DC machines and universal motors

Transcription

1 Chapter 8 machines and universal motors Table of Contents 8.1 Introduction The basic idea and generation of quasi constant voltage Operation of a generator under load Different types of machines Generators and motors Independent, shunt, PM and series excitation motors Universal motors electric drives Historical notes Bibliography Proposed exercises For the teacher

2 8.2 Chapter 8: machines and universal motors 8.1 Introduction Electric machines traditionally considered to be composed by the transformers and the rotating machines. This is justified by the fact that the transformers have a lot of physical similarities with rotating machines: both are based on magnetic circuits made of ferromagnetic materials in which variable flux is produced, so that EMFs are generated by Faraday s law action. Rotating machines are based also on the principles of electromechanical conversion discussed in the previous chapter. The majority of machines used in practice are AC machines. However it is traditional to discuss in basic textbooks also the machines. This approach is followed in this book for the following reasons: the reader has already studied circuits and may want to see how this type of electric energy can be used to generate mechanical power machines structurally very similar to simple machines are used very frequently as singlephase AC motors (the universal motors) small machines are still used whenever a is easily and cheaply available, e.g. onboard vehicles, where the basic energy is an electrochemical battery. Nevertheless, because of the reduced usage machines have today, the topic is dealt with in a particularly simplified way. 8.2 The basic idea and generation of quasiconstant voltage The general arrangement of a machine is shown in fig It is composed by a stationary part, the stator, a rotating part, the rotor, and a space of air, called airgap between the two. The stator contains a winding made by connecting in series al the conductors shown in their crosssection in figure. A simplified idea on how these connections are made can be inferred from the small sketch reported in the lower part of fig This winding, called for reasons that will be presently clear, field winding, during the machine operation is connected to a and therefore is traversed by a current. To analyse what happens in the machine the field winding currents can be imagined to be flowing according to the signs indicated in fig. 8.1, i.e. they exit the page surface when a small dot is reported in the conductors, and enter it when a cross is reported (to understand the rationale beyond this graphical code the reader is suggested to revert to chapter 7). By effect of these currents, a magnetic field is created, whose force lines that flow in the stator, traverse the small air gap, enter the rotor, traverse it then returns to the stator as qualitatively indicated in figure 8.1. In basic analyses of machines, and in this book also, the field H produced in the airgap, and therefore the corresponding flux density B, can be considered to be uniform. B will be proportional to the current I f circulating in the field winding

3 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.3 stator rotor pole tip pole core S S B N N B B N airgap S N I Fig The basic structure of a machine. Left: two polemachine; right fourpole machine; bottom: conductor connections to obtain stator windings. The two machines represented in figure 8.1 differ from eachother from their polepairs: the machine at the left has one pole pair, while the one at the right has two; machines can have even more pole pairs. In all cases, all the windings are connected in series, and are thus traversed by the same current. The analysis of the machine behaviour that will be presently propose is based on the interaction rotorconductors and flux density produced by the stator, and is therefore applicable to either types of machine. Some differences between the two will be later be discussed. The rotor of a machine contains conductors at its periphery, as represented in fig These conductors are inserted into the slots, that are spaces from which iron has been moved, and are able to keep the conductors mechanically tightly connected to the basic iron structure of the rotor. In the electric machines terminology, the armature is the part of the machine in which an EMF is induced by the Faraday s law. Therefore in a machine the armature is the rotor. In other machines (such as the synchronous machine, as will be seen in a further chapter) is located in the stator.

