Transcript 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.

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
The speed of the rotating field (Synchronous speed ) is :
120 f
ns 
rpm
P
ms
sl  ms  m  sms
Slip speed is
Slip, s is :
m
2 *  120 f 4f


rad / sec .
60 P
P
sl ms  m
s

ms
ms
m  1  s ms
sl
Rotor frequency in Rad/Sec. is :
r 
2f s  2sf s
ms
Is the rotor speed ,
Rotor frequency in Hz is :
f r  sf s
I r
Xs
Rs
Xr
Rr
n s nr
Is
Im
V
Xm
Ir
sE
aT 1
EE
Ideal
Transformer
Stator
Rotor
Per Phase Equivalent Circuit
Is
V
X r
Xs
Rs
Im
Xm
Rr / s
I r
E
Per Phase Equivalent Circuit Referred to Stator
X r
Xs
Rs
Is
Rt
I r
Im
Xm
V
Rr / s
E
X r
Xt
Rr / s
I r
Vt  r
E
Per Phase Simplified Equivalent Circuit Referred to Stator
V Xm
Vt 
Rs2   X s  X m 2
t 

1  X s
 tan 
2

jX m  Rs  jX s 
Rt  jX t 
Rs  j  X s  X m 
Is
V
X r
Xs
Rs
Im
Xm
Rr / s
I r
E
Per Phase Equivalent Circuit Referred to Stator
Rr / s  j  X r  X m I r
Is 
jX m
 Xm 

Rs

Xt
Rt
Rr / s
Xr
I r
Vt  r
E
Per Phase Simplified Equivalent Circuit Referred to Stator
Vt t
I r 
 R  Rr   j  X  X  
 t

t
r
s 


R
2
r
Pg  3I r 



s


2

Pm  Pg  3I r Rr
2  Rr
Pm  3I r 

  3I r2 Rr  3I r2 Rr

s


 1  s   1  s P


g
 s 
2  Rr
Pm  3I r 
Rt

  3I r2 Rr  3I r2 Rr

s


Xt
 1  s   1  s P


g
 s 
Rr / s
Xr
I r
Vt  r
E
Rt
X r
Xt
Rr
I r
Vt  r
E
1 s  

 Rr
 s 
T
Pm
m
Pm  1  s Pg
m  1  s ms
Vt t
I r 
 R  Rr   j  X  X  
 t

t
r
s 

T
T 
Pg
ms
3
ms

sPg
sms
 Rr 
2

I r 






Vt2 Rr / s
3 

T
2
ms  

Rr 
2
  j  X t  X r  
  Rt 
s 


 s 
3I r2 Rr

sms



3 
Vt2 Rr / s


T
ms  
Rr 2
2
  Rt  s   j  X t  X r  






Vt2 Rr / s
d 3 
0
ds ms  

Rr  2
2
  j  X t  X r  
  Rt 
s 


dT
0
ds
sm  
Tmax



2ms  Rt 
3
Rt 2   X t  X r 2


2
2
Rt   X t  X r  
Vt2
Rr
A 3-phase, star connected, 6 pole, 60 Hz induction motor has the following
constants:Vt  231 V,
Rt  Rr  1 ,
X t  X r  2
1. the motor is used for regenerative braking,
(i) Determine the range of load torque it can hold and the corresponding
range of speed.
(ii) Calculate the speed and current for a load torque of 150 N.m.
2. If the motor is used for plugging, determine the breaking torque and current
for a speed of 1200 rpm.
N s  120 f / P  120 * 60 / 6  1200 rpm
4f 4 *  * 60
ms 

 125.7 rad / sec .
P
6
For regenerative breaking
sm  

Rr
Rt2  X r  X t
1

 0.24
2
1  16


3 
Vt2
Tmax 
2ms 
2
R

R
t  X r  X t
 t


3
 231 * 231   203.9 N .m

2  2 * 125.7 1  1  16 


(i) the range of active load torque the motor can hold: 0 to 203.9 N.m. The
speed
in
rpm
at
the
maximum
torque:
N  1  sm N s  1   0.24  * 1200  1488 rpm
Therefore the range of speed is 1200 to 1488 rpm.
(ii)








2



Vt * Rr / s
3
3  231 * 231 * 1 / s 
T

 150 N .m


2
2
  1

 ms  
125
.
7


2
Rr 
 1    16 

 X r  X t 
  Rt 
s




s
 


1
Let  x , then x 2  10.49 x  17  0
s
Then, x  2 or  8.5 then s  . 5 or  .118
Since stable operation is usually obtained only up to sm which is -0.24,
the feasible value of s is -0.118.

Then motor speed

N  1  s N s  1.118 *1200  1342 rpm
I r 
Vt
2


231

2

2
 
R
r
 Rt 
  X r  X t

s 

2. For the motor speed N rpm
Then for plugging
1  1   16


  0.118 
Ns  N
s
Ns
N s  1200 rpm
 1200  1200
s
2
 1200
Vt
I r 

2

2
 
R
r
 Rt 
  X r  X t


s




N  1200rpm
231
1  0.52  16
I r2 Rr
3

T
.

