Some Special-Purpose Electric Machines

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Intro

This section is devoted to the study of special-purpose electric machines.

Although all electric machines have the same basic principle of operation, special-purpose machines have some features that distinguish them from conventional machines.

It’s not our intention to discuss all kinds of special-purpose machines in one section; rather, an attempt is made to introduce the basic operating principles of some special-purpose machines that are being used extensively in home, recreational, and industrial applications.

With the proliferation of power electronic circuits and digital control systems, precise speed and position control can be achieved in conjunction with special-purpose electric machines such as permanent-magnet (PM) motors, step motors, switched-reluctance motors, brushless direct-current (dc) motors, hysteresis motors, and linear motors. Some of these devices find applications in computer peripheral equipment or in process-control systems whereas others can be used in devices such as home appliances. For example, step motors are employed extensively in computers where precise positioning is required, as in the case of a magnetic head for a disk drive. For applications that demand constant-speed drives, brushless dc motors offer excellent characteristics. Switched-reluctance motors, on the other hand, find applications where we traditionally use dc or induction motors.

In the following sections we discuss the construction, operating principles, and characteristics of each of the above-mentioned special-purpose electric machines.


===Sectional view of a brushless dc motor. (from Bodine Electric Co.)

Permanent-Magnet Motors

The development of new permanent-magnet materials has made PM motors a viable substitute for a shunt (dc) motor. In a PM motor the poles are made of permanent magnets. Although dc motors up to 75 hp have been designed with permanent magnets, the major application of permanent magnets is confined to fractional-horsepower motors for economic reasons. In a conventional dc motor with a wound-field circuit, flux per pole depends on the current through the field winding and can be controlled. However, flux in a PM motor is essentially constant and depends on the point of operation.

For the same power output, a PM motor has higher efficiency and requires less material than a wound dc motor of the same ratings. However, the design of a PM motor should be such that the effect of demagnetization due to armature reaction, which is maximum at standstill, is as small as economically possible.

Since the flux in a PM motor is fixed, the speed-and current-torque characteristics are basically straight lines. Mathematically, these relations can be expressed as ... and ... where K, V, Op, R, and Td are the machine constant, supply voltage, flux per pole, resistance of the armature winding, and developed torque.

+++1 (a) Cross-sectional view of a PM motor; (b) equivalent circuit.

+++2 Speed-and current-torque characteristics of a PM motor.

+++3 Operating characteristics for different supply voltages.

The speed-torque characteristic of a PM motor can be controlled by changing either the supply voltage or the effective resistance of the armature circuit. The change in the supply voltage varies the no-load speed of the motor without affecting the slope of the characteristic. Thus for different supply voltages, a set of parallel speed-torque characteristics can be obtained.

On the other hand, with the change in the effective resistance of the armature circuit, the slope of the curve is controlled and the no-load speed of the motor remains the same. Using magnets with different flux densities and the same cross-sectional areas, or vice versa, there are almost infinite possibilities for designing a PM motor for a given operating condition. From the same figure we can also conclude that an increase in blocked-rotor torque can be achieved only at the expense of a lower no-load speed.

+++4 Operating characteristics for different resistances of armature circuit.

+++5 Operating characteristics for different fluxes in a PM motor.

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EXAMPLE 1: A PM motor operates at a magnetic flux of 4 mWb. The armature resistance is 0.8 R and the applied voltage is 40 V. If the motor load is 1.2 N-m, determine (a) the speed of the motor and (b) the torque developed under a blocked-rotor condition. The motor constant K, is 95.

EXAMPLE 2: Calculate the magnetic flux in a 200-W, 100-V PM motor operating at 1500 rpm.

The motor constant is 85, the armature resistance is 2 R, and the rotational loss is 15 W.

Since the power developed is pd = 200 + 15 = 215 W, the developed torque becomes...

-----

+++6 Demagnetization effect in a PM motor.

+++7 Intrinsic operating point of the trailing tip of the magnet during start or stall.

+++8 Recoil effect on the operation of a PM motor.

+++9 Temperature effect on the intrinsic curve. Low temperature.