4 8.4 Chapter 8: machines and universal motors brush (to get voltage and current) S here dots indicate induced electric field direction B N here dots indicate field current direction Fig. 8:2. Generation of EMFs in the rotor, and total EMF collected by the brushes. Each conductor moves in the uniform flux density distribution B, and has a local speed v, whose amplitude is R (R being the rotor s radius), but whose direction is different from one conductor to another. On each conductor, an electric field will be generated (see sect. 7.2): Ec vb that will produce a potential difference on the conductor: e () t vb l vbˆ sin( t ) (8.1) c c This varies sinusoidally with time, and the angle c is specific for each conductor. The connections of the rotor s conductors are such that the voltage appearing on the two conductive elements represented in black, called brushes, that are virtually connected to the two conductors present in the direction orthogonal to the uniform field direction is the sum of the voltages generated on half the conductors, as shown in the right part of the figure 8.2. As far as the rotor rotates, the brushes become connected to different conductors, so that the total voltage E reported in the right part of figure 8.2, and corresponding to the instantaneous sum the voltages generated on half conductors (the left half or the right one can be equivalently considered) maintains roughly always the same amplitude. More precisely the amplitude slightly fluctuates, since it is always equal to the vertical distance connecting the upper and lower points of the polygon shown in the right part of figure 8.2. Considering all the conductors, and neglecting the small fluctuation of the voltage around its average value therefore it is: E k vb 1 ˆ or, considering the proportionality of v with and ˆB with I f : and with the total flux produced by the winding coil: E k (8.2) with L I (8.3) f f and k is a dimensionless constant. The actual connection of the brushes to the rotor is not, indeed, made directly on the rotor conductors, but indeed these conductors are linked to a commutator, that contains conductive segments separated by small layers of insulating material, as depicted in fig, 8.3. The conductive E 3 E 4 E 2 E 5 E 1 E E 5 E 1 E 4 E 3 E 2

5 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.5 elements are connected to the active conductor in the rotor periphery, creating a situation that reproduce the principle scheme of fig The collector is mounted on the machine shaft, the conductive sectors connected with the rotor active conductors, and rotates with the rotor; the brushes do not rotate, are pressed using suitable springs against the commutator. The brushes are constituted by a material based on carbon, that is much softer than copper in such a way that the sliding between them and commutator causes some wearing of the brushes instead of the commutator, This causes the need of periodically substituting the brushes, but reduces dramatically wearing the commutator, much more expensive to substitute. brush conductive sector insulating material shaft Fig The basic construction of the machine commutator and its brushes. In a reference integral with the rotor, the flux produced by the stator poles rotates, and therefore the rotor of the machine must be laminated. the stator, instead, is subject (except for the tips) to constant flux and does not need to be laminated. However it often happens that manufacturers laminate all parts of their machines, just because if facilitates the machine assembly or simply because eases the manufacturing process as a whole. 8.3 Operation of a generator under load In the previous section it was seen that a machine is composed by a stator, that contains a coil to be fed in and a rotor, to which two brushes are connected to gather the voltage produced by the interaction of the stator field and the rotor motion. This is summarised in figure 8.4 a), in which the brushes, in turn feed a load. The system reported in figure 8.4 a) can be sufficiently modelled using the circuital equivalent reported in fig. 8.4 b). In fact, being the field coil fed by a voltage, it is constituted by a simple circuit, and the field current I f is simply I f =E f /R f, if E f is the voltage generated by the. The armature of the machine can be modelled by an electromotive force whose value is given by (8.2), and an inner resistance due to the resistance of all the conductors that constitute the armature winding (active conductors, connections among them). Therefore the current flowing in the armature and the load will be: I E/( R R ) k /( R R ) a a f a f

6 8.6 Chapter 8: machines and universal motors I f Load I f E f R f E R a U R l field coil a) b) E U=ER a c) Fig The generator with its field coil, their equivalent circuit and UI characteristic sign chosen to better describe generator operation. From the equivalent circuit of the machine also the UI characteristic can be easily determined, that has ideally the shape shown in fig. 8.4 c). Indeed, in case of real machines, the non linear behaviour of the iron that constitutes stator and rotor makes the actual characteristic to be non linear, but this is beyond the scope of this book. For small armature currents the actual machine characteristic can be assumed to be linear. The current flowing in the machine armature (i.e. in the rotor) will flow in the active conductors in the rotor periphery shown in fig This will produce on the rotor conductors following forces generated according to Lorentz s law: F IB l where I has as amplitude the current and direction that of conductors, and l is the length of the active conductors. These forces will have tangential components F t whose amplitudes will depend on their position (fig. 8.5): B B F t F FF t Fig Forces on the rotor conductors and their contributions F t to torque The set of all these forces will create globally a torque that will be proportional to the current i and the stator flux :