* 54.1 *1 / 2  34.9 N .m
ms
s
 125.7
3
 27.2 A
 54.1A
Speed Control of Induction Motor
1- Variable Terminal Voltage Control
2- Variable Frequency Control
3- Rotor Resistance Control
4- Injecting Voltage in Rotor Circuit
m
1- Variable Terminal Voltage Control
ms
TL
V decreasing
T
variable terminal voltage control
variable frequency control
Low speed range
Wide speed range
Lower rated speed
Lower & higher rated speed
2- Variable Frequency Control
a  f / f rated
Per-Unit Frequency
1- Operation Below the Rated Frequency a <1
Erated Erated
1
Im 

*
Xm
f rated 2 Lm
E
E
1
Im 

*
aX m a * f rated 2 Lm
Comparing of the above equations,
value if
E  a Erated 
f
f rated
Erated
Im
At rated frequency
At any frequency, f
will stay constant at a value equal to its rated
E Erated

f
f rated
The above equation suggests that the flux will remain constant if the
back emf changes in the same ratio as the frequency, in other ward,
when E/f ratio is maintained constant.
The rotor current at any frequency f can be obtained from the following equation:
I r 
aErated
2

 Rr   aX  2
 
r
s
 
ams  m
sl
s

ams
ams
ams  m sl
sa 

ms
T
3
a ms

E rated
2

 Rr    X  2
 
r
as
 
ms Synchronous Speed at rated frequency f rated
m
Angular Speed at frequency f
ms


 2

* Rr / as 
3  Erated
2  Rr 
 
I r 
2


s


 Rr 
2


ms
   X r  

  as 

At a given f and E
Rr
sm  
aX r
sl  ams  m
Torque at frequency f
2
 E rated

Tmax 



2ms  X r 
3
T
3
a ms


 2





E
*
R
/
as
R
3
r
 rated

I r 2  r  
2

2
 s   ms   Rr 
   X r  

  as 

2
3E rated
as   consta nt sl 
T
ms Rr
Rr
  X r
as
(6-51)
m
ms
ams
T
Braking
Motoring
V/f Control
Is
Im
Xm
V
V
Xm
Rr / s
I r
E
X r
Xs
Rs
Im
X r
Xs
Rs
Rr / s
I r
Is
E
V/f Control
At rated frequency
2




 /s

V
*
R
R
3
2
rated
r
r

T 
I r 
 s  ms   R  R  / s 2   X  X  2 
ms


 s
r
s
r 
2

Vrated
3 
Tmax 


2
2
2ms  R  R   X  X   
s
s
r 
 s
3
(6.52)
(6.53)
At any frequency, f,
2

Vrated
* Rr / as
3 
T 


2
2
ms   Rs / a  Rr / as    X s  X r  
2

Vrated
3 
Tmax 


2
2
2ms  R / a   R / a    X  X   
s
s
r 
 s
a 1
(6.54)
(6.55)
V/f Control
m
ms
ams
T
Operation above the rated frequency a>1
The terminal voltage has o be constant = Rated Volatge=
V  consta nt
Vrated
 Flux  when a 
At any frequency, f,
2

Vrated
* Rr / as
3 
T

2
2
2
ms  Rs  Rr / s   a  X s  X r  
Tmax
2


Vrated



2
2

2ms a  Rs   Rs   a  X s  X r  
3
a 1
(6.56)
(6.57)
m
Constant torque
locus
ms
f rated
ams
Constant torque
locus
T
Ex6-3 A 3-phase, Y-connected, 60 Hz, 4 pole induction motor has the following parameters:
Rs  Rr  0.024 
and
X s  X r  0.12
The motor is controlled by the variable frequency control with a constant V/f ratio. For
the operating frequency of 12 Hz, calculate:
1-The breakdown torque as a ratio of its value at the rated frequency for both motoring
and breaking.
2- The starting torque and rotor current in terms of their values at rated frequency.
a
f
f rated
From (6.55)
12

 0.2
60
Tmax
3 


2ms  Rs / a 



2
2
Rs / a    X s  X r  
2
Vrated
0.024 / 1  0.024 / 12  0.24r 2
Tmax a  0.2

 0.68
Tmax a  1.0 0.024 / 0.2  0.024 / 0.22  0.242
Motoring
0.024 / 1  0.024 / 12  0.24r 2
Tmax a  0.2

 1.46
2
2
Tmax a  1.0 0.024 / 0.2  0.024 / 0.2  0.24
Breaking
At starting torque s=1
From (6.54)
2

Vrated
* Rr / as
3 
T


ms  Rs / a  Rr / as 2   X s  X r 2 


0.024 / 0.2

2
2
Tstart a  0.2   0.048 / 0.2   0.24 

 2.6
Tstart a  1.0


0.024 / 1

2
2
 0.048 / 1  0.24 
Ir 
From Fig. 6.1(d)
I r , start a  0.2
I r , start a  1

Vrated
Rs / a  Rr / as2   X s  X r 2
0.024 / 1  0.024 / 12  0.24 2
0.024 / 0.2  0.024 / 0.22  0.24 2
 0.72
Ex6-4 If the rated slip of induction motor of EX6.3 is 0.04, Calculate the motor speed for
rated torque and f=30Hz. The motor is controlled with a constant V/f ratio
a
f
f rated
30

