since the wires will carry current from the same phase in the
same direction, although they are different sections of the length of wire. The tape would
be impregnated with epoxy during potting, increasing the effectiveness of the insulation.
The dowels were removed from the G1O cylinder so that the cylinder could be slid in and
out, making it easier to tape the straight sections.
A metal cylinder about the same length as the straight section of the winding, and of a
very slightly smaller outer diameter compared to the mold core, was made. This served as
a fixture over which the end turns could be bent while holding the straight sections firmly
in place. Rounded, smooth edges allowed bending with minimal abrasion of the insulating
glass cloth tape.
The end turns at the bottom were bent inwards. Oppositely facing end turns were
cinched tightly together with string, pulling them more closely inwards. The windings
were put into the potting mold to bend the end turns at the top. These were bent outwards
and pressed down against the flange of the mold by a ring of spacers attached to the top
plate, which was screwed firmly into place. The spacers maintained a gap between the
winding and the top of the mold, to be occupied by the electrical connectors. The resulting
assembly was baked for three and a half hours at 315 F, to achieve thermosetting of the
adhesive in the glass cloth tape wrapped around the wires. This set the winding in the
76
correct shape. Figures 5-1 and 5-2 show the winding after removal from the oven.
Figure 5-1: The armature winding in the potting mold, with the mold core and cover re-moved, after thermosetting of the tape adhesive has set the end turns in shape.
The lead wires were cut to the appropriate lengths so as to fit into the connecters. The
ends were dipped, 2-3 litz bundles at a time, in a high temperature stripping salt solution
for a few seconds, with the sub-bundles separated slightly so as to increase exposure to
the solution. About 1 cm of insulation was removed. The stripped ends were then sloshed
briefly in a cleaning solution of concentrated citric acid, which dissolved away some of the
residual debris from the reaction with the stripping solution.
After all the lead wires were treated in this manner, the entire stator winding was sup-
ported over a tray of citric acid cleaning solution such that the leads were submerged. This
was left overnight for further cleaning to occur. The stator was then washed with water and
placed in a vacuum chamber. With the vaporization of grease and other contaminants, the
pressure was brought down to about 500 torr.
The exposed ends of the individual litz bundles were tinned using a soldering iron.
After this, wire brushes were used to remove flux and other debris, and further cleaning
77
Figure 5-2: Armature winding with end turns set in shape. A wrap of dark Kapton tape atthe ends of the straight sections is visible against the light-colored glass cloth tape.
was carried out by immersing the leads in a tray of alcohol in an ultrasonic cleaner.
The bending process had stretched the glass cloth tape insulation in the end turns, and
the tape had also been worn through in some places where different bundles had been forced
against each other. To ensure the integrity of the insulation, appropriately shaped sheets of
thin polyetherimide plastic (Ultem) were inserted between end turns of different phases.
The stripped and tinned leads were wrapped with aluminium foil and connected to high
voltage test equipment to check for any weaknesses in insulation between phases. Testing
was done with the winding bound tightly with tie wraps over the mold core, so that the
wires would be as close together as in the eventual machine. The machine passed at a test
voltage of 2100 V.
The lead wires were inserted into electrical connecters, which were filled with solder
78
for good electrical contact. Openings in the glass cloth tape were cut, one for each bundle,
on the inside surface of the straight section, as can be seen in Figure 5-3. This was done
to facilitate penetration of the epoxy among the wires during potting, and reduce thermal
resistance to radial heat flow. Cleaning and high potential testing were carried out once
again.
Figure 5-3: Armature winding with electrical connectors. Windows in the glass cloth tapecan be seen on the inside surfaces of the straight sections of the winding.
Thermistors were placed at various positions along one phase belt, and in neighboring
end turns, to examine temperature distribution during testing. The armature with thermistor
leads is shown in Figure 5-4, and the thermistor locations are diagrammed in Figure 5-5.
Thermistors were also placed at two other points around the circumference, to check for
uniform cooling all around. Although the thermistor leads should eventually emerge from
the connector end of the stator, this end is at the bottom of the potting mold, and having
holes in the bottom plate of the mold might result in epoxy leakage. Thus for potting the
79
thermistor leads were taken out of the top of the mold instead, with the intention of later
running them through the hollow inside of the stator and out the connector end.
Figure 5-4: Armature with thermistor leads
Provisions were made to ensure that the armature could be easily removed from the
mold after potting. Teflon tape was used to cover the inner and outer curved surfaces of
the mold, which would be the largest areas in contact with the epoxy. Mold release was
smeared on the top and bottom plates, including the inner surfaces of various holes for
screws, thermocouples and epoxy.
The epoxy and catalyst were mixed thoroughly using an electric drill, and then placed
in a vacuum chamber for de-gassing. This was done so that air bubbles in the potting would
be minimized.
A strip of fiberglass cloth was painted with epoxy, and wet wound tightly onto the mold
core. This layer will serve as the outer wall of the cooling channel, separating the water
80
from the windings, as shown in Figure 4-9. The winding was lowered carefully over the
core, and bound tightly on the outside with several turns of string. The rest of the mold was
assembled around it. Two concentric cylinders formed an annular channel above the epoxy
holes on the top plate. This served to contain the epoxy while it dripped slowly through the
holes into the winding. The complete mold assembly is shown in Figure 5-6. Silicone was
used as a sealant between parts of the mold, to prevent leakage of epoxy during potting.
The mold assembly was then placed in an oven at 150'F for two hours, to cure the silicone,
and to bring the mold and stator up to an appropriate temperature for potting.
Potting was carried out with the mold assembly in a vacuum chamber. The vacuum
drew epoxy from a container outside the chamber to the annular channel above the mold,
via a length of bent copper pipe. Partway through the process, the mold was taken out of
the chamber, heated slightly to increase fluidity of the epoxy, and tilted all around to allow
escape of air bubbles trapped under the main flange.
Curing of the epoxy took place in an oven at 250 F. Jacking screws and a deadblow
hammer were sufficient to remove the top and bottom plates and the outer housing, while
removal of the core required the application of substantial sustained force from a press.The
potted stator, shown in Figure 5-7, turned out well. There were hardly any air bubbles
except along the edge which had been under the main flange of the mold, which would not
pose any problems. A thin layer of polyurethane was coated on the inside surface, to fill
slight imperfections on this surface and waterproof it. The polyurethane was drawn down
in a vacuum, and excess wiped off the surface, except around the region to be in contact
with the 0 ring. Here a slightly thicker layer of polyurethane was allowed to dry, before
being sanded down to the desired diameter. The 0 rings were covered with lubricating
grease and assembled with the cooling jacket and potted winding.
The potted stator was assembled with the rotor, shaft and bearings. according to the
assembly drawing shown in Figure E-6. The parts of the machine prior to assembly are
shown in Figure 5-8. The machine was first put together with an aluminium ring in place
of the rotor magnets for initial spin-down tests as described in Chapter 6. Figure 5-9 shows
81
the rotor with the aluminium ring installed. The machine was then taken apart, and the
aluminium ring replaced with magnets and spacers mounted on a G1O ring, as shown in
Figure E-4. After each assembly, the machine was sent out for commercial balancing.
Some material was removed from the aluminium ring in the first case and from the rotor in
the second, to ensure that the weight of the rotor was even all around.
82
Figure 5-5: Diagram of armature winding showing locations of thermistors along one phasebelt.
