Download induction motor

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
Construction
Stator
Construction
Squirrel Cage Rotor
Construction
Performance of Three-Phase Induction Motor
n
2
P
f * 60 
s
120 f
p
Slip rpm  ns  n  sns
f2 
p
120
ns  n
ns
ns  n  
p
120
sns  sf1
120 f 2 120sf1
n2 

 sns
p
p
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
120 f
120 * 60
ns 

 1800 rpm
p
4
n  1  s ns  1  0.05 *1800  1710 rpm
(b) 1800 (same as synchronous speed)
Equivalent Circuit of the Induction Motor
Equivalent Circuit of the Induction Motor
sE2
I2 
R2  jsX 2
P2 
E2
I2 
R2 / s   jX 2
2
I 2 R2
P  Pag 
Pmech 
Pmech
2
I 2  R2
I 22

R2

2 R2
 1  s   I 2
s
s

Pmech  1 s  * Pag
R2
1  s 
s

1  s

P
s
2
Pag : P2 : Pmech  1 : s : 1  s 
P2  I 22 R2  sPag
IEEE-Recommended Equivalent Circuit
IEEE recommended equivalent circuit.
Theveninequivalent circuit
Vth 
Xm
R12   X 1  X m 2
V1
jX m R1  jX 1 
Z th 
 Rth  jX th
R1  j  X 1  X m 
Xm
Vth 
V1  K thV1
X1  X m
If ,
R12   X1  X m 2
If R12   X1  X m 2
2
then,
 Xm 
 R1  K th2 R1
Rth  

X

X
 1
m
X th  X 1
Tests To Determine The Equivalent Circuit
R1  RLL / 2
No-load test
VLL
V1 
V / Phase
3
2
2
X NL  Z NL
 RNL
Z NL
V1

I1
RNL
PNL
 2
3I1
X 1  X m  X NL
Locked-rotor test
PBL
 2
3I1
RBL
BL
Z BL
X BL
fBL
fBL


V1 BL
I1 BL
Z
X BL  X BL
2
BL fBL
2
 RBL

Rated Frequency
*
fBL
Frequency at blocked rotor test
X BL  X 1  X 2
X 1  X m  X NL
X m  X NL  X 1
Blocked-rotor equivalent circuit for improved value for
R2
X m2
R 2
R
2 2
R2   X 2  X m 
2
 X 2  X m 
 R
R2  
 Xm 
R  RBL  R1
2
 Xm 
 R2
R  
 X 2  X m 
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 voltage, 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 
From the no-load test:
VLL 440
V1 

 254V / Phase
3
3
Z NL
V1 254
 
 18.143
I1 14
RNL 
PNL
3I12
1470

3 *14
2
 2.5 
2
2
X NL  Z NL
 RNL
 18.1432  2.52  17.97
X1  X m  X NL  17.97 
From the blocked-rotor test
RBL 
PBL
3I12

7200
3 * 60
2
 0.6667 
BL
The blocked-rotor reactance is:
X BL 
Z
2
BL

2
 RBL
 1.5685 2  0.6667 2  1.42 
X BL  X1  X 2  1.42 
 X1  X 2  0.71 
X m  X NL  X1  17.97  0.71  17.26 
R  RBL  R1  0.6667  0.25  0.4167 
2
2
 X 2  X m 
0
.
71

17
.
26





 R2  
R
 * 0.4167  0.4517 

 17 .26 
 Xm 
• 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.
2200
V1 
 1270.2 V / Phase
3
RNL
Z NL
V1 1270.2
 
 282.27 
I1
4.5
PNL
1600
 2
 26.34 
2
3I1 3 * 4.5
(a) No-Load equivalent Circuit
(b) Locked rotor equivalent circuit
2
2
X NL  Z NL
 RNL
 282.27 2  26.342  281
X 1  X m  X NL  281 
PBL
9000
RBL  2 
 4.8 
2
3I1 3 * 25
R2  RBL  R1  4.8  2.8  2
. 281.0 =
:impedance at 15 Hz is
Z BL
V1
270
 