7 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.7 T k (8.4) It would be easy to demonstrate that the constant k of equation (8.4) is exactly the same that appears in eq. (8.2). Assuming this the two equations can be multiplied to each other obtaining: T ke/( k) Ia T E (8.5) Eq. (8.8) is very important, since it states the basic electromechanical conversion in the machine: the power inserted in the system as mechanical power T is converted into electrical form, generating the power E. To summarise, the machine operation can be analysed using the following equations: E k (8.6) T k Ia and the equation of the electric circuit: U=ER a. (8.7) Example 1 A generator runs at n=1000 RPM, supplying 200 V to an electric load. If the flux per pole is 25 mwb, the constant k equals 90 and the armature copper losses are 300 W, calculate: a) the electromotive force E b) the armature current c) the armature resistance d) the electric power P e provided to the load e) the torque T required to move the generator f) the efficiency, neglecting the field losses E R a R l U The angular speed can be easily calculated from n: 2n/ / rad/s Hence: E k V I P / ( EU ) 300 / ( ) = A a cu 2 2 a cu a R = P / I 300 / P UI W e a P P P 1985 W (also EI ) m e cu a T = P / Nm(also ki ) m = P/ P e m a

8 8.8 Chapter 8: machines and universal motors 8.4 Different types of machines In the previous description of the machine principle of operation, the operation of the machine as a generator was taken as reference. Indeed this usage of machines, common in the past, has nowadays been completely abandoned. It was proposed here just because it is simpler to introduce at first the generator than the motor. In the next paragraph, however, a description of the operation as motor of the machine is discussed. Moreover, in the previous description the field current I f was generated using a independent on the armature circuit. This corresponds to the Independent excitation machine, that is not very frequently used, since it requires the availability of an independent. In the following paragraph other important ways to create the flux density B in the machine are rapidly introduced, taking as reference the operation as a motor, that, as just said is the only one today used in practice Generators and motors The machine is intrinsically reversible, and can therefore operate as a generator or as a motor. In the two cases there are obvious differences in the connection of armature and field coils and of the mechanical shaft as shown in fig In the upper part of the figure, where generator operation is reported, the shaft is connected to a mechanical of power, and the load is connected to the armature. Some means of generating the field must be provided, and this can be done in several ways, as will be discussed in the next section. Here, for simplicity sake, an that is totally independent from the armature circuit is considered. In the lower part of the figure, instead, motor operation is considered. The armature is to be connected to a, and some of current for the field coil is as earlier to be provided. The motor generates mechanical power that is transferred to the mechanical load through the machine shaft. In the right part o f the figure equivalent circuit of the arrangements shown at the left part are reported. The only difference, indeed is in the reference sign assumed for current., that is combined with the voltage reference signs in such a way that for the machine generator and load conventions are used for the cases of generator and motor operation respectively. T, P=T field U load R a E U E f I f R f field coil Main U field T, P=T U R a E E f I f R f Fig machine connections and equivalent circuits: up: generator operation; down: motor operation.

9 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.9 To understand what happens in the machine when it is operating as motor, it is simply a matter of repeating the reasoning already made for the operation as generator, just with a slightly different order. This will be rapidly done in this section. Consider the machine to be initially still. Its connection to the s makes a flux density B to be created inside the machine, and an armature current to flow in the rotor active conductors. The two interact and Lorentz s forces are generated: F IB l that globally produce a torque T that will make the rotor start rotating: T k When the rotor rotates, an EMF is produced in the active conductors according to Faraday s law for sliding conductors: E kvb ˆ 1 that can also be expressed in more convenient terms using the (8.2): E k (8.2) As per the operation as a generator, the electric power E is converted into the mechanical power T: E =T (8.8) The general equations of the machine (8.6) are therefore still valid. Also the circuit equation (8.7) is valid, but is more conveniently written using the load sign convention: U E RaIa (load sign convention) It can be concluded that the machine equations are exactly the same in generator and motor operation, although they are usually written in a slightly different way in the two cases, as summarised by the following fundamental machine equations RESULT box. Not evidenced in the box there is also another convention: in the generator convention the reference signs for the mechanical quantities are such that positive values of T and imply mechanical power entering or exiting the machine in the cases of generator and load convention respectively. RESULT: Fundamental machine equations The machine equations are: E k T k Ia U ERaIa (generator convention) E k T k Ia U ERaIa (load convention) (8.8) When generator convention is used positive values of U, imply power going from the mechanical system into the electric one; when a load convention is used they imply opposite power flows. Therefore generator convention is commonly used for generators and load convention for motors. Example 2 A 240 V machine with independent excitation has an armature resistance R a = 0.2. If the armature current is 20 A, calculate the electromotive force: a) for generator operation b) for motor operation.

