Electrical Machines: Part 2b (ii): Induction Motors, slip and equivalent circuit
This is part 2b (ii) of topic 2 in the series of "Electric Machines” and in this part, we will be
discussing the induction motor slip
and equivalent
circuit.
We started this series with introduction to AC machines which then led to Induction motors. In Part 2b (i)
we talked about the construction and working principles of induction motors and in this part we will be talking about the concepts of
slip and introduce you to the equivalent circuit of induction motors. We have a long way to go with this
topic so let us get started!
So let’s start with slip: As discussed earlier, an induction
motor has a rotating magnetic field which rotates at a speed known as synchronous speed or
‘n-sync’, the rotor however, rotates at a speed slightly less than the synchronous speed. This difference in the synchronous
speed and rotor speed is known as slip … (pause) … speed.
Another way of representing this slip speed is by comparing it to the synchronous
speed in fractional form i.e. slip speed divided by synchronous speed. This quantity is known as slip …
(pause) … only. And this is a good numerical indicator of the
difference between slip speed and synchronous speed.
This equation can also be written as:
Let’s see a solved example i.e. Electric Machinery Fundamentals by Chapman’s example#7-1
Consider a 4 pole, 60 Hz motor with speed of 1710 rpm (nrotor). What will be the value of slip for this motor?
As we know the formula for slip of motor is: First we will find out value of synchronous
speed , then by putting it in the above formula we will get the value of slip. So synchronous speed is given by
120 times the electric frequency divided by the number of poles of the motor:
As given frequency is 60 Hz and poles are 4, so by putting these values in formula of nsync we will get magnitude of synchronous speed
i.e. 1800 rpm.
Now putting this value of nsync and also given value of Rotor speed nrotor in formula of slip: Finally we have value of slip which is 5% or we can say 0.05.
Moving on, there is another term that is used frequently in induction motor that is the rotor frequency which is the product of the electrical
supply frequency at stator ‘fs’ and slip ‘s’
You may ask why the rotor speed is always less than the synchronous speed, this is because same values of synchronous speed
and rotor speed will result in zero relative speed, this means there is no emf induced in the rotor and therefore no rotor currents exist.
So, the rotor slows down. Upon slowing down, rotor current is induced again and the relative speed is attained again causing the rotor to
rotate again
smoothly. Now before we move on to the discussion about induction motor equivalent circuit. Here is the sponsor’s pitch: Returning Back:
The equivalent circuit of an induction motor is similar to that of a transformer. Due to the fact that both machines use the same phenomena
of induction that is, a changing magnetic field created by primary winding induces voltage in secondary, this is conceptually similar to the
induction of voltage in a rotor through the stator windings the only difference being that the ratio of primary and secondary sides is
1..So, we can say that transformer primary winding is similar to stator and secondary winding similar to rotor.
Looking at the per phase equivalent circuit of induction motor, as current flows through the stator and rotor windings, heating effect
through resistive losses in motor winding and self-induction in iron core exist in stator. In the figure of equivalent circuit for induction motor,
this resistance is taken as ‘R1’ and the inductive reactance is taken ‘X1’ in series, …(pause)… meanwhile in the rotor, resistance is
‘Rr’ and a reactance ‘Xr’ in series. The core resistance ‘Rc’ corresponds to resistive core losses and inductive reactance ‘Xm’
corresponds to the mutual field between the stator and rotor. ‘Rc’ and ‘Xm’ are placed in parallel.
Next we have ‘aeff’ or the effective turns ratio, this is the ratio between per phase number of turns of stator winding to that of rotor windings
per phase . Now if the rotor speeds up, the slip changes
from 1 to some lesser value, this change of speed in rotor changes the impedance of rotor circuit, which is expressed in equivalent rotor
circuit as rotor resistance ‘Rr’ divided by slip. As shown in figure, total rotor impedance can
be series summed as Rr/s + jXRo
Now for the final equivalent circuit, all we have to do, is to refer or shift the rotor circuit to the stator side and eliminate the aeff air gap
between the stator and rotor, as we do in transformer analysis … i.e. refer secondary circuit to the primary circuit. This can be done
by: • First multiplying the effective turn ratio ‘aeff’ with rotor voltage ‘ERO’. This product can
be denoted as ER’. hence: ER’ = aeffERO and
ER’ will be represented in final equivalent circuit as E1 • Second thing is multiplying the square of
‘aeff’ with the rotor impedance to refer it, we can express the obtained result as Z2. Z2 = aeff2(Rr/s + j XRo) or Z2
= aeff2Rr/s + j aeff2XRo
This can be simplified as: X2 = aeff2XRo R2 = aeff2 * Rr R2 = aeff2*Rr And then
We will model the series sum of rotor impedance i.e. Z2 in final equivalent circuit simply as R2/s and jX2 which represents the
reffered value of Rotar resistance and reactance, that may be directly measured or calculated from Rr and XRo if aeff is known. •
And finally dividing rotor current ‘IR’ by effective turns ratio ‘aeff’ I2 in final equivalent circuit can be expressed as:I2 = IR/aeff So
here is the final equivalent circuit for an induction motor which is modelled after referring the rotor side to stator side.
That’s all from this part. In the next video we will be discussing about torque speed characteristic of induction motor. References: [1] S. J. Chapman, Electric Machinery Fundamentals, 4th Edition, WCB/McGraw-Hill, 1999.