1. Introduction: The Induction motor is a three phase AC motor and is the most widely used machine. Its characteristic features are-

An Induction motor has basically two parts – Stator and Rotor

The Stator is made up of a number of stampings with slots to carry three phase windings. It is wound for a definite number of poles. The windings are geometrically spaced 120 degrees apart. Two types of rotors are used in Induction motors - Squirrel-cage rotor and Wound rotor.
 
 

A squirrel-cage rotor consists of thick conducting bars embedded in parallel slots. These bars are short-circuited at both ends by means of short-circuiting rings. A wound rotor has three-phase, double-layer, distributed winding. It is wound for as many poles as the stator. The three phases are wyed internally and the other ends are connected to slip-rings mounted on shaft with brushes resting on them. The brushes are connected to an external resistance that does not rotate with the rotor and can be varied to change the N-T characteristics. In fact an Induction motor can be compared with a transformer because of the fact that just like a transformer it is a singly energized device which involves changing flux linkages with respect to a primary(stator) winding and secondary(rotor) winding.
 
 

2. Production of rotating magnetic field : An Induction motor operates on the principle of induction .The rotor receives power due to Induction from stator rather than direct conduction of electrical power. It is important to understand the principle of rotating magnetic field in order to understand the operation of an Induction motor. When a three phase voltage is applied to the stator winding , a rotating magnetic field of constant magnitude is produced. This rotating field is produced by the contributions of space-displaced phase windings carrying appropriate time displaced currents. These currents which are time displaced by 120 electrical degrees are shown below-



We will now consider a stator structure depicted along with three phase windings. For convenience, each phase is represented by a single coil (though the winding is distributed in practice).The coil a-a’ represents the entire phase winding for phase a. Similarly b-b’ and c-c’ represent the coils for phases b and c. Each phase winding produces flux along its own flux axis and these axes are separated by 120 electrical degrees. We will now determine the magnitude and direction of resultant flux corresponding to three time instants say t1, t3 and t5.We know that the instantaneous flux of each phase is given by-
 
 
 
 
 
 
 
 




 

 


 
 
 
 


Therefore from the diagram we have-
 
 
 
 




 
 





 
 
 
 
 
 
 
 


Thus we see that at all time instants, the magnitude of resultant vector is constant though the vector rotates at a speed of Ns=120*(f/P) rpm. i.e., w s = 2*p * Ns rad/sec

For three phase supply, this constant magnitude is 1.5 times the maximum value. The speed w s (rad/sec) is called Synchronous Speed. The rotating field induces an emf. In the rotor-circuit .Current flows through the short-circuited rotor windings to produce a flux. The rotor tries to catch up with the stator field but in an Induction this is not possible (since this can be done only by using another starting motor ). If say w m (rad/sec) is the rotor speed, then the difference between stator and rotor flux speeds is (w s - wm) (rad/sec). The slip is now defined as-
 
 



When the rotor is stationary, the frequency of rotor current is same as the supply frequency. But when the rotor starts revolving, the rotor current frequency becomes dependent on the relative speed. Let f’ be the rotor current frequency at any slip say s.
 
 

We have seen that when a three phase supply is given to stator windings, a rotating field is produced. This flux of constant magnitude cuts the stationary conductors of rotor. This induces an emf in the conductors, according to Faradays law of electro-magnetic induction. Since the rotor conductors form a closed circuit, a current is induced in them whose direction as per Lenz’s law is to oppose its very own cause i.e., the relative speed between the rotating flux and the stationary rotor conductors. This tendency to reduce the relative speed results in a torque that rotates the rotor in the direction of the magnetic field.
 
 
 
 


 

3. Equivalent Circuit of an Induction motor: The exact equivalent circuit model of an Induction motor is-


 
 
 
 

where R1 is the stator resistance per phase

X1 is the stator reactance per phase

R2' is the equivalent rotor resistance referred to stator

per phase

X2' is the equivalent rotor reactance referred to stator

per phase

Rc is the resistance representing core losses

Xm is the magnetizing reactance per phase

V1 is the per phasesupply voltage to the stator

s is the slip of the motor
 
 

4. Power flow in an Induction motor :


From the circuit, we see that the total power input to the rotor Pg is

The power flow diagram is-

 

 

 

5. Torque considerations in an Induction motor :

For obtaining the expression for torque we consider the Thevenin’s equivalent of the actual circuit-

From the circuit we see that Zth=Rth+jXth is the Thevenin’s equivalent impedance.

Now Zth is the parallel combination of the shunt branch and Z1 i.e., the stator impedance. So from the circuit we have-



The torque developed T is given-


To obtain the starting torque we put s=1 in the above equation.

