Power Analysis and Efficiency Calculations for Three-Phase Induction Motors
THREE-PHASE INDUCTION
MOTORS
An
Approximate
Equivalent
Circuit
Power
Relations
For
3-Phase
IM
2
E
X
A
M
P
L
E
1
-
2
A
480V,
60Hz, 50hp,
Y-connected,
3 phase induction motor is
drawing
60A
at
0.85
PF
lagging.
The stator
copper losses are
2kW,
and
the rotor
copper losses are
700W.
The
friction and
windage
losses are
600W,
the core
losses are
1800W,
and
the stray
losses are negligible. Using
the
power
flow diagram in Figure 1-6a,
find:
a)
The
air gap
power
P
ag
.
b)
The
power
converted
P
d
.
c)
The
output
power
P
o
.
d)
The
efficiency
of
the
motor.
S
O
L
U
T
I
O
N
a)
The
phase voltage
V
1
=
V
L
/
3
=
480/
3
= 277
V
b)
c)
d)
P
in
P
ag
P
scl
P
in
P
m
42400
2000
1800
38600
W
3
V
1
I
1
cos
3
277
60
0
.
85
42400
W
or
42.4kW
o
r
38
.
6
k
W
P
d
P
ag
P
rcl
38600
700
37900
W
or
37.9kW
P
0
P
d
P
r
P
d
P
fw
P
st
37900
600
0
37300
W
o
r
37
.3k
W
42400
P
in
%
P
o
100
37300
100
88
%
Power
Relations
For
3-Phase
IM
3
E
X
A
M
P
L
E
1
-
3
A
6-pole,
230-V,
60-Hz,
Y-connected,
three-phase induction motor has
the
following
parameters on a per-phase
basis:
R
1
=
0
.
5
,
R
2
=
0
.
2
5
,
X
1
=
0
.
7
5
,
X
2
=
0
.
5
,
X
m
=
1
0
0
,
a
n
d
R
c
=
5
0
0
.
The
friction and
windage
loss is 150
W.
Determine
the
efficiency
of
the
motor at its
rated slip
of
2.5%.
S
O
L
U
T
I
O
N
The
synchronous
speed
of the motor
is
The
effective
rotor
impedance as referred
to
the stator
is
s
P
6
N
120
f
120
60
1200
rpm
or
s
125
.
66
rad /
s
The
per-phase applied voltage is
1
3
3
V
V
L
230
132
.
8
V
2
2
R
2
0.25
Z
j
X
j
0.5
10
j
0.5
s
0.025
Power
Relations
For
3-Phase
IM
4
S
O
L
U
T
I
O
N
(
1
-
3
)
c
o
n
t
.
Since
R
c
,
jX
m
,
and
are in parallel,
we
can
compute
the
equivalent impedance
as:
The stator
winding
impedance
is
Hence,
the
total input impedance
is
The stator
current:
The
power
factor:
Power
input:
R
c
j
X
m
1
1
1
Z
ˆ
e
1
1
1
1
Z
ˆ
2
0
.
102
j
0
.
015
S
500
j
100
10
j
0
.
5
Z
e
9.619
j
1.417
and
1
Z
R
j
X
0.5
j
0.75
1
1
Z
e
i
n
1
Z
Z
10.119
j
2.167
0
V
132
.
8
0
0
12
.
83
2
12
.
1
A
I
1
ˆ
1
10
.
119
j
2
.
167
Z
in
0
v
i
pf
cos(
)
cos(
)
cos(12.1
)
0.978
lagging
P
in
3
V
1
I
1
cos
3
132
.
8
12
.
832
0
.
978
4998
.
54
W
Power
Relations
For
3-Phase
IM
5
The
Rotor copper
loss:
Power
developed:
E
f
f
i
c
i
e
n
c
y
:
S
O
L
U
T
I
O
N
(
1
-
3
)
c
o
n
t
.
Stator
copper
loss:
P
3
I
2
R
3
12
.
832
2
0
.
5
246
.
