Synchronous Machines

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Synchronous Machines

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Synchronous machines come in a variety of different constructions and designs. The differences occur in the physical outlay of the rotor and in the way in which excitation flux is provided (if it is provided at all) in the machine. Regardless of the type however, all the synchronous machines have the same construction of the stator. Stator is of cylindrical crosssection, manufactured from laminated sheets of steel, and it carries a three-phase winding that is supplied with (in the motor case) or that produces (in the generator case) a system of threephase voltages....

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  1. ENGNG 2024 Electrical Engineering SYNCHRONOUS MACHINES 1. INTRODUCTION Synchronous machines come in a variety of different constructions and designs. The differences occur in the physical outlay of the rotor and in the way in which excitation flux is provided (if it is provided at all) in the machine. Regardless of the type however, all the synchronous machines have the same construction of the stator. Stator is of cylindrical cross- section, manufactured from laminated sheets of steel, and it carries a three-phase winding that is supplied with (in the motor case) or that produces (in the generator case) a system of three- phase voltages. The three windings that constitute the stator three-phase winding are displaced in space by 120 degrees around the circumference of the machine. Three-phase voltages have the phase displacement of 120 degrees. Synchronous machines are the main work-horse of the electricity generation industry. They are used as generators in all the hydro, nuclear, coal-fired, gas-fired and oil-fired power plants. This means that a synchronous generator is a standard machine used for conversion of mechanical energy into electric energy in all the power plants that rely on conventional energy sources. Rated powers of synchronous generators are typically from a few megawatts up to a few tens of megawatts, or even a several hundreds of megawatts. Synchronous machines are used as motors as well. In this case rated power of a synchronous motor is either relatively very low (up to few kilowatts) or is in the high power region, from around 150 kW to 15 MW. In between, induction motors are used as a rule, due to their numerous advantages in this power region. The two types of synchronous machines that are relevant for the discussion that follows are: 1. Machines with uniform air-gap and excitation winding on rotor. 2. Machines with non-uniform air-gap and excitation winding on rotor. Rotor cross-section in these two types differs and leads to different mechanisms of torque production. Both machine types are used for both generating and motoring application. In the machine with uniform air-gap torque is produced solely due to interaction between the rotor and stator windings (fundamental torque component). In the machine with non-uniform air-gap, there are two torque components: fundamental torque, produced by the interaction between the stator and rotor windings, and reluctance torque component that is the consequence of the non-cylindrical rotor cross-section. Since the rotor is not cylindrical, stator windings see an air-gap of variable length as rotor rotates. Consequently, magnetic reluctance is variable and all the inductances of the stator windings are functions of the rotor position. Reluctance torque component is therefore produced, in addition to the fundamental torque component. Synchronous machine with uniform air-gap and an excitation winding on rotor: Rotor is of cylindrical construction, with an excitation winding placed on rotor. The excitation winding is supplied with a controllable DC current. Since the rotor rotates, this DC  E Levi, 2000 1
  2. ENGNG 2024 Electrical Engineering current has to be supplied somehow from the stationary outside world. In the past, slip rings and brushes were used to connect a stationary DC source to the rotating rotor winding. Nowadays, so-called brushless excitation systems (or just brushless exciters) are used (these are beyond the scope of interest here). Synchronous machines with uniform air-gap and excitation winding on rotor are often termed turbo-machines. They are used both for motoring application and for generation in coal-, oil-, and gas-fired plants and nuclear plants. Typical number of magnetic pole pairs is one or two, so that the so-called synchronous speed of rotation is 3000 rpm or 1500 rpm for 50 Hz stator frequency. Synchronous speed is defined as n s = 60 f / P ω s = 2πf / P (1) and this is the so-called mechanical synchronous speed of rotation (that is, actual speed of rotation). Symbol P stands for the number of magnetic pole pairs (each pole pair