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MIT6_685F13_chapter5

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Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.685 Electric MachineryClass Notes 5: Winding Inductances
c 2005 James L. Kirtley Jr.
1 Introduction
The purpose of this document is to show how the inductances of windings in round- rotor machineswith narrow air gaps may be calculated. We deal only with the idealized air- gap magnetic ﬁelds,and do not consider slot, end winding, peripheral or skew reactances. We do, however, considerthe space harmonics of winding magneto-motive force (MMF).
2 Description of Stators
Back IronSlotsTeethSlotDepression
Figure 1: Stator Cross-SectionFigure 1 shows a cartoon view of an axial cross-section of a twelve-slot stator. Actually, whatis shown is the shape of a thin sheet of steel, or
lamination
that is used to make up the magneticcircuit. The iron is made of thin sheets to control eddy current losses. Thickness varies accordingto freuqency of operation, but in machines for 60 Hz (the vast bulk of machines made for industrial1
use), lamination thickness is typically .014” (.355 mm). These are stacked to make the magneticcircuit of the appropriate length. Windings are carried in the slots of this structure.Figure 1 shows trapezoidal slots with teeth of approximately uniform cross-section over most of their length but wider extent near the air-gap. The tooth ends, in combination with the relativelynarrow slot depression region, help control certain parasitic losses in the rotor of many machinesby improving uniformity of the air-gap ﬁelds, increase the air-gap permeance and help hold thewindings in the slots. It should be noted that large machines, with what are called “form wound”coils, have straight-sided rectangular slots and consequently teeth of non-uniform cross-section.The description that follows will hold for both types of machine.
1 2 3 4 5 6 7 8 9 10 11 12A A A’ A’B B B’ B’C’ C’ C C
Figure 2: Full-Pitched WindingTo simplify the discussion, imagine the slot/tooth region to be “straightened out” as shown inFigure 2. This shows a three-phase, two-pole winding in the twelve slots. Such a winding wouldhave two slots per pole per phase. One of the two coils of phase A would be wound in slots 1 and 7(six slots apart).
1 2 3 4 5 6 7 8 9 10 11 12A A C’ C’ B B A’ A’ C C B’ B’A C’ C’ B B A’ A’ C C B’ B’ A
Figure 3: Five-Sixths-Pitched WindingMachines are seldom wound as shown in Figure 2 for a variety of reasons. It is usually advanta-geous in reducing the length of the end turns and to reducing space harmonic eﬀects in the machine(usually bad eﬀects!) to wind the machine with “short-pitched” windings as shown in Figure 3.Each phase in this case consists of four coils (two per slot). The four coils of Phase A would spanbetween slots 1 and 6, slots 2 and 7, slots 7 and 12 and slots 8 and 1. Each of these coil spans isﬁve slots, so this choice of winding pattern is referred to as “Five-Sixths” pitch.So this cartoon-ﬁgure machine stator (which could represent either a synchronous or inductionmotor or generator) has both
breadth
because there are more than one slots per pole per phase,and it may have the need for accounting for winding
pitch
. What follows in this note is a simpleprotocol for estimating the important air-gap ﬁelds and inductances.2
3 Winding MMF
To start, consider the MMF of a full- pitch, concentrated winding as shown in schematic form inFigure 4. Assuming that the winding has a total of
N
turns over
p
pole- pairs, and is carryingcurrent
I
the MMF is:
∞
F
=
−
4
NI
sin
npθ
(1)
nπ
2
pn
= 1
nodd
This distribution is shown, as a function of angle
θ
in Figure 5.This leads directly to magnetic ﬂux density in the air- gap:
∞
µ
0
4
NI B
r
=
−
sin
npθ
(2)
g nπ
2
pn
= 1
nodd
Note that a real winding, which will most likely not be full- pitched and concentrated, will have a
winding factor
which is the product of pitch and breadth factors, to be discussed later.
µµ
NIpRgMagneticCircuit: Stator RotorAir-Gaprz
θ
Figure 4: Primitive Geometry ProblemNow, suppose that there is a polyphase winding, consisting of more than one phase (we will usethree phases), driven with one of two types of current. The ﬁrst of these is
balanced
, current:
I
a
=
I
cos(
ωt
)2
πI
b
=
I
cos(
ωt
−
)32
πI
c
=
I
cos(
ωt
+ ) (3)3Conversely, we might consider
Zero Sequence
currents:
I
a
=
I
b
=
I
c
=
I
cos
ωt
(4)3
π
2 p
π
p
π
32 p
π
p2F( )
θθ
NIp
Figure 5: Air-Gap MMFThen it is possible to express magnetic ﬂux density for the two distinct cases. For the
balanced
case:
∞
B
r
=
B
rn
sin(
npθ
∓
ωt
) (5)
n
=1
where
ã
The upper sign holds for
n
= 1
,
7
,...
ã
The lower sign holds for
n
= 5
,
11
,...
ã
all other terms are zeroand3
µ
0
4
NI B
rn
= (6)2
g nπ
2
p
The zero- sequence case is simpler: it is nonzero only for the
triplen
harmonics:
∞
B
r
=
µ
0
4
NI
3(sin(
npθ
−
ωt
) + sin(
npθ
+
ωt
)) (7)
g nπ
2
p
2
n
=3
,
9
,...
Next, consider the ﬂux from a winding on the rotor: that will have the same form as the ﬂuxproduced by a single armature winding, but will be referred to the rotor position:
∞
4
B
=
µ
0
NI
sin
npθ
′
rf
(8)
g nπ
2
pn
= 1
nodd
which is, substituting
θ
′
=
θ
−
ωt p
,
∞
B
rf
=
µ
0
4
NI
sin
n
(
pθ
−
ωt
) (9)
g nπ
2
pn
= 1
nodd
The next step here is to ﬁnd the ﬂux linked if we have some air- gap ﬂux density of the form:
∞
B
r
=
n
B
rn
sin(
npθ
±
ωt
) (10)
=1
4

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