High-Speed Permanent Magnet Motor Generator for
Flywheel Energy Storage
by
Tracey Chui Ping Ho
Submitted to the Department of Electrical Engineering and Computer
Science
in partial fulfillment of the requirements for the degrees of
Bachelor of Science in Electrical Engineering
and
Master of Engineering in Electrical Engineering and Computer Science
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
May 1999 L>
@ Tracey Chui Ping Ho, MCMXCIX. All rights reserved.
The author hereby grants to MIT permission to reproduce and distribute
publicly paper and electronic copies of this thesis document in whole or in
part, and to grant others the right to do so.
MASSACHUSETTS INc
1~~'
OF TECHNOLOG
Author.....
D~partment of lectrical Engineering and
May 20, 1999
Certified by
Jeffrey H. Lang
Professor of Electrical Engineering and Computer Science
Thesis Supervisor
7
-Certified by
c
--
---
-
--
--
James L. Kirtley Jr.
ProfessorfEfectrical Engineering and Computer Science
is S uer vis
Accepted by
........
....
Arthur C. Smith
Chairman, 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 Science
on May 20, 1999, in partial fulfillment of the
requirements for the degrees of
Bachelor of Science in Electrical Engineering
and
Master of Engineering in Electrical Engineering and Computer Science
Abstract
This thesis is part of a joint project between MIT and SatCon Technology Corporation to
develop a high-speed motor-generator for a flywheel energy storage system. Such systems
offer environmental and performance advantages over chemical batteries, with potential
applications in hybrid electric vehicles and uninterruptible power supplies. The development of high-energy Neodymium Iron Boron magnets, as well as advances in composites,
electric drives and magnetic bearings, has contributed towards making flywheel systems
more 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 Kirtley, while much of the mechanical design was done by engineers at SatCon. This thesis
focused primarily on: the development of theoretical models for various loss mechanisms,
with particular interest in the modelling of eddy currents in azimuthally segmented rotor
magnets; the development of theoretical models for thermal performance; the design of a
cooling system; and construction details. Finally, several quantities predicted by the electromagnetic analysis and loss models were experimentally measured, to evaluate the validity of the theory. On the basis of this work it is believed that compact permanent magnet
synchronous motor-generators for flywheel energy storage systems can exhibit efficiencies
near 95%, and can operate with idle losses as low as 12 W.
Thesis Supervisor: Jeffrey H. Lang
Title: 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
7
1
Introduction
2
Machine Design
9
2.1 Existing Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Electromagnetic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Modifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3
Magnet Loss Models
3.1 Stator Current Space Harmonics . . . . . . . . . . . . . . . . . . . . . . .
3.2 Eddy Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Known Magnetic Field and Thin Magnets mounted on Infinitely
Permeable Surface . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 Known Stator Excitation Current and Thin Magnets with Infinitely
Permeable Boundaries . . . . . . . . . . . . . . . . . . . . . . . .
3.2.3 Known Stator Excitation Current and Magnets with Significant Thickness with Infinitely Permeable Boundaries . . . . . . . . . . . . . .
3.2.4 Known Stator Excitation Current and Magnets with Significant Thickness Without Infinitely Permeable Boundaries . . . . . .
3.2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Loss Calculation . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Application of Model to Rotor Magnet Loss Problem . . . . . .
39
48
49
50
Stator Loss Models, Cooling System Design and Thermal Analysis
4.1 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 . . . . . . . . . . . . . . . . . . . . . . . . .
53
53
53
54
57
61
62
63
64
65
4
5
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19
19
22
22
26
31
4.3.1
4.3.2
4.3.3
4.3.4
Thermal Conductivity Experiments . . . .
Film Coefficient for Cooling Channel . . .
Effective Conductivity of Armature Region
Temperature Calculation . . . . . . . . . .
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65
67
68
69
5 Fabrication of the Experiment
75
6
89
89
92
92
95
96
7
Testing
6.1 Resistance and Inductance . .
6.2 Spin-down Tests. . . . . . . .
6.2.1 Loss estim ation . . . .
6.2.2 Back em f . . . . . . .
6.3 Magnetic Field Measurements
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99
Summary and Conclusions
A Inductance Calculation
101
B Matlab code for Rotor Loss Calculation
105
C Thermal Analysis Spreadsheet and Matlab Calculations
C. 1 Thermal Analysis Spreadsheet . . . . . . . . . . . .
C.2 Matlab code for windage calculation . . . . . . . . .
C.3 Matlab code for plotting graphs of loss vs speed . . .
C.4 Matlab code for plotting graphs of loss vs power . . .
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111
111
114
117
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 vehicles. 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 computers, 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 batteries. 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 expensive than the latter. However, over the last decade, technological advances in areas such
as composites, electric drives and magnetic bearings have contributed towards making flywheel 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 machine 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 impedes 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 synchronous 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 nonzero 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. Experimental 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 Professor 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
No. of pole pairs
No. of phases
Wire diameter
Active length
Armature inner radius
Armature thickness
Armature outer radius
Rotational gap width
Magnet inner radius
Magnet thickness
Magnet outer radius
Electrical angle
Symbol
p
q
dw
1
Rai
ta
Rao
g
Rmi
tm
Rmo
Owe
Original Design
4
3
0.254 mm
10.16 cm
6.99 cm
9.53 mm
7.94 cm
0.508 mm
7.99 cm
0.95 cm
8.94 cm
-r/3 = 1.047
Modified Dimensions
4
3
0.254 mm
10.01 cm
6.73 cm
12.0 mm
7.93 cm
1.32 mm
8.06 cm
0.95 cm
9.01 cm
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 flywheel 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
0
m
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
s .(pon
-Brem s
wr
11
terminaL
Figure 2-2: Winding pattern for one phase. The end turns which are bent outward are at
the 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:
./p~m'
4
-Bremkm smn Pr
2
7r
where
km =
jln
: ifp=1
()
(R
P - RI,;;P) Rg;O
(2.1)
: otherwise
Multiplying by 2 to obtain the combined field from both sets of magnets, and dividing by
12
Table 2.2: Machine Quantities at 15,000 rpm
Quantity
Symbol
Rated power
Rms magnetic field at Rao
Effective rms magnetic field
Internal voltage per turn
Ampere-turns
P
Synchronous inductance/N
2
Normalized reactance
Terminal voltage per turn
v
Bia
B1
Ean
NIa
2
Ld/N
Original
Design
30 kW
0.1624 T
0.1278 T
3.09 V
3234 A
With Modified
Dimensions
30kW
0.1564 T
0.1157 T
2.80 V
3573 A
As-built,
Without Halbach
30 kW
0.0958 T
0.0709 T
1.71 V
5835 A
2.36 x 10-8 H
2.30 x 10-8 H
2.30 x 10-8 H
0.155
3.13 V
0.184
2.85 V
0.491
1.91 V
Xa
Vn
to obtain the rms value, we have
A
Bla =Vf2Bremkm sin
=
2/
7r
0.1624 T
To account for the variation in magnetic flux density across the thickness of the armature, 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 radius 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
1
Biao Rao
-
jRao
Rai
Bia
Rai
1
R ao
- XP+1
rdr
(2.2)
(1l-x)(p +)
