1. Induction Motor
•Why induction motor (IM)?
–Robust; No brushes. No contacts on rotor shaft
–High Power/Weight ratio compared to Dc motor
–Lower Cost/Power
–Easy to manufacture
–Almost maintenance-free, except for bearing and
other mechanical parts
•Disadvantages
–Essentially a “fixed-speed” machine
–Speed is determined by the supply frequency
–To vary its speed need a variable frequency
supply
6. Example 5.1 A 3-phase, 460 V, 100 hp, 60 Hz, four pole induction machine
‑
delivers rated output power at a slip of 0.05. Determine the:
(a) Synchronous speed and motor speed.
(b) Speed of the rotating air gap field.
(c) Frequency of the rotor circuit.
(d) Slip rpm.
(e) Speed of the rotor field relative to the
(i) rotor structure.
(ii) Stator structure.
(iii) Stator rotating field.
(f) Rotor induced voltage at the operating speed, if the stator to rotor turns ratio is
‑ ‑
1 : 0.5.
Solution
:
ns=
120f
p
=
120∗60
4
=1800 rpm
n =(1− s ) ns =(1 − 0 . 05 )∗1800 =1710 rpm
(b) 1800 (same as synchronous speed)
12. V th=
Xm
√R1
2
+( X1+ Xm )
2
V 1 If R1
2
<< (X1+ X m)
2
Vth≈
Xm
X1+Xm
V1=Kth V1 Z th=
jX m (R1 + jX 1 )
R1+ j( X1+ X m)
=Rth+ jX th
If , then,
R1
2
<< ( X1+ X m)
2
Rth≃
(
Xm
X1 +Xm
)
2
R1=K th
2
R1
X th≈ X 1
13. Tests To Determine The Equivalent Circuit
No load test
‑
R1= R LL /2
V1=
VLL
√3
V /Phase ZNL=
V1
I1
RNL=
PNL
3I1
2
X NL =√Z NL
2
− R NL
2
X 1 + X m= X NL
15. X BL≃ X1 +X 2
'
X 1 + X m= X NL X m= X NL − X 1
Blocked rotor equivalent circuit for improved value for
‑ R2
16. 2
R=
X m
2
R2
¿
( X2
'
+ X m )
2
R2
'
R2
'
=
(X 2
'
+ X m
Xm
)
2
R R≃
(
X m
X2
'
+ Xm
)
2
R2
'
R = R BL− R1
17. Example 4.2 A no load test conducted on a 30 hp, 835 r/min, 440 V, 3 phase, 60 Hz
‑ ‑
squirrel cage induction motor yielded the following results:
‑
No load voltage (line to line): 440 V
‑ ‑ ‑
No load current: 14 A
‑
No load power: 1470 W
‑
Resistance measured between two terminals: 0.5
The locked rotor test, conducted at reduced volt
age, gave the following results:
‑
Locked rotor voltage (line to line): 163 V
‑ ‑ ‑
Locked rotor power: 7200 W
‑
Locked rotor current: 60 A
‑
Determine the equivalent circuit of the motor.
Solution
:
Assuming the stator windings are connected in way, the resistance per phase is:
R1=0 . 5/2=0 . 25 Ω
From the no load test:
‑
V 1=
V LL
√3
=
440
√3
=254 V / Phase
ZNL=
V1
I1
=
254
14
=18.143 Ω
18. RNL=
PNL
3I1
2
=
1470
3∗14
2
=2 .5 Ω
X NL =√Z NL
2
− R NL
2
= √18 . 1432
−2 . 52
=17 . 97
X 1 + X m= X NL=17 . 97 Ω
RBL=
PBL
3I1
2
∣BL
=
7200
3∗60
2
=0.6667 Ω
From the blocked rotor test
‑
The blocked rotor reactance is
‑
:
XBL=√(ZBL
2
− RBL
2
)=√1.56852
−0.66672
=1.42 Ω
X BL≃ X1 +X 2
'
=1.42 Ω
19. ∴ X1 = X2
'
=0.71 Ω
X m= X NL − X 1= 17 . 97−0 .71=17 . 26 Ω
R = R BL− R1=0 . 6667−0 . 25=0 . 4167 Ω
∴ R2
'
=
(X2
'
+ X m
X m
)
2
R=(0 .71+17 .26
17 .26 )
2
∗0 .4167=0 .4517 Ω
20. • Example 5.3 The following test results are obtained from a three-phase 60
hp, 2200 V, six pole, 60 Hz squirrel cage induction motor.