To obtain the maximum torque produced, the condition is dT/ds=0.

Applying this condition we get smaxT that is the slip at maximum torque Tmax.


We then put in this value of torque in the expression for torque to get the value of maximum torque.


Now to simplify the equation we neglect Rth and the equation for torque is obtained as-


We now get the relationship between the starting torque Tstart, maximum torque Tmax and T i.e., the torque developed at a slip say s.


The torque equation obtained above can be expressed as-


 
 
 
 
 
 

Initially when the motor starts, the slip is high. So k2/s=0. Hence the torque produced is proportional to the speed Nm. However when the motor attains stable speed, slip is negligible.Hence k3.s =0 and the torque is inversely proportional to the speed Nm . From these relationships, the general shape of speed -torque characteristics of Induction motor can be obtained.
 
 

6. Factors affecting the speed-torque characteristics of an Induction motor : The speed-torque characteristics are affected by various factors like applied voltage, R2’ and frequency.

(a) Applied voltage : We know that T µ V2. Thus not only the stationary torque but also the torque under running conditions changes with change in supply voltage.
 
 
 
 

(b) Supply frequency : The major effect of change in supply frequency is on motor speed. The starting torque is reduced with increase in frequency.

(c) Rotor resistance : The maximum torque produced does not depend on R2’. However, with increase in R2’, the starting torque increases. The slip at which Tmax is reached increases too which means that Tmax is obtained at lower motor speeds.
 
 


7. Dynamics of motor-load system : Any motor-load system can be described by the equation-

The load torque TL can be further divided as-

(a) Friction torque : Friction is present at the motor shaft and also in various parts of the load. The friction torque is equivalent value of various friction torques referred to the motor shaft.

(b) Windage torque : The opposing torque generated by wind when the motor runs is called Windage torque.

(c) Torque required to do useful work : The nature of this torque depends on the type of load. It may or may not be a function of speed, it may or may not be time invariant.

The friction torque itself can be resolved into three components – friction at zero speed i.e., static friction Tc . The other component is Tv i.e., viscous friction.

and Tc is called Coulomb friction.

The third component is Ts that accounts for the additional torque present at stand-still. Since Ts is present only at stand-still, it is not considered for dynamic analysis.

The speed-torque requirements of various types of loads are –
 
 





Load torques are of two types- Active and Passive. Active torques have the potential to drive the motor under equilibrium conditions. They retain their sign even when the direction of the drive rotation is changed. Torque due to gravity is an example of this type of torque.

Load torques which oppose motion and change their sign on reversal are Passive. An example of this is the torque due to friction.

Thus the N-T characteristics of an Induction motor are modified due to the type of load.

8. NEMA specifications : The National Electrical Manufacturers Association (NEMA) standardized four basic design categories of Induction motors to match the torque-speed requirements of the most common types of mechanical loads. These basic design categories are Design A, Design B, Design C and Design D.

The Design B motor serves as the basis for comparison of motor performance with other designs. It has the broadest field of application and is used to drive centrifugal pumps, fans, blowers and machine tools. It has a relatively high efficiency, even at light loads, and a relatively high power factor at full load.

The Design A motor has essentially the same characteristics as the Design B, except for somewhat higher break-down torque. Since its starting current is higher, however, its field of application is limited.

The Design C motor has a higher locked-rotor torque, but a lower break-down torque than the Design B. The higher starting torque makes it suitable for driving plunger pumps, vibrating screens and compressors without unloading devices. The starting current and slip at rated torque are essentially the same as that for the Design B.

The Design D motor has a very high locked-rotor torque and a high slip. Its principal field of application is in high-inertia loads such as fly-wheel equipped punch presses, elevators and hoists.

The Design D rotor has relatively high-resistance, low-reactance rotor bars close to the surface. Design B and Design A have low-resistance with high reactance at the deeper bars.
 
 

9. Examples :

  1. A three phase Y connected, 60 Hz, 2 pole induction motor is operating at its rated voltage and frequency. It develops a starting torque of 1.6 times the full-load torque, and a maximum torque of 2 times the full load torque. Determine
  1. The synchronous speed
  2. The slip at maximum torque
  3. The speed at maximum torque
  4. The slip at full load
  5. The speed at full load
Solution:

(a) Synchronous speed Ns = (120*f)/P
= (120*60)/4

Thus Ns= 3600 rpm


(b) Tstart =1.6(Td) and Tdmax=2(Td)

Hence 0.8(sTmax2)-2(sTmax)+1=0

STmax = 2 or 0.5

Since slip for a normal induction motor ranges between 0 and 1, sTmax=0.5
 
 

