40
Induction Motors (Three Phase)
OBJECTIVES
In this chapter you will learn about:
 Rotating magnetic fields
 Direction of rotation
 Slip and its effect
 Construction of a threephase induction motor
 WRIM and SCIM
 Losses and efficiency
 Maximum power
 Induction motor characteristics
 Starting techniques for induction motors
 Induction motor parameters
 Simple problems on the above topics
 Induction motor with squirrel cage motor
40.1 INTRODUCTION
The threephase induction motor has a simple yet exceedingly robust construction which is comparatively cheap to manufacture. It also has good operating characteristics that make it a suitable drive for many production machines such as lathes or fans. These features are reflected by the fact that the induction motor is the most commonly used type of a.c. motor. The induction motor is chosen for its simplicity, reliability and low cost. These features are combined with good efficiency, good overload capacity and minimal or no service requirement. When the facts of very wide availability and simple installation by relatively little trained personnel are added, the choice of an induction motor seems well founded.
The motor is supplied from a threephase source and, therefore, it requires three supply conductors that are connected to three windings attached to the stator. The rotor also has a conductor system but this is not connected to its supply. Instead, the current in the rotor conductors is induced by a transformer action from the stator windings. As a result of this transformer action, the stator windings are sometimes termed the primary windings and the rotor windings may be termed the secondary windings. Thus, the threephase induction motor is a sort of transformer with a rotating secondary winding.
There are two principal methods of connecting the rotor conductors together and this gives rise to two classes of induction motor. These are: the cagerotor induction motor and the woundrotor inductor motor. In either case, the principle of operation is essentially the same and this depends on the action of the stator windings which gives rise to a rotating magnetic field.
40.2 THE ROTATING MAGNETIC FIELD
The most simple form of threephase stator winding is embedded in the inner surface of the cylindrical stator. This form of construction is illustrated is Figure 40.1.
Figure 40.1 Simple Threephase Stator Construction
The windings are spaced at 120° intervals around the stator surface. One may be tempted to think that there is only 60° between the windings but when describing the angle, we consider the difference to be between the start of each winding the start being the point at which the supply is connected. We can see that there is 120° between the external connection of the Rphase and that of the Yphase, whilst there is a further 120° to the connection of the Bphase. A further 120° remains between the Bphase and Rphase, thus, completing a revolution of the stator surface.
The windings are inserted into the stator surface in a manner similar to that already described with the case of armature winding in the d.c. machine. For convenience, each phase winding is shown to be distinctly separate from the winding supplied by the next phase. This makes it easy to observe the different windings, but in practice, the windings would be spread out to cover the entire surface of the stator.
Each winding has a distinct beginning and an end. Such windings are called phase windings, and are not to be confused with the commutator windings which form closed loops tapped by means of commutators. The start terminals of the phase windings are connected to the supply lines. In the case shown in Figure 40.2, the finish terminals are connected together. Thus, the windings are star connected but there is no reason why the windings should not have been delta connected instead.
Figure 40.2 The Pulsating Flux Due to Alternating Current in a Single Coil
Each winding sets up a flux which acts along the axis of the winding. This action is shown in Figure 40.2, showing the field arrangement for only one winding. By considering different instants during a cycle of alternating current flowing in the coil, we can see that the flux always acts in the same axis but with varying magnitude and with alternating direction.
In the threephase system there are three windings, each producing a flux and each acting in a different direction within the stator. If we combine the three fluxes, we find that a similar magnetic field is set up within the stator except that the magnitude of the total field is bigger than that associated with any one winding. Actually, it is 50 per cent larger and for the arrangement shown in Figure 40.1, we can see in Figure 40.3 the manner in which the three fluxes add together to give the larger total flux.
This diagram has been drawn for the instant at which the Rphase current is at its maximum value. It follows that the currents in other two phases at the same instant are half the maximum value and flowing in opposite directions. By applying our observations of Figure 40.3, the Rphase winding produces its maximum flux and the axis of the field is shown by an arrow which indicates the direction of the field. The fields of the other two windings act in relatively opposite directions to that of the first winding because the currents in these other windings are flowing in ‘negative’ directions. Effectively, the three separate fields are similar, although not in identical directions and their combined effect is to produce a field greater than that of the Rphase winding and acting in the direction of its field. We can use the Right Hand Grip Rule to confirm the direction of the separate phase fields and, hence, the direction of the combined field.
Figure 40.3 Field Distribution in a Threephase Stator with Threephase Windings
With one winding we could obtain a certain field. With three windings, we have only increased the resultant field by 50 per cent, which is not a particularly substantial increase. The importance of this arrangement only becomes apparent when we consider subsequent instants.
In Figure 40.4, for example, we can once again look at the flux arrangement as was considered in Figure 40.3, but at an instant 30° later in supply cycle. At this instant, the current in the Rphase has fallen to 0.87 of its maximum value, while that in the Yphase is zero. The current in the Bphase has risen in value to 0.87 of its maximum value and it is again flowing in the negative direction. Figure 40.4 shows the separate phase windings fields and the resultant combined field.
Figure 40.4 Field Distribution in a Threephase Stator with Threephase Windings at an Instant Later than that of Figure 40.3
In this case, we again produce a somewhat bigger resultant field than is produced by one winding but the important observation is that the axis of the resultant field has shifted from that seen in Figure 40.3, the field axis has been shifted by 30°. We have seen that a field set up by three windings fixed in space produces a magnetic field whose axis has shifted relative to the windings.
