Single-Phase Motors--Polyphase Induction Motors (part 3)

9. Some Important Conclusions

Prior to proceeding further, let us pause and make the following observations regarding the torque developed, the rotor current, and the motor efficiency.

The Torque Developed:

When the motor is operating at or near its rated slip, which is usually less than 10%, the hypothetical rotor resistance is considerably greater than the rotor leakage reactance. That is, R,/s >> X,. We can, therefore, approximate Eq. (34) as (40) 1. The torque developed by the motor is proportional to the slip when the applied voltage and rotor resistance are held constant. In this linear range, the ratio of torques developed is equal to the ratio of the slips.

The torque developed is inversely proportional to the rotor resistance at a given slip when the applied voltage is kept the same. In other words, the torque developed at any slip can be adjusted by varying the rotor resistance.

This can be easily accomplished in a wound-rotor motor.

At a definite value of slip and rotor resistance, the torque developed by the motor is directly proportional to the square of the applied voltage.

For a constant-torque operation under fixed applied voltage, the motor slip is directly proportional to the rotor resistance.

The Rotor Current:

From Eq. (33) it’s obvious that the rotor current is directly proportional to the applied voltage as long as the rotor resistance and the slip are held the same.

When R,/s >> X,, the rotor current can be approximated as ...

The rotor current varies linearly with the slip when the motor operates at low slip.

The rotor current varies inversely with the rotor resistance. ...

The Motor Efficiency:

For an ideal motor we can assume that (a) the stator copper loss is negligible and (b) the rotational loss is zero. In this case, the air-gap power is equal to the power input. That is, Pi, = P,. However, the power developed is Pd = (1 -s) Pag = SPag, where S is the per-unit speed. Since the rotational loss is zero, the power output is equal to the power developed. Hence, the motor efficiency under the ideal conditions is …

The above equation places a maximum limit on the efficiency of a three-phase induction motor. This equation highlights the fact that if a motor is operating at 60% of its synchronous speed, the maximum efficiency under the ideal conditions (theoretically possible) is 60 %. Thus, the higher the speed of operation, the higher the efficiency. For example, a motor operating at 5% slip can theoretically have an efficiency of 95%.

EXAMPLE 7:

A 230-V, 60-Hz, 4-pole, A-connected, three-phase induction motor operates at a full-load speed of 1710 rpm. The power developed at this speed is 2 hp and the rotor current is 4.5 A. If the supply voltage fluctuates k lo%, determine (a) the torque range and (b) the current range.

SOLUTION

The torque developed at the rated voltage of 230 V is ...

When the supply voltage is down by lo%, the torque developed by the motor is ...

The corresponding rotor current is ...

Similarly, when the supply voltage is up by lo%, the torque developed by the motor is ...and the rotor current is

12H = 4.5 x 1.1 = 4.95 A

Hence, the torque varies from 6.75 N-m to 10.08 N-m, and the rotor current fluctuates between 4.05 A and 4.95 A.

Exercises:

14. A 440-V, 50-Hz, 4-pole, Y-connected, three-phase induction motor runs at a slip of 5% on full load and develops a power of 10 hp. The slip falls to 2% when the motor is lightly loaded. Determine the torque developed and the power output of the motor when it runs at (a) full load and (b) light load.

A 660-V, 60-Hz, Y-connected, 6-pole, three-phase induction motor runs at 1125 rpm on full load and develops a power of 10 hp. The unregulated supply voltage fluctuates from 600 to 720 V. What is the torque range of the motor if the speed is held constant? What is the speed range for the motor to develop the same torque?

10. Equivalent Circuit Parameters

The equivalent circuit parameters and the performance of a three-phase induction motor can be determined by performing four tests. These are (a) the stator-resistance test, (b) the blocked-rotor test, (c) the no-load test, and (d) the load test.

The Stator-Resistance Test:

This test is performed to determine the resistance of each phase winding of the stator. Let R be the dc value of the resistance between any two terminals of the motor; then the per-phase resistance is ...

The measured value of the resistance may be multiplied by a factor ranging from 1.05 to 1.25 in order to convert it from its dc value to its ac value. This is done to account for the skin effect. The multiplying factor may be debatable at power frequencies of 50 or 60 Hz, but it does become significant for a motor operating at a frequency of 400 Hz.

Fgr. 9 Typical connections to perform test on a three-phase induction motor.

