Chapter 40. Induction Motors (Three Phase) – Electrical Technology, Vol2: Machines and Measurements, 1/e

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 three-phase 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 three-phase 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 three-phase 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 three-phase 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 cage-rotor induction motor and the wound-rotor 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 three-phase 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 Three-phase 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 R-phase and that of the Y-phase, whilst there is a further 120° to the connection of the B-phase. A further 120° remains between the B-phase and R-phase, 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 three-phase 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 R-phase 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 R-phase 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 R-phase 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 Three-phase Stator with Three-phase 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 R-phase has fallen to 0.87 of its maximum value, while that in the Y-phase is zero. The current in the B-phase 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 Three-phase Stator with Three-phase 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 three-phase a.c. supply produces a rotating field.

Figure 40.5 Rotating Magnetic Field

Note: It has been assumed that the construction of the three-phase 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 Three-phase, 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 N-pole at one side and terminates in a S-pole at the other. The same observations may be made of the resultant field, and for this reason, the winding arrangement is described as a two-pole winding.

Figure 40.7 (a) Elementary Three-phase Induction Motor (b) Three-phase Flux Waves and (c) Instantaneous Direction of Resultant Stator Flux

By using six windings instead of three, it is possible to have a four-pole machine, while nine windings give a six-pole machine, and so on; although unusual, it is quite practical to have a machine with about a 100 poles. The winding arrangements of four-pole and six-pole machines are shown in Figure 40.8.

Figure 40.8 Four- and Six-pole, Three-phase Stator Windings

One cycle of the supply causes the field to rotate through one revolution. The field axis starts half way between the R-phase conductors and after half a cycle rotates past one set of R-phase conductors, i.e., one side of R-phase windings to reach a position again half way between the R-phase 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 four-pole 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 four-pole machine requires a further cycle of the supply to rotate the field back to its original position. Thus, a four-pole machine requires two cycles of the supply in order that the field rotates through one revolution. The six-pole 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

 

N = f/P        (40.1)

 

where, ‘n’ is the rotational speed in revolution per second. since ‘n’ revolutions per second is equivalent to N revolutions per minutes.

 

N = 60 f/P        (40.2)

 

Example 40.1

A six-pole, three-phase, 50 Hz induction motor sets up a rotating field. At what speed does it rotate?

Solution:

 

n = ƒ/P = 50/3 = 16.7 rev/s

 

N = 60 ƒ/P = (60 × 50)/3 = 1000 rev/min

 

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 four-pole motor is twice that of an eight-pole motor, assuming the same frequency and the same time period; the centre line of flux rotates 60° (phase A to phase C) for a four-pole winding and 30° (A to C) for an eight-pole winding, as illustrated in Figures [40.9(a) and (b)], respectively.

Figure 40.9 Coil Spans for the Four-pole Winding (a) and Eight-pole 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.

Example 40.2

Determine the synchronous speed of a six-pole, 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 three-phase 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

 

N = nsnr        (40.3)

where,

n = slip speed (rev/min)
ns = synchronous speed (rev/min)
nr = rotor speed (rev/min)
s = slip (pu)

 

The slip, as expressed in Eq. 40.4, is called per-unit 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 nr = 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 reluctance-synchronous motor. However, the application of a small shaft load will cause it to pull out of synchronism, and the induction-motor action will again occur. Solving Eq. (40.4) for nr expresses the rotor speed in terms of slip:

 

nr = ns (1 – s)        (40.6)

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,  ƒr = 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)

 

ns – nr = sns

 

Substituting in Eq. (40.7)

If the rotor is blocked so that it cannot turn, s = 1, and Eq. (40.9) becomes

where, fBR = 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,

 

fr = sfBR        (40.11)

 

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

 

fBR = fstator        (40.12)

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 Rotor-bar Current

 

Er = 4.44 Nfr ϕmax        (40.13)

 

Substituting Eq. (40.11) into Eq. (40.13)

 

Er = 4.44 NsfBR ϕmax        (40.14)

 

At blocked rotor, s = 1, Eq. (40.14) becomes

 

EBR = 4.44 NfBR ϕmax        (40.15)

 

Substituting Eq. (40.15) into Eq. (40.13)

 

Er = sEBR        (40.16)

 

Equation (40.16) is the general expression for voltage induced in a rotor loop at any rotor speed in terms of blocked-rotor voltage and slip.