10 8.10 Chapter 8: machines and universal motors a) E=UR a = =244 V b) E=UR a = =236 V From the previous equations a torquespeed relationship for the machine is easily found, that is of big importance for evaluating the mechanical behaviour of the machine. Consider the equations written with the load convention. Substituting the first of them into the third yields: U k RaIa and Ia U k (8.9) Then, substituting the torque equation:: T U k Ra k T k( U k) A B (8.10) Therefore both the armature current and generated torque are linearly decreasing curves as a function of the angular speed, as reported graphically in fig To evaluate the startup transient of a machine consider the arrangement shown in the right part of figure 8.7 in which we consider the switch S 1 to be held constantly closed, in such a way that the resistor R s does not influence the system behaviour. Imagine that at a given time the switch S, initially open, is closed. Let the machine and its mechanical load have a combined moment of inertia J, and the load torque be of the type: T l =ab 2. The transient that will be activated is governed by the mechanical equation: T Tl J The rotor will thus accelerate from 0 to the final speed. In theory the final speed will be reached in an infinite time, but the system will reach the vicinity of that point in a more reasonable time, e.g. a few seconds. At the beginning of this starting transient, torque and absorbed currents will be at their maximums T 0 and I 0, respectively. the current I 0 can be thought to be as a short circuit current (cf. section ), since at the very starting of the machine the two terminals of the are connected to each other and share the same potential. Indeed often the machines cannot withstand currents as high values as I 0 even for the small period needed for acceleration, without damage. For these machines the starting operation must be made at a reduced voltage, e.g. interposing the resistor R s between the and the motor, just for the first part of the starting transient. In this case, in the first part of the transient switch S 1 will be held open, current and torques will have reduced values, up to the instant t*, which corresponds to speed *, in which it is closed and the machine is fed at full voltage. In some cases, the process is more staged: the resistor interposed is progressively reduced in value, as far as the machine speed increases and with it also increases also the EMF E. Some small motors, however, are designed to be started without the interposition of the starting resistor, thus simplifying the starting process and eliminating the energy losses in the starting resistor: Today, large motors are more and more frequently started without starting resistor, and without making the motor absorb the current I 0, feeding they at starting with a reduced voltage: this will be discussed in sect. 8.6.

11 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.11 I 0 T 0 T,I Main S R s S 1 T T l () R a U S R s S 1 E 0 Fig. 8.7: Currentspeed and torquespeed characteristics, and of starting of an independently excited motor ( 1 ). E knee saturated part linear part I f The analysis performed here was based on a linear model of the machine. This is correct only as an approximation, since due to the nonlinear behaviour of stator and rotor iron, actual machines significantly deviate from linearity, when saturation occurs. The main deviation from linearity in the relationship between the field current I f and the generated EMF, as indicated in the figure aside. In real machines, however, the actual excitation operating point is held in the linear zone, or at most at the knee, since going further would increase excitation losses without increasing the EMF. Example 3 A motor with independent excitation is supplied with 240 V, the flux per pole being 30 mwb and the armature resistance 0.5. If the motor develops a torque T=45 Nm and the constant k= 85, calculate: a) the armature current b) the electromotive force c) the motor speed, in RPM d) the mechanical power provided e) the efficiency T A k E U Ra V E = rad/s RPM k Pm T 4080 W Pel U Ia 4235 W Pm = P el 1 It will be soon shown that approximately the same curves apply also in the cases of shunt and PNM excited machines.