As explained, the operating point of a permanent magnet depends on the permeance of the magnetic circuit. The point of intersection of the operating line and the demagnetization curve determines the flux density in the magnetic circuit. The same situation takes place also in PM motors as long as the demagnetization effect of the armature reaction is neglected.

Let us assume that the point of operation for a PM motor is marked by X when the effect of armature reaction is not considered. However, no matter what, the demagnetization effect of armature reaction should be included to determine the proper operating point of the magnet even though PM motors are designed with relatively large air-gaps to minimize the armature reaction. In this case the operating line moves to the left, where Har corresponds to the magnetic field intensity due to the armature. Thus, the actual operating point of the motor moves to point Y. From the figure, we can conclude that the useful magnetic flux density decreases with the increase in armature reaction.

The demagnetization effect in a PM motor due to the armature reaction is maximum under blocked-rotor condition. In order to study its effect, we consider the intrinsic demagnetization curve. This curve can be extracted from the normal demagnetization curve using Bi = B, + k&, where Bi is the intrinsic flux density. Here, B, and H, are the normal flux density and the corresponding field intensity, respectively. +++ highlights that if Ha, exceeds the intrinsic coercive force Hci, the magnet in the PM motor is completely demagnetized.

Let us assume that the load line of a PM motor with no armature current intersects the demagnetization curve at point X. With an increase in armature current, the point of operation shifts to Y owing to armature reaction. We would expect the operating point to move back to X again as soon as the armature current is switched off. This, in fact, is not so, and the new operating point will be Z on the original operating line. The line from Y to Z is known as the recoil line. The recoil line is approximately parallel to the slope of the demagnetization curve at point B,. The overall influence of the armature reaction is a reduction in the operating flux density in the motor. However, if ceramic magnets are used, the reduction is insignificant, as the demagnetization curve is essentially a straight line.

The effect of temperature should also be taken into consideration when designing a PM motor. +++ the changes in the demagnetization characteristic at two different temperatures. As the temperature increases, the residual flux density in the magnet decreases and the intrinsic coercive force increases. On the other hand, the lower the temperature, the more pronounced the demagnetization effect of the armature reaction.

---- Exercises

===

1. A 0.5-hp, 120-V, 2-pole, 70% efficient PM motor utilizes samarium-based permanent magnets. Its ideal no-load speed is 1000 rpm, and armature resistance is 1.5 Q. The pole length and the average radius of the motor are 55 mm and 45 mm, respectively. Determine the operating line of the motor if all the losses except the copper loss are negligible. Consider the motor constant to be 80. A 100-V, 2-pole, PM dc motor with Alnico magnets drives a load of 0.25 hp

2. at an efficiency of 72%. When the motor operates at no load, its speed is 1000 rpm. The pole length and the average radius of the magnets are 65 mm and 55 mm, respectively, and the resistance of the armature circuit is 1.2 R. Determine the operating line of the motor. The demagnetization characteristic of the Alnico magnet is given. K, is 65.

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+++10 (a) Magnetic circuit with a freely rotating member, (b) reluctance as a function of position, and (c) minimum reluctance, equilibrium, or no rotation position.

Step Motors

Step motors, also known as stepping or stepper motors, are essentially incremental motion devices. A step motor receives a rectangular pulse train and responds by rotating its shaft a certain number of degrees as dictated by the number of pulses in the pulse train. Usually the pulse train is controlled by means of a computer or an electronic circuit. As a result, a step motor is very much compatible with digital electronic circuits and may form an interface between a computer and a mechanical system. Since the motion in a step motor is generally governed by counting the number of pulses, no feedback loops and sensors are needed for its control. Therefore, step motors are suitable for position control in an open loop system. They are relatively inexpensive and simple in construction and can be made to step in equal increments in either direction. Step motors are excellent candidates for such applications as printers, XY plotters, electric typewriters, control of floppy disk drives, robots, and numerical control of machine tools. Some of the drawbacks of step motors are that they don’t offer the flexibility of adjusting the angle of advance, and their step response may be oscillatory in nature with a considerable overshoot. Step motors can be classified into three broad categories-variable-reluctance, permanent-magnet, and hybrid.