The Blocked-Rotor Test:

This test, also called the locked-rotor test, is very similar to the short-circuit test of a transformer. In this case, the rotor is held stationary by applying external torque to the shaft. The stator field winding is connected to a variable three-phase supply. The voltage is carefully increased from zero to a level at which the motor draws the rated current. At this time, the readings of the line current, the applied line voltage, and the power input are taken by using the two-wattmeter method, as illustrated in Fgr. 9.

Since the rotor-circuit impedance is relatively small under blocked-rotor condition (s = l), the applied voltage is considerably lower than the rated voltage of the motor. Thus, the excitation current is quite small and can be neglected. Under this assumption, the approximate equivalent circuit of the motor is given in Fgr. 10 on a per-phase basis. The total series impedance is

Z, = R, + R, + j(Xl + X,) = Re + jX,

Let vb,, Ibr, and Pb, be the applied voltage, the rated current, and the power input on a per-phase basis under the blocked-rotor condition; then

Fgr. 10 An approximate per-phase equivalent circuit of an induction motor under blocked-rotor condition.

Since R, is already known from the stator-resistance test, the equivalent rotor resistance is ...

It’s rather difficult to isolate the leakage reactances XI and X2. For all practical purposes, these reactances are usually assumed to be equal. That is

Xi = X2 = 0.5Xe

Fgr. 11 An approximate per-phase equivalent circuit of an induction motor under no-load condition.

The No-Load Test:

In this case the rated voltage is impressed upon the stator windings and the motor operates freely without any load. This test, therefore, is similar to the open-circuit test on the transformer except that friction and windage loss is associated with an induction motor. Since the slip is nearly zero, the impedance of the rotor circuit is almost infinite. The per-phase approximate equivalent circuit of the motor with the rotor circuit open is shown in Fgr. 11.

Let W,, I,, and V, be the power input, the input current, and the rated applied voltage on a per-phase basis under no-load condition. In order to represent the core loss by an equivalent resistance R,, we must subtract the friction and windage loss from the power input.

The friction and windage loss can be measured by coupling the motor under test to another motor with a calibrated output and running it at the no-load speed of the induction motor. Let P__ be the friction and windage loss on a per-phase basis. Then, the power loss in 4, is ...

Hence, the core-loss resistance is ...

The power factor under no load is ...

The magnetization reactance is ...

The magnetization reactance can also be computed as follows:

When using the two-wattmeter method to measure the power under no load, the reading on one wattmeter may actually be negative because the power factor of the motor under no load may be less than 0.5. If this is the case, the total power input is simply the difference of the two wattmeter readings.

The Load Test:

To experimentally determine the speed-torque characteristics and the efficiency of an induction motor, couple the motor to a dynamometer and connect the three-phase stator windings to a balanced three-phase power source. If need be, the direction of rotation may be reversed by interchanging any two supply terminals.

Starting from the no-load condition, the load is slowly increased and the corresponding readings for the motor speed, the shaft torque, the power input, the applied voltage, and the line current are recorded. From these data, the motor performance as a function of motor speed (or slip) can be computed. Using analog-to-digital converters, the data can be stored on a magnetic disk for further manipulations.

EXAMPLE 8

The test data on a 208-V, 60-Hz, 4-pole, Y-connected, three-phase induction motor rated at 1710 rpm are as follows:

The stator resistance (dc) between any two terminals = 2.4 Q No-Load Test Blocked-Rotor Test Power input 450 W 59.4 w Line current 1.562 A 2.77 A Line voltage 208 v 27 V Friction and windage loss = 18 W Compute the equivalent circuit parameters of the motor.

SOLUTION

For a Y-connected motor, the per-phase resistance of the stator winding is

Using the equivalent circuit parameters of Example 9.8, determine the shaft torque and the efficiency of the motor at its rated speed.

The following data were obtained on a 230-V, 60-Hz, 4-pole, Y-connected, three-phase induction motor: No-load test: power input = 130 W, line current = 0.45 A at the rated voltage.

Blocked-rotor test: power input = 65 W, line current = 1.2 A at a reduced voltage of 47 V. The friction and windage loss is 15 W, and the stator winding resistance between any two lines is 4.1 Q. Calculate the equivalent circuit parameters of the motor.

Fgr. 12 (a) Deep-bar and (b) double-cage rotors.

11. Starting of Induction Motors

At the time of starting, the rotor speed is zero and the per-unit slip is unity. Therefore, the starting current, from the approximate equivalent circuit given in ...