Example 40.3

The frequency and induced voltage in the rotor of a certain six-pole wound-rotor 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:

fr = sfBR = 0.0833 × 60 = 5.0 Hz
Er = sEBR = 0.0833 × 100 = 8.33 V

Example 40.4

A four-pole induction motor operating at a frequency of 60 Hz has a full-load rotor slip of 5 per cent. Calculate the rotor frequency:

(1) At the instant starting; and (2) At full load.

Solution:

  1. At the instant of starting

    where, nr is the rotor speed. Since the rotor speed ns at that instant is zero, s = 1, or unity slip. The rotor frequency is fr = sfBR = 1.0 × 60 = 60 Hz

  2. At full load the slip is 5 per cent (given) and, therefore,

     

    s = 0.05   and  ƒr = s × ƒ = 0.05 × 60

     

    Rotor 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:

 

ns = 50/1 rev/s
= 50 × 60 rev/min = 3000 rev/min

 

Example 40.6

A stator winding supplied from a three-phase 60 Hz system is required to produce a magnetic flux rotating at 900 rev/min. Determine the number of poles.

Solution:

Since,

ns = f/p then p = f/ns = 60/15 = 4

 

Example 40.7

A three-phase two pole motor is to have a synchronous speed of 6000 rev/min. Calculate the frequency of the supply voltage.

Solution:

 

since,

ns = f/p, then  f = ns p
40.6  CONSTRUCTION OF A THREE-PHASE 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 wound-rotor induction motor, as shown in Figure 40.12, has a regular three-phase winding similar to that of the stator and is wound with the same number of poles. The phases are usually wye-connected and terminate at the slip rings. A wye-connected 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 Wound-rotor Induction Motor

The stator core is an assembly of thin laminations stamped from silicon-alloy 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 eddy-current 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 squirrel-cage 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 Squirrel-cage Rotor

A wound-rotor 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 wye-connected rheostat. The three-phase 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 inductor-motor rotor. The power and torque developed by a three-phase 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 Er. The rotor voltage is generated at a frequency fr 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, cross-sectional 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 Xr 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

 

Xr = 2πfrLr        (40.17)

 

Substituting Eq. (40.11) into Eq. (40.17) and simplifying

 

Xr = 2π (sfBR) Lr = s (2π fBRLr) = sXBR        (40.18)

 

Replacing Xr, Er and fr 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

 

Zr = Rr + jsXBR        (40.19)

 

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 blocked-rotor voltage in Figure 40.14(d), combined with an equivalent rotor resistance that varies with slip provides a convenient tool for analysis of induction-motor behaviour.

Expressing the rotor current in terms of magnitude and phase angle,

The magnitude of the rotor current is

Expressing Zr/s and θr in terms of their components, as shown in Figure 40.14(e)

Example 40.7

The frequency of the supply of an eight-pole induction motor is 50 Hz and the rotor frequency is 3 Hz. Determine (1) the slip, and (2) the rotor speed.

Solution:

  1. fr = sf
    Hence, 3 = s × 50 = 0.06 or 6 per cent
  2. Synchronous speed, ns = ƒ/p = 50/4 = 12.5 rev/s

    12.5 × 60 = 75 rev/min

    (0.06) (12.5) = 12.5 – nr

    Rotor speed nr = 12.5 – 0.75

    = 11.75 rev/s or 705 rev/min

Example 40.8

The rotor of a certain 25 hp, six-pole, 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 (EBR) 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:

  1. nr = ns(1– s) = 1200(1 – 0.0124) = 1185 rev/min
40.8  LOCUS OF THE CURRENT

The changes that take place in rotor impedance angle θr and rotor-current magnitude Ir 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 Zr/s in terms of sin θr 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 Rotor-current 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 EBR/XBR.