99
W
scl
1
1
The
induced
emf
in
the
stator
winding
is
E
V
I
Z
132.8
(12.832
12.1
0
)(0.5
j
0.75)
124.763
3.71
0
V
1
1
1
1
Hence,
the
Core loss
is
1
c
P
m
3
E
2
/
R
3(124.76)
2
/ 500
93.75
W
The
Air-gap
power:
P
ag
P
scl
P
in
P
m
4657.8
W
P
rcl
3
I
2
R
or
P
sP
0.25
4657.8
116.46
W
2
2
rcl
ag
P
rcl
P
d
P
ag
4541.34
W
Power
output:
P
0
P
d
P
r
4541.34
150
4391.34
W
P
in
%
P
0
100
87.9%
An
Approximate
Equivalent
Circuit
21
•
A
w
e
l
l
-
d
e
s
i
g
n
e
d
t
h
r
e
e
-
p
h
a
s
e
i
n
d
u
c
t
i
o
n
m
o
t
o
r
u
s
u
a
l
l
y
m
e
e
t
s
m
o
s
t
o
f
t
h
e
f
o
l
l
o
w
i
n
g
g
u
i
d
e
l
i
n
e
s
:
(
r
e
f
e
r
t
o
t
h
e
p
e
r
-
p
h
a
s
e
e
q
u
i
v
a
l
e
n
t
c
i
r
c
u
i
t
s
h
o
w
n
i
n
F
i
g
u
r
e
1
-
4
)
1.
The
stator winding resistance
is kept small in order to reduce the
stator
copper
loss.
2.
The
stator winding leakage reactance
is minimized by reducing the
mean-
turn length of each
coil.
3.
Thin laminations of low-loss steel are used to cut down the core loss.
Thus,
the equivalent
core-loss resistance
is usually
high.
4.
The permeability of steel selected for laminations is high, and the
operating
flux density in
the
motor is kept below
the
knee of the magnetization curve.
Thus, the
magnetization reactance
is usually
high.
Figure
1-4
An
Approximate
Equivalent
Circuit
7
–
An induction motor conforming to the above stipulations can be represented
by an
approximate equivalent circuit
, as shown in Figure 1-7, where we
have
placed the parallel branch (the excitation circuit)
across
the power
source.
–
We
admit that the analysis of an induction motor using the approximate
equivalent circuit is somewhat inaccurate, but the inaccuracy is negligible
for
a well-designed
motor.
–
On the other hand, the approximate equivalent circuit not only simplifies the
analysis but also aids in comprehending various characteristics of the
motor.
–
For instance, we
use
the approximate equivalent circuit to determine
the
speed at which
(a)
the torque developed is maximum
,
(b)
the power
developed is maximum
, and
(c)
the motor
efficiency
is
maximum
.
Figure 1-7
An
approximate
equivalent
circuit
on
a per-phase
basis
of
a
balanced three-phase induction
motor.
An
Approximate
Equivalent
Circuit
8
E
X
A
M
P
L
E
1
-
4
A
6-pole,
230-V,
60-Hz,
Y-connected,
three-phase induction motor has
the
following
parameters on a per-phase
basis:
R
1
=
0
.
5
,
R
2
=
0
.
2
5
,
X
1
=
0
.
7
5
,
X
2
=
0
.
5
,
X
m
=
1
0
0
,
a
n
d
R
c
=
5
0
0
.
T
h
e
f
r
i
c
t
i
o
n
a
n
d
w
i
n
d
a
g
e
l
o
s
s
i
s
1
5
0
W
.
D
e
t
e
r
m
i
n
e
t
h
e
e
f
f
i
c
i
e
n
c
y
o
f
t
h
e
m
o
t
o
r
a
t
i
t
s
r
a
t
e
d
s
l
i
p
o
f
2
.
5
%
.
U
s
i
n
g
t
h
e
a
p
p
r
o
x
i
m
a
t
e
e
q
u
i
v
a
l
e
n
t
c
i
r
c
u
i
t
s
h
o
w
n
i
n
F
i
g
u
r
e
1
-
6
S
O
L
U
T
I
O
N
The
synchronous
speed
of the motor
is
The
effective
rotor
impedance as referred
to
the stator
is
s
P
6
N
120
f
120
60
1200
rpm
or
s
125
.