83
Figure 5-6: Mold assembly for potting stator winding in epoxy
84
Figure 5-7: Armature potted in epoxy
85
S.
A
Figure 5-8: Parts of the machine pictured prior to assembly
86
0
Figure 5-9: Rotor with aluminium ring in place of magnets for initial spindown tests
87
88
Chapter 6
Testing
6.1 Resistance and Inductance
An automatic R-L bridge was used to measure the resistance and self inductance of each
phase, at various frequencies in the range of 200 Hz to 20 kHz, as well as at 20 Hz. The
results are shown in Table 6.1.
The mutual inductance between two phases was found by putting alternating current
through one phase winding, and observing the voltage induced across the open terminals
of the other phases. The voltages across the driven and open phases were measured using
Table 6.1: Resistance and Self Inductance of Individual Phases
Frequency/Hz Ra/mQ Rb/mQ Re/mQ La/lpH Lb /pH Le/pH20 2.07 2.07 2.07 2.10 2.10 2.10
200 2.10 2.11 2.11 2.16 2.14 2.10500 2.30 2.30 2.30 2.39 2.37 2.33
1000 2.65 2.62 2.63 2.36 2.33 2.352000 3.03 2.97 2.99 2.08 2.09 2.065000 4.18 4.17 4.16 2.28 2.28 2.27
10000 4.60 4.63 4.58 2.22 2.21 2.2220000 4.78 4.79 4.79 2.21 2.20 2.21
89
an oscilloscope, this being more convenient than measuring the alternating current directly.
If alternating current is applied to phase A and phase B is open-circuited, the terminal
relations are
diaVa = aRa+ La dt
dtdiaVb = Lo'a d
Writing v2 Re {Le"wt} for x = a, b, and sa = Re {Laej't}, we have
Va =LaRa+jwLaLa
b = jWLbala
Eliminating La from the two equations gives
Vb = jw Lba -- aRa + jWLa
Knowing that Lba is real and negative, we can find Lba from
L|a l_ i(Ra + jwLa)
jo |_Va l
where |b I and |_V I are the measured amplitudes of the voltages across phase A and phase
B respectively.
At low frequencies the induced voltages were too small to measure with an oscillo-
scope, so readings were taken starting from a frequency of 5 kHz. Table 6.2 gives the mea-
sured voltages and the corresponding mutual inductances, calculated as described above.
The dc resistance of a phase was predicted to be R = let~/(uA) = 1.2 mQ, where
the estimated length of the conductor 1tot = 1.63 m, the cross-sectional area A = Npa, X
7r (d,/2)2 - 3.51 x 10-5 M 2 , and the electrical conductivity of copper o = 3.9 x 107
90
Table 6.2: Mutual Inductance Measurements
Frequency / kHz V(driven phase) / V V(open-circuited phase) / V Inductance / pHVa V Lba
5 0.026 0.012 1.05410 0.035 0.016 1.01520 0.065 0.026 0.884
Vb Va Lab5 0.026 0.012 1.054
10 0.035 0.016 1.01120 0.070 0.026 0.817
Va V Lca5 0.026 0.012 1.054
10 0.035 0.014 0.88820 0.065 0.022 0.748
V Va Lac5 0.026 0.012 1.049
10 0.038 0.016 0.93520 0.065 0.024 0.816
Vb V Leb5 0.026 0.012 1.054
10 0.035 0.016 1.01120 0.070 0.028 0.948
V V Lbc5 0.026 0.012 1.049
10 0.038 0.016 0.93520 0.065 0.026 0.884
91
S/m. The measured resistance is somewhat higher, mainly because the end turns had to be
made longer than projected, owing to the springiness of the compacted litz bundles which
required larger bends.
The measured inductances also turned out higher than the predicted values, primar-
ily because the inductance calculations, as discussed in Chapter 2, do not include the
contribution of the end turns. The predicted values of self and mutual inductance were
82 x 1.56 = 0.998 pH and 82 x 0.78 = 0.499 puH respectively, about half of the corre-
sponding measured quantities.
6.2 Spin-down Tests
Spin-down tests were used to estimate various loss mechanisms and to examine the vari-
ation of back emf with rotor speed. These tests involved bringing the motor up to speed
with a hand-held router, and then removing the router and observing the spin-down rate.
Figures 6-1 and 6-2 are pictures of the experimental setup.
6.2.1 Loss estimation
Losses from bearing friction, fluid friction (windage) and eddy currents contribute towards
the gradual slowing of the rotor. Bearing friction typically has a torque component indepen-
dent of rotor speed w and a torque component proportional to w. Since eddy current loss, as
described in Chapter 4, increases with speed as w2 , the associated retarding torque is also
proportional to w. The speed dependence of windage torque is more complicated, since it
changes slightly as different flow regimes are encountered during spin-down. From [5], the
windage torque for the cylindrical surfaces is proportional to W 9 for turbulent air flow and
w"5 for vortex flow. For the disk surfaces, the windage torque is proportional to o 1.8 for
turbulent flow and w1 5 for laminar flow. Therefore
do-J = Co + CiW + T
dt
92
Figure 6-1: Test stand for spin-down tests. The top end of the machine with the couplingfor the router can be seen.
where J is the inertia of the rotor about the spin axis, Co is the coulomb bearing friction
torque, Cjw is the combined torque from viscous bearing friction and eddy currents, and
Ta, the windage torque, is some combination of terms of the form Cwf, with yi around
1.5 to 1.9.
In order to estimate the effects of windage, eddy currents and bearing friction separately,
two sets of spin-down tests were performed, one with magnets on the rotor and one with
an aluminium ring in place of the magnets. The aluminium ring was designed with the
same radial width as the magnets plus the G10 ring on which they are mounted. Since the
rotational gap width is the same in both cases, windage should be about equal for both.
However, eddy currents in the stator winding are present only when there are magnets.
The inertia J for the rotor with the aluminium ring is 0.126 kgm 2, while the complete
assembly with GlO ring, magnets and spacers has an inertia of 0.127 kgm 2 . These val-
ues are from the Pro-Engineer design software used to create the detailed manufacturing
93
Figure 6-2: Back view of machine mounted on test stand.
drawings.
As the rotor spun down, speed measurements were taken every five seconds using an
automatic data logger, over a range of about 12000 rpm to 1000 rpm. Four sets of data were
taken for each rotor configuration. Plots of the results are shown in Appendix F. From the
data, vectors of -J- versus w were made, and fitted using matlab to the equation
dw-J = Co + CiW + Cw'? (6.1)
dt
for y = 1.8 and -y = 1.9. The results are presented in Table 6.3.
The change in Co between the experiments with the aluminium ring and with the mag-
nets is probably due to differences in bearing preload and alignment - important factors in
bearing performance, as the machine had to be taken apart and reassembled between the
two sets of tests.
For the setup with the aluminium ring, the value of coefficient C1 was found to vary
94
Table 6.3: Average values of coefficients from spin-down tests
Coefficient Rotor with Rotor withaluminium ring magnets
1.8 CO / Nm 0.0149 0.0182C1 /104 Nms 0.0591 1.68C1.8 /10- 7 Nms1.8 9.04 5.66
1.9 CO / Nm 0.0131 0.0172C1 /104 Nms 0.303 1.83C1.9/10-7 Nms 1.9 4.08 2.56
a lot over the four data sets. From the plots and results in Appendix F, it can be seen
that the measurement noise is much larger than the average value of the coefficient being
estimated. As a result, no definite conclusions about the magnitude of the eddy current loss
can be drawn from the data. It should be noted at least that C1 increases in the presence
of the magnets but that this increase corresponds to an additional power loss of 400 W at
15,000 rpm, which greatly exceeds the 12 W prediction. Measurement error, fitting error
and inconsistency in bearing performance, as reflected in CO, could have contributed to the
discrepency.