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 eddy-current losses in stator and rotor (called core losses), I2R 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

 

Pgap = PinPcorePscl        (40.28)

where,

Pin = total three-phase power input to stator
Pcore = core loss
Pscl = stator conductor loss or stator copper loss

 

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 P2 is the power input to the rotor from the rotating field, and Pm is the mechanical power (including friction losses), then

From which

Hence,

But P2 – Pm 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 three-phase 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.

Solution:

  1. input power to rotor = stator input power-stator losses

    32 kw – 1.2 kw = 30.8 kw

    Rotor copper loss = (0.05) (30.8) = 1.54 kw

     

  2. Mechanical power developed by the rotor = rotor input power – rotor losses

     

    = 30.8 – 1.54 = 29.26 kw

     

  3. 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:

  1. Slip

    Input power to rotor = 30.8 kw

    Since,

  2. 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 rotor-friction 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 air-gap power or gap-power.

 

Sgap = EBRIrcos θr + j EBRIr sin θr        (40.34)

 

The active and reactive components of gap power are

 

(Active power) Pgap = EBRIr cos θr        (40.35)
(Reactive power) Qgap = EBRIr sin θr        (40.36)

where,

EBR = blocked rotor voltage
Ir = magnitude of rotor current
θr = rotor impedance angle
cos θr = power factor of rotor

 

Active component Pgap: Supplies the shaft power output as well as friction, windage, as well as heat losses in the output.

Reactive component Qgap supplies the reactive power for the alternating magnetic field about the rotor current. Component Qgap is not dissipated. It follows a sinusoidal pattern as it see-saws across the gap between the stator and the rotor.

Components Pgap and Qgap may be represented in a power diagram as the two sides of a right triangle whose diagonal is Sgap as illustrated in Figure 40.18

 

Sgap = Pgap + j Qgap        (40.37)

 

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 three-phase apparent power crossing the air gap its active and reactive components, and the rotor power factor.

Solution:

Converting to rectangular form

 

Sgap = (19732 + j 3197) VA
Pgap = 19732W, Qgap = 3196 Var
PF = cos 9.2° = 0.99
40.11  MAXIMUM TORQUE

Torque

from Eq. (40.31)

Hence, torque per phase

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

R2 = sX2 = Xr         (40.45)

Thus, the maximum torque occurs when rotor resistance and rotor reactance are equal

i.e.,

R2 = X2        (40.46)

 

Example 40.12

A 415 V, three-phase, 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 rotor-stator 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:

  1. Synchronous speed,
  2. Slip,

  3. Phase voltage
  4. Output power, including friction losses, Pm = 2πnrT

     

    (2π) (24) (78.05) = 1170 W

     

    Hence, power output = Pm mechanical losses

     

    1170 – 770 = 11000 W = 11 kW

     

  5. Maximum torque occurs when R2 = Xt = 0.35 Ω

    Slip,

    Hence, maximum torque,

  6. For maximum torque, slip s = 0.1

    Hence, (0.1) (25) = 25 – nr and nr = 25 – (0.1) (25)

    The speed at which maximum torque occurs nr = 22.5 rev/s = 1350 rev/min

     

  7. At the start (at stand still), slip s = 1

    Hence, starting torque

    Starting torque = 22.41 Nm

40.12  INDUCTION MOTOR TORQUE-SPEED 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 torque-speed and torque-slip 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 full-load 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 squirrel-cage 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 wound-rotor induction motor (WRIM) would follow characteristic P when the slip rings are short circuited, which is the normal running condition. However, the slip-rings 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) Torque-speed Characteristics (b) Figure 40.19(a) Redrawn

In general, for three-phase induction motors, the power factor is usually between about 0.8 and 0.9 lagging and the full-load 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.

  1. They have a much higher starting toque.
  2. They have a much lower starting current.
  3. They have a means of varying speed by use of external rotor resistance.

The advantages of the SCIM compared with WRIM are as follows.

