Solved Examples for Three - Phase Induction Motors

1. Tutorial
Electrical Engineering Department
College of Engineering
University of Kerbala
2019/2020
By
Asst. Prof. Dr. Ali Altahir
http://learning.uokerbala.edu.iq/moodle
EEE-32II

2. Academic Example No.1
A two-pole, 50Hz, three–phase design B I.M (for thermal
insulation) supplies 15kW a mechanical load at rotor
speed of 2950 rpm.
Do as required for the followings:
1. What is the percentage motor’s slip at specified rotor
speed ?
2. What is the corresponding induced torque in N.m
under these conditions?
3. What will be the operating speed of I.M, if its induced
torque is doubled in this case?
4. How much the gross power (i.e, convereted) that will
be given by I.M when the induced torque is doubled ?
I.M Load
ώ

3. Solution
1.
2.
120 120 50
3000 rpm
2 2
3000 2950
0.0167*100 1.67% *
3000
100 %sync m
sync
e
sync
n n
s
n
f
n
p

  

  


3
Note: Since, friction and windage losses are not given in question:
Let us consider that is,
2
15 10
48.6 N.m
295
60
0
ind loadconv
conv
in
l ad
m
o
d
f W
P P
P
rpm






 





4. Solution: Cont.
3. In low-slip region, the torque - speed curve is linear (R2>> X2) that
is the induced torque is direct proportional with slip. So, if the
induced torque is doubled, the new slip will, also be doubled, that is
2 * 0.0167= 0.0334 i.e., 3.34% , and the new motor speed is:
(1 ) (1 0.0334) 3000 2900 rpm < rpm2950newm new syncn s n     
4 ( 48.6) ( ) kW, also
2
doubl2 2900 29.5
60
edconv ind m rpP m

     

5. Academic Example No.2
A 460 V, 25hp, 60Hz, four-pole, star connected WRIM has the
following parameters measured in  / ph. referred to the stator
side circuit :
R1 = 0.641 / ph. R2 = 0.332  / ph. RM = 0  / ph
X1 = 1.106  / ph. X2 = 0.464  / ph XM = 26.3  / ph.
1. Compute the maximum torque of this motor? what are
corresponding motor speed and the slip does it occur?
2. Calculate the starting torque of the specified motor?
3. If the rotor resistance is doubled, state the motor speed
corresponding to the maximum torque now occur? What will be
the new starting torque?
4. Sketch Torque -slip curve for both cases shown above.
5. Check the stability of the motor at 1750 and 1300 rpm?

6. Solution
2 22 2
1 1
460
26.3
3 255.2 V/ph.
(0.641) (1.106 26.3)( )
th
M
MX
V
R X X
V
 

  
 
1
2
2
1
26.3
(0.641) 0.590
1.106 26.
.
3
/
M
th
M
X
R
X X
ph
R
 
  

 
 


 


1, 1.106 / .thX pA hnd X 
Note: To compute the maximum torque, it is necessary to prepare Thevenin’s
equivalent circuit :

7. Solution
 The slip corresponding to maximum torque is:
1. R2 /s = ZT = Rth + j (Xth + X2)
 The corresponding motor speed for maximum torque is:
max
2
2 2
2
2 2
( )
0.332
(0.59) (1.106 0.4
0.198 1
64)
th th
R
s
R X X
 
 
 



maxmax. (1 ) (1 0.198) 1444 rpm1800m old syncn s n rpm     

8. Solution
 The maximum induced torque at this motor speed is:
2
2
2
m
2
max
ax 2 2
2
3 (255.2)
2
2 (18
31
(prove that?
00 )[0.59 (0.59) (1.106 0
)
2 ( )
.464) ]
60
229 N.m
th
s th th th
rpm
V
R R X X




 
 
  


 

 
  

2. The starting torque can be found from the induced torque
equation by substituting, s = 1.

9. Solution
 
2 2
21
22
2
1
2
2
2 2
2 2
2
2 2
max.
3
1
( )
3
[ ( ) ]
3 (255.2) (0.332)
1800 [(0.59 0.332) (1.106 0.464) ]
104 N.m
2
0
<
6
th
start ind s
s
th th
s
th
s th th
start
R
V
s
R
R X X
s
V R
R R X X
rpm









 
 
  
 
   
 

  
 

    

5- At motor speed, 1750rpm,s =0.028 < 𝑠𝜏max
= 0.396 I.M is stable
At motor speed,1000rpm,s = 0.444 > 𝑠𝜏max
= 0.396 I.M is unstable

10. Solution
3. If the rotor resistance is doubled, consequently the slip at maximum
torque is doubled for linear operating region.
 Specify type of electrical connection related to insertion external rotor resistance?
4- The corresponding motor speed is:
 The maximum torque is still as it is: max = 229 N.m
max
2
2 2
2
0.396 1 unitles
2
s
( )
*
th th
s
R X X
R
   
 
maxmax. (1 ) (1 0.39 < 14446) 1800 1087 rpm rpmm new syncn s n rpm     

11. Now, the new starting torque could be computed, in case the rotor
resistance is doubled :
2
,,
, 1
2 2
(0.664)
0.664
3 (255.2)
1800 [(0.59 ) (1.106
170 N.m
2
60
0.464) ]
= 104 N.m>st
start new in
art ne
d s
start oldw
rpm
 




 
 
    

4- Sketch Torque -slip curve for both cases shown above.

12. Review Questions
Q1: What is the relation between torque and speed in an induction motor?
Q2: How torque is developed in a three phase induction motor?
Q3: How can we increase the starting torque of an induction motor?
Q4: How do you control the speed of a 3 phase induction motor?
Q5: State maximum torque condition of a three – phase Induction Motor?
Q6: What is full load torque in induction motor?
Q7: Why a three phase induction motor is self starting?
Q8: Why the slip is never zero in an induction motor?
Q9: What are the effects of increasing rotor resistance on starting current
and starting torque?
Q10: Why induction motor has high starting torque?
Q11: Which motor has a high starting torque?