Lecture 4
RELUCTANCE MACHINES
The inductances in the 2x2 inductance matrix are:
)]
(
2
cos
[
x
d
B
A
xx
L
L
L
θ
θ
−
+
=
)]
(
2
cos
[
y
d
B
A
yy
L
L
L
θ
θ
−
+
=
)]
2
cos(
)
(cos(
[
y
x
d
B
y
x
A
yx
xy
L
L
L
L
θ
θ
θ
θ
θ
−
−
+
−
=
=
Single-Phase Reluctance Machine
Consider the x-winding as the a-phase with
θ
x
=0. Then,
]
2
cos
[
d
B
A
a
L
L
L
θ
+
=
. If the a-
phase current is
i
a
, the magnetic co-energy is:
W
m
1
=0.5L
a
i
a
2
.
The torque is
2
5
.
0
a
d
a
e
i
L
T
θ
∂
∂
=
which is
(1)
2
]
2
sin
2
[
5
.
0
a
d
B
e
i
L
T
θ
−
=
How does one make this machine produce a non-zero time invariant torque, so that as a
motor it can drive a constant load torque
T
L
? Newton’s Law of Motion governing the
rotational motion is:
L
e
m
T
T
dt
d
J
−
=
ω
(2)
where
J
is the moment of inertia of the shaft system and
T
L
is the load torque.
So far, most of the class notes are on analysis. Answering the above question requires
synthesis, which is an art. Trial and error is applied most of the time. Often, we use
results from known experience.
For example, one can propose using the ac mains where
cos(
)
a
i
I
t
ω
δ
=
+
and assume
that
d
t
θ
ω
=
. Substituting in (1)
2
0.5[ 2
sin(2
)][0.5
(1
cos(2
2 )]
e
B
T
L
t
I
t
ω
ω
δ
=
−
+
+
Therefore
2
0.5
[sin(2
)
0.5sin(4
2 )
0.5sin 2 ]
e
B
T
I L
t
t
ω
ω
δ
δ
= −
+
+
+
(3)
Substituting (3) in (2)

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*