  1. They are cheaper and more robust.
  2. They have slightly higher efficiency and power factor.
  3. 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 full-voltage 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 in-rush current are reduced voltage starting using auto transformers current limiting through wye-delta connections of stator windings, part winding connections, series impedance and solid-state 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 double-throw 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

Wye-delta Starting Method: This method is used for starting the connections to the stator that are wye-connected, 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 Wye-delta Starting of Induction Motors

Part Winding Method: The part-winding method uses a stator with two identical three-phase windings, each capable of supplying one-half of the rated power. The power circuit for starting a part-winding 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 dual-voltage motors that are operated on the low-voltage connections.

Figure 40.23 Part-winding 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 start-up. The running contacts (R) are open when starting, to limit the in-rush 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

Solid-state Starters: A solid-state starter, shown in Figure 40.25, uses back-to-back 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. Solid-state 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 Solid-state 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 no-load test and a blocked-rotor test.

40.15.1  D.c. Test

The purpose of the d.c. test is to determine R1. 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 R1

If the stator is wye-connected, as shown in Figure 40.26(b),

 

Rd.c.= 2R1 wye and R1 wye = Rd.c./2        (40.48a)

 

If the stator is delta connected, as shown in Figure 40.26(c)

40.15.2  No-load Test

The no-load test is used to determine the magnetizing reactance XM and the combined core, windage, and friction losses. These losses are essentially constant for all load conditions.

The connections for the no-load test are shown in Figure 40.27. The rotor is unblocked and allowed to run unloaded at both rated voltage and rated frequency. At no-load, the operating speed is very close to synchronous speed and the slip = 0, causing the current in R2/s branch to be very small. For this reason, the R2/s branch is drawn with dotted lines as shown in Figure 40.27, and omitted from the no-load current calculations. Since IM >> Ife, I0IM; thus, the Rfe branch is also drawn with dotted lines and omitted from the no-load current calculations.

Figure 40.27 Basic Circuit for No-load Test and Blocked-rotor Test

Referring to the approximate equivalent circuit shown in Figure 40.28 for the no-load test the apparent power input per phase is

Figure 40.28 Equivalent Circuit Per Phase for No-load Test

 

SNL = VNL.INL        (40.49)

 

The reactive power per phase is determined from

Solving for QNL,

Expressing the reactive power in terms of current and reactance, and solving for the equivalent reactance at no-load

 

QNL = I2NL.XNL   and    XNL = QNL/I2NL        (40.52)

 

where as indicated in Figure 40.28

 

XNL = X1 + XM         (40.53)

 

Substituting X1 as determined from blocked rotor test, into Eq. (40.53) permits the determination of XM.

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

 

PNL = I2NLR1 + Pcore + PW, f         (40.54)

Separation of friction and windage losses from the no-load loss may be accomplished by plotting the no-load power versus voltage squared for low values of voltage and extrapolating to zero voltage.

40.15.3  Blocked-Rotor Test

The blocked-rotor test is used to determine X1 and X2. When combined with data from the d.c. test, it also determines R2. The test is performed by blocking the rotor so that it cannot turn, and measuring the line voltage, line current, and three-phase 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 blocked-rotor current to approximately rated current. If instrument transformers and single-phase 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 (I0) at blocked rotor is considerably less than the rotor current (I2) the exciting current may be neglected, enabling a simplification of the equivalent circuit shown in Figure 40.29, where Xm and Rfe are drawn with dotted lines and omitted when making blocked-rotor calculations.

Figure 40.29 Equivalent Circuit Per Phase for Blocked-rotor Test

The IEEE test code recommends that the blocked-rotor 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

 

R1 + R2 = RBR.15        (40.55)

Resistance R2 is obtained from RBR.15 by substituting R1 from the d.c. test into Eq. 40.55

 

Thus

R2 = RBR.15R1        (40.57)