Since the effective rotor resistance, R,, is very small at the time of starting compared with its value at rated slip, R,/s, the starting current may be as much as 400% to 800% of the full-load current. On the other hand, the starting torque may only be 200% to 350% of the full-load torque. Such a high starting current is usually unacceptable because it results in an excessive line voltage drop which, in turn, may affect the operation of other machines operating on the same power source.

Since the starting current is directly proportional to the applied voltage, Eq. (53) suggests that the starting current can be reduced by impressing a low voltage across motor terminals at the time of starting. However, it’s evident from Eq. (54) that a decrease in the applied voltage results in a decline in the starting torque. Therefore, we can employ the low-voltage starting only for those applications that don’t require high starting torques. For instance, a fan load requires almost no starting torque except for the loss due to friction. The induction motor driving a fan load can be started using low-voltage starting.

The starting current can also be decreased by increasing the rotor resistance.

As mentioned earlier, an increase in the rotor resistance also results in an increase in the starting torque which, of course, is desired for those loads requiring high starting torques. However, a high rotor resistance (a) reduces the torque developed at full load, (b) produces high rotor copper loss, and (c) causes a reduction in motor efficiency at full load. These drawbacks, however, don’t represent a problem for wound-rotor motors. For these motors, we can easily incorporate high external resistance in series with the rotor windings at the time of starting and remove it when the motor operates at full load.

For rotors using squirrel cage winding (die-cast rotors), the change in resistance from a high value at starting to a low value at full load is accomplished by using quite a few different designs, as shown in Fgr. 12. In each design, the underlying principle is to achieve a high rotor resistance at starting and a low rotor resistance at the rated speed. At starting, the frequency of the rotor is the same as the frequency of the applied source. At full load, however, the rotor frequency is very low (usually less than 10 Hz). Thus, the skin effect is more pronounced at starting than at full load. Hence, the rotor resistance is higher at starting than at full load owing to the skin effect alone. Also, as the currents are induced in the rotor bars, they produce a secondary magnetic field. Part of the secondary magnetic field links only the rotor conductor and manifests itself as the leakage flux. The leakage flux increases as we move radially away from the air-gap toward the shaft, and is quite significant at starting. Thus, in a multicage rotor at starting, the inner cage presents a high leakage reactance compared with the outer cage. Owing to the high leakage reactance of the inner cage, the rotor current tends to confine itself in the outer cage. If the cross-sectional area of the outer cage is smaller than that of the inner cage, it presents a comparatively high resistance at starting. When the motor operates at full load, the rotor frequency is low. Thereby, the leakage flux is low. In this case, the current tends to distribute equally among all the cages. As a result, the rotor resistance is low.

The approximate equivalent circuit of a double-cage induction motor is given in Fgr. 13. The subscripts o and i correspond to the outer and the inner cage of a double-cage rotor shown in Fgr. 12b. The change in the speed-torque curve for a single-cage to a double-cage rotor is depicted in Fgr. 14.

Fgr. 13 Per-phase equivalent circuit of a double-cage induction motor.

Fgr. 14 Speed-torque characteristics for single-cage and double-cage rotors. Double-cage; Per-unit slip

Fgr. 15 (a) Unskewed and (b) skewed rotor bars. Laminations, Rotor bars rings, Shafrs

Fgr. 16 Equivalent circuit For example 9.

Another technique that is commonly used to increase the rotor resistance and lessen the effects of harmonics in an induction motor is called skewing. In this case, the rotor bars are skewed with respect to the rotor shaft, as illustrated in Fgr. 15. Skew is usually given in terms of bars. The minimum skew must be one bar to avoid cogging. Skews of more than one bar are commonly used.

EXAMPLE 9

The impedances of the inner cage and the outer cage rotors of a double-cage, three-phase, 4-pole induction motor are 0.2 + j0.8 O/phase and 0.6 + j0.2 fkl_phase, respectively. Determine the ratio of the torques developed by the two cages (a) at standstill and (b) at 2% slip.

SOLUTION

Since the stator winding impedance, the core-loss resistance, and the magnetization reactance are not given, the approximate circuit neglecting these impedances is given in Fgr. 16. The rotor currents in the inner and the outer cages are ...

The torque developed by a three-phase induction motor is ...

Thus, the ratio of the torques developed by the outer and inner cages is ...

Substituting s = 1 in Eq. (55), we obtain To in terms of Ti as ....

Thus, the torque developed by the outer cage is five times the torque developed by the inner cage.

However, substituting s = 0.02 in Eq. (55), we get ...

The inner cage is most effective at a slip of 2% because it develops nearly three times as much torque as the outer cage.