12 8.12 Chapter 8: machines and universal motors Independent, shunt, PM and series excitation motors In the previous sections the fundamental equations (8.8) have been demonstrated in both generator and motor operation. They conten the stator flux. The only way earlier considered to generate this flux was by means of an external. As already discussed, this is just one option that is normally referred to as independent excitation. Other ways are possible and commonly used. Some of them are reported in figure 8.8, and rapidly discussed here. Main I f field I f N S a) field coil b) c) =I f d) R Th E Th U f e) Fig Some options for excitation of machines: a) independent (or separate); b) shunt; c) permanent magnet; d) series. In figure a) the excitation is independent (or separate,) as discussed in the previous section. In figure b) the supply to the field coil is the same that feeds the armature. This is called shunt excitation. In case the is ideal, i.e. with zero internal resistance, this solution is perfectly equivalent to the one of independent excitation. In case of a nonzero inner (or Thevenin) resistance of the, a (often minor) difference comes from the fact that when the machine absorbs a very high current, the voltage U f feeding the field coil I f is significantly lower than E Th. thus, with shunt excitation a high armature current tends to reduce the field current. Therefore, except for minor differences, the voltagecurrent and torquecurrent characteristics are the same as reported in the previous section for the case of independent excitation. In figure c) the field is not created by an actual coil, but by the insertion of permanent magnets in the stator circuit. Their effect is, as discussed in chapter 7, to create a fixed field flux, exactly in the same way as if a fixed current flowed in an excitation coil. Therefore PM excitation is equivalent to independent excitation, and as the latter it is nearly equivalent to shunt excitation. Some important, though obvious, differences about these types of excitations are: independent excitation requires an independent ; it allows easy modification of I f, acting on the value of the voltage applied to the field coil; shunt excitation is cheaper than independent excitation, since does not require an external energy ; modification of I f is possible inserting in series with the coil an additional resistance, so that at equal voltage applied to the series the total resistance is increased and I f reduces; PM excitation allows easy and robust machine construction, but does not allow flux modifications. It is typically used in small motors. Finally, in figure c), the field winding is connected in series with the machine armature.

13 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.13 The voltagecurrent and torquecurrent characteristics are easily determined combining the machine equations (8.8) and the excitation equation LfIf LfIa. If the symbol L m is used to indicate the product kl f, it therefore will be: E k klfia LmIa 2 T k Ia klfiaia LmIa and: U ERaIakLf Ia RaIa( Lfa Ra) Ia Thus T L 2 U m 2 Lm Ra The resulting shape, drawn considering U m =const, is summarised in fig Starting from the value T 0 =L m U 2 /R a 2 at standstill, degreases continuously as increases, always being positive. At large speeds the term R a in the denominator can be neglected and the curve decreases proportional to 1/ 2. The standstill current I 0 (and the standstill torque T 0 ) is normally too high and can cause damage to the machine. Therefore, the machine is started at a reduced voltage, and the reduction is obtained at low speeds by interposition of a starting resistor R s, that is then bypassed when the current has become acceptable, in correspondence to the speed * (the curve of I 0 when S 1 is open is not reported in the current plot just to avoid too many curves on the same plot). T,I S S 1 =I f I 0 T 0 R s 1/ 2 Fig. 8.9: Currentspeed and torquespeed characteristics, and of starting of a seriesexcited motor. Series excited machines are used when the load might have sudden peaks of torque: in this chase this machine slows down, and rises its torque by large amounts, thus overcoming the difficulty. It was widely used in the past onboard trains, mainly because of its high torque capability at low speeds, even though starting of trains was made using the temporary insertion of series resistances (such as the R s if figure 8.9). Today they are used as small motors, often sized to be able to start without R s, and as the Universal motor, that will be discussed in section 8.5.

14 8.14 Chapter 8: machines and universal motors Example 4 A 300 V shunt motor absorbs 35 kw from the electric grid. The field and armature resistances are respectively 120 and Calculate: a) the line, armature and field current b) the electromotive force c) the efficiency d) the motor speed, if the flux per pole is 30 mwb and the constant k equals 80 e) the torque developed Pel I A (line) U 300 U 300 I f 25A(field). R f 120 Ia I I f A (armature) E U Ra V Pm EIa W Pm = Pel E rad/s 1122 RPM k T P m / 274 Nm (also k ) The reader can verify that losses (P el P m ) equal R a I 2 a R f I 2 f. Example 5 A 240 V series motor has an armature resistance R a =0.4 and a field resistance R f =0.12. If the motor runs at 900 RPM absorbing 30 A, calculate torque and efficiency. E U ( Rf Ra) I = V n = rad/s 60 Pm EI 6732 W T P m / rad/s Pm = UI 8.5 Universal motors Consider a seriesexcited motor, and its equivalent circuit (fig. 8.10).