‑ ‑
• (1) No load test:
‑
• Supply frequency = 60 Hz, Line voltage = 2200 V
• Line current = 4.5 A, Input power = 1600 W
• (2) Blocked rotor test:
‑
• Frequency = 15 Hz, Line voltage = 270 V
• Line current = 25 A, Input power = 9000 W
• (3) Average DC resistance per stator phase: 2.8
• (a) Determine the no load rotational loss.
‑
• (b) Determine the parameters of the IEEE recommended equivalent
‑
circuit
• (c) Determine the parameters (Vth, Rth, Xth) for the Thevenin
equivalent circuit of Fig.5.16.
25. PERFORMANCE CHARACTERISTICS
2
T mech=
1
ωsyn
I2
¿
R2
'
s
¿
T mech=
1
ωsyn
∗
V th
2
(Rth+ R2
'
/s)
2
+(X th+ X2
'
)
2
∗
R2
'
s
Pmech=Tmech ωmech=I 2
2
R2
s
(1−s) ωmech=
2πn
60
ωmech =(1− s ) ωsyn
ωmech=
nsyn
60
2π (1−s)
ωsyn=
120 f
P
∗
2π
60
=
4π f 1
P
T mech ωsyn=I2
2
R2
s
=Pag
Tmech=
1
ωsyn
Pag
Where
n mech= (1− s) n syn
or
26. At low values of slip
,
Rth +
R2
'
s
>> Xth+ X 2
'
and
R2
'
s
>> Rth
Tmech≃
1
ωsyn
∗
Vth
2
R2
'
∗s
T mech=
1
ωsyn
∗
V th
2
(Rth+ R2
'
/s)
2
+(X th+ X2
'
)
2
∗
R2
'
s
27. At larger values of slip
,
Rth +
R2
'
s
<< Xth + X 2
'
Tmech=
1
ωsyn
∗
V th
2
(Xth +X 2
'
)
2
∗
R2
'
s
T mech=
1
ωsyn
∗
V th
2
(Rth+ R2
'
/s)
2
+(X th+ X2
'
)
2
∗
R2
'
s
31. R1
If is small (hence Rth is negligibly small)
STmax
=
R2
'
√Rth
2
+(Xth + X2
'
)
2
T max =
1
2ωsyn
∗
V th
2
Rth+ √Rth
2
+(X th+ X2
'
)
2
sTmax
≃
R2
'
X th +X 2
'
Tmax=
1
2ωsyn
∗
V th
2
Xth+ X2
'
Then
Then
32. T max
T
=
(Rth +R2
'
/s)
2
+(X th+ X2
'
)
2
(Rth +R2
'
/ sTmax
)
2
+( X th +X 2
'
)
2
¿
s
sT
max
R1
If is small (hence Rth is negligibly small)
T max
T
=
(R2
'
/s)
2
+(X th+ X2
'
)
2
(R2
'
/sTmax
)
2
+( Xth +X 2
'
)
2
¿
s
sT
max
T max
T
=
(R2
'
/s )
2
+(R2
'
/ sT max
)
2
2(R2
'
¿ sT
max )
2
¿
s
sTmax
Tmax
T
=
sT max
2
+s
2
2∗sT max
¿ s
33. Efficiency
Power flow in an induction motor.
Pin =3V1 I1 cos θ1
The power loss in the stator winding is
: P1=3I1
2
R1
P2=3 I 2
2
R2
η=
Pout
Pin
34. Pag = Pin P2= sP ag
Pout = P mecj= P ag (1− s )
ηideal=
Pout
Pin
=(1−s)
Ideal Efficiency
35. Example 4.4 A three-phase, 460 V, 1740 rpm, 60 Hz, four-pole
wound-rotor induction motor has the following parameters per
phase:
1
R =0.25 , 2
.
0
2
R , 5
.
