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116
( ) ( )
re
76.
A threephase 60Hz twopole induction motor runs at a noload speed of 3580 r/min and a fullload
speed of 3440 r/min.
Calculate the slip and the electrical frequency of the rotor at noload and fullload
conditions.
What is the speed regulation of this motor [Equation (457)]?
S
OLUTION
The synchronous speed of this machine is 3600 r/min.
The slip and electrical frequency at no
load conditions is
%
56
.
0
%
100
3600
3580
3600
%
100
sync
nl
sync
nl
=
×
−
=
×
−
=
n
n
n
s
( ) ( )
,nl
0.0056 60 Hz
0.33 Hz
fs
f
==
=
The slip and electrical frequency at full load conditions is
%
44
.
4
%
100
3600
3440
3600
%
100
sync
nl
sync
fl
=
×
−
=
×
−
=
n
n
n
s
( ) ( )
,fl
0.0444 60 Hz
2.67 Hz
f
=
The speed regulation is
%
1
.
4
%
100
3440
3440
3580
%
100
SR
fl
fl
nl
=
×
−
=
×
−
=
n
n
n
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()
M
M
X
X
j
R
jX
R
jX
Z
+
+
+
=
1
1
1
1
TH
(
)
[]
M
M
M
M
X
X
j
R
X
X
j
R
X
X
j
R
jX
R
jX
Z
+
−
+
+
+
−
+
=
1
1
1
1
1
1
1
1
TH
[ ] [ ]
2
1
2
1
2
1
2
1
2
1
2
1
1
1
1
1
TH
M
M
M
M
M
M
M
X
X
R
X
X
X
X
X
R
j
X
R
X
X
R
X
X
R
Z
+
+
+
+
+
+
+
−
=
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
TH
TH
TH
M
M
M
M
M
M
X
X
R
X
X
X
X
X
R
j
X
X
R
X
R
jX
R
Z
+
+
+
+
+
+
+
=
+
=
The Thevenin resistance is
2
1
2
1
2
1
TH
M
M
X
X
R
X
R
R
+
+
=
.
If
1
R
X
M
>>
, then
2
1
2
1
2
1
M
M
X
X
X
X
R
+
≈
+
+
, so
2
1
1
TH
¸
¸
¹
·
¨
¨
©
§
+
≈
M
M
X
X
X
R
R
The Thevenin reactance is
2
1
2
1
2
1
2
1
2
1
TH
M
M
M
M
X
X
R
X
X
X
X
X
R
X
+
+
+
+
=
.
If
1
R
X
M
>>
and
1
X
X
M
>>
then
M
M
M
X
X
X
R
X
X
2
1
2
1
2
1
+
>>
and
2
1
2
2
1
R
X
X
X
M
M
>>
≈
+
, so
1
2
2
1
TH
X
X
X
X
X
M
M
=
≈
713.
Figure P71 shows a simple circuit consisting of a voltage source, two resistors, and two reactances in
series with each other.
If the resistor
R
L
is allowed to vary but all the other components are constant, at
what value of
R
L
will the maximum possible power be supplied to it?
Prove
your answer. (
Hint:
Derive
an expression for load power in terms of
V
,
R
S
,
X
S
,
R
L
and
X
L
and take the partial derivative of that
expression with respect to
R
L
.)
Use this result to derive the expression for the pullout torque [Equation
(752)].
S
OLUTION
The current flowing in this circuit is given by the equation
L
L
S
S
L
jX
R
jX
R
+
+
+
=
V
I
127
()
(
)
2
2
L
S
L
S
L
X
X
R
R
V
I
+
+
+
=
The power supplied to the load is
(
)
2
2
2
2
L
S
L
S
L
L
L
X
X
R
R
R
V
R
I
P
+
+
+
=
=
(
)
(
)
22
2
2
SL
S
L
L
L
S
L
RR
X
X
VV
R
P
R
X
X
ªº
ª
º
++
+
−
+
∂
¬
¼
¬¼
=
∂
+
To find the point of maximum power supplied to the load, set
L
R
P
∂
∂
/
= 0 and solve for
L
R
.
(
)
20
S
L
L
X
X
R
+
−
+=
(
)
2
S
L
L
R
RX
X
R
R
R
+
=
+
2
2
2
SS
L
LS
L
S
L
L
R
RR R
X X
R
+
+
=+
2
2
2
L
R
X
R
+
=
2
LL
R
XX
R
=
Therefore, for maximum power transfer, the load resistor should be
2
2
L
S
S
L
X
X
R
R
+
+
=
714.
A 440V 50Hz sixpole Yconnected induction motor is rated at 75 kW.
The equivalent circuit
parameters are
R
1
= 0.082
Ω
R
2
= 0.070
Ω
X
M
= 7.2
Ω
X
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This note was uploaded on 11/10/2011 for the course EEL 3216 taught by Professor Brooks during the Fall '08 term at FSU.
 Fall '08
 BROOKS
 Frequency

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