66
rad /
s
The
per-phase applied voltage is
1
3
3
V
V
L
230
132
.
8
V
Figure
1-7
2
0.025
2
e
R
Z
ˆ
R
j
X
jX
0.5
j
0.75
0.25
1
1
s
j
0.5
10.5
j
1.25
An
Approximate
Equivalent
Circuit
9
S
O
L
U
T
I
O
N
(
1
-
4
)
c
o
n
t
.
Hence,
the rotor
current:
Core-loss
current:
Magnetization current is
The stator
current:
Power
input:
2
e
V
1
132.8
I
12.558
6.79
0
A
Z
10.5
j
1.25
c
c
V
1
I
132.8
0.266
A
R
500
m
m
V
1
I
132.8
j
1.328
A
j
X
j
100
I
1
I
2
I
c
I
m
12.558
6.79
0.266
j
1.328
13.043
12.45
A
0
0
The
power
factor:
0
v
i
pf
cos(
)
cos(
)
cos(12.45
)
0.977
lagging
P
in
3
V
1
I
1
cos
3
132
.
8
13
.
043
0
.
977
5074
W
Compare
to
4998.57
W
Figure
1-7
An
Approximate
Equivalent
Circuit
10
The
Rotor
copper
loss:
The
Air-gap
power:
P
ag
P
scl
P
in
P
m
4731.3
W
P
rcl
4731.3
118.3
4613
W
P
0
P
d
P
r
4613
150
4463
W
E
f
f
i
c
i
e
n
c
y
:
S
O
L
U
T
I
O
N
(
1
-
4
)
c
o
n
t
.
Stator
copper
loss:
P
3
I
2
R
3
12
.
558
2
0
.
5
236
.
6
W
scl
2
1
Hence,
the
Core loss
is
m
c
c
P
3
I
2
R
3
0.266
2
500
106.1
W
2
2
r
c
l
Power
developed:
P
d
P
ag
Power
output:
P
3
I
2
R
3
12.558
2
0.25
118.3
W
5074
P
in
%
P
0
100
4463
100
87.96%
Compare to
4657.8
W
Compare to
87.9%
Figure
1-7
An
Approximate
Equivalent
Circuit
11
E
X
A
M
P
L
E
1
-
5
A
460V,
25hp, 60Hz, 4 pole,
Y-connected
induction motor has
the
following
impedances
in ohms per phase referred
to the
stator
circuit:
R
1
=
0
.
6
4
1
,
R
2
=
0
.
3
3
2
,
X
1
=
1
.
1
0
6
,
X
2
=
0
.
4
6
4
,
a
n
d
X
m
=
2
6
.
3
.
The total
rotational losses are
1100W
and are assumed
to
be constant.
The
core loss is
lumped in
with
the
rotational
losses.
For
a
rotor
slip
of
2.2%
at the
rated voltage and
rated
frequency,
using
the
approximate equivalent circuit in Figure 1-8, find
the
motor’s
a)
stator
current, b)
power factor,
c)
P
d
and P
o
, d)
the
efficiency,
and e)
T
d
and
T
o
.
S
O
L
U
T
I
O
N
The
synchronous
speed
of the motor
is
a)
The
per-phase applied voltage
is
The
effective
rotor
impedance as referred
to
the stator
is
s
N
120
f
120
60
1800
rpm
or
P
4
s
188
.
5
rad /
s
1
3
3
V
V
L
460
265
.
6
V
2
0.022
2
e
1
R
Z
ˆ
R
j
X
1
s
jX
0.641
j
1.106
0.332
j
0.464
15.732
j
1.57
Figure
1-8
A
modified
approximate
equivalent
circuit.
An
Approximate
Equivalent
Circuit
12
S
O
L
U
T
I
O
N
(
1
-
5
)
c
o
n
t
.
Hence,
the rotor
current:
Magnetization current is
o
r
2
e
V
1
265.6
I
16.8
5.7
0
A
Z
15.732
j
1.57
m
The stator
current:
m
V
1
I
265.6
j
10.1
10.1
90
0
A
j
X
j
26.3
I
1
I
2
b)
The
power
factor:
I
m
16.8
5.7
j
10.1
20.44
35.14
A
0
0
pf
cos(
)
cos(
)
cos(35.14
0
)
0.8177
lagging
1
v
1
i
1
c)
Power
input:
P
in
3
V
1
I
1
cos
1
3
265
.
