This document summarizes key points from a lecture on induction motors: 1. The lecture objectives were to discuss Thevenin's theorem and its applications to induction motor circuits, derive expressions for torque as a function of slip, and examine torque-slip characteristics with variations in rotor resistance. 2. Thevenin's theorem was used to simplify induction motor circuits into equivalent circuits to analyze how torque varies with slip. Maximum torque occurs at a specific slip and decreases to zero as slip approaches zero. 3. Increasing the rotor resistance shifts the maximum torque point to lower slip but does not change the maximum torque value. This allows control of motor speed at maximum torque.

1. Lecture 3: Third class
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. Lecture Objectives
1- Thevenin’s theorem and it’s applications.
2- Derivation of the expression for gross torque developed as a
function of slip ( or speed) of a three – phase induction motor.
3-Sketch the torque-slip (speed), explaining the various features.
4- Derive the expression of maximum torque and the slip (speed)
at which it occurs.
5- Draw the above characteristics with the variation in rotor
resistance.

3. Simple Review
To understand Thevenin’s theorem and simplify I.M circuit into simple
equivalent circuits using.
➢ Thevenin’s Theorem
➢ Academic Examples
➢ Some of Applications
Remember the three steps process:
1- Find the Thevenin Resistance by removing all voltage sources and load.
2- Find the Thevenin Voltage by reconnecting the voltage sources.
3- Use the Thevenin Resistance and Voltage to find the total current flowing
through the load.

4. What is the Purpose of Thevenin Theorem?
Q: Why Thevenin’s theorem is used?
➢ Thevenin's Theorem provides an easy method for analyzing power circuits,
which typically has a load that changes value during the analysis process. This
theorem provides an efficient way to calculate the voltage and current flowing
across a load without having to recalculate your entire circuit.
➢ Thevenin’s theorem can be used to transform the network that is seen between
points ‘a’ and ‘b’ into an equivalent voltage source Vth in series with equivalent
impedance, i.e., Zth = Rth +jXth
Approximate Equivalent Circuit for I.MThevenin’s Equivalent Circuit for I.M

5. Thevenin’s Theorem
➢ Let us consider XM >> X1 and XM >> R1
That is,
(Voltage Divider Rule)
➢ Exact Thevenin’s Impedance
Approximate Thevenin’s Impedance
➢ Because, XM + X1 >> R1
and, XM >> X1
1
M
TH
M
X
V V
X X

+
2
1
1
1
M
TH
M
TH
X
R R
X X
X X
 
  
+ 

1 1( )/ /TH TH TH MZ R jX R jX jX= + = +
1 1( )
M
TH
M
jX
V V
R j X X
=
+ +

6. New Equivalent Circuit
1 1( )
M
TH
M
jX
V V
R j X X
=
+ +
1 1( ) / /TH TH TH MZ R jX R jX jX= + = +
2 2
1 1
| | | |
( )
M
TH
M
X
V V
R X X
=
+ +
Applying voltage divider rule, it gives: Mechanical Load
I2

7. Torque, Power and Thevenin’s Theorem
➢ The power converted to mechanical form (Pconv) measured in (kW) is:
2
2 2
2
2( )
TH TH
T
TH TH
V V
Z R
R X X
s
I = =
 
+ + + 
 
2 2
2
(1 1
3
) ( )
c Rn Lv Co
s s
P R
s
R
I
s
− −
= =
And, the induced mechanical torque (𝜏𝑖𝑛𝑑) measured in (N.m) is:
conv
ind
m
P


=
(1 )
conv
s
P
s 
=
−
2 2
2 (1 )
(
3
1 )
(1 ) (1 )s s
AG AG
s
s
s
s
R
I
Ps
s
P
  
−
−
= = =
− −
➢ Rotor Current measured in (A) is:

8. Derive mechanical induce Torque
2
2
2
22
2
3
( )
th
ind
s
th th
V R
sR
R X X
s


 
 
   
=   
   + + +    
2 2
2
22
2
3
1
(N.m)
( )
th
ind
th th
s
R
V
s
R
R X X
s


 
 
 =
 
+ + + 
 
, s = 00 as r sind  → →
Or,
Prove That ?
1- At no-load
2- At blocked rotor 1
0 , s = 1asstart ind s r  =
==
3- 𝜏 𝑚𝑎𝑥 =? maxmax , sas r r →

9. Torque-Speed Characteristics
Typical torque-speed characteristics of three – phase induction motor

10. Short Comments
1. The mechanical induced torque is zero at corresponding
synchronous speed (Why?).
2. The torque-speed curve is nearly linear varying between
no-load and full load (Why?).
Ans: In this range, the rotor resistance is much greater than
the rotor reactance, R2 >> X2, so the rotor current and induced
torque increase linearly with the slip value.
3. There is a maximum mechanical torque that can’t be
exceeded. This torque is called breakdown torque (pullout) and
it is amplitude bounded (2 – 3) times of full-load torque.
2
2 2
2
2( )
TH TH
T
TH TH
V V
Z R
R X X
s
I = =
 