where x = Rai/Rao. Thus B 1 = Biakt
0.1278 T.
The internal voltage induced across one turn of the armature winding is given by the
13
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 0
km
=
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
NIa = -3Ea
P
En
= 3234 A
(2.4)
The synchronous inductance Ld of the 3-phase armature winding is the apparent in14
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 similar 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
Ld =
(31Npo\ sin
(sNn 2
7r
p_2x
(1_P+ X 2 (1±p)-2xP+l
(1 - x)2(i _ p2)p
21-p+x2
=
N 2 x 2.36 x 10-8 H
The internally normalized per-unit synchronous reactance is obtained by dividing the
voltage drop across the armature winding by the internal voltage:
xa
WLdIa
===
Ea
_w
(Ld/N 2 )NIa
0.155
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
15
E2 (1 + x)
Thus the terminal voltage per turn is
Vin= En /
Vt nEan
2.3
+
X2a+~.3
=
3.13 V
Modifications
A number of parameters in the existing design were altered slightly because of manufacturing 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 armature 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
A,,
6
x
72
x
77
x
(0.0254/2)2/ (7.942 - 6.992)
to be 77,
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 calculating 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 0 m
=
-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 arrangement 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 machine 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
E
n
-
odd ?l
nw
f
J cos (npo)
sin
For a balanced q-phase source of amplitude J and frequency w, total current density is
nOwe
q 4
_.-2mr- sin ( 2
n
for n = 2kq ± 1, integer k
) Jcos(wt -FnpO)
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 = J 0 /r, where
NI
Owe (Rao
-
(3.1)
Rai)
So the overall current density at radius r is
Z
n
" cos(wt - npO)
for n = 2kq ± 1, integer k
T
where
q 4 .nwe
- -sin
2 n7
( 2)
Jn
(3.2)
Jo
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)2
n
gives rise to magnetic fields that can be expressed as the negative gradient of scalar potentials
'is
=
Aln
n
Ra
sin (Lot - npO)
20
for r < Rao
T O,
=
A2n
Rao)psin
r
n
Substituting His = -VTi, and Hos
=
for
fo r > Rao
(wt - npO)
-V To, into the boundary conditions Hos, - Hi,, =
K, and Ho,, = Hjs, at r = Rao, we have
KnRao
=
A
2np
KnRao
2np
Thus, for r > Rao,
E Kn (Rao
Ho
n
Hr
=
2
np+1
cos (wt - np6)
(3.3)
np+1
sin (wt - npO)
(3.4)
r
Kn (Rao
r
2
n
Equating fields from the armature and the equivalent current sheet, we obtain
Kn (Rao
2
np+1
r )
JndR (R)
JRao
2R
r
R ai
np+1
Jn (R np+1 _ Ra+
2(np + 1)rnp+l
=> Kn,
Jn
-Rai
\np+1)
Rao
np+1
(3.5)
Since the windings and magnets have a finite axial length 1, we introduce another
Fourier summation in m:
K=
Kn COS(Wtt - np)
n
004
--- sin
m=1 MT
(7WZ)
modd
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
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 -FnpO'. So the rotor
sees the current distribution
K
004
- sin
Kncos ((1 T n)wt -Fnp')
=
M=1
n
m7rz
MIT
modd
for n = 2kq i 1, integer k
3.2
(3.6)
Eddy Currents
This section presents an electromagnetic analysis of eddy currents in azimuthally segmented 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
Y
(
~00
Bn, sin (wnt - nkx) sin
~
M1
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
OR
at
the y-component of which yields the relation
aE
Ez
- --.
ax
az
23
- BY
at
(3.7)
y
B
magnets
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 ('
h
m
Z
Then
K
=
Kz
=
- O9Z
-Z
h~r
7
lk sin d
m Cm
IkCosm
sin
Chncos
Thus
Amo
: for modes satisfying Ex
0
: otherwise
24
=
0 at x = 0, d; Ez = 0 at z =0, 1
(3.8)
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
z
n mB1
modd
BnmWn
=
n
(COSwUtcosnkx ± sin wt sinrtkx) sin(m
)
m=
modd
The functions cos nkx and sin nkx can in turn be expressed as
z
(27rx)
EU ainu COS
cos nkx
sin nkx
u
sin 2u7rx)
2u7rx)
(2u7x) + a 4 n. sin
(a 3 u COS
=
+ a2
(3.9)
(3.10)
U
where
=sin nkd
a 1 ""
I
+
I
s nkd + 2uir nkd - 2u7r)
a2n""
(1 - cos nkd)
a3n""
=(1 -- cos nk d)
a4n""
=
sinrnkd
(nkd + 2uw + nkd - 2u7)
1
(nkd +2uxr
1
nkdd
-
2uir
-
+
1
nkd - 2nx
1
+
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
at those boundaries. If nkd
=
=
d, which imply the existence of artificial current sheets
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
ie.
1
: foru=mi
0
: otherwise
1
:
0
: otherwise
alau
a2nu
-0
a3nu
-0
foru=mi
(3.11)
a4nu
Since only the terms in sin (2ur) give rise to eddy currents, Equation 3.7 becomes
-E
SBnmn
n
a2,,Sin
(cos wn
2ndr ))
+ sinwnt
(E
a 4 . sin (2wr )
sin (m7rz
modd
OK
OKz)
EE
m
d )
1
'AmoU
-
=1
Am
+ M72
2
Chm
sin (h7rx) sin m7rz
(
Comparing the expressions termwise, we have
En
BnmLn (a2nh/2 cOS wat+a4nh/2 sin
AmoO
Ch0
d
Lot)
: when h even, m odd
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 Infinitely 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
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.
Y
inFinitely permeable material
stator current sheet K
A~
air gap
magnets
Am
0
x
3d
2d
infinitely permeable 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
ac
BC
mir
-- 55
h
Kz
aC
ax
E
E
m
h7r
h7rx
Chm sin
I
ChmCOS
h m
27
h7rx)
d
(
z )z
co
(my)
(3.12)
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 superposition 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
Knm cos (wnt - nkx) sin(m7z
Kz =
Hz
=-K
M7
m=1
modd
fl
= 0
Now
|_a=
Hxdx = E
-
n
nk
Knm
sin (wnt - nkx) sin (m7z)
m=1
modd
-&89/Oz = 0 at y = A for 0 < z < 1, since the summation in m yields a
satisfies Hz
constant in z for 0 < z < 1. Thus the scalar potential
001
E
n
1
k sinh
(nt
in
(k3sinh(#1am
m=1
modd
which is zero at y
0, matches K at y
2
Qx
H
=
0,
+ (rl)2
H=K,ie.
=
-Kz
=K
-
hwx
h7F
d Chm COS(d
=-ZZ
h
Hz
sinh (#nmy)
- nkx) sin
A. Since V 2 91
012m = (nk)
At y = 0,
mn_z
_
Knm
sin
M71z
m
EE
h
m
rn
h'rx
Ch,
28
sin
)
m7z
cos
Since
IF|,l
Hdx =
=-
()
hwrx
d
Chm sin
Z
sin (mlrz)
satisfies H2 = -849/Dz at y = 0, the scalar potential
h7rx)
sin
h
sinh(-#2hmA)
m
sinh
1'mw
) sin (mirz
(d
( 3 2hm (Y -
A))
matches K at y = 0, and is zero at y = A.
#2m =d
2
(h r
+ M72
)
1
since V 2 4' 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
D42
_8DI91
H
ay + aay
=-
Km
_E
."
sin (wt - nkx) sin
mJrz
cosh (#Anmy)
I)
modd
-- Y
E
h
Chm-
hwx
-sin
m
d )
sinh (-#2hmA)
m
cosh (02hm (Y - A))
sin
As in the previous section, the electric field induced by the time-varying magnetic field
satisfies
z
Dx
aB
Now
OB"
Dt
1Z
-Po
00
!nm
Z
Knmk.
nk smh#A
m1
+ po E E
h
m
dChm
dt
#hm
AWn cos (wnt
.
sin
sinh(#aA)
29
h7rx)
d )
-
sin.
nkx) sin
(7)
mIZ
(mwz
cosh (2hmA)
-0
Il
Z/n
po
3
nm
- w (cos
(o w\Wt cos nkx + Sin wnt sin nkx) sin
*
sinh1m
n
nmenk
Knm
modd
#3hm
dChm
dt
m
h
h7rx)
(
- ) s(
s(
sinh (32hm)
"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
Knm nk snnm
rik sinh /3inm A on
=
n
modd
(coszt
a
+
-Po
E
h
m
1
dt
(Kx
sinh(/ 2hmA)
Oz
Amo-
1
a4,, sin (2udrx)
sinont (
. h-ix sin ( MFZ ) cosh (#2hmA)
d
#2hm
dChm
sin (2ndrx
2 nu
OK
x )
h7 2
rn) 2]
h7rx
Cam Sin
m3
h
si1
(m7rz
m
Comparing the expressions termwise, we have, for h = 2u and m odd,
dCh
k1 d
1
k 2 Ch=
(k 3 n cos wnt + k 4 n sin wnt)
n2
where
ki
=
3 2hmcoth
( 3 2hmA)
h7
Amct1
k2
AMUo-
k3
=
k4
=
2 + (m7r)2
( d )
Knm#1inmWn
nk sinh (/31nm)
a2
2/2
Knm/ 3 ino
a
n
4
nk sinh (#1/ma)
30
/2
sin (m7rz)
The solution to the differential equation is of the form
Chm = E (C1, cos wt
n
+ C2, sin wt)
Substituting this into the differential equation, we obtain
k2k3, -
C1"
C 2,"
k 1 k 4 ,wn
(kiWn) 2 + k2
k 2 k 4 n + k 1 k3swn
(kin)2 + k
Thus
(k2k3, -kik4w.) cos wt+(k2k4n+kIk3,W.)sin wt
:
2
(kW" ) +k2
Chm{ =
when h even and m odd
otherwise
0
Expressions for the eddy currents can be obtained by substituting this result into Equations 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,,,, cos (wt - nkx)
fl
at y
=
m=1
modd
m7rz
sin
I
)
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
Y
inFinitely permeable
material
stator current sheet K
!) k,-n) -
". I 7j 7-) 77 V- -, 77 7- '
Region 2
air gap
t
0
Region 1
3d
magnets
x
-T
infinitely permeable
material
Figure 3-4: Diagram for problem with known stator current and magnets with significant
thickness
is given by V x (Ce), where C is of the form
C
Z E Cm(y, t) Sin
=
h m
hyrx
d )
sin
(7)
so as to match the magnet dimensions and the assumption of no radial current. Thus
Jx
aC
m7r
lCm(Y, t)insi
Bz
-
h
Jz
-
OC
Xh
m
h~c
cos ( d
=CdZChm(y,t)
m
()
hirx
d
Cos
(mrz
1)
J sin (7)
From Ampere's Law, the magnetic field induced by the eddy currents satisfies V x H = J,
ie.