It should be sufficient to draw the resultant field only for a selection of instants throughout the supply cycle. This has been illustrated in Figure 40.5 and, by taking instants throughout the complete cycle, it will be seen that the resultant field rotates one complete rotation. We conclude that a system of windings fixed in space and excited from a threephase a.c. supply produces a rotating field.
Figure 40.5 Rotating Magnetic Field
Note: It has been assumed that the construction of the threephase machine is symmetrical, i.e., the windings are identical and displaced at equal angles from one another. It has also been assumed that the supply is symmetrical and, subsequently, the phase currents have all been of equal maximum instantaneous values. This symmetry is implicit to all rotating machine studies.
40.3 SPEED OF THE ROTATING MAGNETIC FIELD
Spaces have been left between the conductors of one phase winding and the conductors of the next phase winding. This failure to use all the surface space of the stator is wasteful and in practice, conductors cover the entire stator surface area, as indicated in Figure 40.6(a). However, drawing the individual conductors is time consuming and it is usual to indicate the winding groups of conductors by shaded areas as shown in Figure 40.6(b).
Figure 40.6 Threephase, Two Pole Winding Arrangement
Each phase winding has been shown to set up its own field, which acts from one side of the stator across to the other. This has been shown in Figures 40.7 and 40.8. Seen from the rotor, it would seem that the field emanates from a Npole at one side and terminates in a Spole at the other. The same observations may be made of the resultant field, and for this reason, the winding arrangement is described as a twopole winding.
Figure 40.7 (a) Elementary Threephase Induction Motor (b) Threephase Flux Waves and (c) Instantaneous Direction of Resultant Stator Flux
By using six windings instead of three, it is possible to have a fourpole machine, while nine windings give a sixpole machine, and so on; although unusual, it is quite practical to have a machine with about a 100 poles. The winding arrangements of fourpole and sixpole machines are shown in Figure 40.8.
Figure 40.8 Four and Sixpole, Threephase Stator Windings
One cycle of the supply causes the field to rotate through one revolution. The field axis starts half way between the Rphase conductors and after half a cycle rotates past one set of Rphase conductors, i.e., one side of Rphase windings to reach a position again half way between the Rphase conductors yet acting in the opposite direction. A further half cycle completes the movement to reach the original relationship in space between the conductor position and the field axis.
In the fourpole machine, the movement of the field past one set of conductors is only sufficient to rotate the field through 90° and the completion of the cycle only rotates the field through 180°. At this instance, the field system appears again to be the same as the field system when it started to move and this is because there are two possible situations which are identical, i.e., when the field system is in its initial position and when it is upside down.
Developing this argument, the fourpole machine requires a further cycle of the supply to rotate the field back to its original position. Thus, a fourpole machine requires two cycles of the supply in order that the field rotates through one revolution. The sixpole machine can, similarly, be shown to require three cycles of the supply for the field to rotate through one revolution. We can observe that the number of cycles of the supply required for one revolution of the magnetic field is always half the number of poles. However, poles in such machines always come in multiples of two, i.e., in pairs of poles. It follows that the number of pole pairs is equal to the number of cycles of the supply required for one complete revolution of the magnetic field.
If the number of pole pairs is p and the supply frequency is f hertz, then the number of revolutions of the magnetic field per second is
where, ‘n’ is the rotational speed in revolution per second. since ‘n’ revolutions per second is equivalent to N revolutions per minutes.
Example 40.1
A sixpole, threephase, 50 Hz induction motor sets up a rotating field. At what speed does it rotate?
Solution:
The speed at which the rotating magnetic field revolves is termed the synchronous speed. In most machines, the supply frequency and the number of poles are fixed and it follows that the synchronous speed is constant for any given machine. However, it is possible to produce variable frequency supplies, and even to change the number of poles, in either of which cases the synchronous speed can be changed, but these cases are relatively unusual.
The relationship between the speed of the rotating flux and the number of stator poles may be visualized by comparing the mechanical degrees of circular arc travelled by the flux in motors with different numbers of stator poles. This is shown in Figure 40.9, where the circular arc travelled by the rotating field of a fourpole motor is twice that of an eightpole motor, assuming the same frequency and the same time period; the centre line of flux rotates 60° (phase A to phase C) for a fourpole winding and 30° (A to C) for an eightpole winding, as illustrated in Figures [40.9(a) and (b)], respectively.
Figure 40.9 Coil Spans for the Fourpole Winding (a) and Eightpole Winding (b)
The synchronous speed of an induction motor operating from a fixed frequency system can be changed by changing the number of poles in the stator, and by using a frequency converter to change the frequency or both.
Determine the synchronous speed of a sixpole, 460 V, 60 Hz induction motor if the frequency is reduced to 85 per cent of its rated value.
Solution:
40.4 DIRECTION OF ROTATION
If the supply connections to any two windings are reversed, then the phase sequence of the stator current is reversed. The result of this change is to cause the field to rotate in the opposite direction, as indicated in Figure 40.10.