6
20
.
44
cos
35
.
14
13318
W
P
in
3
V
1
I
2
cos(
v
1
i
2
)
3
V
1
I
2
cos
2
3
265
.
6
16
.
8
cos
5
.
7
13320
W
Figure
1-8
An
Approximate
Equivalent
Circuit
13
S
O
L
U
T
I
O
N
(
1
-
5
)
c
o
n
t
.
The
Air-gap
power:
Power
developed:
Shaft (load)
torque:
a
g
2
P
3
I
2
R
2
3
16.8
2
0.332
12777.8
W
s
0.022
P
d
1
s
P
ag
1
0.022
12777.8
12496.7
W
d)
Power
output:
P
0
P
d
P
r
12496.7
1100
11396.7
W
13318
P
in
E
f
f
i
c
i
e
n
c
y
:
%
P
0
100
11396.7
100
85.57%
Figure
1-6(b)
T
P
d
12496.7
67.8
N.m
or
P
ag
12777.8
T
67.8
N.m
d
d
m
184.35
s
188.50
1
8
4
.
35
T
P
o
11396.7
61.8
N.m
o
m
e) Developed
torque:
m
1
s
s
1
.
02
2
18
8
.
5
18
4
.
35
rad
/
s
Three-Phase
Induction
Motors
Problems
14
•
P
o
w
e
r
R
e
l
a
t
i
o
n
s
:
7.
A
10-hp, 4-pole,
440-V,
60-Hz,
Y-connected,
three-phase induction motor runs at
1725
rpm
on full load.
The stator
copper loss is 212
W,
and
the
rotational loss is
340
W.
Determine (a) the
power
developed,
(b) the airgap
power,
(c) the
rotor
copper loss, (d) the
total
power
input, and (e) the
efficiency
of the
motor.
What
is
the
shaft
torque?
8.
A
2-hp,
120-V,
60-Hz, 4-pole,
Y-connected,
three-phase induction motor operates
at
1650
rpm
on full load.
The
rotor impedance
at
standstill is 0.02
+
j0.06
/phase.
Determine
the
rotor current
if the
rotational loss
is
160
W.
What
is
the magnitude of
the induced
emf
in
the
rotor?
9.
A
three-phase induction motor operates
at
a slip
of
3% and has a
rotor
copper loss
of
300
W.
The
rotational loss is 1500
W.
Determine (a)
the
air gap
power
and (b)
the
power
output.
If the rotor
impedance is
0.2 +
j0.8
/phase,
what
is
the
magnitude
of the
induced
emf
per phase in
the
rotor?
10.
A
10-hp, 6-pole,
440-V,
60-Hz,
-connected, three-phase induction motor is
designed
to
operate
at
3% slip
on
full load.
The
rotational loss is 4% of
the
power
output.
When the motor
operates
at
full load, determine (a) the
rotor
copper
loss,
(b) the air-gap
power,
(c) the
power
developed, and (d)
the
shaft
torque.
Three-Phase
Induction
Motors
Problems
15
•
D
e
v
e
l
o
p
m
e
n
t
o
f
a
n
E
q
u
i
v
a
l
e
n
t
C
i
r
c
u
i
t
:
11.
A
208-V,
50-Hz, 12-pole,
Y-connected,
three-phase induction motor has a stator
impedance
of 0.1 +
j0.3
/phase and a rotor impedance
of
0.06
+
j0.8
/phase
at
standstill.
The
core-loss resistance
is
150
/phase,
and
the magnetization reactance
is 750
/phase.
The
friction and
windage
loss is 2
kW.
When
the motor
operates
at its
full-load slip of 5%, determine (a)
the
power
input, (b)
the
stator copper loss,
(c) the
rotor copper loss, (d) the
air-gap
power,
(e)
the
power
developed,
(f) the
power
output,
(g)
the
efficiency,
(h) the shaft torque, and
(i)
the
horsepower
rating
of
the
motor.
