The End Effect in Short Secondary Linear Induction Motors
Prof. James L. Kirtley Jr., Lcdr Andrew P. Johnson
Abstract
One of the reasons linear motors, a
technology nearly a century old, have not
been adopted for a large number of linear
motion applications is that they have
historically had poor efficiencies. This is
due to the open-ended geometry of a linear
motor, and has restricted the progress of
linear motor development [1], [2], [3], [4].
The end effect is a direct consequence of
this open-ended geometry. This paper will
address the issue of whether or not the end
effect phenomenon of a short secondary
Linear Induction Motor (LIM) could be
responsible for any reduction in average
thrust and consequently, a reduction in
motor efficiency.
Introduction
A short secondary Double Sided LIM
(DSLIM) is shown in Figure 1.
Secondary
This primary field will induce poles in the
shuttle (Figure 2) that will oppose the
motion of the field according to Lenz’ Law.
Figure 2 Induced Shuttle Poles [5]
Thus the induced poles will create a
reactionary force in the shuttle, and hence
create motion.
The issue here is that the currents induced in
the shuttle have no simple return path. In
order to satisfy conservation of charge, all of
the induced current loops must sum to
exactly zero.
DSLIM End Effect Model
Primary
A mathematical relation was formulated
based on an elemental model of the DSLIM
(see Figure 3).
Figure 1 Short Secondary DSLIM
In motors of this type, the primary coils are
generally segmented to try to reduce total
and end leakage flux. However, it is
unavoidable that the flux immediately ahead
and behind the shuttle will be lost to useful
work because it fails to couple with the
shuttle. As the shuttle moves forward, it
will continually cover and uncover new
sections of track, while the primary applied
magnetic field wave will sweep over the
secondary from back to front at slip speed.
y
x
z
Figure 3 DSLIM Elemental Model
Ampere’s Law in integral form was used
around the integration path to derive the
governing differential equation. The first
order equation assumes a perfectly
sinusoidal stator surface current, no flux
fringing, and magnetic field variations in
only the x direction (Equation 1).
Equation 1 Ampere's Law Relation
1 g dB y
⋅ ⋅
2 μ dx
Ks + Kr
Given the fact that the stator surface current
density is assumed to be perfectly
sinusoidal, it can further be assumed that the
y-directed magnetic field will also be
perfectly sinusoidal and be of the form:
Equation 2 Magnetic Field
By
B o⋅ e
π
− j⋅ ⋅ V s − V t
τ
(
)
It is also known that the rotor current can be
calculated using:
Equation 3 Rotor Current Relation
dE z
dB y
dx
dt
of integration. This model solved for the
magnetic field around the shuttle by
assuming the shuttle was not there. Since
Ampere’s Law requires that the total sum of
current through the surface is accounted for
in the magnetic field formulation, and since
the total sum of the shuttle current over the
entire surface of the shuttle must be exactly
zero, the magnetic field must be a function
of the stator current only. This model was
then used to solve for the magnetic field at
both ends of the shuttle. These solutions
were then equated to the solutions just inside
the ends of the shuttle, and the unknown
constants of integration were solved for.
The airgap field then becomes a sinusoidal
function with two spatial exponential decays
at either end of the shuttle. Figure 4 shows
the magnetic field over a shuttle that is 2
pole pitches long (tau = pole pitch).
When all of these various relations are put
together, the final relation for the magnetic
field becomes second order (Equation 4).
Equation 4 Full Magnetic Field Relation
2
d Bo
2
dx
π
⎡
⎤
j⋅ ⋅ x
⎢ π
⎥
π
τ
⋅ ⎢ j⋅ ⋅ K ⋅ e
− j⋅ σ r⋅ ⋅ ( V s − V) ⋅ B o⎥
g ⎣ τ o
τ
⎦
2⋅ μ
This is known as the airgap field equation.
The solution to this equation will not be
presented in its entirety. In order to solve for
the unknown constants of integration of the
homogeneous solution, boundary conditions
at the ends of the shuttle had to be
established. Laithwaite suggests that the
magnetic field must be a continuous
function across the shuttle-airgap boundary
[3], [6]. Implicit in this statement is that the
shuttle current must sum to zero across the
full length of the shuttle, which is precisely
the physical condition that must exist in
order to preserve conservation of charge.
Thus, a second model had to be formulated
in order to solve for the unknown constants
Figure 4 Magnetic Field Strength
DSLIM Average Force
Using Poynting’s Theorem, the timeaverage value of the thrust of the motor was
calculated by taking one-half the integral of
the real part of the complex conjugate
product of the rotor surface current times the
magnetic field over the entire length of the
rotor (see Equation 5) [1].