Figure 40.10 Direction of Rotation of the Magnetic Field
As a common application of the threephase induction machine is to serve as a motor and since the rotor turns in the same direction as the rotating flux, it follows that the direction of rotation of a motor can be reversed simply by interchanging any two of the three supply conductors.
40.5 SLIP AND ITS EFFECT ON ROTOR FREQUENCY AND VOLTAGE
The difference between the speeds of the rotating flux and the speed of the rotor is called slip speed and the ratio of slip speed to synchronous speed is called slip. Expressed mathematically
where,
The slip, as expressed in Eq. 40.4, is called perunit slip. The slip depends on the mechanical load connected to the rotor shaft (assuming a constant supply voltage and a constant supply frequency). Increasing the shaft load decreases the rotor speed, thus, increasing the slip. If the rotor is blocked to prevent turning n_{r} = 0, and Eq. (40.4) reduces to
Releasing the brake allows the rotor to accelerate. The slip decreases with acceleration and approaches zero when all mechanical load is removed.
If operating with the no shaft load and if the windage and friction are sufficiently small, the very low relative motion between the rotor and the rotating flux of the stator may cause the rotor to become magnetized along an axis of minimum reluctance. If this occurs, the rotor will lock in synchronization with the rotating flux of the stator; the slip will be zero, no induction motor torque will be developed and the motor will act as a reluctancesynchronous motor. However, the application of a small shaft load will cause it to pull out of synchronism, and the inductionmotor action will again occur. Solving Eq. (40.4) for n_{r} expresses the rotor speed in terms of slip:
40.5.1 Effect of Slip on Rotor Frequency
The frequency of the voltage induced in a rotor loop by a magnetic field is given by
where, ƒ_{r} = rotor frequency (Hz)
P = number of stator poles
n = slip speed (rev/min)
substituting Eq. (40.3) into Eq. (40.7)
From Eq. (40.4)
Substituting in Eq. (40.7)
If the rotor is blocked so that it cannot turn, s = 1, and Eq. (40.9) becomes
where, f_{BR} = frequency of voltage generated in the blocked rotor (BR)
Substituting Eq. (40.6) into Eq. (40.9) results in the general expression for rotor frequency.
Thus,
At blocked rotor, also called locked rotor, there is no relative motion between rotor and stator, the slip is 1.0, and the frequency of the voltage generated in the rotor is identical to the frequency of the applied stator voltage. That is
40.5.2 Effect of Slip on Rotor Voltage
The voltage generated in a rotor loop, as illustrated in Figure 40.11 (formed by two rotor bars and end connections), as it is swept by the rotating stator flux is given by
Figure 40.11 (a) Rotating Field Sweeping a Rotor Bar (b) Direction of Flux Generated Around Rotor Bar and (c) Direction of Rotorbar Current
Substituting Eq. (40.11) into Eq. (40.13)
At blocked rotor, s = 1, Eq. (40.14) becomes
Substituting Eq. (40.15) into Eq. (40.13)
Equation (40.16) is the general expression for voltage induced in a rotor loop at any rotor speed in terms of blockedrotor voltage and slip.
Example 40.3
The frequency and induced voltage in the rotor of a certain sixpole woundrotor induction motor, whose shaft is blocked, are 60 Hz and 100 Hz, respectively. Determine the corresponding values when the rotor is running at 1100 rev/min
Solution:
Example 40.4
A fourpole induction motor operating at a frequency of 60 Hz has a fullload rotor slip of 5 per cent. Calculate the rotor frequency:
(1) At the instant starting; and (2) At full load.
Solution:
 At the instant of starting
where, n_{r} is the rotor speed. Since the rotor speed n_{s} at that instant is zero, s = 1, or unity slip. The rotor frequency is f_{r }= sf_{BR} = 1.0 × 60 = 60 Hz
 At full load the slip is 5 per cent (given) and, therefore,
s = 0.05 and ƒ_{r} = s × ƒ = 0.05 × 60Rotor frequency = 3 Hz
Example 40.5
A three phase two pole induction motor is connected to a 50 Hz supply. Determine the synchronous speed of the motor in rev/min.
Solution:
Example 40.6
A stator winding supplied from a threephase 60 Hz system is required to produce a magnetic flux rotating at 900 rev/min. Determine the number of poles.
Solution:
Since,
Example 40.7
A threephase two pole motor is to have a synchronous speed of 6000 rev/min. Calculate the frequency of the supply voltage.
Solution:
since,
40.6 CONSTRUCTION OF A THREEPHASE INDUCTION MOTOR
Many a.c. motors are classified as induction motors. An induction motor is a motor that has no electric connections between the power sources and the rotor, yet the rotor has conductors which carry current. The current in the rotor conductors is an induced current. It is induced by the magnetic field created by the stator windings.
A woundrotor induction motor, as shown in Figure 40.12, has a regular threephase winding similar to that of the stator and is wound with the same number of poles. The phases are usually wyeconnected and terminate at the slip rings. A wyeconnected rheostat with a common lever is used to adjust the resistance of the rotor circuit. The rheostat provides speed control, torque adjustment at locked rotor and current limiting during starting and acceleration.
Figure 40.12 Cutaway View of Woundrotor Induction Motor
The stator core is an assembly of thin laminations stamped from siliconalloy sheet steel; the use of silicon steel for the magnetic material minimizes hysteresis losses. The laminations are coated with oxide or varnish in order to minimize eddycurrent losses.