Variable-Reluctance Step Motors:

Variable-reluctance step motors operate on the same principle as a reluctance motor. The principle involves the minimization of the reluctance along the path of the applied magnetic field.

The stator of a variable-reluctance step motor consists of a single stack of steel laminations with phase windings wound on each stator tooth. The rotor, which is also made of a stack of steel laminations, does not carry any winding. In order to make only one set of stator and rotor teeth align, the number of teeth in the rotor is made different from that of the stator.

The step motor has six stator teeth and four rotor teeth. The stator windings are excited at different times, leading to a multiphase stator winding. The stator of the step motor has three phases-A, B, and C, with teeth 1 and 4,2 and 5, and 3 and 6, respectively.

Rotor teeth 1 and 2 are aligned with stator teeth 1 and 4 when phase-A winding is excited by a constant current. As long as phase-A is energized while all the other phases are not, the rotor is stationary and counteracts the torque caused by the mechanical load on the shaft. Since the angle between the magnetic axis of phase-B or -C and the axis of rotor teeth 3 and 4 is 30", if phase-A is switched off and phase43 winding is excited, this time rotor teeth 4 and 3 align under stator teeth 3 and 6, leading to 30" of displacement of the rotor. Finally, if we excite phase-C winding after de-energizing phases, the rotor rotates another 30" and aligns with phase-C. The rotor can be made to rotate continuously in the clockwise direction by following the switching sequence described above. To achieve a counterclockwise rotation, however, phases should be sequentially switched in the order of A, C, B. ---the phase voltages applied to the variable-reluctance step motor discussed, and Table 1 shows the proper switching sequence for a clockwise rotation. For this particular motor, applied voltage must have at least five cycles for one revolution.

The step angle, 6, for a variable-reluctance step motor is determined by ....

...where n and p are the number of phases and the number of rotor teeth (poles), respectively.

--- ===+++11 Variable-reluctance step motor.

+++12 Phase-voltage waveforms for a variable-reluctance motor.

Table 1: Switching Sequence for a Variable-Reluctance Motor. "1" and "0" correspond to positive and zero current in a phase winding, respectively.

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Permanent-Magnet (PM) Step Motors:

A PM step motor differs from its variable-reluctance counterpart in that its rotor is made of permanent magnets. The stator construction of a PM step motor is the same as that of a variable-reluctance step motor. A two-phase, 2-pole-rotor PM step motor. In this motor the rotor is radially magnetized so that the rotor poles align with the appropriate stator teeth.

When phase-A winding is excited by a constant current, tooth 1 acts as a south pole. This makes the north pole of the PM rotor align with the south pole of the stator. Later in time, phase-A is de-energized while the phase4 winding is activated causing a 90" displacement in the counterclockwise direction to align the rotor's north pole with stator tooth 2. If we reverse the polarity of the applied current and start exciting phase-A again, the rotor will further rotate 90" along the counterclockwise direction, this time to align the rotor's north pole with the stator tooth 3. So far the motor has completed one-half revolution, and with the continuation of the appropriate switching, the rotor continues to rotate and completes its revolution.

+++ the input waveforms to phase-A and phase-B of a two-phase step motor, while Table ===2 describes the switching sequence for one full revolution of the motor.

Table -2: Switching Sequence for a Two-Phase PM Step Motor. "1," "-1," and "0" correspond to positive, negative, and zero current in a phase winding, respectively.

+++13 Two-phase permanent magnet step motor.

+++14 Applied voltage waveforms for a two-phase PM step motor.

+++15 Various views of a hybrid step motor.

Hybrid Step Motors:

The stator construction of a hybrid step motor is no different from that of a variable-reluctance or a PM step motor. However, the rotor construction integrates the design of the rotors of a variable-reluctance and a PM step motor. The rotor of a hybrid step motor consists of two identical stacks of soft iron as well as an axially magnetized round permanent magnet. Soft iron stacks are attached to the north and south poles of the permanent magnet. The rotor teeth are machined on the soft iron stacks. Thus the rotor teeth on one end become the north pole, while those at the other end become the south pole. The rotor teeth at both north and south poles are displaced in angle for the proper alignment of the rotor pole with that of the stator. The operating mode of the hybrid step motor is very similar to that of a PM step motor.