12.
The
per-phase equivalent circuit parameters
of
a
208-V,
4-pole, 50-Hz, three-phase,
Y-connected,
induction motor are R
1
= 0.4
,
X
1
= 0.8
, R
2
= 0.3
,
X
2
= 0.9
, and
X
m
=
40
.
The
core loss is 45
W,
and
the
friction and
windage
loss is 160
W.
When
the
motor
operates
at
a slip
of
5%, determine (a)
the
input current, (b)
the
power
input,
(c) the
air-gap
power,
(d)
the
power
developed, (e)
the
power
output,
( f )
the
shaft torque, and
(9)
the
efficiency
of
the
motor.
Draw
its
power-flow
diagram.
13.
Calculate
the
starting torque developed by
the motor of
Problem
12.
Three-Phase
Induction
Motors
Problems
16
A.
A
4-pole,
230-V,
60-Hz,
Y-connected,
three-phase induction motor has
the
following
parameters on a per-phase basis: R
1
=
10.12
,
X
1
=
38.61
, R
2
=
21.97
,
X
2
=
11.56
, and
X
m
=
432.48
.
The
core loss is 10.72
W,
and
the
friction and
windage
loss is
5.9
W.
When
the motor
operates
at its
rated speed
of
1550
rpm,
determine
(a)
the
stator
current, (b)
the
magnetization current,
(c) the rotor current,
(d) the
power
input, (e) the
stator
copper loss,
( f ) the rotor
copper loss, (g) the
power
output, (h)
the shaft
torque, and
(i)
the
efficiency.
•
A
n
A
p
p
r
o
x
i
m
a
t
e
E
q
u
i
v
a
l
e
n
t
C
i
r
c
u
i
t
:
15.
Redo Problem 12 using
the
approximate
equivalent
circuit. Also, solve Problem
13.
16.
The
equivalent circuit parameters
of
a
208-V,
60-Hz, 6-pole,
Y-connected,
three-
phase induction
motor
in ohms/phase are
R
1
=
0.21, R
2
=
0.33,
X
1
=
0.6,
X
2
=
0.6,
R
c
=
210, and
X
m
=
450. When
the motor
runs
at
a slip
of
5% on full load, determine
the torque developed by
the motor
using
the
approximate equivalent circuit.
What
is
the
starting torque developed
by the
motor?
The provided content discusses the power relations and efficiency calculations for two examples of three-phase induction motors. It covers topics such as air gap power, power converted, output power, stator copper losses, rotor copper losses, core losses, stray losses, and efficiency calculations. Detailed solutions and relevant formulas are provided for a better understanding.
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
E N D
Presentation Transcript
THREE-PHASE INDUCTION MOTORS An Approximate Equivalent Circuit
Power Relations For 3-Phase IM EXAMPLE 1-2 A 480V, 60Hz, 50hp, Y-connected, 3 phase induction motor is drawing 60A at 0.85 PF lagging. The stator copper losses are 2kW, and the rotor copper losses are 700W. The friction and windage losses are 600W, the core losses are 1800W, and the stray losses are negligible. Using the power flow diagram in Figure 1-6a, find: a) The air gap power Pag. b) The power converted Pd. c) The output power Po. d) The efficiency of the motor. SOLUTION a) The phase voltage V1 = VL/ 3 = 480/ 3 = 277V Pin Pag Pscl = Pin Pm=42400 2000 1800=38600W = 3V1I1cos = 3 277 60 0.85 = 42400W or 42.4kW or 38.6kW Pd = Pag Prcl = 38600 700 = 37900W P0 = Pd Pr= Pd Pfw Pst= 37900 600 0 =37300W % =Po 100 =37300 100 =88% or 37.9kW b) c) or 37.3kW d) 42400 Pin 2
Power Relations For 3-Phase IM EXAMPLE 1-3 A 6-pole, 230-V, 60-Hz, Y-connected, three-phase induction motor has the following parameters on a per-phase basis: R1 = 0.5 , R2 = 0.25 , X1 = 0.75 , X2 = 0.5 , Xm = 100 , and Rc = 500 . The friction and windage loss is 150 W. Determine the efficiency of the motor at its rated slip of 2.5%. SOLUTION The synchronous speed of the motor is N =120 f=120 60 = 1200 rpm or s 6 P s= 125.66 rad / s The per-phase applied voltage is V =VL=230 = 132.8V 1 3 3 The effective rotor impedance as referred to the stator is =0.25 0.025 =R2 + jX + j 0.5 = 10 + j 0.5 Z 2 2 s 3