 6.24 
I1
3 * 25
The blocked-rotor reactance at 15 Hz is
Its value at 60 Hz is
X BL  3.98 *
X BL 
6.24
2
60
 15.92 
15
X BL  X 1  X 2
15 .92
 X 1  X 2 
 7.96 
2
X m  281  7.96  273.04 
R  RBL  R1  4.8  2.8  2 
7.96  273.04 

R2  
 2  2.12 
 273.04 
2

 4.82  3.98 
at 60 Hz
)c(
273.04
Vth 
V1  0.97 V1
7.96  273.04
Rth  0.97 R1  0.97 * 2.8  2.63 
2
2
X th  X 1  7.96 
PERFORMANCE CHARACTERISTICS
Pmech  Tmech mech
 mech  1  s  syn
syn
Where
R2
1  s 
I
s
2
2
nmech  1  s  nsyn
or
120 f 2 4 f1

*

P
60
P
Tmech 
1
syn
I 22
 mech
 mech 
R2
s
Tmech 
nsyn
60
1
syn
2 R2
Tmech syn  I 2
 Pag
s
R2
Tmech 
*
*
2
2
syn  Rth  R2 / s    X th  X 2 
s
1
2n

60
Vth2
2 1  s 
Pag
At low values of slip,
R2
R2
Rth 
 X th  X 2 and
 Rth
s
s
R2
Tmech 
*
*
2
2
syn  Rth  R2 / s    X th  X 2 
s
1
Tmech
Vth2

* *s
 syn R2
1
Vth2
At larger values of slip,
R2
Rth 
 X th  X 2
s
R2
Tmech 
*
*
2
2
syn  Rth  R2 / s    X th  X 2 
s
1
Tmech
Vth2
Vth2
R2

*
*
2
 syn  X th  X 2  s
1
Torque-speed profile at different voltages.
Maximum Torque
dT / ds  0
R2
Tmech 
*
*
2
2
syn  Rth  R2 / s    X th  X 2 
s
Then
R2
2
 Rth2   X th  X 2 
STmax
Tmax 
1
2 syn
*
Vth2
1
Vth2
Rth 
Rth2
2

  X th  X 2 
STmax 
R2
Rth2   X th  X 2 2
Tmax 
1
2 syn
*
Vth2
2
Rth  Rth2   X th  X 2 
.
Torque speed characteristics for varying
R2
If
R1
STmax 
Tmax 
is small (hence
Rth
is negligibly small)
R2
Rth2   X th  X 2 2
1
2 syn
*
Vth2
Rth 
Rth2
sTmax
Then
2

  X th  X 2 
R2

X th  X 2
Then
Tmax
Vth2

*
2 syn X th  X 2
1
2
2

Tmax
Rth  R2 / s    X th  X 2 
s

*
2
2
T
Rth  R2 / sTmax   X th  X 2  sTmax

If
R1

Rth
is small (hence
is negligibly small)

Tmax
R2 / s    X th  X 2 
s

*
2
2
T
R2 / sTmax   X th  X 2  sTmax
2

Tmax

T
2

R2 / s 2  R2 / sT

2 R2 / sTmax
s
Tmax

T
2 * sTmax * s
2
sTmax
2

2

2
max
*
s
sTmax
Efficiency
Power flow in an induction motor.
Pin  3V1I1 cos1
The power loss in the stator winding is:
P2  3 I 22 R2
P1  3I R
Pout

Pin
2
1 1
Ideal Efficiency
Pag  Pin
ideal
P2  sPag
Pout

 1  s 
Pin
Pout  Pmecj  Pag 1  s 
Example 4.4 A three-phase, 460 V, 1740 rpm, 60 Hz, four-pole
wound-rotor induction motor has the following parameters per
phase:
R1 = 0.25 , R2  0.2 , X 1  X 2  0.5 , X m  30 
The rotational losses are 1700 watts. With the rotor terminals
short-circuited, find
(a)
(i) Starting current when started direct on full voltage.
RZ
ohm
(ii) Starting torque.
(b)
(i) Full-load slip.
(ii) Full-load current.
(iii) Ratio of starting current to full-load current.
(iv) Full-load power factor.
(v) Full-load torque.
(iv) Internal efficiency and motor efficiency at full load.
(c)
(i) Slip at which maximum torque is developed.
(ii) Maximum torque developed.
(d)
How much external resistance per phase should be
connected in the rotor circuit so that maximum torque occurs at
start?
Xm
Ohm
=163.11 N.m
28022.3
motor 
*100  87.5%
32022.4
int ernal  1  s  *100  1  0.0333 *100  96.7%
(c) (i)
(c) (ii)
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:
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.
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
Solution:
From no load test:
a  Z NL 
RNL 
460 / 3
 6.64 
40
PNL
2
3 * I1
4200

