CHAPTER 17
Solutions for Exercises
E17.1
From Equation 17.5, we have
)
240
cos(
)
(
)
120
cos(
)
(
)
cos(
)
(
gap
°
°

+

+
=
θ
θ
θ
t
Ki
t
Ki
t
Ki
B
c
b
a
Using the expressionsgiven in the Exercise statement for the currents, we have
)
240
cos(
)
120
cos(
)
120
cos(
)
240
cos(
)
cos(
)
cos(
gap
°
°
°
°


+


+
=
θ
ϖ
θ
ϖ
θ
ϖ
t
KI
t
KI
t
KI
B
m
m
m
Then using the identity for the products of cosines, we obtain
)]
360
cos(
)
120
cos(
)
360
cos(
)
120
cos(
)
cos(
)
[cos(
2
1
gap
°
°
°
°

+
+
+

+

+
+


+
+
+

=
θ
ϖ
θ
ϖ
θ
ϖ
θ
ϖ
θ
ϖ
θ
ϖ
t
t
t
t
t
t
KI
B
m
However we can write
0
)
120
cos(
)
120
cos(
)
cos(
=
+

+


+

°
°
θ
ω
θ
ω
θ
ω
t
t
t
)
cos(
)
360
cos(
θ
ω
θ
ω
+
=

+
°
t
t
)
cos(
)
360
cos(
θ
ω
θ
ω
+
=

+
°
t
t
Thus we have
)
cos(
2
3
gap
θ
ω
+
=
t
KI
B
m
which can be recognized as flux pattern that rotates clockwise.
E17.2
At 60 Hz, synchronous speed for a fourpole machine is:
(
29
rpm
1800
4
60
120
120
=
=
=
P
f
n
s
The slip is given by:
%
778
.
2
1800
1750
1800
=

=

=
s
m
s
n
n
n
s
The frequency of the rotor currents is the slip frequency.
From Equation 17.17, we have
ω
ω
s
=
slip
.
For frequencies in the Hz, this becomes:
1
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Hz
667
.
1
60
02778
.
0
slip
=
×
=
=
sf
f
In the normal range of operation, slip is approximately proportional to output power and torque.
Thus at half power,
we estimate that
%
389
.
1
2
2.778
=
=
s
.
This correspondsto a speed
of 1775 rpm.
E17.3
Following the solution to Example 17.1, we have:
rpm
1800
=
s
n
02
.
0
1800
1764
1800
=

=

=
s
m
s
n
n
n
s
The per phase equivalent circuit is:
(
29
8
.
0
4
.
29
6
.
0
50
8
.
0
4
.
29
6
.
0
50
2
2
.
1
j
j
j
j
j
Z
s
+
+
+
+
+
+
+
=
15.51
+
22.75
j
=
29
.
34
53
.
27
∠
=
(
29
lagging
%
62
.
82
34.29
cos
factor
power
=
=
29
.
34
98
.
15
29
.
34
53
.
27
0
440

∠
=
∠
∠
=
=
s
s
s
Z
V
I
A rms
For a deltaconnected machine, the magnitude of the line current is
rms
A
68
.
27
3
98
.
15
3
=
=
=
s
line
I
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 Spring '08
 Briggs
 Electrical Engineering, Cos, Electric motor, Bgap

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