aHz
aHy
ay
az
32
iJx
OH_ OHx
Ox
Oy
0H
OHz
Ox
Oz
0
-
Thus Hz, Hy and Hz are of the form
H
Ahm
-
Hy
=
Ayh(Y, t) sin
Z
sin
(hwx)
Azhm (y, t) sin
S:
m
=
h
7Z
(hdx) sin (M7FZ)
m
h
Hz
(h7rx)
t) Cos
(,
m
h
Cos (1)
Now H satisfies the diffusion equation
= V2)7
pLoa O
within region 1. Therefore,
OAxm
ILOU
h m
at
sin (
cos
(
=5IE
h m
7IZ
2
(d)
-Axh
-
OA
o
y
-
Axh
-
2
(7)
Axh
hw
± a2xh
Cos
(
hdx) sin
)
2
-
Axhm
Tr2
+
2
Axhm
Similarly,
- A Ym
at
IL a
z
_
[U at
_
-
h
- AYhm
+
-Azhm
±
hm \d)
(hwN)2
A h
dh
a 2 AYm
g2Ah
Since the excitation is a sum of sinusoids with frequencies w, we can write Ajhm (Y)
33
in the form E Re
AZhmn (y)ewnt}I
for i = x, y, z. Substituting this into the differential
equation, we obtain
(
+2 (
h7
E I-10Or)WnA
E
Ahmn
A
n
02 A ihmn
A ihmn
+±E
n
+(rnlr)2
Azh
-
Oy 2
w2)
dT
ihmn(
±
2hmn
Oy
2
jn/l0)
a -hmn+-hmnY
ai+hmn
where
7hmn
Chmn+
d)2
+M72
dhmn
h7r2
WnP;OU
T
4
(h)2
M,)2
2
2
(Wn0 )2
(
2
+
d)
(3.13)
)
±
Chmn
MT2
+
dhmnJ =
2)
)
2
2
+ (Wn0IoC)2)
Thus
h
m
n
m
n
Re (ay+ e
Hy zz55
h
h
{
Re (ax+eO" + ax e-7) eiwnt} cos
m
E Re {(az+ey
+ aye-Y)
eiW
sin
+ az-e YT) eint sin
n
where the subscripts hmn have been omitted from
pactness.
34
ai±hmn
(h7x
J
(
d
h7rx)
( d'rx
sin
TT
sin
Cos (1
(3.14)
and 7hmn for notational com-
Applying the boundary condition that H,
=
0, Hz = 0 at y = -T, we have
yT
ax+e-T + axe?
0
S ax-_
az+e~-T + az-e T
0
az-
-
(3.15)
ax+
-e -2-yr
(3.16)
Thus
HX
=
Re { ax+ (eY - e -2-yTe
1:1
19Y) ej"t} Cos
h
Re {az+ (ey - e -2-Te -YY)
h
ejUn}
sin
m n
(
sin (m1rZ)
(
Cos
hurx)
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
F
Iy
=-
Jkt
fx
Hxdx = 1
W
n
Knk
sin (wnt - nkx) sin (n7z
m=1
modd
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
nk
n
sin(ot
- nkx) sin
i,2m =
(nk) 2 + (7)2
mrz
J
sinh (#1nmY)
sinh (AinmA)
modd
where
matches K at y = A, and is zero at y = 0.
AF2Z=
h7rx) sin
EDhm(t) nsi d
h m
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
=
q1
+
X2.
0, tangential H is continuous since the current density is finite, and normal H is
At y
also continuous since we assume that the permeability y of the magnet is close to yo. Now
HX(2 , lyzz
h7rx)
sin(
d )
ZDhyh7rCOS
d
h m
-
H|"2 ",--o =-
Z)
EEoddKnm-n k (sin watcos nkx -
#1""
cos wat sin nkx) sin
sinh (i1nmA)
m =1
modd
hmw
H( 2 )
jY=
h7rx)
EDhm sin
+
Z
=-
m7r
Dhm
h
si(")
h7rx)
s in
(
m
coth (/ 3 2hmA)
32,m
m7rz
COS
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
-E h
h7r
Dhm d COS
m
()
h7rx
= EEE Re ax+ (I
m
h
(7)
sin
d
e-2yT) ejwnt cos (h7rx
-
1 {
EKnmk (sinwnt
rik
m_=
00
n
sin (mrz
n
(
modd
sin 2u7rx)
-cos
wnt
(E
a 4 nu sin
a 2 nu
2urx
d
I
sin (n1Fz)
/ 3 inm
sinh (#1inmA)
+
EE
h
h7rx
d
Dhm sin
m
=E
h
E
E
m n
Re
((ay+
mr
-
/2hm
#Z)
sin
+ ay_)
h
Writing
m
Dhm =
n
Re
jaz+ (I
En Re
(7)
COS (
m
-EE
h7rx) sin
si d )
sin
m~rz
h7rx
E EDhm 17 sin hdx
h
ejwnt}
coth (#2hmA)
-
e-2yT) ejwt}
{fhmneWnt}I,
sin h~x
(m7rz
a termwise comparison of the expressions yields, for
36
h = 2u and m odd,
ax+ (I
=
e-2yT)
-
an
d--bhmn
-Knm! 3 inm
ay+ + ay-
(j a2nh, 2 +
nk sinh #1nm
a2+ (I - e 2-T)
a4h/ 2 )
A
+
Dhmn 3 2hm
coth
( 3 2hmA)
Dmn
= -
which gives the following expressions for ax+ and az+ in terms of Dhmn:
ax+
hir
-
=
bhmn
2
e
m = -hmn
2
I I 1- e-
(3.17)
r
(3.18)
y
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:
+
OHx
OH(1)
09 Y
Oz
- E E E
-
h
m
E E E
+
h
E
-
h
m
e~Y7)
(7 (ay+ey
ej'nt} sin
ejw" 11 sin
n
Re
- ay-e~7Y)
n
SE
m
hiRe {ax+ (eY - e2T
n
n7 Re
I
{az+ (eYY
- e 2,Te-YY)
ejwnt}
h7rx)
d )
Re
ey (
+e-7y9(
dL ax+ + -yay+ ax+e 2-
-
M7r
\
i az+)
-yay_ +
-az+e
- 0
h7r
- -dax++
-ax+e-r
mir
-yay+
az+
-
- -yay-
+
0
-
-az+e-r=
37
0
2r
sin
sin (7)
m7rz
sin h7rx sin
-0
-
(
hux)
eiwt
7
Substituting in the expressions for ax+ and az+ in terms of Dhmn obtained earlier, we have
ax(a++
-
az+
((h7 2
Dhmn
+ (Mnr)2
d /)
ax+ +
-
d
rraz+
e27T
()
Dhmne2r
)2)
± (TnT
We can substitute these expressions into the expression for a,+ + ay
Knm/'3 nm
ja2.h/2
nk sinh (#1nA)(
ay+ + ay_
h7 2
d )
+
(3.19)
-e-2-yr) -Y
a4h/ 2 )
(3.20)
to solve for hmn:
+ bhmn2hm coth (/32hm/A)
bmn (1 + e--Y)
S(7r)2
(1-
\} l
-,)
-y
Dhmn coth (yT)
(7r)2
d )
Knkinm n
:::
Dhmn
flk sifh(O 1innA)
-
#2hm
kj
a/
2
nh/2
+a
2
coth (2hmA)
coth(yT)
+
Knm 3 1nm7
nk sinh (#1nmA)
nh/2,
()2)
)
(ja2
+ (m))
h/2
+
a4fh/2 )
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
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 magnet 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
sin
npo)
(wut
K ( K cos
m,,rz
(
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.