+ + + 
 

11. Short Comments, Continue
4. The starting torque of the motor is slightly higher
than its full-load torque, so the motor will start
carrying any load, (How?), it can supply at full
load.
5. The induced torque of the I.M for a given slip
varies as the square of the applied voltage,
Discuss this phrase?
6. If the rotor is driven faster than synchronous
speed it will run as a induction generator I.G, that
is converting mechanical power to electric power.

12. Torque - Slip Characteristics of an Induction Motor
Ns > Nr Ns < Nr

13. Derive Maximum Slip?
➢ Maximum torque occurs when the power transferred to R2 /s is
maximum (How?).
➢ As you know, the maximum torque condition occurs , iff
R2 /s = ZT = Rth + j (Xth + X2)
max
2 22
2( )th th
T
R
R X X
s
= + +
max
2
2 2
2
Prove that?)(
( )
T
th th
R
s
R X X
= 
+ +
➢ So, the slip factor corresponding to maximum torque is:
Or,
𝒅𝝉
𝒅𝒔
=0 𝒔 𝑻 𝐦𝐚𝐱
➢ Use + ve slip for motoring mode
➢ Use - ve slip for generating mode
I2
ZT

14. ( ) ( )22max
rtt
r
XXR
R
s
++

=
0=
ds
dT
( )
0
/3
2
2
2
=












++




 
+

rt
r
t
rt
ms
XXj
s
R
R
sRV
ds
d

( ) 











++




 
+

=
2
2
2
/3
rt
r
t
rt
ms
XXj
s
R
R
sRV
T

( ) 







++
=
22
2
max
2
3
rttt
t
ms XXRR
V
T

Derive Maximum Torque?
Induced torque

15. Maximum Torque for Wound Rotor WRIM
➢ Rotor resistance can be increased by inserting external
resistance in the rotor of a wound-rotor WRIM type.
➢ The value of the maximum torque remains unaffected.
➢ The operating speed corresponding to maximum torque
can be controlled.
➢ What about the starting torque of three – phase WRIM?

16. Maximum Torque for a WRIM Type
Effect of increasing rotor resistance on torque-speed characteristic
Q1: How torque – slip Curve vary when adding resistance to the rotor circuit?
Increasing rotor resistance

17. Maximum Starting Torque of a three – Phase IM
𝑘2 =
3 𝑉𝑡ℎ
2
2𝜔𝑠
Q2: What is the condition for maximum starting torque of a 3 phase WRIM?
Approximately:

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

19. Solution
1.
2.
120 120 50
3000 rpm
2
3000 295
1.6
0
*100 7% *
3000
%100sync m
sync
e
sync
f
n
n
P
n
s
n
− −
= ==

= = =
3
Note: Since, friction and windage losses are not given in this question:
Let us consider that is,
2
6
15 10
48.6 N.m
2950
0
ind loacon d
conv
i
v loa
nd
m
d
f W
P
P
rpm
P  



+

=
= =
=
=


20. Solution
3. In low-slip region, the torque - speed curve is linear (R2>> X2)
and the induced torque is direct proportional to slip. So, if the
induced torque is doubled, the new slip will also be doubled, that
is 2 * 0.167= 0.333 i.e., 3.33% , and the new motor speed is:
(1 ) (1 0.0333) 3000 2900 rpmnewm new syncn s n= − = −  =
4 ( 48.6) ( ) kW, also
2
doubl2 2900 29.5
60
edconv ind m rpP m

− = =   =

21. Academic Example No.2
A 460V, 25hp, 60Hz, four-pole, star connected wound-rotor three –
phase induction motor WRIM has the following impedances
measured in  / ph. referred to the stator side circuit :
R1 = 0.641 / ph. R2 = 0.332  / ph.
X1 = 1.106  / ph. X2 = 0.464  / ph XM = 26.3  / ph.
1. Compute the maximum torque of this motor? At what
corresponding motor speed and slip factor does it occur?
2. Calculate the starting torque of this 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?

22. 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 prepare Thevenin’s
equivalent circuit :

23. Solution
➢ The slip factor corresponding to maximum mechanical 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= − = −  =

24. Solution
➢ The maximum mechanical induced torque at this 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.

25. 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

26. Solution
3. If the rotor resistance is doubled, consequently the slip at maximum
torque is doubled for linear operation region.
➢ Note: 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= − = −  =

27. Solution
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.

28. 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?