The values for C were more consistent over the four experimental runs. The windage
calculations of Chapter 4 predict a windage loss of 791.4 W at 15,000 rpm, or Q = 1570.8
rad/s. This corresponds to the predictions C1.8 = 791.4/1570.82.8 = 8.90 x 10~7 Nms 1.8
and C1.9 = 791.4/1570.82.9 = 4.26 x 10-7 Nms 1'9 . These values are close to the experi-
mental values for the rotor with the aluminium ring. The corresponding values for the rotor
with magnets are a bit further from the predictions. This is not surprising, given that the
aluminium ring has a smooth cylindrical surface similar to that assumed in the windage cal-
culations, while the magnet surface is azimuthally segmented, with narrow gaps between
adjacent magnets and spacers.
95
6.2.2 Back emf
Back emf was also measured during spin-down tests of the rotor with magnets installed.
The voltage across the open terminals of each phase was measured simultaneously with the
rotational speed, with readings taken every five seconds. As expected from Equation 2.3,
the back emf E, was found to have a linear relationship with speed w. The data was fitted to
the equation E, = Cw, and the results and plots from 2 experimental runs are presented in
Appendix F. The average value of the ratio C = Ea/W was found to be 0.0 150 Vs for phase
A, 0.0147 Vs for phase B and 0.0145 Vs for phase C. This is a bit higher than the predicted
value of 8 x 1.71/1571 = 0.0087 Vs from Chapter 2. The results, however, fit better with
the magnetic field measurements given in the following section, where the fundamental
component of field at the outer radius of the armature is found to be Bia = 0.1309 T. The
corresponding value of the ratio C = EaIw is 0.0127 Vs.
6.3 Magnetic Field Measurements
With the magnets installed on the rotor, but before assembling the machine with the stator,
the magnetic field produced by the permanent magnets was measured using a Hall probe.
The probe coil was placed 1.32 mm from the magnet inner surface, a location correspond-
ing to the outer radius Rao of the armature. It was mounted on a height gauge so that it
could be moved vertically while preserving its horizontal position. Measurements were
taken at eight heights. At each height, the rotor was spun slowly, and the output of the
Hall probe for one revolution captured on a digital oscilloscope. Two such readings were
taken at each height. The Hall probe was calibrated by taking readings at the surface of a
permanent magnet, and then measuring magnetic flux at the same locations with a guass-
meter. A plot from one of the experimental runs is shown in Figure 6-3. The signals were
fourier analyzed in matlab, and the average values of the fundamental and the 3rd, 5th and
7th harmonics are given in Table 6.4. The measured fundamental magnetic field strength
was somewhat higher than the predicted value of 0.0958 T.
96
Table 6.4: Magnetic flux density harmonics at Rao
Height / mm Magnetic flux density / TFundamental 3rd harmonic 5th harmonic 7th harmonic
1.00 0.1413 0.0179 0.0282 0.02500.1413 0.0184 0.0280 0.0251
15.5 0.1293 0.0243 0.0279 0.02760.1296 0.0244 0.0279 0.0278
30.0 0.1232 0.0246 0.0280 0.02780.1232 0.0246 0.0279 0.0279
44.5 0.1212 0.0248 0.0278 0.02780.1211 0.0247 0.0280 0.0278
59.0 0.1221 0.0249 0.0279 0.02800.1221 0.0248 0.0279 0.0281
73.5 0.1251 0.0250 0.0281 0.02860.1248 0.0243 0.0285 0.0281
88.0 0.1394 0.0240 0.0287 0.02900.1391 0.0233 0.0291 0.0288
96.0 0.1460 0.0183 0.0293 0.02640.1460 0.0182 0.0293 0.0267
mean 0.1309 0.0229 0.0283 0.0275
97
run 1
-~0. 05-
Time.- s
Figure 6-3: Magnetic flux pattern at Rao over one revolution of the rotor
98
Chapter 7
Summary and Conclusions
A high-speed permanent magnet synchronous motor-generator for flywheel energy storage
was designed, built and experimentally evaluated. It was based largely on an existing elec-
tromagnetic design developed by Professor Kirtley. This design was discussed in Chapter 2,
along with modifications to it which are present in the actual machine. These modifications
included increasing the armature thickness and the rotational gap width to make manufac-
turing easier. In addition, the magnet arrangement was changed from a Halbach array to
one involving only radially magnetized magnets. The stator cooling system, presented in
Chapter 4, was part of this thesis, while much of the mechanical design was performed by
engineers at SatCon.
Since low-loss and high-efficiency are major design goals in flywheel energy storage
systems, this project aimed to investigate various loss mechanisms. Theoretical loss mod-
els were developed in Chapters 3 and 4, with particular interest in the modelling of eddy
current losses in segmented rotor magnets and in the stator windings. At 15,000 rpm, the
predicted conduction loss is 3895 W, eddy current loss is 12 W and windage loss is 224 W,
with a corresponding efficiency of 87%. The conduction loss could be decreased by using
a Halbach array to provide a stronger magnetic field, which would reduce the required cur-
rent and improve efficiency to 94.5%. A machine for actual use would have a vacuum to
eliminate windage loss and magnetic bearings to reduce bearing friction loss. Thus, at idle,
99
its main loss mechanism would be eddy currents in the windings, which are projected to be
around 12 W. The eddy current loss could be further reduced by using thinner wire.
The fabrication process of the machine was documented in Chapter 5. The machine
was originally intended to have 72 turns, but was incorrectly constructed with 9 of the
conductors in parallel, resulting in an 8-turn machine. As a result, the machine operates
at much lower voltage and higher current than anticipated. This does not affect most of
the intended experimental work in measuring losses and other machine quantities. The
machine cannot however be run as a practical motor/generator at appreciable power because
of its high current and low voltage requirements.
Testing was described in Chapter 6. The fundamental machine parameters such as re-
sistances, inductances and magnet flux were measured, as were several quantities predicted
by the loss models. The measured resistances and inductances of the armature winding
were close to what was expected, but the magnets turned out to be stronger than antici-
pated, resulting in a higher back emf. Spin-down tests were carried out to determine losses
from windage and eddy currents in the magnets. The measured windage loss was close
to its predicted value, but the much smaller eddy current loss could not be distinguished
owing to measurement noise.
More experimental work remains to be done. Tests yet to be carried out on this machine
include a locked rotor test to measure eddy current losses in the magnets, and a generator
test in which the machine, driven by another motor, generates power into a resistor bank.
Future experimental machines, besides having more turns, might also evaluate the perfor-
mance improvements from having a vacuum, magnetic bearings, thinner wire strands, and
a Halbach magnet array which produces a stronger magnetic field at the armature, and thus
reduces conduction loss. The inclusion of these features would hopefully provide con-
crete evidence that this is a practical design for a highly efficient, low-loss flywheel energy
storage machine.