Hod
8
(
=
Kn
np+1
oo
cos (wnt - npO) (
m
n
4
--- sin
mrz
1
modd
Kn
HoSr
Rao
Rao
np+1
=
sin (wnt - np6) (
o
4
sin
m~sz
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
E E
Hx =
h
Hy=
-
m
§
Re
(ax+e"
+ axe~-Y) ent} cos (hx
sin (m7zN
d
n
Re
((ay+eYY
+ aye-Y7) eiwn'} sin hdx sin (7r z)
39
EE
Hz
where
'Yhmn = Chmn
+ dhmnj =
rx
d )
f
sin
e~(
n
m
h
j
--YY) e
Re Iaz+e?"+ az-e
\t()
+ (7)2
Cos (
)
(3.21)
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
=
h
Tor
3hm.
EEEA
m
n
F3hm (r) sin
(hTO)
hvO)
= EEE A 4 m F4h (r) sin
h
m
hi0
sin
sin
for r < Rmi
(1UZ
m7Z)
for r > Ro
n
which satisfy Laplace's equation
V2T = 0
or
I
1 &2T
a (rO)
=0
r2 002 +
in cylindrical coordinates. Substituting Tir and Wo, into Laplace's equation, we obtain
82 F
Or2
1 F
r
h7r
2
(' m
2~
r +
Or
F =0
for i = 1, 2
The form of this differential equation matches a version of Bessel's Equation, which has
as solutions the hyperbolic Bessel Functions I
Kh
(r),
(TjZMr),
which grows with radius, and
which decreases with radius and is singular at the origin. Thus
F 3h.(r)
=
Ih
om
F 4hm(r)
mT r
i<
r
r
In the air gap between the stator current and the magnets, the total field is the superpo40
sition of H,, = -V4'
and Hir = -VWTr.
0
0K
f
Ho =
n
2
m=1
(Rao
4
nP+1
cos (wnt - npO)
ao
(
At the inner surface of the magnets, r
Rmi
=
Rmi,
m7z
7
mr
sin
modd
Rmi h m n
Hr
ooK
(
=
(
E
np+1
/Ra
2
-
h7r
1mr
Ihir
h
R ZEA3
-Rmi
R
1 t
4
sin (wnt - npO) -
mr
mirz
hr
Cos
m
sin
m)
\
1
mr
sin("~z
modd
- E
E Y
Hz
h
m
As
Rm
Ih
z
(1
m7r
(
=-EE(
/sin
j-(1(0)
n1
m
h
sin
R
I/
A
sin
n
h7rO
6
0
hirO
m7
m7rz
- Cos
m1
(
Assuming a magnet permeability close to po, the normal and tangential components of
magnetic field are continuous across the magnet boundary. However, only some components 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)
sin(npO)
cos
+ a 2 . Sin
a 3nu cos
+ a 4 ,. sin
(a1n.
=
(3.22)
(2ur)
(3.23)
where
+
sin (npom)
a1u=
np6m + 2u7r
a2n
=
(3.24)
up6,
(1 - cos (npO))
(3.25)
np6,
(np6m + 2ugr
a
=(1 - Cos (npOm))
(
npm
a4n
=
2u7r)
sin (np~m)
I
(npom - 2u-c
41
2u~r)
(3.26)
+
+ 27
np6 , - 27
)(
1(3.27)
np6m
+ 2ugr)
unless npOm is an even multiple of 7, of the form np0m2 m17, in which case
1
: foru=mi
0
: otherwise
1
: forun=mi
0
: otherwise
alnu
a2nu
-0
a3nu
-0
(3.28)
a4nu
For those components of magnetic field that couple to eddy currents, we have, noting
that Hr = -Hy,
np+1
0Kn(a
modd
MSin
(cos
t1
n
m
np+1
a
(
_0K
n
modd
4.
m7r
-
h7r
- cos
Rn
a 2nu sin
(sin wnt (E
h
m0
=E
h
I' W
n
(E
a3 u
cos (2urO)))
( m wz
(n7IZ
- cos wnt (1a
m2
n
m
mi
h76
sin (0)
(him sin (M17Z)
I R(
Re (ay+ + ay_) en't I sin
3
hmi
r
/Rmi
E E Re( (az+ +
m
(hrO) sin
(,h7r
m
n
I sin (
az- ) ej''t
(hm ) sin
h7
MTr
- EEE5A
miM
)0
mir
A3
E E
h
mr
1
m
- E
sin Wnt
mr z
sin
-EE
h
+
(a-+ + ax) ejW't cos (hurO) sin
= EEERe
E
Cos (2uyr6
( M"Z
F1
REEEAshmnnIh
Rmi h m n
h
ainu
)
c(
m)
h7rO)
sin (0)
42
(n7z
os
)
CS(m7rz
(mrz
4 nu sin
2zurO
(
Orl))
Comparing these expressions termwise, with
ax+ + ax_ ay+ + ay_ -
Rao np+l1
2Kn
m7
Rmi
2Kn
Rao )np+1 (.a
2n
-A
3hm
--
+ a4,) +
eiWn},
m7r
hir
Rmi
-m
mir
I'
A3hm
we have
Rmi
h7
(3.29)
OM
(T Rmi)
hir
)7
Rmi
I h
{A3hm
Z3h
jaa,)
nu
mm~r
az+ + az- =
= Re
A3hmn
(3.30)
(3.31)
yT
T
for h = 2u and m odd. A similar calculation at the outer surface of the magnets yields
ax+e y + axe
2Kn
Rao) np+l
mr
Rmo
( a1nu - ja3nu )
(
I
RAo A4hn
ay+e
+yTay-eT
2Kn
m7r
Z
az+e
yT+
ae
yT
-
4A
h7r
mo
mR
I
(3.32)
m
Rao) np+1
(ja
(Rmo
hK'
1
lh
OM(
n
A 4 hKh, (7
2
u + a 4 n.)
( '7 Rmo)
(3.33)
Rmo)
(3.34)
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
+±yay+-
mir
1az+
m7
-az_
- -yay_
1
= 0
(3.35)
(3.36)
The eight equations 3.29- 3.36 can be solved simultaneously for the eight unknowns
43
ax+, ax_, ay+, ay_, a2+, az-,
A3hm
and
A4hm,
in terms of K,:
-
A-ly
(3.37)
az-
A4hm
where
1
1
0
0
0
0
0
0
1
1
0
0
"'"I'h7,
0
0
0
0
1
1
" I
e-yT
e-T
0
0
0
0
1
om
0
( mrRmi)
I
h^*
0
mRir
("'fni
(
0
0
hr
Rmo m
K
mirRo)
h,
m
0
0
e-7T
eT
0
0
0
0
0
0
0
e-T
eyT
0
0
0
0
ME
0
0
y0
0
-h
0
-^lr
0
-~
0
44
""~rK'
"Kh,,
)
m~rRmo
h~r
(mirRmo)
and
2K, (Rao \np+l
,mir
\Rmi)
(ah/2
2K, (Rao" )np-frni7r kRmt)
-
(ja2h/2
)
+ a4fh/2 )
ja3nh/2
0
2K,
m.7r
(Rao
(
")np+1
lnh/
Rmo)
2Kn (Rao )n~
mir \Rmo)
2
-
a2lh/ 2
3/2)
+ a4fh/ )
2
0
0
0
The values of the Bessel functions Ih- (m''m")
m
and Kh -
Tm
(mRmi
) vary greatly with h,
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 _
(3.38)
az+
az-
Ahm n
4
hK
,
(miRmi
45
where
1
1
0
0
0
0
0
0
1
1
0
0
0
0
0
0
1
1
mit
0
e -T
eyT
0
0
0
0
0
hitr
Rmo~m
0
0
e-?T
eT
0
0
0
0
0
0
0
eyT
0
mr
0
0Y
0
0
0
0
mt
0
0
0
Rr
hir
07
0
hit
0
- I
-0
0
0
myram
K hr
(mrRmo)
and y is, as before,
2Kn
mir
(\Rmi)J
SRaonP+1
(
2K.