Insulated coils are set in slots within the stator core and the overlapping coils are connected in series or parallel arrangements in order to form phase groups and the groups are connected in wye or delta. The connections wye or delta, series or parallel, are dictated by voltage and current requirements.
The rotors are of two basic types (squirrel cage and wound rotor). Small squirrelcage rotors use a slotted core of laminated steel into which the molten aluminium is cast to make the conductors, end rings, and fan blades. Large squirrel rotors, as shown in Figure 40.13, use brass bars and brass end rings that are brazed together to form the squirrel cage. There is no insulation between the iron core and the conductors, and none is needed; the current induced in the rotor is contained within the circuit formed by the conductors and end rings also called end connections. Skewing the rotor slots helps avoid crawling and reduces vibration.
Figure 40.13 Conductor Arrangement in a Squirrelcage Rotor
A woundrotor induction motor uses insulated coils that are set in slots and connected in wye arrangement. The rotor circuit is completed through a set of slip rings, carbon brushes and a wyeconnected rheostat. The threephase rheostat is composed of three rheostats connected in a wye; a common lever is used to simultaneously adjust all three rheostat arms.
The transfer of energy occurs in a manner similar to that in a transformer. The stator is often referred to as the primary and the rotor as the secondary. The air gap is made quite small so as to offer minimum reluctance.
Each coil of an induction motor stator spans a portion of the stator circumference equal to or slightly less than the pole pitch. The pole pitch is equal to the stator circumference divided by the number of stator poles, it may be expressed in terms of stator slots or arc. If the span is less than a pole pitch, it is called a fractional pitch winding.
40.7 ROTOR IMPEDANCE AND CURRENT
When an induction motor is stationary, the stator and rotor windings form the equivalent of a transformer, as illustrated in Figure 40.14(a), which represent one phase of an inductormotor rotor. The power and torque developed by a threephase motor is three times that developed by one of its phases. It will be assumed that the stator is an ideal stator in that it produces a rotating magnetic field of constant amplitude and constant speed, and that it has no core losses, no copper losses and no voltage drops. The rotor is represented by an electrically isolated closed circuit containing resistance and reactance acted on by an induced rotor voltage E_{r}. The rotor voltage is generated at a frequency f_{r} by the rotating flux of the stator. The model shows one phase of a wound rotor, or one phase of an equivalent wound rotor, if the rotor is squirrel cage.
Figure 40.14 Equivalent Circuits and Corresponding Impedance Diagrams for an Induction Motor with an Ideal Stator and a Real Rotor
The rotor resistance is dependent on the length, crosssectional area, resistivity and the skin effect of the rotor conductors as well as external rheostat resistance if it is a wound rotor like that shown in Figure 40.12.
The inductive reactance X_{r} of the rotor, called leakage reactance, is caused by leakage flux and is dependent on the shape of the rotor conductors, its depth in the iron core, the frequency of the rotor voltage and the length of the air gap between the rotor and the stator iron.
The leakage reactance of the rotor expressed in terms of rotor frequency and rotor inductance is
Substituting Eq. (40.11) into Eq. (40.17) and simplifying
Replacing X_{r}, E_{r} and f_{r} in Figure 40.14(a) with their equivalent values in terms of slip results in Figure 40.14(b): the rotor impedance as determined from the associated impedance diagram in Figure 40.14(c) is
Applying Ohm’s law to the rotor circuit in Figure 40.14(b)
Dividing both numerator and denominator by s,
A modified equivalent circuit and associated impedance diagram corresponding to Eq. (40.19) are shown in Figures [40.14(d) and (e)], respectively. The constant blockedrotor voltage in Figure 40.14(d), combined with an equivalent rotor resistance that varies with slip provides a convenient tool for analysis of inductionmotor behaviour.
Expressing the rotor current in terms of magnitude and phase angle,
The magnitude of the rotor current is
Expressing Z_{r}/s and θ_{r} in terms of their components, as shown in Figure 40.14(e)
Example 40.7
The frequency of the supply of an eightpole induction motor is 50 Hz and the rotor frequency is 3 Hz. Determine (1) the slip, and (2) the rotor speed.
 f_{r} = sf
Hence, 3 = s × 50 = 0.06 or 6 per cent  Synchronous speed, n_{s} = ƒ/p = 50/4 = 12.5 rev/s
12.5 × 60 = 75 rev/min
(0.06) (12.5) = 12.5 – n_{r}
Rotor speed n_{r} = 12.5 – 0.75
= 11.75 rev/s or 705 rev/min
Example 40.8
The rotor of a certain 25 hp, sixpole, 60 Hz induction motor has equivalent resistance and equivalent reactance per phase of 0.10 Ω and 0.54 Ω, respectively. The blocked rotor voltage/phase (E_{BR}_{}) is 150 V. If the rotor is running at 1164 rev/min, determine (1) synchronous speed, (2) slip, (3) rotor impedance, (4) rotor current, (5) rotor impedance if changing the shaft load resulted in 1.24 per cent slip, (6) speed for the conditions in (5).