Torque-Speed Characteristic:

Step motors are generally used in a range from 1 W to about 3 hp, and their step sizes vary from approximately 0.72" to 90". However, the most common step sizes are 1.8", 7.5", and 15".

Since a step motor rotates when a series of pulses is applied to its phase windings, the duration of each pulse should be sufficiently long to accurately rotate the motor at the desired speed. If the pulse duration is too short, the rotor will miss the steps and be unable to follow the applied pulses accurately. Thus either the motor won’t rotate or the required speed won’t be achieved. To avoid such an undesirable operation, usually the pulse duration is selected so that it’s greater than the inertial time-constant of the combination of the rotor and the mechanical load. Therefore, it’s expected that a large motor with a high moment of inertia requires a slower pulse rate for accurate operation.

The pull-in torque characteristic illustrates the permissible range of rate of steps for a given load and a motor in order not to miss a step. When the motor achieves its steady-state operation, the speed is uniform and no starting and stopping take place at each step. We can load the motor up to a limit, which is defined by the pull-out torque characteristic. Above that torque level the motor starts missing the steps, thereby losing speed.

+++16 Speed-torque characteristic of a step motor.

+++17 Schematic diagram of a switched-reluctance motor in its simplest form.

Switched-Reluctance Motors

In principle, a switched-reluctance motor operates like a Variable-reluctance step motor discussed in the previous section. However, the operation differs mainly in the complicated control mechanism of the motor. In order to develop torque in the motor, the rotor position should be determined by sensors so that the excitation timing of the phase windings is precise. Although its construction is one of the simplest possible among electric machines, because of the complexities involved in the control and electric drive circuitry, switched-reluctance motors have not been able to find widespread applications for a long time.

However, with the introduction of new power electronic and microelectronic switching circuits, the control and drive circuitry of a switched reluctance motor have become economically justifiable for many applications where traditionally dc or induction motors have been used.

A switched-reluctance motor has a wound stator but has no windings on its rotor, which is made of soft magnetic material. The change in reluctance around the periphery of the stator forces the rotor poles to align with those of the stator. Consequently, torque develops in the motor and rotation takes place.

The total flux linkages of phase-A is 1, = L,(O)i, and of phase-B is hb = Lb(0)ib, with the assumption that the magnetic materials are infinitely permeable. Since the magnetic axes of both windings are orthogonal, no mutual flux linkages are expected between them.

The co-energy in the motor is ... and the developed torque is ...

We can conclude that the developed torque in the motor is independent of the direction of the supply current because it’s proportional to the square of the phase currents. However, the initial rotor position has a significant impact on the developed torque and rotation. Thus, a reliable rotor position sensor and a control circuitry are necessary to energize the motor at the proper instant to rotate it in the desired direction.

Brushless DC Motors

+++18 Schematic of a brushless dc motor that indicates the operating principle.

+++19 Speed-torque characteristic of a brushless dc motor.

Owing to their inherent characteristics, as discussed in previous sections, dc motors find considerable applications where controlling a system is a primary objective. However, electric arcs produced by the mechanical commutator-brush arrangement are a major disadvantage and limit the operating speed and voltage.

A motor that retains the characteristics of a dc motor but eliminates the commutator and the brushes is called a brushless dc motor.