m7r
a2nh/
(i,,
2Kn
mit
2Kn
mitr
nh/2
(Rao
)flP+1
Ja3nh/2)
2
+ a4nh/2 )
0
np+1
Rnp+1
-
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 cylindrical 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
boundaries, assuming a rectilinear geometry gives a very good approximation.
h
E
Z
m n
hwx
sin
A4hn e Amy
sin
d
m
h
or
(hwrx
J
e-AmY sin
A3hm
Wir
m7z
sin ( 7Fz)
for r < Rmi
for r > Rmo
d)
Matching fields at the boundaries of the magnets then yields, for h = 2u and m odd,
ax+ + a
2Kn
=
np+1
Rao
m7
Rmi
(ai1n
-
ja 3
A
-
u)
hr
3
h-
(3.39)
(3.40)
ay+ + ay-
2Kn
Rao
m7r
Rmi
np+l
(ja 2n + a4nu) +
(3.41)
A3hm
m7r
az+ + az-
-A3hm
(3.42)
1
at the inner surface, and
ax+e-T
ay+e-T
+a
2Kn
eyT
+
Ro)np+1
Z4hm e~OTh7
(ain - jas.) -- Z
h~4d
-
mrnz
eyT =
2Kn
mrn
az+e-T + a_ eI =
-A4hm
np+ 1
Rao
(ja 2n + a 4 nu)
RmoI
-
A4hm
I3e- T
e~TmT
1
1
0
0
0
o
0
0
1
1
0
0
-#
0
0
0
0
0
1
1
T
0
e~yT
eyT
0
0
0
0
0
e-,T
0
0
e -1T
eT
0
0
0
13e-OT
0
0
0
0
e-?T
e?yT
0
Te- OT
0
0
0 -T
0
0
0
0y
-I0
Tt
0
0
hr
d7
0
-j
47
(3.44)
(3.45)
1
at the outer surface. Matrix A in Equation 3.37 then becomes
Ac=
(3.43)
4
-
hir
d1
0
The cartesian version Ac also becomes poorly conditioned for larger h and m, owing to
the term e-OT. The following equation works better:
ax+
ax-
ay+
ay~
(3.46)
=-A'- y
az+
azA3hmn
where
hit
1
1
0
0
0
0
0
0
1
1
0
0
-#13 0
0
0
0
0
1
1
0
e-yT
eyT
0
0
0
0
0
it
0
0
e--T
e-yT
0
0
0
/3
0
0
0
0
e-?yT
eyT
0
hit
d
0
7
0
-T
0
0
0
0
h-i
0
-7
0
mt
0
0
d
0
d
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
dissipation.
Loss Calculation
3.3
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)
-
h
Sin
=~
Jz
+ ay_
ey- (az-7
I7 ) e
eiwnt
n
m
hx
cos
(
1Z)
OHX
ax
19y
E(ay+
hir
-
Re
- ax+7 e" + (ay
)
I( d
hm
+
axj)
eiwnt
e-]
nz
sin
cos (hx
hir
The time-average power dissipation in one magnet is thus
jd
- J
dzdydx
o U
-Tf
dl E
+
IT
(az+7 - ay+ Mi
h7r
ay+ -
d
-
ax+,l
dl
(IC12
e"
+
ay
)
+ |C3 2 (1
±az-7+
ay_)
e?" -
2
i + axY) e-t 2dy
-
e-(-y+y)T)
8rh m n
49
+
(C12
+ C30 4) (I
-
e(--y+)T)
(C201
+
403)
(i
-
(IC2|2 +C4
e(--)T)
12) (i
-y+y)r
where the constants Cihmf, whose subscripts have been omitted from the notation above,
are
m7r
Clhmn
=
az+7 - ay+ ,
m7r
C2hmn
-az--7 - ay _
hir
3.n
=
C4hmn
3.4
-y
y--
ax±_y
+ ax~
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
otherwise
0
(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
No. of pole pairs
No. of phases
Armature inner radius
Armature outer radius
Magnet inner radius
Symbol
p
q
Ri
Rao
Rmi
Magnet outer radius
Magnet conductivity
Rmo
Value
4
3
0.0673 m
0.0793 m
0.0806 m
0.0901 m
7 x 104 W/mOC
Magnet length
a
1
Magnet angle
6m
ir/6
Electrical angle
Ampere-turns
owe
0.856
5835 A
NI
0.1001 m
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
La
41
0.2
1
1
0.4
0.5
I
0.8
1
thetam -rad
I
1.2
Figure 3-5: Graph of eddy current loss vs magnet angle
52
1.4
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
=
strands is Npa,
1.037 A, where the number of turns N = 8 and the number of parallel
=
693. The resulting power loss per unit length of wire from winding
resistance is
PiR
-
N
o7rr2
53
0.544 Wm
1
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 df Conduction Loss with Speed at 30 kW
Variationof ConductionLosswithRatedPowerat 15000 rpmn
3500
3000
02500
5D
1000
1000DO
0
0.5
1
15
Spead. rpm
2
2.5
3
x
Rated Power W
10'
, 10
10
Figure 4-1: Graphs showing variation of conduction loss with rated power and speed
C in Figure 4-2. Accordingly,
c
B.ds
L
-d
2EL
E
=
-
d (2xLB,,sin wt)
dt
-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
21
o x 2 ( Bow cos wt) 2 2fr2
_ X2
10
= 4cr(Bow cos wt) 2 J(rW cos 6) 2 rW sin Orw(- sin )dO
4ur(Bowcoswt) 2r
- sidn
4L
2
55
B0 sir
wt
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
=
8
-B2W2 r
0
(4.1)
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
< B2 >
={
Rao
Rai
(2BIa)2
Rao,
2B1a
Rai
r
)P-1)2
Rao
d
1 -x(p
= 0.0237 T 2
(1 - z)(2p - 1)
2
1_X(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
Speed rprn
2
2.5
3
, 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.
cooling
channel
plena
armature
winding
Figure 4-4: Cross section of the machine showing major parts and indicating regions of
windage 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 as
Cf =
T
wR 4 lpQ2
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
P
where the density of air p
=
/2
go)
= 1330
(Rao)
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
1.2V2 7 (1 + 0.5go/Rai)
P
(
2 (1 - go/Rai)
858
For go/Rao << 1, which is the case here, this expression approximates to
pG~ango-0.136
Cf
= 0.00655
pQRaogo
=
0.001864
p
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 R 1 and that on a disk of radius
R 2 , where R 1 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 compared 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
Cmi =
0.051(gc/Rao)0 .1
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, -
Td 2
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--
0,
I 800BDD--
600400200
0
0
0.5
1
1.5
Speedirpm
2
_j
3
2.5
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). Calculations 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
Speedi rpm
2
2.5
3
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
Rated Power ' W
2
2.5
3
1a"
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 between 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
=
vDHP
P
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 5 0 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
i
le
AP= f
2DH
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 channel. 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, epoxyimpregnated 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
the copper. Thus,
dQ
dt
TW -- Te _
Rfilm + Rmateriai
_______________=
65
mc
dTe
dt
Q of
hot
water
bath
Lagging
styrofoam
insulation
test
materiaL
copper
cyindler
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
1
hA
Rfilm
where h is the film coefficient and A the surface area, while the layer of test material has
thermal resistance
Rmaterial
=
66
ln (r,/ri)
2,rlk
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/m 0 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 fiberglass 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 determined. 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 C
DH
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
(
+ 9 2 )(u1 + < O- >) -
12
( +±
(Ui
2
)(Or2+ < & >) - 42 (1(0 -
9 1(071 + 072)(022+ < Or >) u1+ < & >) (os +
2 )(
OrL
1)2
#2(02-
0r)2
#1 2 (o-2 - oTI)2#1
2(07
-
03)
2
and
#
1
and
#2
=
1-
#1
respectively. (1 and
< o > =
0-141 + U202
< Or >
Or201 + 91#2
=
are the volume fractions of the cylinders and insulating material
c2
model the cylinder-to-cylinder interactions. From [8], they are
found to be
2=
3
-0.0570742
The thermal conductivity 92 of copper wire is 390 W/m C, and that of the epoxy, or, is
1.8 W/m 0 C. For a copper volume fraction of 0.373, the effective conductivity is calculated
to lie between o-L = 1.54 W/m 0 C and o-u = 14.4 W/m 0 C. As a compromise, a conservative
68
estimate of conductivity is taken to be the geometric mean
4.3.4
oLoyU = 4.71 W/m 0 C.