0
2
1
X
X , 30
m
X
Therotationallossesare 1700watts.Withtherotorterminals
short-circuited,find
(a) (i)Starting currentwhenstarted directonfullvoltage.
(ii) Startingtorque.
(b) (i)Full-loadslip.
(ii) Full-loadcurrent.
(iii)Ratioofstartingcurrenttofull-loadcurrent.
(iv)Full-loadpowerfactor.
(v) Full-loadtorque.
(iv)Internalefficiencyandmotorefficiencyatfullload.
(c) (i)Slipat whichmaximumtorqueisdeveloped.
(ii) Maximumtorquedeveloped.
(d) How muchexternalresistanceperphaseshouldbe
connected intherotorcircuitsothatmaximumtorqueoccurs at
start?
RZ=0.2
ohms
Xm -30
Ohms
41. Note that for parts (a) and (b) it is not necessary to use Thevenin
equivalent circuit. Calculation can be based on the equivalent
circuit of Fig.5.15 as follows:
42. A three-phase, 460 V, 60 Hz, six-pole wound-rotor induction motor
drives a constant load of 100 N - m at a speed of 1140 rpm when
the rotor terminals are short-circuited. It is required to reduce the
speed of the motor to 1000 rpm by inserting resistances in the
rotor circuit. Determine the value of the resistance if the rotor
winding resistance per phase is 0.2 ohms. Neglect rotational
losses. The stator-to-rotor turns ratio is unity.
44. Example The following test results are obtained from three
phase 100hp,460 V, eight pole star connected induction machine
No-load test : 460 V, 60 Hz, 40 A, 4.2 kW. Blocked rotor test is
100V, 60Hz, 140A 8kW. Average DC resistor between two stator
terminals is 0.152
(a)Determine the parameters of the equivalent circuit.
(b)The motor is connected to 3 , 460 V, 60 Hz supply and runs
at 873 rpm. Determine the input current, input power, air
gap power, rotor cupper loss, mechanical power developed,
output power and efficiency of the motor.
(c) Determine the speed of the rotor field relative to stator
structure and stator rotating field
45. Solution: From no load test:
(a ) Z NL=
460 /√3
40
=6 .64 Ω
RNL=
PNL
3∗I1
2
=
4200
3∗40
2
=0 . 875 Ω
X NL =√6 . 642
− 0 . 8752
=6 .58 Ω
X 1 + X m= 6 . 58 Ω
From blocked rotor test
:
RBL=
8000
3∗1402
=0 .136 Ω R1=
0.152
2
=0.076Ω
Z BL=
100/√3
140
=0 .412Ω
48. Stator CU losses
:
Pst=3∗125.22
2
∗0.076=3.575 kW
Air gap power
Pag =88 . 767−3 . 575= 85 .192 kW
Rotor CU losses
P2= sP ag= 0. 03∗85 .192=2 .556 kW
Mechanical power developed
:
Pmech =(1− s) Pag =(1− 0 . 03)∗85 . 192=82 . 636 kW
Pout = P mech − Prot
From no load test
: Prot=PNL−3I1
2
∗R1=4200−3∗40
2
∗0 .076=3835.2 W
Pout=82.636∗10
3
−3835.2=78.8 kW
η=
Pout
Pin
∗100=
78.8
88.767
∗100=88.77
49. Example A three phase, 460 V 1450 rpm, 50 Hz, four pole
wound rotor induction motor has the following parameters per
phase ( 1
R =0.2, 2
R=0.18 , 2
1 X
X
=0.2, m
X =40). The
rotational losses are 1500 W. Find,
(a) Starting current when started direct on full load voltage.
Also find starting torque.
(b) (b) Slip, current, power factor, load torque and efficiency
at full load conditions.
(c) Maximum torque and slip at which maximum torque will
be developed.
(d) How much external resistance per phase should be
connected in the rotor circuit so that maximum torque
occurs at start?
50. V 1=
460
√3
=265.6 V / phase
Z1=0.2+ j0 .2+
j 40∗(0.18+0.2)
0.18+ j 40.2
=0.55∠46 .59
o
Ω
Ist =
V1
I1
=
265.6
0.55∠46.59
o
=482.91 ∠−46.3o
Ω
s=
1500−1450
1500
=0.0333
R2
'
s
=
0.18
0 .0333
=5 .4
Z1=0.2+ j0 .2+
j 40∗(5.4+ j0.2)
5.4+ j 45.4
=4.959 ∠10.83o
Ω
51. I1∣FL=
265.6
4.959∠10.83o
=53.56∠−10.83o
A
Then the power factor is: 9822
.