100
Appendix A
Inductance Calculation
A calculation of armature inductance was referenced in Section 2.2, and is presented in
detail here. This calculation parallels the approach used in [4], which finds the inductances
of an air gap armature winding that has uniform current density in each phase belt. There
the inductances are found to be reasonably well approximated by the first space harmonic
term, the next being two orders of magnitude lower.
In this machine, the number of turns does not increase with radius. So the current
density J varies inversely with radius, that is, J = Jo/r. Integrating J over one phase belt,
we obtain J, in terms of the terminal current Ia, the number of turns N and the number of
poles p:Rao f 2JIRa -rdOdr = JOw (Rao - Rai) = NIa/pat z 2Ow r
NIa NIa
p0w (Rao - Rai) Owe (Rao - Rai)
where Ow = Owe/P is the angle subtended by one phase belt.
As in [4], we find the magnetic field produced by a shell of surface current and integrate
over shells between Rai and Rao to obtain the total field. At radius R, the surface current is
101
KR = odR, which can be expanded as a fourier series KR = EKR. cos npO, where
4JdR sin ("l-)
0
for n odd
: for n even
: for n odd
: for n even
[4] gives the fundamental component of radial magnetic field due to a surface current
dH KR 1 .d HiS, = 2
KR,d~o~r - 2
(o)P sin p0
( p sin pO
In the region of the windings, Rai < r < Rao, the fundamental component of total magnetic
field is then
H rHr = )Rai
Raod HoS, + I
1 Er= -- sin(pO)II
2 s LORai
2- -- sin(pO)Jo sin
7r
4JodR .irR
Owe
2 )
(Owe) (-R)+1S2 r 7
12i- 2
f Rao
r
4JodR,R
I-2p - (1
. Owe r P -2 \RJ
+ (1 + p) (r)PRao
If the winding has a significantly large turns density, the number of turns in a differential
area element can be expressed as
Nd2N = rdrd@
r
where No is given in terms of the total number of turns N by
No- N/p N(Rao - Rai) Ow (Rao - Rai) Owe
102
KR,,sin f"lw :
nir 2
0
shell at R:: for r < R
: for ft r
dHiSr
IRai P*+P)(r )
The flux linked by this element and its full-pitched complementary element is
d2 A = d2 Nl poHrdO
The total flux linked by p pole pairs is then
Rao N O
fRai (Rao - Rai) 0 e P-
pioHrrd] drdt
Total flux linked by one phase due to excitation of that same phase is
8l p Josi2 Rai 2A = 8Ngo sin 2p) R 20 [1-p+ (Ri)2 (1+p) - 2
r(Rao - Rai) Owe (1 - p 2 )p aRao
12N2 PoI )( 1 - p + x2(1 + p) - 2x)+
7r Owe (1 - x)2(i - p2)p
Rai )~]Rao)P1
The self inductance of each phase winding is thus
La -21 N2 pof 2 sin 1 - p+ x2(1 + p) _ - 2x+
7r Ow) (1 - x)2(i - p2)p
For a three-phase machine, the mutual inductance between phases is
.~22lN 2 O w in 1 - px 2 (1 +p)-2x+ 1 2rLw? = gr (1 - x)(- 2) )cos2IN 2 po /sin- (1- p +x 2 (1±+p) -2xv+
- r Ow) (1 - x)2(1 _ p2 )p
The actual machine, however, ended up being built with many parallel strands and few
turns. Each phase belt has only two turns, and the winding pattern is such that the wires
alternately occupy the inner and outer layers. As such, flux linked by each turn can be
calculated using an average value of Hr.
< Rao
< H, > = aRa
Hdr
103
A = p1OW-2
2 .(Owe\ 1= - sin(pO) Jo sin -- 1IT 2 i-p 2 I p1 - X)
Total flux linked by one phase due to excitation of that phase becomes
A - II f 2
we 2 J- P< H, > Rao+Rai dd
4N 1pouJo (Rao + Rai)
irwe (1 -p2 ) psin 2 (we
(2 )
12
1-- p2(1 - p)x(xP - 1) + (1 + p)(l - xP)
p(i - )
and the self inductance of each phase is
La _ lpoN 2 (Rao + Rai)7L (Rao - Rai)
)2 I (1I(1 - p2), I
The mutual inductance between phases for a three-phase machine is
Lab - lyoN 2 (Rao + Rai)27c (Rao - Rai)
wsin 2
2
1 (1-p)x(xP -1) + (1 +p)(l - xP)
(1 - P2)p p2(1 -X)
104
(1 -p)x(xP - 1) + 1 - xP1I
- p)x(xP - 1) + (1 + p)( - xP)
p2 (i - x)- 2
-2
Appendix B
Matlab code for Rotor Loss Calculation
This code implements the equations in Chapter 3 to calculate rotor magnet eddy current
loss.
clear all
total=O;
p=4; q=3;
n=[2*q-1 2*q+l 4*q-1 4*q+l 6*q-1 6*q+l 8*q-1 8*q+l];
nn=[2*q 2*q 4*q 4*q 6*q 6*q 8*q 8*q];
ome = 2*pi*15000/60;
omegan = ome*nn;
muO = 4e-7*pi;
sigma = 7e4;
I = 5835 /8 *sqrt(2);
Rai = 0.0673; Rao = Rai+0.012;
Rmi = Rao+1.32e-3; Rmo = Rmi + 9.53e-3;
1= 0.1001; T = Rmo-Rmi; Rm = (Rmi+Rmo)/2;
thetawe = 0.856;
h=2*(1:8);
105
thetamv [ pi/6];
%thetamv = linspace(pi/8,pi/2,25);
for ith= 1:length(thetamv)
total(ith) = 0; total2(ith)=O;
thetam = thetamv(ith);
d=thetam*(Rmi+Rmo)/2;
%nummags = floor(2*pi/thetam)
%nummags = 8;
nummags = 1;
for im = 1:38
m=2*im-1;
for in = 1:4
ni = n(in);
for ih = 1:length(h)
hi = h(ih);