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 winding from the water is
Rf
-0.9160CW-tf
kgwle
where tj = 0.41 mm is the thickness of this layer, and k1 = 0.31 W/m 0 C is the experimentally 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
Lass cLoth tape
outer Layer
wires and
epoxy
nd epoxy
x=d_
inner Layer
/x=d
epoxy
window
fioerglass
and epoxy
water
channet
Figure 4-9: Cross section of a phase belt with adjacent cooling channel
The water film at the fiberglass-epoxy surface has resistance
Rfilm
1
= .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 =
ti k
k
= 0.4680CW 1
(0.45kg + 0.55ke) wie
where the thermal conductivity of epoxy ke = 0.66 W/m 0 C and that of epoxy-impregnated
glass cloth tape, k9 = 0.31 W/m 0 C. The corresponding temperature drop is AT 1 =
P
x
R1 = 76.0 0 C.
Within the region occupied by the wires, the temperature distribution satisfies the dif70
ferential equation
d2 T
dx
2
<}
+
=0
kce
where
P
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
-
+ C 1 x + C2
2
2ke
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 is
dT
dx
Thus C1
=
-
P/48
kcewl
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 C 2 , which is equal to the temperature at x = 0:
T(x =0)=C
2
=T(x = d)+
q d2 -C
2ke
1
d
Then the temperature drop across the inner winding layer is
ATi =
q d 2 - Cid = 33.5 0 C
2kee
The epoxy impregnated glass cloth tape layer between the two wire layers has thermal
resistance
R2_
t2
-
0.759 CW-
where thickness t 2 = 0.34 mm and conductivity kg = 0.31 W/m 0 C. The temperature drop
71
across this layer is AT 2 =
P
x R2
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
-
2kce
30.8 0 C
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
d2 T
dx 2
+R
PiR
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,:
Tend
To+
1 PR
2 Ao-cu,
lend
2
2
2
+lxj
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 upper and lower bounds for conductivity of the epoxy-copper composite, and their geometric
mean.
Temperature drop across
Symbol
Water film
Epoxy/fiberglass wall
Layer with epoxy window
Inner winding layer
Epoxy/glass layer
Outer winding layer
End turn
Total
ATfilm
ATf
AT 1
AT
AT 2
ATo
kce = VU/oL
U
10.8 0 C
148.6 0 C
76.0 0 C
92.4 0 C
61.6 0 C
30.8 0 C
43.3 0 C
463.4 0 C
ATend
EAT
o-U
kce = o-L
10.8 0 C
148.6 0 C
76.0 0 C
30.3 0 C
61.6 0 C
10.1 C
43.3 0 C
380.6 0 C
10.8 0 C
148.6 0 C
76.0 0 C
281.8 0 C
61.6 0 C
93.9 0 C
43.3 0 C
715.9 0 C
kce
=
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 manufacture 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 consisted 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 removed, 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 supported 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 at
the 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 tape
can 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 phase
belt.
83
Figure 5-6: Mold assembly for potting stator winding in epoxy
84
Figure 5-7: Armature potted in epoxy
85
S
.
A
0
Figure 5-8: Parts of the machine pictured prior to assembly
86
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
20
200
500
1000
2000
5000
10000
20000
Ra/mQ
2.07
2.10
2.30
2.65
3.03
4.18
4.60
4.78
Rb/mQ
2.07
2.11
2.30
2.62
2.97
4.17
4.63
4.79
Re/mQ
2.07
2.11
2.30
2.63
2.99
4.16
4.58
4.79
89
La/lpH Lb /pH
2.10
2.10
2.16
2.14
2.39
2.37
2.36
2.33
2.08
2.09
2.28
2.28
2.22
2.21
2.21
2.20
Le/pH
2.10
2.10
2.33
2.35
2.06
2.27
2.22
2.21
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
dia
Re {Le"wt} for x
Writing v2
Va
=
Vb
=
=
dt
dia
dt
Lo'a d
a, b, and
Va
b
aRa+ La
sa =
Re {Laej't}, we have
=LaRa+jwLaLa
=
jWLbala
Eliminating La from the two equations gives
Vb =
jw Lba
-- a
Ra + jWLa
Knowing that Lba is real and negative, we can find Lba from
L|a
+ jwLa)
l_
i(Ra
jo |_Va l
where |b I and |_V Iare 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 oscilloscope, so readings were taken starting from a frequency of 5 kHz. Table 6.2 gives the measured 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
7r (d,/2)2
-
3.51 x 10-5 M 2 , and the electrical conductivity of copper o
90
=
= Npa, X
3.9 x 107
Table 6.2: Mutual Inductance Measurements
Frequency / kHz
V(driven phase) / V V(open-circuited phase) / V
Inductance / pH
Va
V
Lba
5
10
20
0.026
0.035
0.065
0.012
0.016
0.026
1.054
1.015
0.884
Vb
Va
Lab
5
10
20
0.026
0.035
0.070
0.012
0.016
0.026
1.054
1.011
0.817
Va
V
Lca
5
10
20
0.026
0.035
0.065
0.012
0.014
0.022
1.054
0.888
0.748
V
Va
Lac
5
10
20
0.026
0.038
0.065
0.012
0.016
0.024
1.049
0.935
0.816
Vb
V
Leb
5
10
20
0.026
0.035
0.070
V
0.026
0.038
0.065
0.012
0.016
0.028
1.054
1.011
0.948
5
10
20
91
V
Lbc
0.012
0.016
0.026
1.049
0.935
0.884
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, primarily 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 variation 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 independent 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
-J
do
dt
= Co + CiW + T
92
Figure 6-1: Test stand for spin-down tests. The top end of the machine with the coupling
for 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 values 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
-J
dw
= Co + CiW + Cw'?
dt
(6.1)
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 magnets 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
1.8
1.9
CO / Nm
C 1 /104 Nms
C 1 .8 /10- 7 Nms 1 .8
CO / Nm
C 1 /104 Nms
C1.9/10-7 Nms 1.9
Rotor with
aluminium ring
0.0149
0.0591
9.04
0.0131
0.303
4.08
Rotor with
magnets
0.0182
1.68
5.66
0.0172
1.83
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 C 1.8
=
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 experimental 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 calculations, 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 corresponding 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 guassmeter. 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
1.00
15.5
30.0
44.5
59.0
73.5
88.0
96.0
mean
Fundamental
0.1413
0.1413
0.1293
0.1296
0.1232
0.1232
0.1212
0.1211
0.1221
0.1221
0.1251
0.1248
0.1394
0.1391
0.1460
0.1460
0.1309
Magnetic flux density / T
3rd harmonic
0.0179
0.0184
0.0243
0.0244
0.0246
0.0246
0.0248
0.0247
0.0249
0.0248
0.0250
0.0243
0.0240
0.0233
0.0183
0.0182
0.0229
97
5th harmonic
0.0282
0.0280
0.0279
0.0279
0.0280
0.0279
0.0278
0.0280
0.0279
0.0279
0.0281
0.0285
0.0287
0.0291
0.0293
0.0293
0.0283
7th harmonic
0.0250
0.0251
0.0276
0.0278
0.0278
0.0279
0.0278
0.0278
0.0280
0.0281
0.0286
0.0281
0.0290
0.0288
0.0264
0.0267
0.0275
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 electromagnetic 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 manufacturing 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 models 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 current 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 resistances, 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 anticipated, 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 performance 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 concrete 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:
IRa
Rao f
atz
2J
-rdOdr = JOw (Rao - Rai) = NIa/p
r
2Ow
NIa
NIa
p 0 w (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
nir
KR,,
sin
:
f"lw
2
for n odd
: for n even
0
4JdR sin ("l-)
: for n odd
0
: for n even
[4] gives the fundamental component of radial magnetic field due to a surface current
shell at R:
dH
dHiS,
=
d~o~r-
:
KR 1 .
2
(o)P
sin p0
KR,
2
( p
sin pO :
for r < R
for ft
r
In the region of the windings, Rai < r < Rao, the fundamental component of total magnetic
field is then
Rao
Hr
= )Rair dHoS, + I
H
1
Er
= -- sin(pO)II
-
--
2s
2
7r
LORai
dHiSr
4JodR .i
rR
sin(pO)Jo sin
Owe
2 )
(Owe)
S2
12
i- 2
(-R)+1
r7
Rao
f
r
I-2p - (1
4JodR
,R
Rai P*+
P)(r )
.