15 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.15 =I f U T, U R a R f E P=T P=E Fig Seriesexcited motor with indication of electric and mechanical reference signs. Moreover, consider and the usual general equations of all motors (8,8): E k U E R T k aia Ia In case of series excitation, being LI f a the same equations become: E LmIa U E R 2 aia T LmIa In case the actual polarity of the supply voltage is opposite to the reference signs reported in fig. 8.10, it will be: E<0 <0 >0 T>0 P=E =T>0 This indicates that in case of reversal of the supply polarity, the machine continues to rotate in the same direction, and the electric circuit will continue to transfer power to the mechanical part of the machine. As a consequence, if a seriesexcited machine is connected to an alternating, that continuously switches its polarity, it will operate as a seriesexcited machine, with a torquespeed relationship similar to that discussed in the previous section. A motor with commutator, designed to be fed but an AC is normally called universal motor 2 Under AC operation the motor will draw a current that can be assumed to be sinusoidal: it () Iˆ sint This will generate a torque: 2 Tt () LI m (1 cos2 t) that will have an average value and a fluctuation around that value at twice the frequency of the supply voltage. Indeed it must be noted that all the equations that have been written in the previous section, did not take into account the possibility of dynamical variation of the quantities, in particular of the current =I f. In case the machine is connected to an AC, instead, this current will continuously change, and this will introduce significant changes in the machine behaviour, since the machine has inner inductances and current flowing into inductors cause electromotive forces to be generated: di UL L. dt 2 Sometimes they are also called AC commutator motors.

16 8.16 Chapter 8: machines and universal motors The higher the frequency applied to the universal motor, the more its operation will be different from the characteristics. For instance the torquespeed relationship of the same machine operating at different frequency is qualitatively represented in fig In a recent past, when electric trains were typically operated using commutator motors, very large AC commutator motors were used: up to hundreds of kilowatts. To have the advantage of AC feeding (that allowed used of transformers to feed the contact lines), and minimise the disadvantages connected to the effect of machine inductances traversed with fast variable currents, a lot of railway contact lines were built that operated at a frequency that is lower than the one typically used for stationary applications: 25 Hz in some American states, 50/3=162/3 Hz in some European countries. In Europe there still exist thousands of kilometres operated at 162/3 Hz, even though modern trains innerly convert electricity first in, and then supply a power trains that typically are based on synchronous or induction motor drives. T Railway freq Railway frequency: 16,67Hz; 25 Hz Stationary frequency: 50 Hz, 60 Hz Stationary freq Fig. 8.11: Universal motor torque behaviour at different frequencies. Small universal motors are today very frequently used in small domestic appliances, where simple construction and high torque at small speeds are requested. therefore, typical application is in drills, food mixers, saws, vacuum cleaners. Characteristic of these motors is a high noise and small arcs in the commutator: both are mainly due to the continuous switching of current between the armature conductors, commanded by the switching of brushes from a commutator sector to another. Needless to say, universal motors must be always laminated in both stator and rotor. 8.6 electric drives Not always the characteristics of a standard supply to an electric machine matches what the machine would need for its optimal behaviour this occurs both when and AC supply systems and machines are involved. One situation in which occurs has already been met in this chapter, when it was seen that normally a constant voltage is not adequate for the starting of a machine, since when ist mechanical speed in null or very low, its electromotive force E is null or very low, and he machine draws a very large current that can damage it. This matching of the characteristics of the supply system and the machine was done in the phase using electromechanical means. As regards the motors it was seen in sects and 0 that one or more starting resistors R s put in series with the armature can do this matching. These electromechanical means, however have significant disadvantages. For instance the resistor R s dissipates significant quantities of energy during starting, and does not easily allow continuous variation of the voltage during startup, or operation at low speeds. In recent years, new devices called electronic converters have appeared that allow very flexible and effective modification of the characteristic of supply, and allow much easier and effective matching of machines to their supply system.