hpd = hi*pi/d; mpl=m*pi/l;
gam = sqrt(hpd.^2+ mpl^2 + j*omegan(in)*muO*sigma);
HPTH = hi*pi/thetam;
egt = exp(gam*T); emgt = exp(-gam*T);
A = [ 1 1 0 0 0 0 1/Rmi*HPTH 0;
0 0 1 1 0 0 -mpl*besip(HPTH, m*pi*Rmi/l)/besseli(HPTH,
m*pi*Rmi/1) 0;
0 0 0 0 1 1 mpl 0;
emgt egt 0 0 0 0 0 1/Rmo*HPTH;
o 0 emgt egt 0 0 0 -mpl*beskp(HPTH, m*pi*Rmo/l)/besselk(HPTH,
m*pi*Rmo/1) ;
0 0 0 0 emgt egt 0 mpl;
-hpd 0 gam 0 -mpl 0 0 0;
0 -hpd 0 -gam 0 -mpl 0 0];
u=hi/2;
npthm = ni*p*thetam;
106
if npthm/(2*pi) == floor(npthm/(2*pi))
if u == npthm/(2*pi)
alu = 1;
a2u = 0;
a3u = 0;
a4u = 1;
else
alu = 0;
a2u = 0;
a3u = 0;
a4u = 0;
end
else
tup = 2*u*pi;
alu = sin(npthm)*(l/(npthm+tup) + 1/(npthm-tup));
a2u = (1-cos(npthm))*(1/(npthm+tup) - 1/(npthm-tup));
a3u = (1-cos(npthm))*(l/(npthm+tup) + 1/(npthm-tup));
a4u = sin(npthm)*(l/(npthm-tup) - 1/(npthm+tup));
end
Kn = 4*p*q*I*sin(ni*thetawe/2) / (ni*pi*thetawe*(Rao-Rai)*(ni*p+l)) *
(1-(Rai/Rao)^(ni*p+l));
ki = 2*Kn/(m*pi) *(Rao/Rmi)^(ni*p+l);
ko = 2*Kn/(m*pi) *(Rao/Rmo)^(ni*p+l);
y = [ki*(alu-j*a3u); ki*(j*a2u+a4u); 0; ko*(alu-j*a3u);
ko*(j*a2u+a4u); 0; 0; 0];
x = inv(A)*y;
axp
axm
ayp
aym
azp
= x(1,1);
= x(2,1);
= x(3,1);
= x(4,1);
= x(5,1);
107
azm = x(6,1);
Cl azp*gam - ayp*mpl;
C2 = -azm*gam - aym*mpl;
C3 = ayp*hpd - axp*gam;
C4 aym*hpd + axm*gam;
cgam= conj(gam);
power = nummags*d*l/(8*sigma) * ( ((abs(C1))^2 + (abs(C3))^2) *
(1-exp(-(gam+cgam)*T)) / (gam + cgam) + (C1*conj(C2) + C3*conj(C4))
* (1-exp((-gam+cgam)*T)) / (gam - cgam) + (C2*conj(C1) +
C4*conj(C3)) * (1-exp((gam-cgam)*T)) / (-gam + cgam) + ((abs(C2))^2
+ (abs(C4))^2) * (1-exp((gam+cgam)*T)) / (-gam - cgam) );
total(ith) = total(ith) + power;
%%%%%%%%%%%%% Cartesian approximation %%%%%%%%%%%%%%%
beta = sqrt(hpd^2+mpl^2);
embt = exp(-beta*T);
A2 [ 1 1 0 0 0 0 hpd 0;
0 0 1 1 0 0 -beta 0;
o 0 0 0 1 1 mpl 0;
emgt egt 0 0 0 0 0 hpd;
0 0 emgt egt 0 0 0 beta;
0 0 0 0 emgt egt 0 mpl ;
-hpd 0 gam 0 -mpl 0 0 0;
0 -hpd 0 -gam 0 -mpl 0 0];
x = inv(A2)*y;
axp = x(1,1);
axm = x(2,1);
108
ayp = x(3,1);
aym = x(4,1);
azp = x(5,1);
azm = x(6,1);
C1 = azp*gam - ayp*mpl;
C2 = -azm*gam - aym*mpl;
C3 = ayp*hpd - axp*gam;
C4 = aym*hpd + axm*gam;
power = nummags*d*l/(8*sigma) * ( ((abs(Cl))^2 + (abs(C3))^2) *
(1-exp(-(gam+cgam)*T)) / (gam + cgam) + (Cl*conj(C2) + C3*conj(C4))
* (1-exp((-gam+cgam)*T)) / (gam - cgam) + (C2*conj(Cl) +
C4*conj(C3)) * (1-exp((gam-cgam)*T)) / (-gam + cgam) + ((abs(C2))^2
+ (abs(C4))^2) * (1-exp((gam+cgam)*T)) / (-gam - cgam) );
total2(ith) = total2(ith) + power;
end
end
end
end
figure(l)
plot(thetamv,total)
title('Eddy current loss in a magnet of angle theta-w');
xlabel('theta-w / rad')
ylabel('Loss / W')
figure(2)
plot(thetamv,total2)
title('Cartesian approximation of eddy current loss in a magnet of
angle theta-w');
xlabel('thetaw / rad')
ylabel('Loss / W')
109
110
Appendix C
Thermal Analysis Spreadsheet and
Matlab Calculations
The spreadsheet and matlab code used to calculate the values quoted in Chapter 4 are
presented here.
C.1 Thermal Analysis Spreadsheet
Quantity
turns
parallel strands
poles
phases
armature outer radius
armature inner radius
wire radius
Ampere turns
current in one strand
elect. conductivity Cu
Power loss / unit length
Symbol geom mean
N 8
Npar 693
p 4
q 3
Rao 0.0793
Rai 0.0673
x=Rai/Rao 0.84867591
r_w 0.000127
N Ia 5749
Ia/Npar 1.0369769
sigma 39000000
P_l_R_15 0.54414702
111
P_l_R_30 0.13603675
Fundamental field at Rao
mean square magnetic field
Speed
Electrical freq 15000 rpm
Eddy cur loss /unit length
active length
end turn length
safety length
total i^2R loss (15)
total eddy cur loss(15)
windage loss (15)
total loss
therm cond. fib+epoxy
thermal cond. epoxy
thermal cond. Cu
winding angle
vol. fraction epoxy
vol. fraction Cu
upper bound conductivity
lower bound conductivity
geom mean cond. Cu+epoxy
length of cooling channel
dist betw channel&end turn
Bla
<Bo^2>
rpms
omega
P_1_ec_15
P_1_ec_30
la
lend
ls
PR
P_ec
P_w (15)
P(15)
sigl+sig2
(s2-sl) ^2
<sig>
<sig bar>
E2
El
k_f,k_g
s igmal
sigma2
thetawe
phil
phi2
sigmaU
sigmaL
k_ce
lc
lx
0.0958
0.023667014
15000
6283.1853
0.0037225675
0.01489027
0.1001
0.088123859
0.0142
3663.9744
12.395131
218 .4
3894.7695
390.66
151585.64
145.92152
244.73848
0.11642418
0.88357582
0.31
0.66
390
0.214
0.62690318
0.37309682
14.363205
1.5433015
4.7081584
0.1003
0.012
112
radial width of channel
channel outer radius
channel inner radius
hydraulic diameter
mass flow rate
velocity of flow
Reynolds no.
Moody friction factor
pressure drop
Nusselt no.
film coeff.