consists of two magnetic poles, one north and one south). Rotor of a turbo-machine carries, apart from the excitation winding, a short-circuited multi-phase winding similar to the squirrel-cage winding of an induction motor. The existence of this winding is crucial for motoring applications since, without it, the machine would not be capable of starting when connected to a three-phase supply on stator side. It is important however to emphasise that this winding has no impact on steady-state operation, since no currents exist in the short-circuited rotor winding (often called damper winding) in steady- state. The damper winding is important in generation as well, since it has an important role during transients. Once again, however, it has no impact on steady-state operation and is therefore omitted from the schematic representation of a synchronous turbo-machine in Fig. 1, where only excitation winding is shown on rotor. Stator three phase winding is shown in Fig. 1 with three concentric turns. It is important to stress once more that an electromagnetic torque in a synchronous machine with uniform air-gap is produced purely due to the interaction between the stator currents and the rotor flux (that stems from the DC current in the rotor winding). The torque consist therefore entirely of the fundamental torque component, very much the same as the case is with other uniform air-gap machines (DC machines, induction machines). According to the condition of average torque existence, zero frequency in rotor means that the stator frequency must equal the frequency of rotation. Rotor flux, produced by rotor DC current, is stationary with respect to rotor. However, as rotor rotates at the frequency of stator supply, rotor flux seen from the stator rotates at the same, synchronous, speed as does the stator rotating field created by the system of stator three-phase currents. Hence all the fields in a synchronous machine rotate at the same, synchronous speed. Moreover, rotor rotates at the synchronous speed as well. Note that for the constant frequency of stator voltages there is a single speed at which the machine can rotate. This means that, regardless of the loading, a synchronous machine rotates at a constant speed. This is in huge contrast to some other types of electric machines (DC machines, induction machines) where the speed of rotation is dependent on the loading of the machine. Synchronous machine with non-uniform air-gap and excitation winding on rotor: As already noted, if the air-gap is not uniform, magnetic reluctance (resistance) seen by windings will vary. In synchronous machines with non-uniform air-gap stator remains to be cylindrical, but the rotor is not. Consequently, magnetic reluctance seen by stator windings will vary as the rotor rotates. This gives rise to dependence of stator inductances on instantaneous rotor position and leads to creation of the second torque component, not present in uniform  E Levi, 2000 2
  3. ENGNG 2024 Electrical Engineering air-gap machines (reluctance torque component). Since the machine contains an excitation winding, then a fundamental torque component exists as well, so that the total torque is a sum of the fundamental torque component and the reluctance torque component. Rotor b -c Stator Air-gap -a a c -b Rotor winding Fig. 1 - Synchronous machine with uniform air-gap and excitation winding on rotor. The structure of the machine is shown in Fig. 2. The rotor is of so-called salient-pole structure and the machine is usually used for very low speed of rotation, so that the number of poles is very high (say, 84 poles - that is, 42 pole pairs, is a rather common structure). The machine of Fig. 2 is shown as being of 2-pole structure for simplicity. This type of synchronous machines is often called hydro-machine or salient-pole synchronous machine, since these machines are used in hydro-powered electric generation plants. The machine is used in motoring application as well, for those cases where a low speed of rotation is required. Rotor of a hydro-machine carries, apart from the excitation winding, a short-circuited multi-phase winding similar to the squirrel-cage winding of an induction motor (this winding is not shown in Fig. 2). This is very much the same as for a turbo-machine. This winding once more has no impact on steady-state operation, since no currents exist in the short-circuited rotor winding in steady-state (there is not any induced emf, since the rotor rotates at the same speed as does the stator rotating field). Rotor b -c Stator Air-gap -a a Rotor c -b winding Fig. 2 - Cross-section of a salient pole synchronous machine with excitation winding.  E Levi, 2000 3