0
83
.
10
cos
o
lag.
.
sec
/
08
.
157
2
*
60
1500
rad
sys
V
j
j
V o
th 285
.
0
275
.
264
2
.
40
2
.
0
40
*
6
.
265
Then,
2
.
0
198
.
0
285
.
45
281432
.
0
2
.
40
2
.
0
2
.
0
2
.
0
*
40
j
j
j
j
Z o
th
∴ T=
3∗(264.275)2
∗5. 4
157.08∗(0.198+5.4)2
+(0.2+0.2)2
=228.68 Nm
52. Then, W
T
P sys
ag 1
.
35921
08
.
157
*
68
.
228
*
Then, W
sP
P ag 1197
1
.
35921
*
0333
.
0
2
And, W
P
s
P ag
m 7
.
34723
1
Then, W
P
P
P rot
m
out 7
.
33223
1500
7
.
34723
W
Pin 41917
9822
.
0
*
56
.
53
*
6
.
265
*
3
Then, %
26
.
79
41914
7
.
33223
in
out
P
P
∴ Tm=
3∗(264 .275)2
2∗188.5 [0.198+(0.1982
+(0.2+0.2)2
)]
1 /2
=862.56 Nm
sTmax
=
0.18
[0.1982
+(0.2+0.2)2
]
1/2
=0.4033
53. (d)
2
/
1
2
2
2
2
.
0
2
.
0
198
.
0
1
max
ext
T
R
R
s
Then, 446323
.
0
2
ext
R
R
Then, 26632
.
0
18
.
0
446323
.
0
ext
R
54. Example 5.6 The rotor current at start of a three-phase, 460 volt,
1710 rpm, 60 Hz, four pole, squirrel-cage induction motor is six
times the rotor current at full load.
(a) Determine the starting torque as percent of full load torque.
(b) Determine the slip and speed at which the motor develops
maximum torque.
(c) Determine the maximum torque developed by the motor as
percent of full load torque.
Note that the equivalent circuit parameters are not given. Therefore equivalent circuit
parameters cannot be used directly for computation.)a) The synchronous speed is
T =
I2
2
R2
sωsyn
α
I2
2
R2
s
57. Example 4.9 A 4 pole 50 Hz 20 hp motor has, at rated voltage
and frequency a starting torque of 150% and a maximum torque of
200 % of full load torque. Determine (i) full load speed (ii) speed
at maximum torque.
Solution:
5
.
1
FL
st
T
T
and 2
max
FL
T
T
then, 75
.
0
2
5
.
1
max
T
Tst
75
.
0
1
2
2
max max
max
T
T
st
s
s
T
T
Then, 0
75
.
0
2
75
.
0 max
max
2
T
T s
s
Then 21525
.
2
max
T
s (unacceptable) Or 451416
.
0
max
T
s
58. 2
*
2 max
max
2
2
max
FL
T
FL
T
FL s
s
s
s
T
T
But 451416
.
0
max
T
s
Then 2
*
451416
.
0
*
2
451416
.
0 2
2
max
FL
FL
FL s
s
T
T
0
451416
.
0
451416
.
0
*
4 2
2
FL
FL s
s
0
203777
.
0
80566
.
1
2
FL
FL s
s
6847
.
1
FL
s (unacceptable) or 120957
.
0
FL
s
59. rpm
ns 1500
4
50
*
120
then (a) s
FL
FL n
s
n *
1
rpm
nFL 1319
1500
*
120957
.
0
1
(b) rpm
n
s
n s
T
T 823
1500
*
451416
.
0
1
*
1 max
max
60. Example 4.10 A 3, 280 V, 60 Hz, 20 hp, four-pole induction
motor has the following equivalent circuit parameters.
12
.
0
1
R , 1
.
0
2
R , 25
.