T
Rco
Rci
DH
m
v
Re
f
deltaP
Nu
h-c
0.00014
0.06689
0.06675
0.00028
0.18590785
3.165417
676.03642
0.094669456
169760.67
5
10446.429
conductivity of epoxy/Cu
Width of a phase belt
Winding layer thickness
Resistance of water film
Temp drop across film
Thickness of fiberglass
Resistance of fiberglass
Temp drop across fiberglass
Thickness glass cloth tape
Resistance of layer 1
Temp drop across layer 1
Power density
Temp drop inner layer
Resistance of layer 2
Temp drop across layer
Temp drop outer layer
temp drop end turn
Total temp drop
2
k_ec
w
d
Rf ilm
dTfilm
tf
Rf
dTf
tg
R1
dTl
q dot
Cl (i)
dTi
R2
dT2
dTo
dTend
sum dT
geom mean
4.7081584
0.0144
0.00516
0.066277899
10.755714
0.00041
0.91571165
148.60357
0.00034
0.46846746
76.023865
10887481
-11932.351
92.356399
0.75937063
61.616116
30.785466
43.271347
463.41248
upper bound
14.363205
0.0144
0.00516
0.066277899
10.755714
0.00041
0.91571165
148.60357
0.00034
0.46846746
76.023865
10887481
-3911.3416
30.273784
0.75937063
61.616116
10.091261
43.271347
380.63566
lower bound
1.5433015
0.0144
0.00516
0.066277899
10.755714
0.00041
0.91571165
148.60357
0.00034
0.46846746
76.023865
10887481
-36402.09
281.75218
0.75937063
61.616116
93.917392
43.271347
715.94018
113
C.2 Matlab code for windage calculation
clear all
Omega = 15000*2*pi/60;
rho = 1.16; go = 1.32e-3; Rao = 0.0793; mu = 1.85e-5;
%%%%%%% Area a %%%%%%%%
1_a = 0.1001;
Ta = rho*Omega*Rao*go/mu * (go/Rao)^0.5
cfv = 0.476*Ta^0.5/(rho*Omega*Rao*go/mu)
cft = 0.00655*(rho*Omega*Rao*go/mu)^(-0.136)
if cfv > cft
cf = cfv;
else cf = cft;
end
tau = cf*pi*Rao^4*1_a*rho*mega^2;
fprintf('cf = %9.6g\n', cf)
fprintf('tau = %9.6g\n', tau)
%%%%%%% Area b %%%%%%%%
1_b = 0.044; R-b = Rao; g-b = 0.017;
Ta = rho*Omega*R-b*g b/mu * (g-b/R-b)0.5
cfv = 0.476*Ta^0.5/(rho*Omega*R b*g-b/mu)
cfts = eval(solve('exp( (1+g-b/R-b)/(1.2*sqrt(2*c)*(1+0.5*gb/Rb)) -
log( sqrt(c/2)/(2*(l-g_b/R-b))) - 8.58 ) = rho*Omega*R-b*gb/mu',
cft = cfts(l)
if cfv > cft
114
cf = cfv;
else cf = cft;
end
taub = cf*pi*Rb^4*1_b*rho*Omega^2;
fprintf('cf = %9.6g\n', cf)
fprintf('taub = %9.6g\n', tau-b)
%%%%%%% Area 3 %%%%%%%%
g_c = 0.012; asr = gc/Rao
Re = rho*Omega*Rao^2/mu
cm1 = 0.051*asr^0.1/Re^0.2;
tau_cl = cml*rho*Omega^2*Rao^5/2;
fprintf('cml = %9.6g\n', cml)
fprintf('tau_ci = %9.6g\n', tau-cl)
R_c2 = 0.020; asr = gc/Rc2
Re rho*Omega*R-c2^2/mu
cml = 0.051*asr^0.1/Re^0.2;
tauc2 = cm1*rho*Omega^2*R-c2^5/2;
fprintf('cm1 = %9.6g\n', cml)
fprintf('tauc2 = %9.6g\n', tauc2)
tauc = tau_cl -tauc2;
fprintf('tauc = %9.6g\n', tau-c)
%%%%%%% Area d %%%%%%%%
R_d = 0.109; gd = 0.0063; asr = g_d/Rd,
Re = rho*Omega*Rd^2/mu % -- > regime IV
cml = 0.051*asr^0.1/Re^0.2;
tau_dl = cm1*rho*0mega^2*R-d^5/2;
fprintf('cm1 = %9.6g\n', cml)
115
fprintf('tau-dl = %9.6g\n', tau_dl)
Rmi = 0.0806; asr = g_d/Rmi,
Re = rho*Omega*Rmi^2/mu % -- > regime IV
cml = 0.051*asr^0.1/Re^0.2;
taud2 = cml*rho*Omega^2*Rmi^5/2;
fprintf('cml = %9.6g\n', cml)
fprintf('tau-d2 = %9.6g\n', taud2)
taud = tau_dl-tau-d2;
fprintf('tau-d = %9.6g\n', tau-d)
%%%%%%% Area e %%%%%%%%
1_e = 0.166; R-e = 0.109; g-e = 0.0127;
Ta = rho*Omega*R-e*ge/mu * (ge/Re)^0.5
cfv = 0.476*Ta^0.5/(rho*Omega*R-e*g-e/mu)
cfts = eval(solve('exp( (1+g-e/R-e)/(1.2*sqrt(2*c)*(1+0.5*g_e/Re)) -
log( sqrt(c/2)/(2*(1-g_e/R-e))) - 8.58 ) rho*Omega*R-e*g_e/mu',
'c'))
cft = cfts(l)
if cfv > cft
cf = cfv;
else cf = cft;
end
taue = cf*pi*R_e^4*1_e*rho*Omega^2;
fprintf('cf = %9.6g\n', cf)
fprintf('tau-e = %9.6g\n', tau-e)
%%%%%%% Area f %%%%%%%%
R_f = 0.109; g-f = 0.0063; asr = gf/R f,
116
Re rho*Omega*R-f^2/mu % -- > regime IV
cml = 0.051*asr^0.1/Re^0.2;
tauf = cml*rho*Omega^2*Rjf^5/2;
fprintf('cml = %9.6g\n', cml)
fprintf('tauf = %9.6g\n', tau-f)
trqSR = tau +taub + tauc + tau_d
Omega = 15000*2*pi/60;
powerSR = trqSR*Omega
powerHR = (tau-e+tau-f)*Omega
C.3 Matlab code for plotting graphs of loss vs speed
clear all
N=8; Npar = 693; rw = 0.127e-3; q=3; p=4;
sigma = 3.9e7;
B1 = 0.0709; % effective radial component of B
msqB = 0.023667; %mean sq amplitude of B,includes radial and
%azimuthal components
actual =1;
if actual == 0
Rao = 0.07938; Rai = 0.06985; Rmi = 0.07988; Rmo = 0.08941;
thetawe = pi/3; la=0.1016;
else
Rai = 0.0673; Rao = Rai+0.012; Rmi = Rao+1.32e-3; Rmo = Rmi + 9.53e-3;
thetawe = 0.856; la=0.1001;
end
kw = sin(thetawe/2)/(thetawe/2);
is = 1.27e-2; lend = 8.81e-2;
rho = 1.16; mu = 1.85e-5;
117
g-a = 1.32e-3; 1_a = la;
kav = 0.476*(ga/Rao)^0.25/(rho*Rao*g-a/mu)^0.5 * pi*Rao^4*1_a*rho;
kat = 0.00655*(rho*Rao*ga/mu)^(-0.136) * pi*Rao^4*1_a*rho;
1_b = 0.044; Rb = Rao; g-b = 0.017;
kbv = 0.476*(gb/Rao)^0.25/(rho*Rao*g-b/mu)^0.5 * pi*Rao^4*1_b*rho;
kbt = 0.00655*(rho*Rao*g-b/mu)^(-0.136) * pi*Rao^4*1_b*rho;
g-c = 0.012; asr-c = gc/Rao;
kc2 = 1.85* (gc/Rao)^O.1 / (Rao^2*rho/mu)^0.5 *rho*Rao'5/2;
kc4 = 0.051*(gc/Rao)^O.1 / (rho*Rao^2/mu)^0.2 *rho*Rao^5/2;
R_d = 0.109; g-d = 0.0063; asrdl = gd/Rd; asr_d2 = g_d/Rmi;
kd12 = 1.85* (g-d/Rd)^0.1 / (R_d^2*rho/mu)^0.5 *rho*R_d^5/2;
kd14 = 0.051*(g-d/R_d)^0.1 / (rho*R_d^2/mu)^0.2 *rho*R_d^5/2;
kd22 = 1.85* (g-d/Rmi)^0.1 / (Rmi^2*rho/mu)^0.5 *rho*Rmi^5/2;
kd24 = 0.051*(g-d/Rmi)^0.1 / (rho*Rmi^2/mu)^0.2 *rho*Rmi^5/2;