Solution:
 n_{r} = n_{s}(1– s) = 1200(1 – 0.0124) = 1185 rev/min
40.8 LOCUS OF THE CURRENT
The changes that take place in rotor impedance angle θ_{r} and rotorcurrent magnitude I_{r} as an unloaded induction motor accelerates from standstill (blocked rotor) to synchronous speed are illustrated in Figure 40.15(a). The curves are plots of Eqs. (40.23) and (40.24), respectively. It can be seen that the rotor current and the rotor impedance angle have their greatest values at blocked rotor both decrease in values as the rotor accelerates and both approach zero as the rotor approaches synchronous speed. For low values of slip (s < 0.05), the rotor current is proportional to the slip.
A phasor diagram representing the magnitude and phase angle of the rotor for values of slip from s = 1 to s = 0 is shown in Figure 40.15(b). The current phasor changes in both magnitude and phase angle as the machine accelerates from blocked rotor (s = 1) to synchronous speed (s = 0). The locus of the current phasor is a semicircle. Proof of its semicircle character is obtained by expressing Z_{r}/s in terms of sin θ_{r} and then substituting into Eq. (40.23). Thus, from Figure 40.14(e),
Figure 40.15 (a) Rotor Current and Rotor Impedance Angle Versus Speed for Representative Induction Motor (b) Locus of Rotorcurrent Phasor
Substituting Eq. (40.26) into Eq. (40.23) and simplifying
Equation (40.27) is the polar equation for a circle that is tangent to the horizontal axis at the origin and whose diameter is E_{BR}/X_{BR}.
40.9 LOSSES AND EFFICIENCY
Calculations involving overall motor efficiency must take into account the losses that occur in both the stator and the rotor. The stator losses include all hysteresis losses and eddycurrent losses in stator and rotor (called core losses), I^{2}R losses in the stator winding (called stator copper losses). Given the input power to the stator, and the stator losses, the net power crossing the air gap is
where,
Figure 40.16 shows the flow of power from stator input to shaft output and accounts for the loss in both stator and rotor. The power flow diagram is a useful adjunct to problem solving. It suggests a convenient method of solution.
Figure 40.16 Shows the Flow of Power from Stator Input to Shaft Output and Accounts for the Losses in Both Stator and Rotor Power
Power P = 2πnT where T is the torque in Newton metres; hence, torque T = P/(2πN). If P_{2 }is the power input to the rotor from the rotating field, and P_{m} is the mechanical power (including friction losses), then
From which
Hence,
But P_{2} – P_{m} is the electrical or copper loss in the rotor, i.e.,
Hence slip
Or power input to the rotor
See Figures 40.16 and 40.17
As has been shown in Figure 40.17(a), when an induction motor is stationary, the stator and rotor windings form the equivalent of a transformers and
As can be seen in Figure 40.17(b), when running, the induced emf is proportional to the slip, s. Hence, when running, the rotor emf per phase
Figure 40.17 Rotor e.m.f., Frequency Impedance and Current (a) When Stationary (b) When Running
Motor Efficiency,
Example 40.9
The power supplied to a threephase induction motor is 32 kw and the stator losses are 1200 w. If the slip is 5 per cent, determine (1) the rotor copper loss, (2) the total mechanical power developed by the rotor, (3) the output power of the motor if friction and windage losses are 150 w and (4) the efficiency of the motor neglecting rotor iron loss.
 input power to rotor = stator input powerstator losses
32 kw – 1.2 kw = 30.8 kw
Rotor copper loss = (0.05) (30.8) = 1.54 kw
 Mechanical power developed by the rotor = rotor input power – rotor losses
= 30.8 – 1.54 = 29.26 kw
 Output power of motor = power developed by the rotor – friction and windage losses
= 29.26 – 0.75 = 28.51 kw
Example 40.10
The speed of induction motor in Example 40.9 is reduced to 35 per cent of its synchronous speed by using external rotor resistance. If the torque and rotor losses are unchanged, determine (1) the rotor copper loss, and (2) the efficiency of motor.
Solution:
 Slip
Input power to rotor = 30.8 kw
Since,
 Power developed by rotor = input power to rotor – rotor copper loss
= 30.8 – 20.02 = 10.78 kw
Output power of motor = power developed by rotorfriction and windage losses
= 10.78 – 0.75 = 10.03 kw
Efficiency,
40.10 AIR GAP POWER
The power transferred electromagnetically across the air gap between the stator and rotor is called airgap power or gappower.
The active and reactive components of gap power are
where,
Active component P_{gap}: Supplies the shaft power output as well as friction, windage, as well as heat losses in the output.
Reactive component Q_{gap} supplies the reactive power for the alternating magnetic field about the rotor current. Component Q_{gap} is not dissipated. It follows a sinusoidal pattern as it seesaws across the gap between the stator and the rotor.
Components P_{gap} and Q_{gap} may be represented in a power diagram as the two sides of a right triangle whose diagonal is S_{gap} as illustrated in Figure 40.18
Figure 40.18 Power Diagram for Air Gap Power
Example 40.11
For the motor operating at 1164 rev/min in Example 40.8, determine the total threephase apparent power crossing the air gap its active and reactive components, and the rotor power factor.