(c) NTmax = (1-s)Ns

= (1-0.5)3600

NTmax = 1800 rpm
 
 


(d) We have the relation
 
 

Hence s2+smax2-4(s)(smax)=0

i.e., s2+2s+0.25=0

s = 0.134
 
 

  1. Nm=(1-0.134).3600 =3118 rpm
    1. A three phase induction motor is selected to drive a particular load in a plant that operates continually during each working day. The load requires a normal running torque of 200N.m. at a motor speed of 1800 rpm. Occasionally the motor must start under heavy load requiring a torque of 280N.m. It is also essential that the motor be capable of starting with supply voltage as low as 80% of normal. The plant has 440 V, three phase, 60 Hz supply.
  1. What must be the continuous horse power rating of the motor?
  2. If you have to select only from Class B, class C, class D motors, which motor satisfy the conditions of the load?
Solution:
  1. To calculate the horse power rating we know that



  2.  
     

    Hence Hp=50.536


    For Class B: from the fig. below Tstart is almost equal to160% of Tfull

It cannot operate the load at 280 N.m.

Class C motor can operate the load.

For Class D: From the figure below, we see that the motor can provide the required starting torque of 280N.m. However the slip at normal operating conditions is very high. The efficiency will be very poor.
 
 

    1. A three phase Y connected, 400 hp, 380 V(l-l), 50 Hz, wound rotor Induction motor operating at rated conditions has a slip of 0.0159. The machine parameters expressed in ohms are: Rth=0.00536 ohms; R2´=0.00613 ohms; Xth=0.0383ohms; X2´=0,0383 ohms. Determine (a) the rotor frequency (b) the slip at which maximum torque occurs (c) rotor speed at one half of the rated torque (d) the external resistance per phase required to operate the machine at 1000 rpm and at one half of the rated torque load(assume the motor is operating on the linear section of the torque curve)
Solution:
  1. This problem can be interpreted in two ways-frequency (i.e., speed ) of rotor or frequency of rotor currents.

  2. According to the first interpretation rotor frequency is wm= (1-s). w s
     
     


    w m= (1-s) ws = (1-0.0159).157.1 =154.6 rad/sec

    So w m= 154.6 rad/sec = 1476.2 rpm

    According to the second interpretation, the frequency of rotor currents f¢ is given



    f¢ =s.f=(0.0159)(50)=0.795 Hz
     
     

  3. Let sTmax be the slip at which maximum torque is obtained .

  4.  

     
     

    (c) We need to know the rotor speed at one half of the rated torque now.

    Hence the torque now is 0.5Trated=0.5(Prated/wrated)

    At Trated we know that w mrated = 154.6 rad/sec


    However since in this case Rc is ¥ and hence Vth=V1.

    Substituting this in the above equation for torque we get,


    To decide the correct value of speed from above values, we look at the following figure -
     
     
     
     

    It is seen that the load curve intersects the Induction motor curve on the right hand side where s < sTmax. Hence this is the actual operating point for the loaded Induction motor. So s=0.00678 and w m=156 rad/sec .
     
     

  5. The speed here is Nm=1000 rpm and torque is half that of the rated value that is T=965.1 Nm.
Let us now add an external resistance say Rex to the armature resistance R1. Let the new armature resistance be Rnew= R1+Rex

    1. A load for which the torque-speed characteristic is constant, is driven by a three phase wound rotor Induction motor. If the rotor resistance of the Induction motor is increased, indicate whether the following quantities will increase, decrease or remain constant:
(a) The rotor speed (b) The slip (c) The starting torque (d) The maximum torque and (e) Synchronous speed Solution:

(a) We know that the torque equation is-
 


 

In this equation we see that rotor resistance appears at several places and hence it is not easy to see how s(or speed) varies with R2’. The analytical method of root locus technique can be used for this. However it is beyond the scope of this module. Here we just change the value of R2’ and see how the value of s or (speed) changes. Hence the value of rotor resistance is changed from 0.00163(from problem 9.3) to say 0.01.We calculate the s(speed) value for this value of rotor resistance from the torque equation.
 
 

Hence s= 1.582 or 0.0110

However for normal mode of the Induction motor s<1.

Thus we take s=0.0110

Thus w m= (1-0.0110)157.1=155.37 rad/sec

We see that when the rotor resistance is increased, the speed decreases.


(b) In case of starting torque s=1.

Here R2’ in denominator is squared and thus the starting torque will decrease as R2

increases.
 

(c) The equation for Tmax is-

Thus we see that Tmax is independent of R2’and hence does no change even if rotor

resistance is changed.

(d) w s is a constant and hence does not change.