Owe
r
2 \RJ
P -
+ (1 + p) (r)P
Rao
I
If the winding has a significantly large turns density, the number of turns in a differential
area element can be expressed as
d 2N =
N
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
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
N
Rao
A = p1
OW fRai
-2
(Rao
O
pioHrrd] drdt
- Rai)
0 e
P-
Total flux linked by one phase due to excitation of that same phase is
8lp Josi2
A =
Rai 2 (1+p)
(Ri)2
8Ngo sin 2p)
R 20 [1-p+
r(Rao - Rai) Owe (1 - p 2 )p aRao
12NPoI)(
2
-
2
Rai )~]
Rao)P1
1 - p + x2(1 + p) - 2x)+
7r
(1 -
Owe
x)2(i
- p2)p
The self inductance of each phase winding is thus
La -21
N2 pof
7r
sin
Ow)
p+
2 1 -
_
x2(1 + p)
- 2x+
(1- x)2(i - p 2)p
For a three-phase machine, the mutual inductance between phases is
2
Lw?
= 2lNgr
-
IN 2 po
r
Oin
.~2
w
/sin-2
1 - px
(1 -
2 (1 +p)-2x+ 1
x)(-
2)
)cos
2r
(1- p +x 2 (1±+p) -2xv+
(1 - x)2(1 _ p 2 )p
Ow)
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.
<
< H, >
=
Rao
a
Hdr
Ra
103
2
.(Owe\
- sin(pO)Jo sin --
=
2
IT
1
i-p
1
2
I
1
1 - xP
(1 - p)x(xP - 1) +
p1 - X)
I
Total flux linked by one phase due to excitation of that phase becomes
A
-
II
f
we
J-
< H, >
2
2
Rao+Rai
dd
P
4N 1pouJo (Rao + Rai)
2
sin 2
12
(2 ) 1-- p2
(we
irwe(1 - p )p
(1 - p)x(xP - 1) + (1 + p)(l - xP)
p(i - )
and the self inductance of each phase is
La
_
lpoN
2
(Rao + Rai)
)2
7L(Rao - Rai)
(1I - p)x(xP - 1) + (1 + p)(
p2 (i - x)
(1 - p2), I
I
-
xP)
-
2
The mutual inductance between phases for a three-phase machine is
Lab
- lyoN 2 (Rao + Rai)
27c (Rao - Rai)
wsin 2
2
1
(1- P2)p
104
(1-p)x(xP -1) + (1+p)(l
p 2(1 -X)
-
xP)
-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
[ pi/6];
thetamv
%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) -
a3u =
(1-cos(npthm))*(l/(npthm+tup) + 1/(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*(j*a2u+a4u); 0;
0;
0];
x = inv(A)*y;
axp
=
x(1,1);
axm
=
x(2,1);
ayp
=
x(3,1);
aym = x(4,1);
azp = x(5,1);
107
ko*(alu-j*a3u);
*
azm =
Cl
x(6,1);
azp*gam -
C2
=
-azm*gam -
C3
=
ayp*hpd -
ayp*mpl;
aym*mpl;
axp*gam;
aym*hpd + axm*gam;
C4
cgam= conj(gam);
power
=
nummags*d*l/(8*sigma)
(1-exp(-(gam+cgam)*T))
*
+
*
(abs(C4))^2)
((abs(C1))^2 + (abs(C3))^2)
(
/ (gam + cgam)
(gam -
(1-exp((-gam+cgam)*T)) /
C4*conj(C3))
*
cgam)
(1-exp((gam+cgam)*T))
total(ith) = total(ith) +
(C1*conj(C2) + C3*conj(C4))
+
(C2*conj(C1) +
/ (-gam + cgam)
(1-exp((gam-cgam)*T))
*
+
/ (-gam - cgam)
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
*
+
((abs(C2))^2
);
ayp = x(3,1);
aym = x(4,1);
azp = x(5,1);
azm = x(6,1);
C1 = azp*gam - ayp*mpl;
C2 = -azm*gam C3 = ayp*hpd -
aym*mpl;
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)
*
(1-exp((-gam+cgam)*T))
C4*conj(C3))
*
/
(gam - cgam)
+
(Cl*conj(C2) + C3*conj(C4))
+
(C2*conj(Cl)
(1-exp((gam-cgam)*T)) / (-gam + cgam)
+ (abs(C4))^2) *
*
(1-exp((gam+cgam)*T)) / (-gam -
+
+
((abs(C2))^2
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
Symbol
geom mean
turns
N
parallel strands
Npar
poles
p
4
phases
q
3
armature outer radius
Rao
0.0793
armature inner radius
Rai
0.0673
8
693
x=Rai/Rao
wire radius
r_w
Ampere turns
N Ia
current in one strand
Ia/Npar
elect. conductivity Cu
sigma
Power loss / unit length
P_l_R_15
0.84867591
0.000127
5749
1.0369769
39000000
111
0.54414702
P_l_R_30
0.13603675
Fundamental field at Rao
Bla
mean square magnetic field
<Bo^2>
Speed
rpms
Electrical freq 15000 rpm
omega
Eddy cur loss /unit length
P_1_ec_15 0.0037225675
0.0958
0.023667014
15000
6283.1853
P_1_ec_30
0.01489027
0.1001
active length
la
end turn length
lend
safety length
ls
0.0142
PR
3663.9744
total eddy cur loss(15)
P_ec
12.395131
windage loss
P_w
total i^2R loss
(15)
(15)
total loss
0.088123859
(15)
218 .4
3894.7695
P(15)
sigl+sig2
390.66
(s2-sl) ^2
151585.64
<sig>
145.92152
<sig
bar>
244.73848
E2
0.11642418
El
0.88357582
therm cond. fib+epoxy
k_f,k_g
0.31
thermal cond. epoxy
s igmal
0.66
thermal cond. Cu
sigma2
390
winding angle
thetawe
vol. fraction epoxy
phil
0.62690318
vol. fraction Cu
phi2
0.37309682
upper bound conductivity
sigmaU
14.363205
lower bound conductivity
sigmaL
1.5433015
geom mean cond. Cu+epoxy
k_ce
4.7081584
length of cooling channel
lc
0.1003
dist betw channel&end turn
lx
0.012
112
0.214
radial width of channel
T
0.00014
channel outer radius
Rco
0.06689
channel inner radius
Rci
0.06675
hydraulic diameter
DH
0.00028
mass flow rate
m
0.18590785
velocity of flow
v
3.165417
Reynolds no.
Re
Moody friction factor
f
pressure drop
deltaP
Nusselt no.
Nu
film coeff.
h-c
676.03642
0.094669456
169760.67
5
10446.429
upper bound
lower bound
4.7081584
14.363205
1.5433015
geom mean
conductivity of epoxy/Cu
k_ec
Width of a phase belt
w
0.0144
0.0144
0.0144
Winding layer thickness
d
0.00516
0.00516
0.00516
Resistance of water film
Rf ilm
0.066277899
0.066277899
0.066277899
Temp drop across film
dTfilm
10.755714
10.755714
10.755714
Thickness of fiberglass
tf
0.00041
0.00041
0.00041
Resistance of fiberglass
Rf
0.91571165
0.91571165
0.91571165
148.60357
148.60357
148.60357
0.00034
0.00034
0.00034
0.46846746
0.46846746
0.46846746
76.023865
76.023865
76.023865
10887481
10887481
10887481
-11932.351
-3911.3416
-36402.09
92.356399
30.273784
281.75218
0.75937063
0.75937063
0.75937063
Temp drop across fiberglass dTf
Thickness glass cloth tape
tg
Resistance of layer 1
R1
Temp drop across layer 1
dTl
Power density
q dot
Cl (i)
Temp drop inner layer
dTi
Resistance of layer 2
R2
Temp drop across layer 2
dT2
61.616116
61.616116
61.616116
Temp drop outer layer
dTo
30.785466
10.091261
93.917392
temp drop end turn
dTend
43.271347
43.271347
43.271347
Total temp drop
sum dT
463.41248
380.63566
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',
fprintf('tau = %9.6g\n',
cf)
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 )
cft = cfts(l)
if cfv > cft
114
=
rho*Omega*R-b*gb/mu',
-
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
rho*Omega*R-c2^2/mu
Re
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
Re
cml
=
0.109;
gd = 0.0063; asr = g_d/Rd,
rho*Omega*Rd^2/mu % -- > regime IV
=
=
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
)
'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
rho*Omega*R-e*g_e/mu',
-
rho*Omega*R-f^2/mu % -- > regime IV
Re
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',
trqSR =
tau-f)
tau +taub + tauc + tau_d
Omega = 15000*2*pi/60;
powerSR = trqSR*Omega
powerHR =
C.3
(tau-e+tau-f)*Omega
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;
thetawe = pi/3;
Rmi
0.07988; Rmo = 0.08941;
=
la=0.1016;
else
Rai = 0.0673;
Rao = Rai+0.012;
thetawe = 0.856;
Rmi
=
Rao+1.32e-3; Rmo = Rmi + 9.53e-3;
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;
kc2 = 1.85*
asr-c
=
gc/Rao;
(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
%rpm
linspace(10,30000,80);
=
=
[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 *
P_ec(i)
=
sigma * msqB *
(p*Omega)^2
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
*
rw^4;
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 ==
Rao =
0
0.07938; Rai = 0.06985;
thetawe = pi/3;
Rmi =
0.07988;
Rmo
0.08941;
la=0.1016;
else
Rai =
0.0673;
Rao =
Rai+0.012;
Rmi = Rao+1.32e-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
Rmo
Rmi
+ 9.53e-3;
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 temperature 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 +