Solution:
Converting to rectangular form
40.11 MAXIMUM TORQUE
Torque
from Eq. (40.31)
Or
If there are m phases, then torque
where, K is a constant for a particular machine. (40.42)
i.e., torque
Under normal conditions, the supply voltage is usually, constant; hence, Eq. (40.43) becomes
The torque will be a maximum, when the denominator is a minimum and this occurs when
Thus, the maximum torque occurs when rotor resistance and rotor reactance are equal
i.e.,
Example 40.12
A 415 V, threephase, 50 Hz, four pole star connected induction motor runs at 24 rev/s on full load. The rotor resistance and reactance per phase are 0.35 Ω and 3.5 Ω, respectively, and the effective rotorstator turns ratio is 0.85:1. Calculate: (1) the synchronous speed; (2) the slip; (3) the full load torque; (4) the power output if mechanical losses amount to 770 W; (5) the maximum torque; (6) the speed at which maximum torque occurs; and (7) the starting torque.
Solution:
 Synchronous speed,

Slip,
 Phase voltage
 Output power, including friction losses, P_{m} = 2πn_{r}T
(2π) (24) (78.05) = 1170 W
Hence, power output = P_{m} – mechanical losses
1170 – 770 = 11000 W = 11 kW  Maximum torque occurs when R_{2} = X_{t} = 0.35 Ω
Slip,
Hence, maximum torque,
 For maximum torque, slip s = 0.1
Hence, (0.1) (25) = 25 – n_{r} and n_{r} = 25 – (0.1) (25)
The speed at which maximum torque occurs n_{r} = 22.5 rev/s = 1350 rev/min
 At the start (at stand still), slip s = 1
Hence, starting torque
Starting torque = 22.41 Nm
40.12 INDUCTION MOTOR TORQUESPEED CHARACTERISTICS
The normal starting torque may be less than the full load torque. The speed at which maximum torque occurs is determined by the value of rotor resistance. At synchronous speed, slip s = 0 and torque is zero. From these observations, the torquespeed and torqueslip characteristics of an induction motor can be drawn (Figure 40.19).
The rotor resistance of an induction motor is usually small compared with its reactance, so that the maximum torque occurs at a high speed, typically about 80 per cent of the synchronous speed. Curve P in Figure 40.19(a) is a typical characteristic of an induction motor. The curve P cuts the fullload torque line at X, showing that at full load the slip is 4–5 per cent. The normal operating conditions are between O and X. It can be seen that for normal operation the speed variation with load is quite small the induction motor is an almost constant speed machine. Redrawing the speed torque characteristic between O and X gives the characteristic shown in Figure 40.19(b), which is similar to that of the d.c. shunt motor.
If maximum torque is required at starting, then a high resistance rotor is necessary, which gives characteristic Q in Figure 40.19(a), however, the motor has a full load slip of over 30 per cent which results in a drop in efficiency. Also, such a motor has a large speed variation with variation of load curves R and S, as seen in Figure 40.19(a), are characteristic for values of rotor resistance between those of P and Q. Better starting torque than for curve P is obtained but with lower efficiency and with speed variations under operating conditions.
A squirrelcage induction motor (SCIM) would normally follow characteristic P. This kind of machine is highly efficient and about constant speed under normal running conditions. However, it has a poor starting torque and must be started off load or very highly loaded. Also, on starting, the current can be four or five times the normal full load current due to the motor acting like a transformer with secondary short circuited.
A woundrotor induction motor (WRIM) would follow characteristic P when the slip rings are short circuited, which is the normal running condition. However, the sliprings allow for the addition of resistance to the rotor circuit externally and as a result for starting, the motor can have a characteristic similar to curve Q in Figure 40.19(a). The high starting current experienced by the squirrel cage induction motor can be overcome.
Figure 40.19 (a) Torquespeed Characteristics (b) Figure 40.19(a) Redrawn
In general, for threephase induction motors, the power factor is usually between about 0.8 and 0.9 lagging and the fullload efficiency is usually about 80–90 per cent. As torque is proportional to the square of the supply voltage, any voltage variations, therefore, would seriously affect the induction motor performance.
40.13 WRIM AND SCIM: A COMPARISON
The advantages of the WRIM compared with the SCIM are as follows.
 They have a much higher starting toque.
 They have a much lower starting current.
 They have a means of varying speed by use of external rotor resistance.
The advantages of the SCIM compared with WRIM are as follows.
 They are cheaper and more robust.
 They have slightly higher efficiency and power factor.
 They are explosion proof, since the risk of sparking is eliminated by the absence of slip rings and brushes.
40.14 STARTING TECHNIQUES FOR INDUCTION MOTORS
Induction motors of almost any horse power may be started by connecting them across full voltages, as shown in Figure 40.20, and most are started that way. In many cases, however, the high in rush current associated with fullvoltage starting can cause large voltage dips in the distribution system. Lights may dip or flicker, and unprotected control systems may drop out due to low voltage. The impact torque that occurs when starting at full voltage can damage the driven equipment.
Figure 40.20 Starting by Connecting Induction Motors Across Line
The methods commonly used for reducing inrush current are reduced voltage starting using auto transformers current limiting through wyedelta connections of stator windings, part winding connections, series impedance and solidstate control.
Auto Transformer Method: Auto transformer reduces the stator voltage and, thus, the starting current. However the the starting torque is seriously reduced, so the voltage is reduced only sufficiently to give the desired reduction of the starting current.
A typical arrangement is shown in Figure 40.21. A doublethrow switch connects the autotransformers in the circuit for starting and when the motor is up to speed the switch is thrown to the run position which connects the supply directly to the motor.