3 * 40
2
 0.875
X NL  6.64  0.875  6.58
2
2
X1  X m  6.58
From blocked rotor test:
RBL 
Z BL 
8000
3 *140 2
 0.136
100 / 3
 0.412
140
0.152
R1 
 0.076
2
X BL  0.4122  0.1362  0.389
X 1  X 2  0.389
0.389
X1  X 2 
 0.1945
2
X m  6.58  0.1945  6.3855
R  RBL  R1  0.136  0.076  0.06
0.1945  6.3855 

R2  
 * 0.06  0.0637
6.3855


2
0.076 
j0.195 
j6.386 
j0.195 
0.0637
s
b 
120 f 120 * 60
ns 

 900rpm
P
8
ns  n 900  873
s

 0.03
ns
900
R2 0.0637

 2.123
s
0.03
Input impedance
Z1  0.076  j 0.195 
 j 6.3862.123  j 0.195
 2.12127.16o 
2.123  j 6.386  0.195
V1
460 / 3
o
I1 

 125.22  27.16
Z1 2.1227.16
Input power:


460
Pin  3 *
*125.22 cos 27.16o  88.767 kW
3
Stator CU losses:
Pst  3 *125.222 * 0.076  3.575 kW
Pag  88.767  3.575  85.192 kW
Air gap power
Rotor CU losses
P2  sPag  0.03* 85.192  2.556 kW
Mechanical power developed:
Pmech  1  s  Pag  1  0.03 * 85.192  82.636 kW
Pout  Pmech  Prot
From no load test:P
rot
 PNL  3I12 * R1  4200  3 * 402 * 0.076  3835.2W
Pout  82.636 *103  3835.2  78.8 kW
Pout
78.8

*100 
*100  88.77 %
Pin
88.767
Example
A three phase, 460 V 1450 rpm, 50 Hz, four pole
wound rotor induction motor has the following parameters per
phase ( R1 =0.2, R2 =0.18 , X 1  X 2 =0.2, X m =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?
460
V1 
 265.6 V / phase
3
j 40 * 0.18  0.2 
o
Z1  0.2  j 0.2 
 0.5546.59 
0.18  j 40.2
V1
265.6
o
I st  
 482.91   46.3 
o
I1 0.5546.59
1500  1450
s
 0.0333
1500
R2
0.18

 5.4
s 0.0333
j 40 * 5.4  j 0.2 
Z1  0.2  j 0.2 
 4.959 10.83o 
5.4  j 45.4
I1 FL
265.6
o

 53.56  10.83 A
o
4.95910.83
Then the power factor is: cos 10.83  0.9822 lag.
o
 sys
1500

* 2  157.08 rad / sec .
60
265.6 *  j 40 
Vth 
 264.275 0.285o V
0.2  j 40.2
Then,
j 40 * 0.2  j 0.2 
Z th 
 0.28143245.285o  0.198  j 0.2 
0.2  j 40.2
T 
3 * 264.275 * 5.4
2
157.08 * 0.198  5.4   0.2  0.2 
2
2
 228.68 Nm
Then, Pag  T *  sys  228.68 *157.08  35921.1W
Then, P2  sPag  0.0333 * 35921.1  1197 W
And, Pm  1  s Pag  34723.7W
Then, Pout  Pm  Prot  34723.7  1500  33223.7W
Pin  3 * 265.6 * 53.56 * 0.9822  41917 W
Then,  
 Tm 
Pout 33223.7