EXERCISES

18. The stator winding resistance of the double-cage induction motor of Example 9 is 2.1 + j8.3 fl/phase. Find the starting torque developed by the motor if the phase voltage is 110 V, 60 Hz.

A double-cage, three-phase, 6-pole, Y-connected induction motor has an inner cage impedance of 0.1 + j0.6 fl/phase and an outer cage impedance of 0.4 + j0.l fl/phase. Determine the ratio of the torque developed by the two cages at (a) standstill and (b) 5% slip. What is the slip at which the torque developed by the two cages is the same? If the motor in Exercise 19 is connected across a 230-V power source and has a stator impedance of 1.5 + j2.5 fl/phase, obtain the torque developed by the motor at standstill.

12. Rotor Impedance Transformation

Thus far we have tacitly assumed for both the squirrel-cage and wound rotors that the rotor circuit impedance can be transformed to the stator side in terms of an a-ratio. The a-ratio was defined on a per-phase basis as the ratio of the effective turns in the stator winding to the effective turns in the rotor winding. That is ....

For a wound rotor having the same number of poles and phases as that of the stator winding, the total turns per phase N2 and the winding factor kw2 can be calculated the same way as that for the stator winding. However, the problem is somewhat more perplexing for the squirrel-cage (die-cast) rotor. Let us suppose that there are P poles in the stator and Q bars on the rotor. Let us assume that one of the bars is under the middle of the north pole of the stator at any given time. Another bar also exists which is in the middle of the adjacent south pole. The induced emf in both bars is maximum but of opposite polarity. Thus, these two bars carry the maximum current and can be visualized as if they form a single turn. Hence, the total number of turns on the rotor is Q/2. The emfs are also induced in other bars. If the flux is distributed sinusoidally, the induced emfs and thereby the induced currents also follow the same pattern. However, the root-mean-square (rms) value of the induced emf in each turn is the same.

Since each turn is offset by one slot on the rotor, the induced emf in each bar is offset by that angle. Therefore, we can assume that each turn is equivalent to a phase group, and there are Q/2 phase groups in all. Since there are Q bars and P poles, the number of bars per pole is Q/P. As each bar identifies a different phase group, the number of bars per pole is then equivalent to the number of phases m^2 on the rotor. That is …

This realization highlights the fact that the number of bars per pole per phase is …

1. Stated differently, the number of turns per pole per phase is 1/2. Now we can determine the total number of turns per phase by multiplying the number of turns per pole per phase by the number of poles. That is, …

Since the two bars that are displaced by 180-degree electrical form a turn, the winding factor is unity because (a) the pitch factor is unity as each turn is a full-pitch turn and (b) the distribution factor is unity as there is only one turn in each phase group.

Since we are trying to transform the rotor circuit elements to the stator side, let m, be the number of phases on the stator side, E, the induced emf, and l2 the equivalent rotor current. For the equivalent representation to be valid, the apparent power associated with the rotor circuit on the rotor side must be the same for the equivalent rotor circuit as referred to the stator side. Thus, …

… where E_bar is the induced emf in the rotor bar and I_bar is the induced current.

Hence, Since the induced emf on the stator side is E1, E2 must be equal to El.

Also the rotor copper loss prior to and after the transformation must be equal.

That is, …

… where R_bar is the resistance of a rotor bar.

Finally, the magnetic energy stored in the rotor leakage inductance before and after the transformation must also be the same. Therefore, …

… where X, is the leakage reactance of each rotor bar.

Equations (62) and (63) outline how the actual rotor parameters for a squirrel-cage rotor can be transformed into equivalent rotor parameters on the stator side. R, and X, are the rotor resistance and leakage reactance that we have used in the equivalent circuit of an induction motor. From the above equations, it’s evident that the a-ratio is ...

Note that for a wound rotor m, = m^2.

EXAMPLE 10

A $-pole, 36-slot, double-layer wound, three-phase induction motor has 10 turns per coil in the stator winding and a squirrel-cage rotor with 48 bars. The resistance and reactance of each bar are 20 pi2 and 2 msZ, respectively. Determine the equivalent rotor impedance as referred to the stator on a per-phase basis.

SOLUTION

The number of coils for a double-layer wound stator is 36. The coils per pole per phase are n = 3 [36/(4 X 3)). The pole span is 36/4 = 9 slots. The slot span is 180/9 = 20 -degree electrical. The coil pitch is 7 slots, 140 -degree electrical, and can be determined from the developed diagram. The pitch factor is kp, = sin (140/2) = 0.94.

The distribution factor is ...