A brushless dc motor consists of a multiphase winding wound on a non-salient stator and a radially magnetized PM rotor. +++ a brushless dc motor. The multiphase winding may be a single coil or distributed over the pole span. Direct voltage or alternating voltage is applied to individual phase windings through a sequential switching operation to achieve the necessary commutation to impart rotation. The switching is done electronically using power transistors or thyristors. For example, if winding 1 is energized, the PM rotor aligns with the magnetic field produced by winding 1. When winding 1 is switched off while winding 2 is turned on, the rotor is made to rotate to line up with the magnetic field of winding 2. As can be seen, the operation of a brushless dc motor is very similar to that of a PM step motor. The major difference is the timing of the switching operation, which is determined by the rotor position to provide the synchronism between the magnetic field of the permanent magnet and the magnetic field produced by the phase windings. The rotor position can be detected by using either Hall-effect or photoelectric devices. The signal generated by the rotor position sensor is sent to a logic circuit to make the decision for the switching, and then an appropriate signal triggers the power circuit to excite the respective phase winding. The control of the magnitude and the rate of switching of the phase currents essentially determine the speed-torque characteristic of a brushless dc motor.

Hysteresis Motors

Hysteresis motors utilize the property of hysteresis of magnetic materials to develop torque. The stator may have a uniformly distributed three-phase or single-phase winding. In a single-phase hysteresis motor, the stator winding is connected as a permanent split capacitor (PSC) motor. The capacitor is selected in such a way that a balanced two-phase condition can be approximately achieved, so that almost a uniform rotating field can be established in the motor. The rotor is a solid hard magnetic material with no teeth or windings. +++ a schematic diagram of a hysteresis motor with two-phase windings.

When the stator winding is excited, a rotating field is set up in the air-gap of the motor that revolves at the synchronous speed. The rotating field magnetizes the rotor and induces as many poles on its periphery as there are in the stator.

Owing to the large hysteresis loss in the rotor, the magnetic flux developed in the rotor lags the stator magnetomotive force (mmf). Thus a rotor angle, 6, takes place between the rotor and stator magnetic axes. +++ illustrates the relative positions of the rotor and the stator magnetic axes for a 2-pole hysteresis motor.

The greater the loss due to hysteresis, the larger the angle between the magnetic axes of the rotor and the stator. Because of the tendency of the magnetic poles of the rotor to align themselves with those of the stator, a finite torque, called hysteresis torque, is produced. This torque is proportional to the product of the rotor flux and the stator mmf and the sine of the rotor angle 6. Thus, it should be noted that a rotor with a large hysteresis loop results in a higher hysteresis torque.

Since the rotor is a solid magnetic material, eddy currents are induced in the rotor by the magnetic field of the stator as long as there is a relative motion between the stator magnetic field and the rotor. These eddy currents produce their own magnetic fields and thereby their own torque, which further enhances the total torque developed by the motor. The torque due to the eddy currents is proportional to the slip of the motor, and it’s maximum at standstill and zero when the synchronous speed is reached.

+++20 (a) Schematic of a hysteresis motor; (b) Flux distribution showing hysteresis effect. Rotor-Stator --Speed-torque characteristic of a hysteresis motor.

When the motor is excited at a certain voltage while it has been at standstill, a constant torque is developed by the motor, and if it’s greater than the torque required by the load, the motor will start rotating. With the fixed applied voltage, the hysteresis torque remains almost constant over the entire speed range of the motor up to the synchronous speed because the rotor angle, 6, essentially depends on the rotor material. However, with the influence of the eddy-current torque, a slight reduction in the total developed torque is observed as the motor speed increases. After synchronous speed is achieved, the motor adjusts its rotor angle, 6, so that the required torque can be developed by the motor. A typical torque-speed characteristic of a hysteresis motor. From this figure it can be observed that the torque developed is maximum at standstill.

Thus the starting torque is never a problem in hysteresis motors. Furthermore, because the developed torque is almost uniform from standstill to synchronous speed, a hysteresis motor can accelerate a high-inertia load.

Linear Induction Motors

So far we have examined the fundamental operating principles of electric machines that produce rotation or circular motion. During the last few decades, extensive research in the area of propulsion has led to the development of linear motors. Theoretically each type of rotating machine may find a linear counterpart. However, it’s the linear induction motor that is being used in a broad spectrum of such industrial applications as high-speed ground transportation, sliding door systems, curtain pullers, and conveyors.

If an induction motor is cut and laid flat, a linear induction motor is obtained.