Power Relations For 3-Phase IM SOLUTION (1-3) cont. Since Rc, jXm, and are in parallel, we can compute the equivalent impedance as: 1=1 Z 2 1 1 + + Z e Rc 1 500 j Xm 1 1 Z e= 9.619 + j 1.417 and = + + = 0.102 j 0.015 S 10 + j0.5 j100 The stator winding impedance is Z = R + jX = 0.5 + j 0.75 1 1 1 Hence, the total input impedance is Z = Z + Z = 10.119 + j 2.167 1 in e The stator current: 132.8 00 V Zin I1 = 1=10.119 + j2.167 =12.832 12.1 A 0 The power factor: pf = cos( ) = cos( ) = cos(12.1 ) = 0.978 lagging 0 v i Power input: Pin= 3V1 I1 cos = 3 132.8 12.832 0.978 = 4998.54 W 4
Power Relations For 3-Phase IM SOLUTION (1-3) cont. Stator copper loss: P = 3I 2 R = 3 12.8322 0.5 =246.99 W scl 1 1 The induced emf in the stator winding is E = V I Z =132.8 (12.832 12.10)(0.5+ j0.75)=124.763 3.710V 1 1 1 1 Hence, the Core loss is Pm = 3E2/ R = 3(124.76)2 / 500 = 93.75W The Air-gappower: Pag Pscl 1 c = Pin Pm = 4657.8 W The Rotor copper loss: Prcl= 3 I2R P = sP rcl = 0.25 4657.8 = 116.46W or 2 2 ag Power developed: Prcl Pd = Pag = 4541.34 W Power output: P0 = Pd Pr= 4541.34 150 = 4391.34 W Efficiency: % =P0 100 =87.9% Pin 5
An Approximate Equivalent Circuit A well-designed three-phase induction motor usually meets most ofthe following guidelines: (refer to the per-phase equivalent circuit shown in Figure 1-4) 1.The stator winding resistance is kept small in order to reduce thestator copper loss. 2.The stator winding leakage reactance is minimized by reducing the mean- turn length of each coil. 3.Thin laminations of low-loss steel are used to cut down the core loss.Thus, the equivalent core-loss resistance is usually high. 4.The permeability of steel selected for laminations is high, and the operating flux density in the motor is kept below the knee of the magnetization curve. Thus, the magnetization reactance is usually high. Figure1-4 21
An Approximate Equivalent Circuit An induction motor conforming to the above stipulations can be represented by an approximate equivalent circuit, as shown in Figure 1-7, where wehave placed the parallel branch (the excitation circuit) across the powersource. We admit that the analysis of an induction motor using the approximate equivalent circuit is somewhat inaccurate, but the inaccuracy is negligible for a well-designed motor. On the other hand, the approximate equivalent circuit not only simplifies the analysis but also aids in comprehending various characteristics of the motor. For instance, we use the approximate equivalent circuit to determinethe speed at which (a) the torque developed is maximum, (b) the power developed is maximum, and (c) the motor efficiency ismaximum. Figure 1-7 An approximate equivalent circuit on a per-phase basis of a balanced three-phase induction motor. 7
An Approximate Equivalent Circuit EXAMPLE 1-4 A 6-pole, 230-V, 60-Hz, Y-connected, three-phase induction motor has the following parameters on a per-phase basis: R1 = 0.5 , R2 = 0.25 , X1 = 0.75 , X2 = 0.5 , Xm = 100 , and Rc = 500 . The friction and windage loss is 150 W. Determine the efficiency of the motor at its rated slip of 2.5%. Using the approximate equivalent circuit shown in Figure 1-6 SOLUTION The synchronous speed of the motor is N =120 f=120 60 = 1200 rpm or s 6 P s= 125.66 rad / s The per-phase applied voltage is V =VL=230 = 132.8V Figure1-7 1 3 3 The effective rotor impedance as referred to the stator is ) s R + jX =(0.5 + j0.75) + 0.25 + j0.5 = 10.5 + j1.25 Z = (R + jX + 2 2 0.025 1 1 e 8