From Figure 40.29

And

Converting XBR.15 to 60 HZ

where,

SUMMARY
  1. The induction motor is the most commonly used type of a.c. motor.
  2. The current in the rotor conductors is induced by transformer action.
  3. The number of cycles of the supply required for one revolution of the magnetic field is always half the number of poles.
  4. The speed at which the rotating magnetic field revolves is termed the synchronous speed.
  5. The synchronous speed is constant for any given machine.
  6. The direction of rotation of a motor can be reversed by interchanging any two of the three supply conductors.
  7. The difference between the speed of the rotating flux and the speed of the rotor is called slip speed.
  8. The ratio of slip speed to synchronous speed is called slip.
  9. The slip decreases with acceleration and approaches zero when all mechanical load is removed.
  10. At blocked rotor the slip is 1.0.
  11. The connections wye or delta, series or parallel are dictated by voltage and current requirements.
  12. The rotors are of two basic types: squirrel cage and wound rotor.
  13. The stator is often referred to as the primary and the rotor as the secondary.
  14. 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.
  15. The stator losses include all hysteresis losses and eddy current losses in the stator and rotor (called core losses).
  16. The power transferred electromagnetically across the air gap between the stator and rotor is called the air-gap power or gap power.
  17. The maximum torque occurs when rotor resistance and rotor reactance are equal.
  18. The induction motor parameters are determined by d.c. test, no-load test and blocked rotor test.
MULTIPLE CHOICE QUESTIONS (MCQ)
  1. The speed at which the rotating magnetic field, produced by the stator current rotates is
    1. Rotor speed
    2. Synchronous speed
    3. Greater than synchronous speed
    4. Less than synchronous speed
  2. The no-load slip of a three-phase induction motor is of the order of
    1. 1 per cent
    2. 2 per cent
    3. 6 per cent
    4. 4 per cent
  3. The rotating field of the rotor rotates relative to the stator core at a speed equal to
    1. sns
    2. ns
    3. n
    4. sn
  4. The stator of a three-phase induction motor is laminated to
    1. Reduce eddy current losses
    2. Reduce copper losses in the stator winding
    3. Reduce hysteresis losses
    4. All of the above
  5. In a three-phase induction motor iron losses occur in
    1. Stator winding
    2. Rotor winding
    3. Stator core and teeth
    4. Rotor core and teeth
  6. The no-load current of a three-phase induction motor in terms of its full-load current is of the order of
    1. 10 per cent
    2. 20 per cent
    3. 50 per cent
    4. 25 per cent
  7. The efficiency of a three-phase induction motor as compared to that of a transformer is
    1. Lesser
    2. Much less
    3. Higher
    4. Much higher
    5. Comparable
  8. The nature of the PF of a three-phase induction motor is
    1. Leading
    2. Unity
    3. Lagging
    4. May be leading or lagging
    5. Always lagging
    6. Always leading
  9. The slip of a three-phase induction motor under blocked rotor test is
    1. 1.0
    2. 0.5
    3. Zero
    4. 0.2
  10. The starting torque of three-phase squirrel cage induction motor is
    1. High
    2. Zero
    3. Low
    4. Equal to full-load torque
ANSWERS (MCQ)
  1. (b)
  2. (b)
  3. (b)
  4. (a)
  5. (c)
  6. (d)
  7. (a)
  8. (e)
  9. (a)
  10. (c)
CONVENTIONAL QUESTIONS (CQ)
  1. What is meant by a doubly excited motor?
  2. How does a uniform strength of a rotating magnetic field induce voltage in a.c. induction motor rotor?
  3. How do voltages that are induced in rotor produce a rotating magnetic field?
  4. What is slip in an induction motor?
  5. Why must some slip be present for motor action?
  6. Why does running an induction motor unloaded enable the rotational losses to be determined?
  7. What is the utility of the blocked rotor test?
  8. Why is it desirable to know the resistance of the stator winding?
  9. Why is induction motor rotor current related to slip?
  10. What is the difference between gross developed torque and net output torque?
  11. Why is maximum torque called breakdown torque?
  12. What is the synchronous speed of an induction motor with six poles operating on 60 HZ?
  13. What is the synchronous speed of an induction motor with four poles operating on 400 HZ?
  14. An induction motor operates at 4.45 per cent slip and has four poles. What is its r.p.m. on 60 HZ?
  15. A three-phase induction motor draws 4.5 A from its lines at 230 V line-to-line 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?
  16. 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
    1. The equivalent resistance reflected to the stator per phase?
    2. The equivalent impedance per phase?
    3. 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.