Figure 40.21 Auto Transformer: Starting of Induction Motors
Wyedelta Starting Method: This method is used for starting the connections to the stator that are wyeconnected, so that the voltage across each phase winding is (i.e., 0.577) of the line voltage for running the windings are switched to delta connections. A typical arrangement is shown in Figure 40.22. This method of starting is less expensive than that by autotransformer.
Figure 40.22 Wyedelta Starting of Induction Motors
Part Winding Method: The partwinding method uses a stator with two identical threephase windings, each capable of supplying onehalf of the rated power. The power circuit for starting a partwinding motor is shown in Figure 40.23. Contacts 1 are closed first, energizing one winding. After a brief time delay, contacts 2 are closed energizing both windings. The part winding starter is one of the least expensive starters but is limited to dualvoltage motors that are operated on the lowvoltage connections.
Figure 40.23 Partwinding Starter
Series Impedance Starter: The series impedance starter shown in Figure 40.24(a) uses a resistor and inductor in series with each phase of the stator windings to limit the current during startup. The running contacts (R) are open when starting, to limit the inrush current and are closed to short out the impedance when the motor is near rated speed. The Ohmic values of the resistor or reactor are selected to provide approximately 70 per cent rated voltage at the motor terminals when starting. The series impedance stator provides smooth acceleration and is the simplest method of starting induction motors.
At each step, the motor operation will transfer from one characteristic to the next so that the overall starting characteristic will be as shown by the bold line in Figure 40.24(b).
Figure 40.24 (a) Series Impedance Start (b) At Each Step the Motor Characteristic Transfers from One to the Other
Solidstate Starters: A solidstate starter, shown in Figure 40.25, uses backtoback thyristors to limit the current. The control circuit (not shown) allows a gradual build up of current. The smooth build up permits a soft start with no impact loading and no significant voltages dips. Solidstate starters can be designed to incorporate many special features such as speed control, power factor control, protection against overload and single phasing.
Figure 40.25 A Solidstate Starter
40.15 DETERMINATION OF INDUCTION MOTOR PARAMETERS
In those cases where induction motor parameters are not readily available from the manufacturers, they can be approximated from a d.c. test, a noload test and a blockedrotor test.
40.15.1 D.c. Test
The purpose of the d.c. test is to determine R_{1}. This is accomplished by connecting any two stator leads to a variable voltage d.c. source as shown in Figure 40.26(a). The d.c. source is adjusted to provide approximately rated stator current and the resistance between the two stator leads is determined from voltmeter and ammeter readings. Thus, from Figure 40.26(a)
Figure 40.26 Circuit for d.c. Test to Determine R_{1}
If the stator is wyeconnected, as shown in Figure 40.26(b),
If the stator is delta connected, as shown in Figure 40.26(c)
40.15.2 Noload Test
The noload test is used to determine the magnetizing reactance X_{M} and the combined core, windage, and friction losses. These losses are essentially constant for all load conditions.
The connections for the noload test are shown in Figure 40.27. The rotor is unblocked and allowed to run unloaded at both rated voltage and rated frequency. At noload, the operating speed is very close to synchronous speed and the slip = 0, causing the current in R_{2}/s branch to be very small. For this reason, the R_{2}/s branch is drawn with dotted lines as shown in Figure 40.27, and omitted from the noload current calculations. Since I_{M} >> I_{f}_{e}, I_{0} ≈ I_{M}; thus, the R_{fe} branch is also drawn with dotted lines and omitted from the noload current calculations.
Figure 40.27 Basic Circuit for Noload Test and Blockedrotor Test
Referring to the approximate equivalent circuit shown in Figure 40.28 for the noload test the apparent power input per phase is
Figure 40.28 Equivalent Circuit Per Phase for Noload Test
The reactive power per phase is determined from
Solving for Q_{NL},
Expressing the reactive power in terms of current and reactance, and solving for the equivalent reactance at noload
where as indicated in Figure 40.28
Substituting X_{1} as determined from blocked rotor test, into Eq. (40.53) permits the determination of X_{M}.
The input power per phase at no load includes the core loss, stator copper loss, windage loss and friction loss (all per phase).
That is
Separation of friction and windage losses from the noload loss may be accomplished by plotting the noload power versus voltage squared for low values of voltage and extrapolating to zero voltage.
40.15.3 BlockedRotor Test
The blockedrotor test is used to determine X_{1} and X_{2}. When combined with data from the d.c. test, it also determines R_{2}. The test is performed by blocking the rotor so that it cannot turn, and measuring the line voltage, line current, and threephase power input to the stator. Connections for the test are shown in Figure 40.29. An adjustable voltage a.c. supply (not shown) is used to adjust the blockedrotor current to approximately rated current. If instrument transformers and singlephase Watt meters are used, the effect of transformer ratios and the direction of Watt meter readings (whether a positive or negative) must be considered.
Since the exciting current (I_{0}) at blocked rotor is considerably less than the rotor current (I_{2}) the exciting current may be neglected, enabling a simplification of the equivalent circuit shown in Figure 40.29, where X_{m} and R_{fe} are drawn with dotted lines and omitted when making blockedrotor calculations.