An Approximate Equivalent Circuit SOLUTION (1-4) cont. Hence, the rotor current: 132.8 =V1 = =12.558 6.790A I 2 10.5 +j1.25 Z e Core-loss current: Figure1-7 =V1 =132.8=0.266 A 500 I c R Magnetization current is c V1 jX =132.8 = j1.328 A j100 = I m The stator current: m I1 =I2 + Ic + Im =12.558 6.79 +0.266 j1.328 =13.043 12.45 0 A 0 The power factor: pf = cos( ) = cos( ) = cos(12.45 ) = 0.977lagging 0 v i Power input: Pin= 3V1 I1 cos = 3 132.8 13.043 0.977 = 5074 W Compare to 4998.57W 9
An Approximate Equivalent Circuit SOLUTION (1-4) cont. Stator copper loss: P = 3I2R = 3 12.5582 0.5= 236.6 W scl 2 1 Hence, the Core loss is P = 3I2R = 3 0.2662 500 = 106.1 W Figure1-7 m c c The Air-gappower: = Pin Pm = 4731.3 W Pscl Pag Compare to 4657.8W The Rotor copper loss: = 3I2R = 3 12.5582 0.25 =118.3 W P rcl 2 2 Power developed: Prcl=4731.3 118.3 =4613 W Pd = Pag Power output: P0 = Pd Pr= 4613 150 = 4463 W Efficiency: % =P0 100 =4463 100 =87.96% Compare to 87.9% 5074 Pin 10
An Approximate Equivalent Circuit EXAMPLE 1-5 A 460V, 25hp, 60Hz, 4 pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: R1 = 0.641 , R2 = 0.332 , X1 = 1.106 , X2 = 0.464 , and Xm = 26.3 . The total rotational losses are 1100W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2% at the rated voltage and rated frequency, using the approximate equivalent circuit in Figure 1-8, find the motor s a) stator current, b) power factor, c) Pd and Po, d) the efficiency, and e) Td andTo. SOLUTION The synchronous speed of the motor is N =120 f=120 60 = 1800 rpm or P 4 s= 188.5 rad / s s a) The per-phase applied voltage is V =VL=460 = 265.6V Figure 1-8 A modified approximate equivalent circuit. 1 3 3 The effective rotor impedance as referred to the stator is ) s R + jX =(0.641 + j1.106) + 0.332 + j0.464 = 15.732 + j1.57 Z = (R + jX + 2 2 0.022 1 1 e 11
An Approximate Equivalent Circuit SOLUTION (1-5) cont. Hence, the rotor current: 265.6 =V1 =16.8 5.70A = I 2 15.732 + j1.57 Z e Magnetization current is Figure1-8 V1 jX =265.6 = j10.1=10.1 900A j26.3 = I m m The stator current: Im =16.8 5.7 j10.1 = 20.44 35.14 0 A I1 = I2+ 0 b) The power factor: pf = cos( ) = cos( ) = cos(35.140 ) = 0.8177 lagging 1 v1 i1 c) Power input: Pin= 3V1I1cos 1= 3 265.6 20.44 cos(35.14) =13318 W or Pin= 3V1I2cos( v1 i2)= 3V1I2cos 2= 3 265.6 16.8 cos(5.7)=13320 W 12
An Approximate Equivalent Circuit SOLUTION (1-5) cont. The Air-gappower: P = 3 I 2 R2 = 3 16.82 0.332 = 12777.8 W s 2 ag 0.022 Power developed: Pd= (1 s)Pag=(1 0.022) 12777.8 =12496.7 W d) Power output: P0 = Pd Pr= 12496.7 1100 = 11396.7 W Figure1-6(b) Efficiency: % =P0 100 =11396.7 100 = 85.57% 13318 Pin e) Developed torque: m =(1 s) s = (1 .022) 188.5=184.35 rad / s Pag s 12777.8 =Pd =12496.7 = 67.8 N.m m 184.35 T = d = = 67.8 N.m T or d 188.50 Shaft (load) torque: T =Po=11396.7 = 61.8 N.m o m 184.35 13