1_e = 0.166; R-e = 0.109; g-e = 0.0127;
kev = 0.476*(ge/R-e)^0.25/(rho*R-e*g-e/mu)^0.5 * pi*R_e^4*1_e*rho;
ket = 0.00655*(rho*Re*ge/mu)^(-0.136) * pi*R-e^4*1_e*rho;
R_f = 0.109; gf = 0.0063; asr-f = g_f/R_f;
kf2 = 1.85* (g-f/R~f)^0.1 / (R_f'2*rho/mu)^0.5 *rho*R_f^5/2;
kf4 = 0.051*(gf/R-f)^O.1 / (rho*R_f^2/mu)^0.2 *rho*R_f^5/2;
%%% Variation with speed at 30 kW %%%
rpm = linspace(10,30000,80);
%rpm = [150001;
for i=1:length(rpm)
Omega = rpm(i)*2*pi/60;
if rpm(i) < 15000
118
P=rpm(i)/15000 * 30e3;
else
P = 30e3;
end
Ean = 2*Rao*la*Bl*kw*Omega;
NIa = P/(3*Ean);
P_1_R = (NIa/(N*Npar))^2/(sigma*pi*rw^2);
P_R(i) = 2*q*N*Npar* P_1_R*(la+ls+lend);
P_lec = pi/8 * sigma * msqB * (p*Omega)^2 * rw^4;
P_ec(i) = 2*q*N*Npar* P_1_ec*la;
if Omega < 1188
taua = kav* Omega^1.5;
else taua = kat*Omega^1.864;
end
if Omega < 534
taub = kbv* Omega^1.5;
else taub = kbt*Omega^1.864;
end
if Omega < 380
tauc = kc2* Omega^1.5;
else tauc = kc4*Omega^1.8;
end
if Omega < 201
taudl = kdl2* Omega^1.5;
else taudl = kdl4*Omega^1.8;
end
if Omega < 368
taud2 = kd22* Omega^1.5;
else taud2 = kd24*Omega^1.8;
end
119
taud = taudl-tau-d2;
if Omega < 342
taue = kev* Omega^1.5;
else taue = ket*Omega^1.864;
end
if Omega < 201
tauf = kf2* Omega^1.5;
else tauf = kf4*Omega^1.8;
end
P-w(i) = (tau-a + taub + tauc + taud)*Omega;
end
figure(l)
plot(rpm,P_R)
title('Variation of Conduction Loss with Speed at 30 kW')
ylabel('Conduction Loss / W')
xlabel('Speed / rpm')
print Rloss.ps
figure(2)
plot (rpm, Pec)
title('Variation of Stator Eddy Current Loss with Speed at 30 kW')
ylabel('Eddy Current Loss / W')
xlabel('Speed / rpm')
print ecloss.ps
figure(3)
plot(rpm,P-w)
title('Variation of Windage Loss with Speed at 30 kW')
ylabel('Windage Loss / W')
xlabel('Speed / rpm')
print windloss.ps
figure(4)
plot(rpmP_R+Pec+P-w)
120
title('Variation of Total Stator Loss with Speed at 30 kW')
ylabel('Total Stator Loss / W')
xlabel('Speed / rpm')
print totloss.ps
C.4 Matlab code for plotting graphs of loss vs power
clear all
N=8; Npar = 693; rw = 0.127e-3; q=3; p=4;
sigma = 3.9e7;
B1 = 0.0709; % effective radial component of B
msqB = 0.023667; %mean sq amplitude of B,includes radial and
%azimuthal components
actual =1;
if actual == 0
Rao = 0.07938; Rai = 0.06985; Rmi = 0.07988; Rmo 0.08941;
thetawe = pi/3; la=0.1016;
else
Rai = 0.0673; Rao = Rai+0.012; Rmi = Rao+1.32e-3; Rmo Rmi + 9.53e-3;
thetawe = 0.856; la=0.1001;
end
kw = sin(thetawe/2)/(thetawe/2);
ls = 1.27e-2; lend = 8.81e-2;
%%% Variation with power at 15,000 rpm %%%
rpm = 15000;
Omega = rpm*2*pi/60;
Ean = 2*Rao*la*Bl*kw*Omega;
121
P=linspace(0,30e3,30);
for i = 1:length(P)
NIa = P(i)/(3*Ean);
P_1_R = (NIa/(N*Npar))^2/(sigma*pi*rw^2);
P_R(i) = 2*q*N*Npar* P_1_R*(la+s+lend);
end
P_ec = 12.39*ones(1,length(P));
P-w = 224.4*ones(1,length(P));
figure(1)
plot(P, PR);
title('Variation of Conduction Loss with Rated Power at 15,000 rpm')
ylabel('Conduction Loss / W')
xlabel('Rated Power / W')
print powRloss.ps
figure(2)
plot(P, PR+Pec+P-w);
title('Variation of Total Stator Loss with Rated Power at 15,000 rpm')
ylabel('Total Stator Loss / W')
xlabel('Rated Power / W')
print powtotloss.ps
122
Appendix D
Thermal Conductivity Experimental
Results
The results of the experiments described in chapter 4 are presented here. Copper tempera-
ture Tc was plotted against time t and fitted to the equation Tc = Ce-t/L + Tw. The thermal
resistances of the materials and the water film are determined from the coefficient L using
the relation L = Rfilm + RmateriaImc, where m is the mass of the copper cylinder, and
c = 393.6 J/kg0 C is the specific heat capacity of copper.
Copper
Diameter of cylinder di
Exposed length 11
Mass mi
Li,
1- (0.02533 + 0.02533 + 0.02533 + 0.02533)
4
= 0.02533 m1
= -(0.08023 + 0.08040 + 0.08029 + 0.08038)4
= 0.08033 m
= 0.470 kg
Rfilmmlc = mic/ (hirdil)
1= (27.1 + 28.0 + 25.6 + 26.9) = 26.9
4
123
Film coefficient h = 1075.7 Wm-2 oC-l
time / S
0 20 40 60 80Time / s
Figure D-1: Plots of temperature of uncoated cylinder vs time
Epoxy
Diameter of cylinder d2
Thickness of epoxy layer t2
Exposed length 12
1= -(0.03826 + 0.03810 + 0.03819 + 0.03817)4
= 0.03818 m
1=- (d2- di)
2
= 6.425 x 10-3 m
1-(0.08126 + 0.08190 + 0.08154 + 0.08168)4
= 0.08160 m
124
0L
E(D
96-
90-
84-
78-
72-
66-
60-
54-
48-
42-
36-
30-
CL)
E
time / s
Time / s
M2 = 0.466 kg
L2 = (Rfjim + Rmaterial) m
1= -(242 + 237 + 241 +
4
2C = (hrd2i2
235) = 238.8
S)n (d2c)27rl2k2 )mc
Conductivity of epoxy ke = 0.663Wm-loClo
Time I s
Time / s
Cylinder with Epoxy layer, run 2
0 100 200 300 400Time s
Wyinder with epoxy layer, run 4
500 600
Time / s
Figure D-2: Plots of temperature of cylinder with epoxy layer vs time
Epoxy-impregnated glass cloth tape
Diameter of cylinder d3
Thickness of epoxy-tape layer t3
1= -(0.02679 + 0.02681 + 0.02673 + 0.02653)4
0.02672 m
1
2
125
Mass
= 6.925 x 10-4 m
1= -(0.07976 + 0.07966 + 0.07993 + 0.08050)
4
= 0.07996 m
Mass m3 - 0.468 kg
L3 = (Rfilm + Rmateriai) m 3 c
1- -(88.1 + 88.9 + 89.0 +
4
=
91.1) = 89.28
Conductivity of epoxy-glass cloth tape composite kg = 0.306 Wm- C -lo
Cylinder with epoxy-impregnated glass cloth tape, run 1
E-
a)
Time / s
E.