The winding factor for the stator is k,, = 0.94 X 0.96 = 0.9. Total turns per phase are 10 X 36/3 = 120. It’s assumed that all coils in a phase are connected in series.

For the rotor, k,, = 1. The number of phases m^2 = Q/P = 12. Turns per phase are N, = P/2 = 2. Thus, the a-ratio, from Eq. (64), is ...

Hence, the rotor parameters as referred to the stator are ...

EXERCISES

21. In a wound-rotor induction motor the stator winding has twice as many turns as the rotor winding. The winding distribution factor for the stator is 0.85 and that for the rotor is 0.8. The actual impedance of the rotor on the rotor side at standstill is 0.18 + j0.25 O/phase. What is the rotor impedance as referred to the stator side? A 6-pole, 36-slot, double-layer wound, three-phase induction motor has 20 turns per coil and a squirrel-cage rotor with 48 bars. Each bar has a resistance of 15 ~LQ and a reactance of 1.2 mR. Determine the equivalent circuit parameters of the rotor as referred to the stator side.

13. Speed Control of Induction Motors

It was pointed out in the preceding sections that the speed of an induction motor for a stable operation must be higher than the speed at which it develops maximum torque. In other words, the slip at full load must be less than the breakdown slip. For an induction motor having a low rotor resistance, the breakdown slip is usually less than 10%. For such a motor, the speed regulation may be within 5%. For all practical purposes, we can refer to a low-resistance induction motor as a constant-speed motor. Therefore, we must devise some methods in order to vary its operating speed.

We already know that the synchronous speed is directly proportional to the frequency of the applied power source and inversely proportional to the number of poles; the motor speed at any slip is

...

From this equation it’s evident that the operating speed of an induction motor can be controlled by changing the frequency of the applied voltage source and/or the number of poles. Speed can also be controlled by either changing the applied voltage, the armature resistance, or introducing an external emf in the rotor circuit.

Some of these methods are discussed below.

Frequency Control:

The operating speed of an induction motor can be increased or decreased by increasing or decreasing the frequency of the applied voltage source. This method enables us to obtain a wide variation in the operating speed of an induction motor.

The only requirement is that we must have a variable-frequency supply. To maintain constant flux density and thereby the maximum torque developed, the applied voltage must be varied in direct proportion to the frequency. This is due to the fact that the induced emf in the stator winding is directly proportional to the frequency. The speed-torque characteristics of an induction motor at four different frequencies are given in Fgr. 17. Also shown in the figure is a typical load curve. At each frequency the motor operates at a speed at which the load line intersects the speed-torque characteristic for that frequency.

Fgr. 17 Speed-torque characteristics for various frequencies and adjusted supply voltages.

Changing Stator Poles:

This method is quite suitable for an induction motor with a squirrel-cage rotor.

In this case, the stator can be wound with two or more entirely independent ...windings. Each winding corresponds to a different number of poles and therefore different synchronous speed. At any time, only one winding is in operation. All other windings are disconnected. For example, an induction motor wound for 4 and 6 poles at a frequency of 60 Hz can operate either at a synchronous speed of 1800 rpm ($-pole operation) or at 1200 rpm (6-pole operation). This method of speed control, although somewhat limited, is very simple, provides good speed regulation, and ensures high efficiency at either speed setting. Use has been made of this method in the design of traction motors, elevator motors, and small motor driving machine tools.

An induction motor is usually wound such that the current in each phase winding produces alternate poles. Thus, the four coils for each phase of a 4-pole induction motor produce two north poles and two south poles, with a south pole located between the two north poles and vice versa. However, if the phase coils are reconnected to produce either four north poles or four south poles, the winding is said to constitute a consequent-pole winding. Between any two like poles an unlike pole is induced by the continuity of the magnetic field lines. Thus, a 4-pole motor when reconnected as a consequent-pole motor behaves like an %pole motor.

Therefore, by simply reconnecting the phase windings on an induction motor, a speed in the ratio of 2:1 can be accomplished by a single winding.

Rotor Resistance Control:

We have already discussed the effect of changes in the rotor resistance on the speed-torque characteristic of an induction motor. This method of speed control is suitable only for wound-rotor induction motors. The operating speed of the motor can be decreased by adding external resistance in the rotor circuit. However, an increase in the rotor resistance causes (a) an increase in the rotor copper loss, (b) an increase in the operating temperature of the motor, and (c) a reduction in the motor efficiency. Because of these drawbacks, this method of speed control can be used only for short periods.