Three-Phase Induction Motors Problems Power Relations: 7. A 10-hp, 4-pole, 440-V, 60-Hz, Y-connected, three-phase induction motor runs at 1725 rpm on full load. The stator copper loss is 212 W, and the rotational loss is 340 W. Determine (a) the power developed, (b) the airgap power, (c) the rotor copper loss, (d) the total power input, and (e) the efficiency of the motor. What is the shaft torque? 8. A 2-hp, 120-V, 60-Hz, 4-pole, Y-connected, three-phase induction motor operates at 1650 rpm on full load. The rotor impedance at standstill is 0.02 + j0.06 /phase. Determine the rotor current if the rotational loss is 160 W. What is the magnitude of the induced emf in the rotor? 9. A three-phase induction motor operates at a slip of 3% and has a rotor copper loss of 300 W. The rotational loss is 1500 W. Determine (a) the air gap power and (b) the power output. If the rotor impedance is 0.2 + j0.8 /phase, what is the magnitude of the induced emf per phase in the rotor? 10.A 10-hp, 6-pole, 440-V, 60-Hz, -connected, three-phase induction motor is designed to operate at 3% slip on full load. The rotational loss is 4% of the power output. When the motor operates at full load, determine (a) the rotor copper loss, (b) the air-gap power, (c) the power developed, and (d) the shaft torque. 14
Three-Phase Induction Motors Problems Development of an Equivalent Circuit: 11.A 208-V, 50-Hz, 12-pole, Y-connected, three-phase induction motor has a stator impedance of 0.1 + j0.3 /phase and a rotor impedance of 0.06 + j0.8 /phase at standstill. The core-loss resistance is 150 /phase, and the magnetization reactance is 750 /phase. The friction and windage loss is 2 kW. When the motor operates at its full-load slip of 5%, determine (a) the power input, (b) the stator copper loss, (c) the rotor copper loss, (d) the air-gap power, (e) the power developed, (f) the power output, (g) the efficiency, (h) the shaft torque, and (i) the horsepower rating of the motor. 12.The per-phase equivalent circuit parameters of a 208-V, 4-pole, 50-Hz, three-phase, Y-connected, induction motor are R1 = 0.4 , X1 = 0.8 , R2 = 0.3 , X2 = 0.9 , and Xm = 40 . The core loss is 45 W, and the friction and windage loss is 160 W. When the motor operates at a slip of 5%, determine (a) the input current, (b) the power input, (c) the air-gap power, (d) the power developed, (e) the power output, ( f ) the shaft torque, and (9) the efficiency of the motor. Draw its power-flow diagram. 13.Calculate the starting torque developed by the motor of Problem 12. 15
Three-Phase Induction Motors Problems A. A 4-pole, 230-V, 60-Hz, Y-connected, three-phase induction motor has the following parameters on a per-phase basis: R1 = 10.12 , X1 = 38.61 , R2 = 21.97 , X2= 11.56 , and Xm = 432.48 . The core loss is 10.72 W, and the friction and windage loss is 5.9 W. When the motor operates at its rated speed of 1550 rpm, determine (a)the stator current, (b) the magnetization current, (c) the rotor current, (d) the power input, (e) the stator copper loss, ( f ) the rotor copper loss, (g) the power output, (h) the shaft torque, and (i) the efficiency. An Approximate Equivalent Circuit: 15.Redo Problem 12 using the approximate equivalent circuit. Also, solve Problem 13. 16.The equivalent circuit parameters of a 208-V, 60-Hz, 6-pole, Y-connected, three- phase induction motor in ohms/phase are R1 = 0.21, R2 = 0.33, X1 = 0.6, X2 = 0.6, Rc = 210, and Xm = 450. When the motor runs at a slip of 5% on full load, determine the torque developed by the motor using the approximate equivalent circuit. What is the starting torque developed by the motor? 16