Figure 40.29 Equivalent Circuit Per Phase for Blockedrotor Test
The IEEE test code recommends that the blockedrotor test be made using 25 per cent rated frequency with the test voltage adjusted to obtain approximately rated current. Thus, a 60 HZ motor would use 15 Hz test voltage. The total reactance calculated from the 15 Hz test is then corrected to 60 Hz by multiplying by 60/15. The total resistance calculated from the 15 Hz test is essentially correct, however, and must not be adjusted Referring to Figure 40.29, when all values are per phase
Resistance R_{2} is obtained from R_{BR}_{.15 }by substituting R_{1} from the d.c. test into Eq. 40.55
Thus
From Figure 40.29
And
Converting X_{BR.15} to 60 HZ
where,
SUMMARY
 The induction motor is the most commonly used type of a.c. motor.
 The current in the rotor conductors is induced by transformer action.
 The number of cycles of the supply required for one revolution of the magnetic field is always half the number of poles.
 The speed at which the rotating magnetic field revolves is termed the synchronous speed.
 The synchronous speed is constant for any given machine.
 The direction of rotation of a motor can be reversed by interchanging any two of the three supply conductors.
 The difference between the speed of the rotating flux and the speed of the rotor is called slip speed.
 The ratio of slip speed to synchronous speed is called slip.
 The slip decreases with acceleration and approaches zero when all mechanical load is removed.
 At blocked rotor the slip is 1.0.
 The connections wye or delta, series or parallel are dictated by voltage and current requirements.
 The rotors are of two basic types: squirrel cage and wound rotor.
 The stator is often referred to as the primary and the rotor as the secondary.
 The rotor current and the rotor impedance have their greatest values at blocked rotor; both decrease in value as the rotor accelerates and both approach zero as the rotor approaches synchronous speed.
 The stator losses include all hysteresis losses and eddy current losses in the stator and rotor (called core losses).
 The power transferred electromagnetically across the air gap between the stator and rotor is called the airgap power or gap power.
 The maximum torque occurs when rotor resistance and rotor reactance are equal.
 The induction motor parameters are determined by d.c. test, noload test and blocked rotor test.
MULTIPLE CHOICE QUESTIONS (MCQ)
 The speed at which the rotating magnetic field, produced by the stator current rotates is
 Rotor speed
 Synchronous speed
 Greater than synchronous speed
 Less than synchronous speed
 The noload slip of a threephase induction motor is of the order of
 1 per cent
 2 per cent
 6 per cent
 4 per cent
 The rotating field of the rotor rotates relative to the stator core at a speed equal to
 sn_{s}
 n_{s}
 n
 sn
 The stator of a threephase induction motor is laminated to
 Reduce eddy current losses
 Reduce copper losses in the stator winding
 Reduce hysteresis losses
 All of the above
 In a threephase induction motor iron losses occur in
 Stator winding
 Rotor winding
 Stator core and teeth
 Rotor core and teeth
 The noload current of a threephase induction motor in terms of its fullload current is of the order of
 10 per cent
 20 per cent
 50 per cent
 25 per cent
 The efficiency of a threephase induction motor as compared to that of a transformer is
 Lesser
 Much less
 Higher
 Much higher
 Comparable
 The nature of the PF of a threephase induction motor is
 Leading
 Unity
 Lagging
 May be leading or lagging
 Always lagging
 Always leading
 The slip of a threephase induction motor under blocked rotor test is
 1.0
 0.5
 Zero
 0.2
 The starting torque of threephase squirrel cage induction motor is
 High
 Zero
 Low
 Equal to fullload torque
ANSWERS (MCQ)
 (b)
 (b)
 (b)
 (a)
 (c)
 (d)
 (a)
 (e)
 (a)
 (c)
CONVENTIONAL QUESTIONS (CQ)
 What is meant by a doubly excited motor?
 How does a uniform strength of a rotating magnetic field induce voltage in a.c. induction motor rotor?
 How do voltages that are induced in rotor produce a rotating magnetic field?
 What is slip in an induction motor?
 Why must some slip be present for motor action?
 Why does running an induction motor unloaded enable the rotational losses to be determined?
 What is the utility of the blocked rotor test?
 Why is it desirable to know the resistance of the stator winding?
 Why is induction motor rotor current related to slip?
 What is the difference between gross developed torque and net output torque?
 Why is maximum torque called breakdown torque?
 What is the synchronous speed of an induction motor with six poles operating on 60 HZ?
 What is the synchronous speed of an induction motor with four poles operating on 400 HZ?
 An induction motor operates at 4.45 per cent slip and has four poles. What is its r.p.m. on 60 HZ?
 A threephase induction motor draws 4.5 A from its lines at 230 V linetoline at a power factor of 0.153 while running at no load. Its d.c. resistances line to line between two phases of the stator is 1.863 Ω. What is its rotational loss?
 An induction motor is tested in the blocked – rotor test. Its rated line current of 8.5 A is drawn when the line voltage is 16.6 V and the total voltage is 48.8 W. Under these conditions what is
 The equivalent resistance reflected to the stator per phase?
 The equivalent impedance per phase?
 The equivalent inductive reactance per phase?
ANSWERS (CQ)
12. 1200 r.p.m.
13. 12000 r.p.m.
14. 1720 r.p.m.
15. 204 W
16. (a) 0.255 Ω (b) 1.13 Ω (c) 1.11 Ω.