  4. ENGNG 2024 Electrical Engineering As already noted, synchronous machines are used as motors either for low power applications or for very high power applications. In the low power range the advantage of synchronous motors is their fixed synchronous speed of rotation that is always governed by the applied frequency in stator winding. In other words, speed of rotation depends only on stator frequency and it is load independent, in contrast to DC motors and induction motors. As far as the high power region is concerned, the advantage of synchronous motors over the induction motors is that they are capable of producing reactive power (while induction motors always have to consume reactive power). The higher capital cost of a synchronous motor, when compared to an induction motor of the same size, is thus offset by the capability of the machine to satisfy its own reactive power needs and, if necessary, to delivery reactive power to the grid (it is important to realise that reactive power is not for free; industrial customers are charged for the consumed reactive power). In what follows, only the synchronous machines with uniform air-gap will be elaborated. The reason is that an analysis of synchronous machines with non-uniform air-gap is more involved, due to the existence of the reluctance torque component. The concept of the rotating fields is reviewed next. This is followed by an explanation of the mechanism of the reactive power production in a synchronous machine. Motoring mode of operation and the generating mode of operation are then discussed in two separate sections. 2. ROTATING FIELDS AND REACTIVE POWER BALANCE OF A SYNCHRONOUS MACHINE Consider the stator winding of a synchronous machine. Suppose that the machine operates as a motor, supplied with a system of three phase currents ia = I m cosωt ib = I m cos( ωt − 2π / 3) ic = I m cos( ωt − 4π / 3) (2) These currents are delivered by the grid and can be regarded as being impressed into the three windings, that are spatially displaced along the circumference of the stator by 120 degrees. Current flow in each of the three phases causes an appropriate m.m.f., that acts along the magnetic axis of the given phase. The situation is illustrated in Fig. 3, where individual phase m.m.f.’s are given with (N is the number of turns per phase): Fa = NI m cosωt Fb = NI m cos(ωt − 2π / 3) (3) Fc = NI m cos(ωt − 4π / 3) a c b Fa Fb Fc b c a Fig. 3 - Individual phase m.m.f.’s of a three-phase winding.  E Levi, 2000 4
  5. ENGNG 2024 Electrical Engineering One observes that in terms of spatial dependence, all the three individual phase magneto- motive forces are stationary and they act along the defined magnetic axis of the winding. From (3) one notices that each of the three m.m.f.’s is varying in time. The values of the three phase m.m.f.’s in the given instant of time correspond to those met in any three phase system. The resultant magneto-motive force that stems from the three phase system of currents flowing through spatially displaced windings is the sum of the individual contributions of the three phases. The summation is done in the cross-section of the machine, and it is necessary to observe the spatial displacement between the three m.m.f.’s. One may regard the cross-section of the machine as a Cartesian co-ordinate system in which phase a magnetic axis corresponds to x-axis, while y-axis is perpendicular to it. For the purposes of calculation this co-ordinate system may be treated as a complex plane, with x-axis corresponding to the real axis, and y- axis corresponding to the imaginary axis. In terms of the complex plane, spatial displacement of the m.m.f. by 120 degrees corresponds to the so-called ‘vector rotator’, a = exp( j 2π / 3) . Hence the resultant magneto-motive force can be written as 2π 4π ( ) j j Fres = Fa + aFb + a Fc 2 a=e 3 , a =e 2 3 (4) Fres = NI m ( cosωt + a cos( ωt − 2π / 3) + a cos( ωt − 4π / 3) ) 2 The expression for the resultant magneto-motive force is most easily found if one recalls the well-known correlation cosδ = 0.5( exp( jδ ) + exp(− jδ )) . Hence Fres = NI 1 2 (e jωt +e − jωt + ae j ( ωt − 2π / 3) + ae − j ( ωt − 2 π / 3) + a 2e j ( ωt − 4 π / 3) + a 2e − j ( ωt − 4π / 3) )= Fres = NI m 1 2 (e ω + e j t − jωt + aa * e jωt + aae − jωt + a 2 a 2*e jωt + a 2 a 2e − jωt ) 1+ a + a 2 = 0 a* = a2 a 2* = a a3 = 1 a4 = a (5) Fres = NI m 1 2 ( e ω (1 + a j t 2 a + aa 2 ) + e − jωt (1+ a 2 + a) ) = Fres = NI m 1 2 ( e ω ( 3) + e j t − jωt ( 0) ) 3 jωt Fres = NI me 2 Symbol * in (5) stands for complex conjugate. The result obtained in (5) is an important one. The equation 3 jωt Fres = NI m e (6) 2 describes a circular trajectory in the complex plane. This means that the resulting magneto- motive force, produced by three stationary, time-varying m.m.f.’s (often called pulsating m.m.f.’s) is a time-independent and rotating magneto-motive force. Thus it follows that the three-phase system creates a rotating field (called also Tesla’s field) in the machine. Figure 4 illustrates the rotating field in different instants of time. The speed at which the rotating field travels equals the angular frequency of the stator three-phase supply.  E Levi, 2000 5