17 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.17 Therefore we can insert between the and the machine of the converter so that the voltage and current the machine sees are different from those at the interface with the (fig. 8.12a) This is normally done when a cheap is available to feed the converter, e.g. in a system in which the main us an electrochemical battery 3. In the majority of cases, however an AC is easily available. For powers larger than just a kilowatt this is typically a threephase, as reported in fig. ## b. U s I s Electronic converter U m I m T, a) Electric drive AC U ab Electronic converter U m I m T, b) Electric drive U c I b U b I c U a U ab Fig An electric drives is composed by a machine and an electronic converter that allows matching of electric and machine. The general purpose of the electronic converter is to modify the parameters of the electricity supply, ideally leaving the power unchanged. In case the main is the following equation will ideally apply: P=U s I s =U m I m (8.11) In case in which the is threephase AC, indicating with the symbols U l and I l the RMS value of the linetoline voltages and line currents respectively, and with cos the power factor (phase displacement between phase voltage and phase current) it will be: P 3UI l lcos UmIm (8.12) Obviously enough, it will not possible to realize the exact equivalence of input and output power, because, to make voltage and current conversion some power loss would unavoidably occur. For the purpose of basic analysis of electric drives, these losses can be initially neglected: they are considered only at a later stage to evaluate global efficiencies. Therefore in this section (8.11) or (8.12) will be assumed to be satisfied. Actual electronic converters ca have very high efficiencies. It is not uncommon to have efficiencies 4 of around 9798%: A converter realising the basic equation (8.11) is called converter or chopper. It performs in a much same way as an ideal transformer: it is like the transformer, but operating in. 3 This is for instance the case of some Italian submarines: submarines are indeed often fed by electrochemical s, because when submerged they cannot consume the oxygen combustionbased engines require. 4 The efficiency of a converter will be computed as the ratio of output and input power: =U m I m /(U s I s ).

18 8.18 Chapter 8: machines and universal motors The ratio between output and input voltage, and between input and output current: =U m /U s =I s /I m can be set by a control action on the inner operation of the converter, differently from the transformer, in which this ratio is fixed because it equals the turns ratio. A converter realising the basic equation (8.12) is called AC converter or rectifier. Rectifiers can be realised in different ways in practice. To obtain an effective drive, a controlled rectifier is need, e.g. a system only is able to transfer power from an AC to a load in a controlled way. So, again, the ration of input and output voltages and currents can be set by a control action on the inner operation of the converter: =U m /U l =I s /I m Some information of the inner structure and operation of electronic converters in general, and converters in particular, will be provided in part III of this book. i A second electronic converter can be used to vary the voltage applied to the field coil, so that the field current I f =U f /R f is modified according to the needs. The arrangements in this case can be of the type reported in fig. 8.13, / converter (1) U m I m I f converter (2) a) AC Controlled rectitier (1) U m I m I f Controlled rectitier (2) b) (1): armature electronic converter (2): field electronic converter Fig Electric drives with controlled armature and field voltages: a) starting from a ; b) starting from an AC. It is important to evaluate the maximum torque and power performance of an electric drive of the type shown in figure this can be easily done considering the fundamental equations of the motor (8.8): E k T k Ia U ERaIa

