1. Introduction: The Induction motor is a three phase AC motor and is the most widely used machine. Its characteristic features are-
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 t_{1}, t_{3} and t_{5}.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 N_{s}=120*(f/P)
rpm. i.e., w _{s} = 2*p
* N_{s }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} - w_{m})
(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 R_{1 }is the stator resistance per phase
X_{1} is the stator reactance per phase
per phase
per phase
R_{c }is the resistance representing core losses
X_{m} is the magnetizing reactance per phase
V_{1} 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 P_{g}
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 Z_{th}=R_{th}+jX_{th} is the Thevenin’s equivalent impedance.
Now Z_{th }is the parallel combination of the shunt branch and Z_{1} 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 s_{maxT} that is the slip at maximum torque T_{max}.
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 R_{th }and the equation
for torque is obtained as-
We now get the relationship between the starting torque T_{start},
maximum torque T_{max} 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 k_{2}/s=0.
Hence the torque produced is proportional to the speed N_{m}. However
when the motor attains stable speed, slip is negligible.Hence k_{3}.s
=0 and the torque is inversely proportional to the speed N_{m }.
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, R_{2}’ and frequency.
(a) Applied voltage : We know that T µ
V^{2}. 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 R_{2}’. However, with increase in R_{2}’,
the starting torque increases. The slip at which T_{max} is reached
increases too which means that T_{max} 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 T_{L }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 T_{c }. The other component is T_{v }i.e., viscous friction.
and T_{c} is called Coulomb friction.
The third component is T_{s} that accounts for the additional torque present at stand-still. Since T_{s} 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 :
(a) Synchronous speed N_{s} = (120*f)/P
= (120*60)/4
Thus N_{s}= 3600 rpm
(b) T_{start} =1.6(T_{d}) and T_{dmax}=2(T_{d})
Hence 0.8(s_{Tmax}^{2})-2(s_{Tmax})+1=0
S_{Tmax} = 2 or 0.5
Since slip for a normal induction motor ranges between 0 and 1, s_{Tmax}=0.5
(c) N_{Tmax} = (1-s)N_{s}
= (1-0.5)3600
N_{Tmax }= 1800 rpm
(d) We have the relation
Hence s^{2}+s_{max}^{2}-4(s)(s_{max})=0
i.e., s^{2}+2s+0.25=0
s = 0.134
Hence Hp=50.536
For Class B: from the fig. below T_{start} is almost equal
to160% of T_{full}
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.
According to the first interpretation rotor frequency is w_{m}=
(1-s). w _{s}
w _{m}= (1-s) w_{s}
= (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
(c) We need to know the rotor speed at one half of the rated torque now.
Hence the torque now is 0.5T_{rated}=0.5(P_{rated}/w_{rated})
At T_{rated} we know that w _{mrated = }154.6 rad/sec
However since in this case R_{c} is ¥
and hence V_{th}=V_{1}.
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 < s_{Tmax}. Hence this is the
actual operating point for the loaded Induction motor. So s=0.00678 and
w
_{m}=156 rad/sec .
(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 R_{2}’.
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
R_{2}’ 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 R_{2}’ in denominator is squared and thus the starting torque will decrease as R_{2}’
increases.
(c) The equation for T_{max} is-
Thus we see that T_{max} is independent of R_{2}’and hence does no change even if rotor
resistance is changed.
(d) w _{s} is a constant and hence does not change.