0
2
1
X
X , and 10
m
X
The rotational loss is 400 W. For 5% slip, determine (a) The
motor speed in rpm and radians per sec. (b) The motor current. (c)
The stator cu-loss. (d) The air gap power. (e) The rotor cu-loss. (f)
The shaft power. (g) The developed torque and the shaft torque.
(h) The efficiency.
rpm
ns 1800
4
60
*
120
, sec
/
5
.
188
2
*
60
1800
rad
s
Solution:
61. 0.12 j0.25
j10
j0.25
2
05
.
0
1
.
0
e
e X
R
j
Z
25
.
0
12
.
0
1
o
j
j
j
j
Z 55
.
23
1314
.
2
25
.
10
2
25
.
0
2
*
10
25
.
0
12
.
0
1
1
.
120
3
208
1
V V
o
o
I 55
.
23
1314
.
2
55
.
23
1314
.
2
1
.
120
1
A
(c) W
P 031
.
1143
12
.
0
*
3479
.
56
*
3 2
1
62. (d) W
P o
s 9794
.
18610
55
.
23
cos
*
3479
.
56
*
1
.
120
*
3
W
P
P
P s
ag 9485
.
17467
1
(e) W
sP
P ag 3974
.
873
9785
.
17467
*
05
.
0
2
(f) W
P
s
P ag
m 5511
.
16594
1
(g) m
N
P
T
ag
.
6682
.
92
5
.
188
9485
.
17467
5
.
188
Nm
P
T
shaft
shaft 9127
.
85
5
.
188
5511
.
16194
5
.
188
(h) %
02
.
87
100
*
s
shaft
P
P
63. Example 4.11 A 30, 100 WA, 460 V, 60 Hz, eight-pole induction
machine has the following
equivalent circuit parameters:
07
.
0
1
R , 05
.
0
2
R , 2
.
0
2
1
X
X , and 5
.
6
m
X
(a) Derive the Thevenin equivalent circuit for the
induction machine.
(b)If the machine is connected to a 30, 460 V, 60 Hz supply,
determine the starting torque, the maximum torque the machine
can develop, and the speed at which the maximum torque is
developed.
(c) If the maximum torque is to occur at start, determine the
external resistance required in each rotor phase. Assume a
turns ratio (stator to rotor) of 1.2.
65. Speed in rpm for which max torque occurs
= rpm
n
s s
T 5
.
787
900
*
1249
.
0
1
*
1 max
(c)
2
2
2
1
2
1
2
max
R
X
X
R
R
sT
or 4
.
0
05
.
0
*
1249
.
0
1
*
1
2
2
max
R
s
s
R
T
start
start
Then 243
.
0
2
.
1
/
05
.
0
4
.
0 2
ext
R
![I2=
sE2
R2+ jsX 2
P2=I2
2
R2
I2=
E2
( R2 /s )+ jX 2
P= Pag =I2
2
[R2+
R2
s
(1−s)]=I2
2 R2
s
Pmech=I2
2
R2
s
(1−s) Pmech =(1− s)∗Pag
Pmech=
(1−s)
s
P2
P2=I2
2
R2=sPag
Pag : P 2 : Pmech =1: s :(1− s)](https://image.slidesharecdn.com/inductionmotor1118ppt-250522054550-e6b0c484/85/induction_motor1_1_1_8space-vector-_ppt-ppt-10-320.jpg)
![Then, W
T
P sys
ag 1
.
35921
08
.
157
*
68
.
228
*
Then, W
sP
P ag 1197
1
.
35921
*
0333
.
0
2
And, W
P
s
P ag
m 7
.
34723
1
Then, W
P
P
P rot
m
out 7
.
33223
1500
7
.
34723
W
Pin 41917
9822
.
0
*
56
.
53
*
6
.
265
*
3
Then, %
26
.
79
41914
7
.
33223
in
out
P
P
∴ Tm=
3∗(264 .275)2
2∗188.5 [0.198+(0.1982
+(0.2+0.2)2
)]
1 /2
=862.56 Nm
sTmax
=
0.18
[0.1982
+(0.2+0.2)2
]
1/2
=0.4033](https://image.slidesharecdn.com/inductionmotor1118ppt-250522054550-e6b0c484/85/induction_motor1_1_1_8space-vector-_ppt-ppt-52-320.jpg)