c
=
RmateriaImc,
where m is the mass of the copper cylinder, and
393.6 J/kg0 C is the specific heat capacity of copper.
Copper
1
(0.02533 + 0.02533 + 0.02533 + 0.02533)
4
= 0.02533 m
1
Exposed length 11 = -(0.08023 + 0.08040 + 0.08029 + 0.08038)
4
= 0.08033 m
Diameter of cylinder di
Mass mi
-
=
0.470 kg
Rfilmmlc = mic/ (hirdil)
Li,
=
1
(27.1 + 28.0 + 25.6 + 26.9)
4
123
=
26.9
Film coefficient h
= 1075.7 Wm-2 oC-l
96908478720L
6660-
E
54-
(D
48423630time / S
time / s
CL)
E
0
20
40
Time/ s
60
80
Time/ s
Figure D-1: Plots of temperature of uncoated cylinder vs time
Epoxy
Diameter of cylinder d2
1
= -(0.03826 + 0.03810 + 0.03819 + 0.03817)
4
=
Thickness of epoxy layer t 2
=
0.03818 m
1
=- (d2- di)
2
6.425 x 10-3 m
1
-(0.08126 + 0.08190 + 0.08154 + 0.08168)
4
Exposed length 12
=
0.08160 m
124
Mass M 2 =
L2
=
0.466 kg
(Rfjim +
Rmaterial) m 2 C =
(hrd2i
S)n
(d2c)
27rl2k2
2
)mc
1
= -(242 + 237 + 241 + 235) = 238.8
4
Conductivity of epoxy ke = 0.663Wm-loClo
Cylinder with Epoxy layer, run 2
0
Time I s
100
200
300
400
Time s
Wyinder with epoxy layer,run 4
500
Time / s
Time / s
Figure D-2: Plots of temperature of cylinder with epoxy layer vs time
Epoxy-impregnated glass cloth tape
Diameter of cylinder d 3
Thickness of epoxy-tape layer t 3
1
= -(0.02679
+ 0.02681 + 0.02673 + 0.02653)
4
0.02672 m
1
2
125
600
= 6.925 x 10-4 m
1
Exposed length 13 = -(0.07976 + 0.07966 + 0.07993 + 0.08050)
4
= 0.07996 m
Mass m3
L3
-
0.468 kg
=
(Rfilm +
Rmateriai) m 3 c
=
1
-
+
-(88.1 + 88.9 + 89.0 + 91.1)
=
2rl3 k 3
)mc
89.28
4
Conductivity of epoxy-glass cloth tape composite kg = 0.306 Wm-
C -lo
Cylinder with epoxy-impregnated glass cloth tape,run 2
Cylinder with epoxy-impregnated glass cloth tape, run 1
C,
In (d3 /di)
E-
E
a)
HD
Time / s
Time / s
Cylinder with layer of epoxy-impregnated glass cloth tape,run 4
95-1
908580-
0
75-
6O
E.
706560555045-
4035-
//5
0
Time / s
O
60
120
180
1
240
2
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
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 reproduced here.
Figure E-1: Stator cooling jacket
129
C
0
z
0
0
t
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
1.8
1.9
Coefficient
CO / Nm
C 1 /10-4 Nms
C2 /10-7 Nms 1 8
CO / Nm
C1/10 4 Nms
C2/10-7 Nms 1 9
Run 1
0.0172
0.0206
9.226
0.0151
0.2842
4.136
Run 2
0.0151
0.0591
9.002
0.0137
0.2732
4.130
135
Run 3
0.0149
0.0122
9.116
0.0127
0.2790
4.074
Run 4
0.0125
0.1444
8.817
0.0110
0.3758
3.979
Mean
0.0149
0.0591
9.040
0.0131
0.3030
4.080
run 1
0
run 2
2
5,
0.30.2
025 -
(00
2
0.2-
0-15
0.15-
0.1
0.1
0.05
0.05200
0
200
400
000
8
Speed(rads)
1000
1200
140
400
800
1000
1200
Speed (rads)
run 3
run 4
0.35
0.350.3 -
0.3
0.2-
L0.2
VOW.,
0.15
0.15 01
0.1
005-
0.05
0
500
00
200
400
S00
800
Speed (rds)
1000
1200
0
1400
200
400
600
800
Speed (rads)
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
1.8
1.9
Coefficient
CO / Nm
C 1 /10-4 Nms
C2 /10- 7 Nms 1 8.
CO / Nm
C1/104 Nms
C2 /10-
7 Nms 1.9
Run 1
0.0180
0.1725
5.541
0.0172
Run 2
0.0178
0.1627
5.759
0.0168
Run 3
0.0192
0.1677
5.788
0.0182
Run 4
0.0177
0.1703
5.572
0.0168
Mean
0.0182
1.683
5.665
0.0172
1.868
1.777
1.831
1.848
1.831
2.506
2.599
2.606
2.515
2.557
136
run1
run2
run3
run4
D.4
1/7
0.35
03
SD0.25
0.2
0. 15
0.1
0.05
1400
0
200
Speed(rads
4D
8
BOD d
Speed(rac~s)
1000
1200
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
Phase A
Phase B
Phase C
Run 1 /Vs
0.0150
0.0147
0.0145
Run 2/Vs
0.0149
0.0147
0.0145
137
Mean/Vs
0.0150
0.0147
0.0145
1400
Phase A, run 1
Phase A run 2
9
8
7
~1
>0
~r
m
m
3
2
K
1
0
200
400
B00
Ebtatona
800
speed 'rads
1000
1200
1400
Eotatonal
Phase B. run 1
speed -rads
Phase B, run2
ca
fotatond
speed.: rads
Phase C run 1
Phase C. run2
9
9-
8
8-
7
7-
>0
>0
a3
4
Go 4-
3
3-
2
21 lD0
200
400
600
8DD
Fbtatonal speed rads
/
1000
1200
1400
0
200
400
6DD
800
Ftationd speed rads
Figure F-3: Plots of Back Emf versus rotor speed
138
1000
1200
140D
Bibliography
[1] Regis Roche, "Magnet Losses in a Flywheel Energy Storage System". Internal report
at MIT, 1997.
[2] James L. Kirtley Jr., Mary Tolikas, Jeffrey H. Lang, Chee We Ng, Regis Roche, "Rotor Loss Models for High Speed PM Motor-Generators". Presented to the International Conference on Electric Machines, Istanbul, Sept. 2-4, 1998.
[3] James L. Kirtley Jr., "Notes for 6.1 Is: Design of Electric Motors, Generators and
Drive Systems", Massachusetts Institute of Technology, Cambridge, MA, 1997.
[4] James L. Kirtley Jr., "Design and Construction of an Armature for an Alternator
with a Superconducting Field Winding", Ph.D Thesis (Electrical Engineering), Massachusetts 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 Associated
Power Electronics for Direct-Drive Vehicle Propulsion", Ph.D Thesis (Electrical Engineering), Massachusetts Institute of Technology, Cambridge, MA, 1996.
[8] S. Torquato and F. Lado, "Bounds on the Conductivity of a Random Array of Cylinders", Proceedings of the Royal Society of London, Series A (Mathematical and Physical Sciences), Vol. 417, No. 1852, May 1998.
139