Equation 5 Average Force Calculation
a
F
⌠
Χ
⎮ .5⋅ Kr⋅ B dx
⌡
0
Χ
B = conjugate of B
This force relation was used to calculate the
total force on a shuttle of a DSLIM with the
following operating parameters.
Table 1 Sample DSLIM Parameters
DSLIM Motor Parameters
Pole Pitch
0.385
Airgap
9
Shuttle Thickness
2
Primary Stack Depth
1.25
Primary Linear Current Density
190,000
m
cm
cm
m
A/m
The force that this DSLIM could create was
calculated using the above model with end
effect corrections and without end effect
corrections for a DSLIM with a shuttle that
was only 2 pole pitches long. Figure 5
shows the difference in the thrust-slip
profiles with the thrust results normalized.
The shuttle length was then increased to 5
pole pitches in length and the simulation
was re-run (Figure 6). This creates much
better tracking between the two models with
only a 10% difference in peak thrust
between the two.
Code was then generated to determine the
percentage difference between the two
models as a function of shuttle length.
Shuttle length was varied from 1 to 25 pole
pitches in length. The peak average force
was calculated for each shuttle length for
each of the models, and a comparison was
made. This is presented in Figure 7.
Figure 7 Thrust versus Shuttle Length
Figure 5 Thrust-Slip at shuttle = 2*tau
The end effect causes a reduction in peak
thrust while increasing thrust in higher slip
regions. Thus, this has the same effect as an
increase in secondary resistance [3], [6].
Figure 7 shows that once the shuttle gets to
be at least 8 pole pitches long, the end effect
only manages to have a 5% effect on the
total thrust of the DSLIM (as compared to
no end effect at all). As the shuttle length
increases to encompass an infinite number
of poles, obviously the end effect will go
away entirely. For the purposes of
designing and building a DSLIM though,
once one makes a shuttle that is at least 8
pole pitches long, most of the lost thrust that
was due to the end effect has been
recovered.
Efficiency
Figure 6 Thrust-Slip at shuttle = 5*tau
The assumption that will be used here is that
this DSLIM is operating under constant flux
in a field-oriented control scheme.
Therefore, a constant amount of current is
being supplied to the stator to maintain the
field. In order to produce thrust, a second
current is injected to the machine in
quadrature with the field-producing current.
It is this current that is used to produce the
power of the machine.
Recognizing that the end effect reduces the
overall thrust of the machine by about 5%,
in order to compensate, an additional 5% of
current must be injected in phase with the
applied voltage, or in quadrature with the
field-producing current. Assuming the two
currents are exactly equal prior to the
compensation, this 5% increase results in a
net 3.6% increase in the total stator current.
This produces a net increase of
approximately 7% in the total resistive
losses of the motor.
Conclusion
While the end effect is a deleterious effect in
short secondary DSLIM, it has been shown
that, provided the secondary (shuttle) of the
motor is at least 8 pole pitches in length, the
total thrust reduction is only about 5%.
Thus, modeling of the DSLIM should only
require that a correction factor of
approximately .95 be used in the thrust
calculation of the transient motor model.
With that .95 thrust correction factor comes
the requirement to boost thrust by increasing
current. This increase in current, at a
minimum, increases resistive losses in the
machine. These resistive losses could
account for a 7% or more loss in efficiency
depending on the relative magnitudes of the
field producing and thrust producing
currents. Thus, the total reduction in
efficiency would be notable, but perhaps not
large compared to other loss mechanisms
such as flux leakage [1], [3], [4].
References
[1] Sakae Yamamura, Theory of Linear
Induction Motors. John Wiley and
Sons, 1972.
[2] E. R. Laithwaite, Transport Without
Wheels, Elek Science, 1977.
[3] E. R. Laithwaite, Induction Machines
for Special Purposes, George Newnes
Limited, 1966.
[4] Ion Boldea and S.A. Nasar, Linear
Motion Electromagnetic Devices, Taylor
and Francis, 2001.
[5] M. A. Plonus, Applied
Electromagnetics, McGraw Hill, 1978.
[6] James L. Kirtley (private
communication), 2004.
Acknowledgements
Dr. James L. Kirtley Jr is a Professor of
Electrical Engineering at MIT and advised
LCDR Johnson on his thesis. Dr Kirtley has
a wealth of knowledge related to linear
motor design, and was indispensable in the
writing of this paper.
LCDR Andrew Johnson, USN obtained
his B.S. in Electrical Engineering from
SUNY Buffalo in 1994. He is a former
nuclear trained surface warfare officer who
transferred to the EDO Community. He
graduated in 2005 from MIT with a Naval
Engineers degree and SM in Electrical
Engineering. His thesis topic was the design
of a linear induction motor for aircraft
launch purposes. He currently works as an
Assistant Project Officer for the Supervisor
of Shipbuilding, Newport News, VA.