  6. ENGNG 2024 Electrical Engineering Im ωt=90° ωt=135° Fres ωt=45° ωt=0° Re (a) 1.5NIm Fig. 4 - Resultant field in the three-phase machine for sinusoidal supply conditions. Since the rotor winding carries excitation current, a field is produced by this current. This field is stationary with respect to rotor. Since the rotor rotates at synchronous speed, then, looking in from stationary stator, this rotor field rotates at synchronous speed. This is always the case in any multi-phase AC machine: regardless of whether the rotor rotates synchronously or asynchronously, all the fields in the machine rotate at synchronous speed. Since the resulting m.m.f. is responsible for the resulting flux density and ultimately resulting flux, this means that apart from the rotating m.m.f., there is a rotating flux density wave and a rotating flux in the machine as well. The term rotating field in general denotes any of the three. In other to show how a synchronous machine can produce reactive power, consider the operation of a synchronous motor. The motor will always consume real power; however, it can either generate or absorb reactive power. Let us further consider a couple of characteristic situations. Let us assume at first that a synchronous motor operates under ideal no-load conditions, so that all the losses in the motor can be neglected. This means that the input real power and the output mechanical power are both zero. Such an ideal situation is useful in explaining the reactive power production and consumption. Suppose at first that the rotor current is zero. Therefore the rotor does not produce any rotating field. Rotating field of the stator is generated due to current flow in stator and the stator rotating field equals the total rotating field of the machine (total field is a vectorial sum of the field generated by the stator currents and the rotor field; it is fixed since the supply voltage is fixed, and does not depend on individual values of the rotor and stator fields). Since rotor current is zero, there is not an induced emf in the stator winding due to rotor flux. Under these conditions the machine behaves like a pure reactance and consumes reactive power. The situation is illustrated in Fig. 5a, where total field and stator rotating field are shown for one instant in time. If the reactance of a stator phase winding is Xs, then the current in one phase is I = V/Xs, where V is the rms of the phase to neutral voltage. Hence the reactive power consumed by the machine is Q = 3 V I = 3V2/Xs. Suppose next that rotor current is now increased to a certain value, say, such that the emf in the stator winding equals one half of the applied phase to neutral voltage. The situation is illustrated in Fig. 5b. The stator field is now just one half of the value for zero rotor current, since one half of the total field is provided by rotor. The current in stator is I = (V - E) /Xs, and  E Levi, 2000 6
  7. ENGNG 2024 Electrical Engineering since E is 0.5V, then I = 0.5V/Xs. Consequently, reactive power consumed by the machine becomes Q = 3 V I = 1.5V2/Xs. The consumed reactive power is just one half of what it was originally. Let the rotor current be now exactly such that the induced emf in stator due to rotor flux equals applied voltage. This means that the total field exactly equals rotor field and, since there is not any current in stator, stator field is zero (Fig. 5c). As the stator current is zero, the machine now neither consumes nor delivers the reactive power. Finally, let the rotor current be further increased, so that the induced emf becomes 1.5V. The corresponding rotor filed is now larger than the total field and consequently stator field has to change the direction (Fig. 5d). This means that the stator current now flows from the machine towards the grid, i.e. I = (V - E) /Xs = -0.5V/Xs. Hence the reactive power is now equal to Q = - 1.5V2/Xs, where negative sign indicates that the reactive power is delivered to the grid rather than being absorbed. In all the cases discussed so far stator current was purely reactive and the reactive power was either taken from or delivered to the grid. If a synchronous machine absorbs reactive power it is said that it operates in under-excited mode. If a synchronous machine delivers reactive power to the grid, it is said that it operates in over-excited mode. These two terms are exclusively related to the reactive power balance of the machine. One notes that for all these cases the angle between the total rotating field and the rotor rotating field in the machine is zero. 3. MOTORING MODE OF OPERATION Suppose now that the synchronous motor drives a load torque. Hence it delivers mechanical power at the shaft and