The stator and rotor of the rotating motor correspond to the primary and secondary sides, respectively, of the linear induction motor. The primary side consists of a magnetic core with a three-phase winding, and the secondary side may be just a metal sheet or a three-phase winding wound around a magnetic core. The basic difference between a linear induction motor and its rotating counterpart is that the latter exhibits endless air-gap and magnetic structure, whereas the former is open-ended owing to the finite lengths of the primary and secondary sides. Also, the angular velocity becomes linear velocity, and the torque becomes the thrust in a linear induction motor. In order to maintain a constant thrust (force) over a considerable distance, one side is kept shorter than the other. For example, in high-speed ground transportation, a short primary and a long secondary are being used. In such a system, the primary is an integral part of the vehicle, whereas the track manifests as the secondary.

A linear induction motor may be single-sided or double-sided. In order to reduce the total reluctance of the magnetic path in a single-sided linear induction motor with a metal sheet as the secondary winding, the metal sheet is backed by a ferromagnetic material such as iron.

When the supply voltage is applied to the primary winding of a three-phase linear induction motor, the magnetic field produced in the air-gap region travels at the synchronous speed. The interaction of the magnetic field with the induced currents in the secondary exerts a thrust on the secondary to move in the same direction if the primary is held stationary. On the other hand, if the secondary side is stationary and the primary is free to move, the primary will move in the direction opposite to that of the magnetic field.

Let us consider the simplified schematic diagram for a linear induction motor. Only one phase winding, say phased, of the three-phase primary winding is shown. The N-turn phase winding experiences an mmf of NIb. If we focus our attention only on the fundamental of the mmf waveform, we get where k, is the winding factor, in is the instantaneous value of the fundamental current in phase-a, A is the wavelength of the field (in essence it’s the winding pitch), n is the number of periods over the length of the motor, and z is an arbitrary location in the linear motor. Each phase winding is displaced from the others by a distance of A/3 and excited by a balanced three-phase supply of angular frequency w. Thus the net mmf in the motor consists of only a forward-traveling wave component as given by ... where ...

+++22 (a) Single-sided and (b) double-sided linear induction motors.

+++23 Schematic diagram of a linear induction motor and its mmf waveform.

The synchronous velocity of the traveling mmf can be determined by setting the argument of the cosine term of Eq. (7) to some constant K ... and then differentiating with respect to t to obtain us as ... where f is the operating frequency of the supply. The equation can also be expressed in terms of the pole pitch T as ....

Both Eqs. suggest that the synchronous velocity is independent of the number of poles in the primary winding. Moreover, the number of poles need not be an even number.

Similar to the rotating induction motors, the slip in a linear induction motor is defined as ....

.... where u,, is the velocity of the motor.

The power and thrust in a linear induction motor can be calculated by using the same equivalent circuit employed for its rotating counterpart. Thus, the air-gap power, Px, is .... the developed power, Pd, is ...and the developed thrust, F_d, is ...

+++24 Typical velocity versus thrust characteristic of a linear induction motor. Velocity.

The speed-torque characteristic of a conventional induction motor is equivalent to the velocity-thrust characteristic of a linear induction motor. The velocity in a linear induction motor decreases rapidly with the increasing thrust. For this reason these motors often operate with low slip, leading to a relatively low efficiency.

A linear induction motor displays a phenomenon known as end effects because of its open-ended construction. The end effects can be classified as static and dynamic. Static end-effect occurs solely because of the asymmetric geometry of the primary. In this case, the mutual inductances of the phase windings are not equal to one another. This results in asymmetric flux distribution in the air-gap region and gives rise to unequal induced voltages in the phase windings. The dynamic end effect occurs as a result of the relative motion of the primary side with respect to the secondary. As the primary moves over the secondary, at every instant a new secondary conductor is brought under the leading edge of the primary, while another secondary conductor is leaving the trailing edge of the primary. The conductor coming under the leading edge opposes the magnetic flux in the air-gap, while the conductor leaving the trailing edge tries to sustain the flux. Therefore, the flux distribution is distorted. It's weaker in the leading edge region as compared to the trailing edge region. Furthermore, the conductor leaving the trailing edge, although still carrying the current and contributing to the losses, does not contribute to the trust. Therefore, the increased losses in the secondary reduce the efficiency of the motor.