 79.26 %
Pin
41914
3 * 264.2752


2 *188.5 0.198  0.1982  0.2  0.2 
sTmax 
0.198
2

2 1/ 2
0.18
 0.2  0.2

2 1/ 2
 862.56 Nm
 0.4033
(d) sTmax  1 
0.198

R2  Rext
2
 0.2  0.2 

2 1/ 2
  0.446323
Then, R2  Rext
  0.446323  0.18  0.26632 
Then, Rext
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
I 22 R2 I 22 R2
T

s syn
s
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:
Tst
Tst
Tmax
1.5
 1.5 and
 2 then,

 0.75
TFL
TFL
Tmax
2
2sTmax
Tst

 0.75
2
Tmax 1  sTmax
Then, 0.75 sTmax  2 sTmax  0.75  0
2
Then sTmax  2.21525 (unacceptable) Or sTmax  0.451416
2
sT2max  sFL
Tmax

2
TFL 2sTmax * s FL
But sTmax  0.451416
2
Tmax
0.4514162  s FL
Then

2
TFL 2 * 0.451416 * s FL
2
s FL
 4 * 0.451416 sFL  0.451416 2  0
2
s FL
 1.80566 s FL  0.203777  0
s FL  1.6847 (unacceptable) or s FL  0.120957
120 * 50
ns 
 1500 rpm
4
then (a) nFL  1 s FL  * ns
nFL  1  0.120957  *1500  1319 rpm


(b) nTmax  1  sTmax * ns  1  0.451416 *1500  823 rpm
Example 4.10 A 3, 280 V, 60 Hz, 20 hp, four-pole induction
motor has the following equivalent circuit parameters.
R1  0.12 , R2  0.1 , X 1  X 2  0.25 , and X m  10 
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.
Solution:
120 * 60
1800
ns 
 1800 rpm ,  s 
* 2  188.5 rad / sec
4
60
0.12 
j0.25 
j0.25 
j10 
0.1
2
0.05
Z1  0.12  j 0.25  Re  X e
j10 * 2  j 0.25
Z1  0.12  j 0.25 
 2.131423.55o 
2  j10.25
V1 
208
 120.1 V
3
120.1
o
I1 

2
.
1314


23
.
55
A
o
2.131423.55
(c) P1  3 * 56.3479 * 0.12  1143.031W
2

(d) Ps  3 *120.1 * 56.3479 * cos  23.55
o
  18610.9794 W
Pag  Ps  P1  17467.9485 W
(e) P2  sPag  0.05 *17467.9785  873.3974 W
(f) Pm  1  s  Pag  16594.5511W
Pag
17467.9485
(g) T 

 92.6682 N .m
188.5
188.5
Pshaft
16194.5511
Tshaft 

 85.9127 Nm
188.5
188.5
Pshaft
(h)  
*100  87.02%
Ps
Example 4.11 A 30, 100 WA, 460 V, 60 Hz, eight-pole induction
machine has the following
equivalent circuit parameters:
R1  0.07 , R2  0.05 , X 1  X 2  0.2 , and X m  6.5 
(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.
Solution:
Vth 
Xm
6.5
* V1 
* 265.6  257.7 V
X1  X m
0.2  6.5
Rth  jX th 
 j 6.5 *  j 0.2  0.07   0.06589 
0.07  j 0.2  j 6.5
0.06589 j0.1947 
j0.2 
257.7V
(b) Tst 
Tmax 
0.05
s
3 * 257.7 2 * 0.05

94.25 0.06589  0.052  0.1947  0.2 2

 624.7 Nm
3 * 257.7 2

2 * 94.25 0.06589  0.06589 2  0.1947  0.2 2
 2267.8 Nm
sTmax 
j 0.1947 
0.05
0.06589 2  0.1947  0.2 2
 0.1249

Speed

in

rpm
for
which
max
torque
= 1  sTmax * ns  1  0.1249 * 900  787.5 rpm
(c) sTmax 
or R2 start 
R2
R12
  X 1  X 2 
sstart  1
sTmax
2
* R2 
1
0.1249
 R2
* 0.05  0.4 
Then Rext  0.4  0.05 / 1.2 2  0.243 
occurs