90-85-80-
75-
70-65-60-
55-50-45-
40-
35-
Time / s0 60 120
Time / s
Figure D-3: Plots of temperature of cylinder with epoxy-glass cloth tape layer vs time
Comments
The fit of the data to the exponential relation assumed was very good, as can be seen
from the graphs. The measured conductivity for the epoxy/glass cloth tape composite was
126
Exposed length 13
In (d3/di)+ 2rl3 k 3 )mc
C,
EHD
Cylinder with epoxy-impregnated glass cloth tape, run 2
0
6O
Time / s
Cylinder with layer of epoxy-impregnated glass cloth tape, run 495-1
1 2180 240
//5 O
lower than that of the epoxy, which was as expected, since the glass fibers have a lower
conductivity than the epoxy.
127
128
Appendix E
Manufacturing Drawings
As mentioned in Chapter 5, machine parts were constructed and assembled according to
manufacturing prints drawn by Mike Amaral of SatCon. Some of the drawings are repro-
duced here.
Figure E-1: Stator cooling jacket
129
C
0
z0
0t
0
0
0
Figure E-3: Cross section of Potting mold
131
/
I/
1'
//
~ >\I
Figure E-4: Rotor with magnets and spacers mounted on GlO ring
132
Figure E-5: Cross section of rotor
133
Figure E-6: Assembly drawing of machine
134
Appendix F
Experimental Results from Spin-Down
Tests
The experimental results from the spin-down tests described in Chapter 6 are presented
here.
Table F. 1: Experimentally determined spin-down coefficients for rotor with aluminium ring
7 Coefficient Run 1 Run 2 Run 3 Run 4 Mean1.8 CO / Nm 0.0172 0.0151 0.0149 0.0125 0.0149
C1 /10-4 Nms 0.0206 0.0591 0.0122 0.1444 0.0591C2 /10-7 Nms1 8 9.226 9.002 9.116 8.817 9.040
1.9 CO / Nm 0.0151 0.0137 0.0127 0.0110 0.0131C1/10 4 Nms 0.2842 0.2732 0.2790 0.3758 0.3030C2/10-7 Nms 1 9 4.136 4.130 4.074 3.979 4.080
135
run 1
0.3 -
025 -
0.2-
0.15 -
0.1
0.05-
0 200 400 000 8 1000 1200 140Speed (rads)
run 3
0.35-
0.3 -
0.2 -
0.15 -
01
005-
0 200 400 S00 800Speed (rds)
1000 1200
run 20
25,
0.35
0.3
L0.2
0.15
0.1
0.05
1400
200 400 500
Speed (rads)
run 4
0 200 400 600 800Speed (rads)
800 1000 1200
1000 1200 1400
Figure F-1: Plots of loss torque versus rotor speed for rotor with aluminium ring
Table F.2: Experimentally determined spin-down coefficients for rotor with magnets
-y Coefficient Run 1 Run 2 Run 3 Run 4 Mean1.8 CO / Nm 0.0180 0.0178 0.0192 0.0177 0.0182
C 1/10-4 Nms 0.1725 0.1627 0.1677 0.1703 1.683C2 /10- 7 Nms1 .8 5.541 5.759 5.788 5.572 5.665
1.9 CO / Nm 0.0172 0.0168 0.0182 0.0168 0.0172C1/104 Nms 1.868 1.777 1.831 1.848 1.831
C2 /10- 7 Nms 1.9 2.506 2.599 2.606 2.515 2.557
136
0.2
2 0-15
0.1
0.05
00
(00
VOW.,
run 2
D.4
0.35
03
SD0.25
0.2
0. 15
0.1
0.05
run 4
1400 0 200 4D BOD d 8 1000 1200Speed (rac~s)
1400
Figure F-2: Plots of loss torque versus rotor speed for rotor with magnets
Table F.3: Experimentally determined ratio of back emf to rotor speed
Run 1 /Vs Run 2/Vs Mean/VsPhase A 0.0150 0.0149 0.0150Phase B 0.0147 0.0147 0.0147Phase C 0.0145 0.0145 0.0145
137
run 3
Speed (rads
1/7
run 1
0 200 400 B00 800 1000 1200 1400Ebtatona speed ' rads
Phase B. run 1
fotatond speed.: rads
Phase C run 1
200 400 600 8DD 1000 1200 1400 0Fbtatonal speed / rads
Phase C. run 2
9-
8-
7-
>0
Go 4-
3-
2-
1 l-
200 400 6DD 800 1000 1200 140DFtationd speed rads
Figure F-3: Plots of Back Emf versus rotor speed
138
9
8
7
>0
~r
3
2
1
Phase A, run 1
~1
K
Phase A run 2
Eotatonal speed -rads
Phase B, run 2
9
8
7
>0
a34
3
2
D0
mm
ca
Bibliography
[1] Regis Roche, "Magnet Losses in a Flywheel Energy Storage System". Internal reportat MIT, 1997.
[2] James L. Kirtley Jr., Mary Tolikas, Jeffrey H. Lang, Chee We Ng, Regis Roche, "Ro-tor Loss Models for High Speed PM Motor-Generators". Presented to the Interna-tional Conference on Electric Machines, Istanbul, Sept. 2-4, 1998.
[3] James L. Kirtley Jr., "Notes for 6.1 Is: Design of Electric Motors, Generators andDrive Systems", Massachusetts Institute of Technology, Cambridge, MA, 1997.
[4] James L. Kirtley Jr., "Design and Construction of an Armature for an Alternatorwith a Superconducting Field Winding", Ph.D Thesis (Electrical Engineering), Mas-sachusetts Institute of Technology, Cambridge, MA, 1971.
[5] "Heat Transfer Data Book", General Electric Company Corporate Research and De-velopment, Schenectady, NY, 1971.
[6] Frank P. Incropera and David P. DeWitt, "Fundamentals of Heat and Mass Transfer",Wiley, New York, 1990.
[7] John Ofori-Tenkorang, "Permanent-Magnet Synchronous Motors and AssociatedPower Electronics for Direct-Drive Vehicle Propulsion", Ph.D Thesis (Electrical En-gineering), Massachusetts Institute of Technology, Cambridge, MA, 1996.
[8] S. Torquato and F. Lado, "Bounds on the Conductivity of a Random Array of Cylin-ders", Proceedings of the Royal Society of London, Series A (Mathematical and Phys-ical Sciences), Vol. 417, No. 1852, May 1998.
139