EXAMPLE 3: The pole pitch of a linear induction motor is 0.5 m, and the frequency of the applied three-phase voltage is 60 Hz. The speed of the primary side of the motor is 200 h/h, and the developed thrust is 100 kN. Calculate the developed power by the motor and the copper loss in the secondary side.

SOLUTION: The speed of the motor ... The developed power ... The synchronous speed of the motor is ... Slip for this operating condition is ... Copper loss in the secondary side is ...

Exercises

A linear induction motor with a pole pitch of 50 cm is used on a trolley that travels a distance of 10 km. The resistance and the current of the secondary side as referred to the primary are determined to be 4 R and 500 A, respectively. Determine the developed thrust by the motor when the slip is 25% while operating at 60 Hz.

A 660-V (line), 50-Hz linear induction motor drives a vehicle a slip of 20%. The motor has 5 poles and the pole pitch is 30 cm. The parameters of its equivalent circuit are:

Primary side: r1 = 0.15 R, x1 = 0.5 R Secondary side: r2 = 0.3 R, x2 = 0.3 R Magnetizing reactance: X, = 3 R

Determine (a) the synchronous speed, (b) the power delivered to the load, (c) the thrust, (d) the input current, and (e) the power factor. Neglect the core losses.

SUMMARY

Recent developments in the area of power electronics technology and digital control systems have facilitated the use of special-purpose motors for the accurate control of speed and/or position. The special-purpose motors studied are: permanent-magnet (PM) motors, step motors, switched-reluctance motors, brushless dc motors, and linear induction motors. These motors find a wide variety of applications ranging from computer peripheral equipment to high-speed ground transportation and process control.

PM motors operate as separately excited dc motors except that they don’t have a field winding. Instead, a PM motor has permanent magnets to set up the necessary magnetic field for the electromechanical energy conversion. However, special care should be taken in order not to demagnetize the magnets by exceeding their coercive force due to the armature reaction.

A step motor is an incremental motion device and is widely used in computer peripherals. Its rotation is dictated by the number of pulses applied to the stator winding. Thus, no feedback loops or sensors are required for its operation. The step response of a step motor may be oscillatory. The three kinds of step motors are variable-reluctance, permanent-magnet, and hybrid.

A switched-reluctance motor operates similarly to the variable-reluctance step motor. However, a sensor accurately detects the position of the rotor to maintain precise timing of the phase windings for a given operating condition. Switched-reluctance motors have found widespread applications where ac or dc motors have traditionally been employed.

Brushless dc motors don’t exhibit the disadvantages of conventional dc motors because they don’t have the commutator-brush arrangement. Yet their performance characteristics are very similar to those of a conventional dc shunt motor. Construction of a brushless dc motor is similar to the construction of a PM step motor. However, in a brushless dc motor, the rotor position is accurately sensed by a sensor in order to properly time the switching of the stator windings.

Hall-effect or optoelectronic devices are used to sense the rotor position in a brushless dc motor.

A hysteresis motor is a synchronous motor that uses the hysteresis property of magnetic materials to develop the torque. The torque developed by a hysteresis motor is inherently higher at any speed other than the synchronous speed.

Consequently, the starting torque is never a problem for a hysteresis motor.

Even though each type of rotating machine can find its linear counterpart, the linear induction motor is the one that covers a wide spectrum of applications such as high-speed ground transportation, conveyors, and sliding doors. A linear induction motor has a primary and a secondary side. The primary consists of a three-phase winding and a magnetic core, and the secondary is either a metal sheet or a three-phase winding wound around a magnetic core. For high-speed ground transportation, a short primary (the vehicle) and a long secondary (the track) are used. The synchronous velocity u, can be calculated from u, = 27f, where T is the pole pitch (m) and f is the frequency (Hz) of the applied three-phase voltage. The developed thrust, on the other hand, is Fd = pd/u, where pd(w) is the developed power and u, (m/s) is the velocity of the motor.