 0.5
60
2

Vrated
* Rr / s
3 
At rated frequency,f=60Hz, From (6.52) T 


ms  Rs  Rr / s 2   X s  X r 2 
Trated
2
2
 3 *Vrated
Vrated
* 0.024 / 0.04
3 
1.34 



 m s  0.024  0.024 / 0.04 2  0.24 2 
m s
At f=30Hz, From (6.54)
Trated
[1]
2

Vrated
* Rr / as
3 
T


ms  Rs / a  Rr / as 2   X s  X r 2 



2
* 0.024 / 0.5 s 
3  Vrated



2
ms   0.024 0.024 
2
  0.5  0.5s   0.24 



[2]
Equating [1] and [2]
s 2  0.98s  1  0
s  0.089
and
s  0.43
 N  ams1  s   0.5 *1800 * 1  0.089   820 rpm
Ex 6-5
A 480V, 60 Hz 4 pole 1710 rpm, Y-connected induction motor has the following parameters
per phase: Rs  0.3 , Rr  0.15 , X s  X r  1.1
X m  40 
The motor is controlled by variable frequency control at a constant flux of rated value.
(i) Calculate the speed and stator current at half the rated torque and 20 Hz supply.
(ii) Solve part (i) assuming the speed-torque curves to be parrallel stright lines for s  s m .
(iii) Calculate the frequency, the stator current, and voltage at rated breaking torque and
1400 rpm.
Solution:
120 f 120 * 60
Ns 

 1800 rpm ms  2 * 1800  188.5 rad / sec
P
4
60
N s  N 1800  1710
s

 0.05
Ns
1800
Rr
0.15
Z r 
 jX r 
 j1.1  3  j1.1  3.195320.14o 
s
0.05
Z s  Rs  jX s  0.3  j1.1  1.1474.75 
o
Z in
Z r * Z m
 Zs 
Z r  Z m
3.195320.14 * 4090
Z in  1.1474.75 
 3.927437.245o 
3.195320.14  4090
Is 
I r 
V ph
Z in
480 / 3

 70.563 A
3.9274
4090
Zm
* Is 
* 70.563  68.57 A
3.195320.14  4090
Z r  Z m
E  I r * Z r  0.971* 3.1953  218.93V
Rated torque=
I r 2 Rr
3
 0.15
T
.

* 68 .517  *
 224 .14 N .m
m s
s
188 .5
0.05
3
(i) At 20 Hz
20 1
a

60 3












2
2
3  E rated * Rr / as 
3  218.93 * 0.15 * 3 / s  224.14
T


N .m


2
2

ms  R  
188.5  
2

2
 r 
  0.15 

r 

X

  as 




1
.
1





  1s 

  3 

 
0.152 * 9
2
 1.1 s  3.063 or 1.21s 2  3.063s  0.2025  0
s
s  2.4635 It is not acceptable value, s  0.0679 resonable value.
1
 m  a ms 1  s   * 188.5 * 1  0.0679  58.567 rad / sec
3
1
or N  a * N s 1  s   * 1800 * 1  0.0679   559.26 rpm
3
At 20 Hz
20
E  * 218.95  72.977V
60
Rr
0.15
j1.1
Z r 
 jaX r 

 2.23949.424o 
s
0.0679
3
E
72.9770
I r 

 32.588  9.424o A
Z r 2.23949.424
E
72.9770
Im 

 5.473  90o A
jaX m
j 40 / 3
I s  I r  I m  33.9167   18.584 A
n
(ii)
1800
2x
60Hz
600
3y
x
20Hz
y
T
TFL
2
TFL
2x
x 3 * 0.05
 S FL 
 0.05
 
 0.075
3y
y
2
 n  600 * 1  s   555 rpm
E
 cons tan t
For constant flux, the
f
x
 Sn   0.075
y
20
At 20 Hz E 
* 218.93  72.977V
60
Rr
0.15
j1.1
Z r 
 jaX r 

 2.03310.39o 
s
0.075
3
E
72.9770
I r 

 35.896  10.39 o A
Z r
2.03310.39
E
72.9770
o
Im 

 5.473  90 A
jaX m
j 40 / 3
I s  I r  I m  37.274  18.694 A
N s  1400  90  1310 rpm
90
s
 0.069
1310
120 * f 120 * f
Ns 

 1310
P
4
 f  43 .66 Hz
43 .66
* 218 .93  158 .6V
60
aErated
158 .6
I r 

 68 .46  159 .78 


Rr / s  ja X r 0.15 /(0.069 )  j 0.73 *1.1
E  aErated 
I m  remais the same  5.473  90o A
I s  I r  I m  68.46 159 .78   5.474   90  70.52 155 .55  A
V  E  Z s I s  158 .60  0.3  j 0.73 *1.1* 70 .52   155 .55
 173 .62   20 .3V