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ECE 321
Exam 3 SolutionSpring 2006
Problem 1
The winding function expresses how many turns of a particular phase are linking the flux at a
specific angle.
Problem 2
P6
=
W
1
1
2
j 1to
N
aslt
P
⋅
=
N
asj
∑
⋅
=
1
2
j 1to2
⋅
=
N
asj
∑
⋅
=
1
2
55
+
()
⋅
=
5
=
W
2
W
1
N
1
−
=
0
=
W
3
W
2
N
2
−
=
5
−
=
W
4
W
3
N
3
−
=
0
=
etc.
W
as
50 5
−
050 5
−
−
0
T
=
Problem 3
P8
=
Note: Check the integral equation below.
Some got full credit since this answer can
essentially be arrived at by inspection, even though the equation they used might have had
errors in it.
w
as
1
2
0
2
π
P
φ
sm
n
as
⌠
⎮
⎮
⌡
d
⋅
0
φ
sm
φ
sm
n
as
⌠
⎮
⌡
d
−
=
25
−
sin 4
φ
sm
25
6
sin 12
φ
sm
⋅
+
=
Problem 4
ω
rm
2
P
ω
r
⋅
=
Machines with more poles rotate more slowly (under the same excitation) than otherwise
similar machines with fewer poles (and vice versa).
Problem 5
Fw
as
i
as
⋅
w
bs
i
bs
⋅
+
=
50 cos 4
φ
sm
cos 30t
(
)
50 sin 4
φ
sm
sin 30t
+
=
F
50 cos 4
φ
sm
30t
−
=
The maximum MMF occurs at
4
φ
sm
30t
−
0
=
Therefore, the speed is
t
φ
sm
d
dt
7.5t
d
d
=
7.5
=
rad
s
⎛
⎝
⎞
⎠
CCW (since positive)
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View Full Document Problem 6
The maximum of the MMF is at
4
φ
sm
100t
−
0
=
φ
sm
100
t
4
=
100
0.1ms
4
⋅
=
0.0025
=
rad
()
B
µ
0
F
g
⋅
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This note was uploaded on 02/19/2012 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Staff
 Flux

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