QUIZ

Why is the efficiency of a PM motor higher than that of a wound dc motor? Why does the useful magnetic flux density decrease in a PM motor while it’s operating under a load? Under what conditions can a PM motor be completely demagnetized? What is the recoil line in a PM motor? How does the temperature affect the operation of a PM motor? How does a step motor operate? What are the different kinds of step motors? What is the expression for the step angle of a variable-reluctance step motor? What is the basic difference between a PM and a variable-reluctance step motor? What is the difference between a variable-reluctance step motor and a switched-reluctance motor? In what kind of applications can one use a switched-reluctance motor? What is the difference between the PM step motor and a brushless dc motor? How does the speed vary in a brushless dc motor with changing torque? Explain the operating principle of a hysteresis motor.

What is the hysteresis torque? Is a hysteresis motor a synchronous motor? How does a linear induction motor operate? How is the synchronous velocity expressed in a linear induction motor? How does the slip vary as a function of thrust in a linear induction motor? Can a backward-traveling field be supported in a three-phase linear induction motor? Give reasons.

EXAMPLE PROBLEMS:

A 20-V, PM dc motor develops a torque of 1 N-m at the rated voltage. The magnetic flux in the motor is 2 mWb. The armature resistance is 0.93 R, and the motor constant is 95. Calculate the operating speed of the motor.

Neglect the rotational losses.

What is the armature current of the motor given in Problem 1 under the blocked-rotor condition? Determine the magnitude of the applied voltage when the motor given in Problem ===1 develops a torque of 10 N-m under blocked-rotor condition.

A 100-V, PM dc motor operates at 1200 rpm and the rated voltage. The flux per pole due to the magnets is 1.5 mWb, and the armature resistance is 0.7 R. Determine the developed torque if the motor constant is 82. Neglect the rotational losses.

A 12-V, 2-pole, PM dc motor manufactured with ceramic magnets drives a load of 0.134 hp with an efficiency of 54%. The ideal no-load speed of the motor is 800 rpm, and the armature resistance is 2 Q. Determine the operating line of the motor if the pole length and the average radius are 35 mm and 25 mm, respectively. The motor constant is 75, and all the losses are assumed to be negligible except the copper loss. The demagnetization curve is given.

Determine the performance of the motor given in Problem ===5 if samarium-based rare-earth magnets are substituted for the ceramic magnets without changing the dimensions of the motor.

Determine the magnetic flux in a 120-V, 1-hp, PM motor operating at a speed of 1500 rpm. The motor constant is 85, the armature resistance is 0.7 R, and the rotational losses are SO W. A 120-V, PM dc motor operates at a speed of 400 rad/s at no load. If the resistance of the armature circuit is 1.3 R, determine the speed of the motor when the load demands 5 N-m at SO V. Draw the speed-torque characteristics for both 50-V and 100-V operations. Assume that the motor maintains constant magnetic flux with no rotational losses.

A three-phase linear induction motor has a pole pitch of 1 m. Determine the velocity of the resultant traveling mmf wave if the motor is excited by a three-phase supply having a frequency of 50 Hz.

The synchronous speed, the operating frequency, and the peak current of a linear induction motor are 10 m/s, 60 Hz, and 10 A, respectively. Determine the net mmf traveling wave if the number of turns per phase is 300 with 3 winding factor of 0.9. Assume that the number of periods over the length of the motor is 2.

A linear induction motor drives a conveyor belt at a speed of 20 km/h with a slip of 20% at 60 Hz and develops a thrust of 200 N. (a) Determine the pole pitch of the motor. (b) Calculate the power developed by the motor. (c) Calculate the amount of copper loss in the secondary side.

A 2300-V, 60-Hz, 10-MW linear induction motor has 0.8 pf lagging while operating with a slip of 30% and an efficiency of 72%. The pole pitch of the motor is 60 cm. If the magnetizing current is 15% of the applied current and lags the applied voltage by an angle of 88", determine the developed thrust by the motor, and the winding resistance of the secondary as referred to the primary. Neglect the core, friction, and windage losses.








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