Stator Voltage Control:

Since the torque developed by the motor is proportional to the square of the applied voltage, the (reduction/augmentation) in operating speed of an induction motor can be achieved by (reducing/augmenting) the applied voltage. The speed-torque characteristics for two values of the applied voltage are depicted in Fgr. 18. The method is very convenient to use but is very limited in its scope because to achieve an appreciable change in speed a relatively large change in the applied voltage is required.

Injecting an EMF in the Rotor Circuit:

The speed of a wound-rotor induction motor can also be changed by injecting an emf in the rotor circuit, as shown in Fgr. 19. For proper operation, the frequency of the injected emf must be equal to the rotor frequency. However, there is no restriction on the phase of the injected emf. If the injected emf is in phase with the induced emf in the rotor, the rotor current increases. In this case, the rotor circuit manifests itself as if it has a low resistance. On the other hand, if the injected emf is in phase opposition to the induced emf in the rotor circuit, the rotor current decreases. The decrease in the rotor current is analogous to the increase in the rotor resistance. Thus, changing the phase of the injected voltage is equivalent to changing the rotor resistance. The change in the rotor resistance is accompanied by the change in the operating speed of the motor. Further control in the speed can also be achieved by varying the magnitude of the injected emf.

Fgr. 18 Speed-torque characteristics as a function of supply voltage.

Fgr. 19 An equivalent circuit of an induction motor with an external source in the rotor circuit.

14. Types of Induction Motors

The National Electrical Manufacturers Association (IWMA) has categorized squirrel-cage induction motors into six different types by assigning them the letter designations A through F. Each letter designation is intended to satisfy the requirements of a certain application.

Class A Motors:

A class A motor is considered a standard motor and is suitable for constant-speed applications. The motor can be started by applying the rated voltage. It develops a starting torque of 125% to 175% of full-load torque. The starting current at the rated voltage is 5 to 7 times the rated current. The full-load slip is usually less than 5% because the rotor resistance is relatively low. The speed regulation is 2% to 4%. The rotor bars are placed close to the surface of the rotor laminations in order to reduce the leakage reactance. These motors drive low-inertia loads and possess high accelerations. They are employed in such applications as fans, blowers, centrifugal type pumps, and machine tools.

Class B Motors:

A class B motor is considered a general-purpose motor and can be started by applying the rated voltage. The rotor resistance for a class B motor is somewhat higher than for a class A motor. The rotor conductors are placed deeper in the slots than for the class A motor. Therefore, the rotor reactance of a class B motor is higher than that of a class A motor. The increase in the rotor reactance reduces the starting torque, whereas an increase in the rotor resistance increases the starting torque. Thus, the starting torque range for a class B motor is almost the same as that of the class A motor. Owing to the increase in reactance, the starting current is about 4.5 to 5.5 times the full-load current. The low starting current and almost the same starting torque make class B motors appropriate for class A applications as well. Therefore, class B motors can be substituted in all applications using class A motors. The speed regulation for class B motors is 3% to 5%. Class C Motors.

A class C motor usually has a double-cage rotor and is designed for full-voltage starting. The high-resistance rotor limits the starting current to 3.5 to 5 times the full-load current. The starting torque is 200% to 275% of the full-load torque. The speed regulation is 4% to 5%. Class C motors are used in applications that require high starting torques, such as compression pumps, crushers, boring mills, conveyor equipment, textile machinery, and wood-working equipment.

Class D Motors:

A class D motor is a high-resistance motor capable of developing a starting torque of 250% to 300% of the rated torque. The high rotor resistance is created by using high-resistance alloys for the rotor bars and by reducing the cross-sectional area of the bar. Depending upon the design, the starting current may be 3 to 8 times the rated current. The efficiency of a class D motor is lower than that of those discussed above. The speed regulation may be as high as 10%. These motors are used in such applications as bulldozers, shearing machines, punch presses, stamping machines, laundry equipment, and hoists.

Class E Motors:

Class E motors in general have low starting torque and operate at low slip at rated load. The starting current is relatively low for motors below 7.5 horsepower. These motors may be started at rated voltage. However, for motors above 7.5 horsepower the starting current may be high enough to require a low-voltage starting circuit.

Class F Motors:

A class F motor is usually a double-cage motor. It’s a low-torque motor and requires the lowest amount of starting current of all motors. The starting torque is usually 1.25 times the rated torque, whereas the starting current is 2 to 4 times the rated current. The speed regulation is over 5%. These motors can be started by applying the rated voltage. They are designed to replace class B motors and are built in sizes above 25 horsepower.