19 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering 8.19 considering that the machine will have maximum values of armature voltage and current U max and I max, that cannot be overcome, and that in normal operation it can be assumed UE. Consider the machine operation starting from standstill, with growing speed. In the beginning the field flux is held equal to its maximum value. The armature voltage will be approximately linearly growing with, because it is assumed to be approximately equal to the machine EMF E (fig. 8.14). Correspondingly the maximum torque available will stay at the fixed value T k Imax. There is a speed, that is called base, at which the maximum voltage has reached its maximum value U max and cannot therefore further increase. The machine speed can further increase over base, maintaining the armature voltage constant. This can be done by reducing the flux : this is the field weakening region, that can continue up to the point in which the maximum machine speed max is reached. Note that this behaviour allow to exploit the maximum torque the machine is able to deliver up to the point in which the machine has reached its maximum power; starting from that point the machine is able to deliver its maximum power up to the maximum allowed. flux torque voltage power a) b) base max base max Fig Flux, voltage, torque, power trends at the maximum current I max of a drive controlled in armature and field voltages. Obviously, the drive is able to make the machine deliver torque and power that are lower than the limit values reported in fig. 8.14: it is just a matter of making the machine absorb currents that are lower than I max. Controlling the drawn current is easy done. Consider the equation: U ERaIak RaIa in which, at given flux, E varies slowly since its variation is connected with the mechanical inertia of the machine rotor and its mechanical load. Any variation in the applied armature voltage U, will therefore result in a corresponding variation of the drawn current: U E Ia Ra and therefore of the torque produced: T k The usual rule for controlling the drive is to keep the flux at its maximum (function of as indicated in figure 8.14) controlling the field current acting on the field electronic converter, and to vary the armature current (and therefore torque) acting on the applied voltage through the armature electronic converter.

20 8.20 Chapter 8: machines and universal motors [1]. For further information on electric drives the reader is recommended to consult the book 8.7 Historical notes 8.8 Bibliography [1] N. Mohan, T. M. Undeland, W. P. Robbins, Power Electronics, 3 rd edition, John Wiley & sons, 2003.

21 M. Ceraolo D. Poli: Fundamentals of Electrical Engineering Proposed exercises 8.1. Calculate the voltage induced in the armature of a machine running at 1500 RPM, if the flux per pole is 30 mwb and the dimensionless constant k equals The generator of the previous exercise has an armature resistance of 0.4 and supplies a load of 12. Calculate: a) the armature current b) the voltage applied to the load c) the electric power provided to the load d) the torque and the mechanical power required to move the generator 8.3. A separately excited fourpole generator runs at 1500 RPM, supplying 240 V to a certain load. The flux per pole is 20 mwb and the constant k equals 85. The armature copper loss is 600 W and the total brushcontact drop is 2 V. Calculate: a) the electromotive force b) the armature current c) the armature resistance d) the electric power provided to the load e) the torque required to move the generator 8.4. A 200V separately excited machine has an armature resistance of 0.3. If the armature current is 24 A, calculate the electromotive force: a) for generator operation b) for motor operation A 250V shunt generator provides 5 kw to a certain load. The armature and field resistances are respectively 0.5 and 200. Calculate the armature and field currents, the electromotive force and the torque required to move the generator, if it runs at 1200 RPM. Nm, where is the angular speed of the motor A 400V shunt motor has field and armature resistances of respectively 100 and 0.1. If the motor absorbs 40 kw from the electric grid, calculate: a) the line, armature and field current b) the electromotive force c) the efficiency 8.9. The 400V motor of the previous exercise is running at 1100 RPM and absorbing a line current of 80 A. Calculate the torque developed and the efficiency A 230V shunt motor, with an armature resistance of 0.4 and a field resistance of 200, absorbs 2.5 A in noload operation. In another loading condition, the same motor runs at 300 RPM while absorbing 30 A. Calculate the noload speed A 400V series motor, with an armature resistance of 0.3 and a field resistance of 0.15, absorbs 40 A. If the motor runs at 800 RPM, calculate the torque and the efficiency A 230V series motor, with a total circuit resistance of 0.45, absorbs 40 A while running at 600 RPM. Calculate the speed when the motor absorbs 30 A and the torque in both conditions. Compare 1 / 2, I 1 /I 2, T 1 /T A 230V series motor, with a total circuit resistance of 0.5, absorbs 20 A while running at 800 RPM. Calculate the motor speed when the torque is reduced by 30% A separately excited motor is supplied with 200 V. The flux per pole is 25 mwb and the constant k equals 70. If the motor develops a torque of 40 Nm and the armature resistance is 0.3, calculate: a) the armature current b) the electromotive force c) the motor speed, in RPM d) the mechanical power provided e) the efficiency 8.7. Repeat the previous exercise, assuming that the mechanical load requires a torque T=50.4

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