consequently consumes real (active) power at stator winding terminals. Let the rotor current be such that the machine is in over-excited mode (Fig. 5d). What now happens is that an angular displacement occurs between the rotor field and total field. This angle is called load angle and it is illustrated in Fig. 6, where motoring operation in over-excited mode is depicted. This load angle will appear in the phasor diagram later on. As will be shown, power delivered by a synchronous motor directly depends on the value of this angle. On the basis of considerations in the previous section, it is easy to derive an equivalent per-phase representation of the synchronous motor. The applied stator phase to neutral voltage (note that the value given is always line to line voltage and that stator winding of a synchronous motor is always connected in star) can be represented with a corresponding phasor. This voltage phasor corresponds to the total field in Fig. 6. Rotor current causes rotor flux, that rotates at synchronous speed and therefore induces an emf in the stationary stator winding. This emf corresponds to the rotor field in the Fig. 6. Hence, in phasor diagram, the angle between the grid voltage and the induced emf will be equal to the load angle of Fig. 6. Thus equivalent circuit contains an internal voltage source (induced emf) and the external applied voltage. In between is the stator reactance of the machine, called synchronous reactance. Stator winding resistance can usually be neglected and it is omitted from all the further considerations. Furthermore, iron loss (that takes place in stator core) and mechanical loss will be neglected as well, so that under these idealised conditions input active power equals output mechanical power. Equivalent circuit of a synchronous motor with excitation winding on rotor and uniform air-gap is shown in Fig. 7.  E Levi, 2000 7
  8. ENGNG 2024 Electrical Engineering Ftotal Fs (= Fres) Ftotal Fr ω ω Phase a axis Phase a axis a. c. Fs Ftotal Fs (= Fres) Ftotal Fr Fr ω Phase a axis ω b. Phase a axis d. Fig. 5 - Illustration of rotating fields in a synchronous motor (for ideal no-load conditions) for four different values of the rotor current. Fs Ftotal Fr δ ω Phase a axis Fig. 6 - Rotating fields in motoring operation in over-excited mode. jXs I + V E Fig. 7 - Equivalent circuit of a synchronous motor.  E Levi, 2000 8
  9. ENGNG 2024 Electrical Engineering The following phasor equation follows directly from Fig. 7: = E + jX s I (7) The phasor diagram, that this equation describes, is drawn in Fig. 8 for two cases: operation in the under-excited mode when the power factor is lagging (since the machine consumes reactive power) and operation in the overexcited mode when the power factor is leading (since the machine consumes active power but simultaneously delivers reactive power to the grid). jXsI V jXsI V E δ E φ δ φ I I lagging power factor leading power factor Fig. 8 - Phasor diagram for motoring operation in under-excited and overexcited modes. Phasor diagrams of Fig. 8 are used most frequently to determine the unknown load angle and the induced emf on the basis of the known motor loading and stator terminal conditions. They simultaneously represent the starting point in deriving the so-called load angle characteristic of a synchronous motor. Consider the phasor diagram of Fig. 8 for overexcited mode, that is for convenience re-drawn in Fig. 9. Due to the correlation that exists, one immediately recognises that the angle BCA is equal to the power factor angle (angle between current and voltage phasors). On the other hand, angles ODA and ABC are right angles. When solving the phasor diagram of Fig. 9, it is much easier to use one of the two triangles (triangle ODC or triangle OBC) and project the phasors on the sides of these two triangles, than to directly solve the complex phasor equation (7). In order to obtain the load angle characteristic of a synchronous motor one considers the triangle OBC in Fig. 9. By projecting the phasors on sides OB and CB one has C B φ A jXs I D V E φ δ I O Fig. 9 - Phasor diagram for derivation of load angle characteristic.  E Levi, 2000 9
  10. ENGNG 2024 Electrical Engineering + X s I sin φ = E cos δ (8) X s I cos φ = E sin δ When terminal voltage, stator current and the power factor are known, these two equations enable simple calculation of the load angle and the induced emf. However, for the purpose of deriving the load angle characteristic one expresses the active stator current component from the second equation of (8) as E sin δ I cos φ = (9) Xs Next, it is necessary to recall that the phasor diagram represents one phase of the machine and that the machine is three-phase. Hence the input active power and the reactive power (that is either absorbed or delivered, as in Fig. 9) are given in terms of phase to neutral voltage and phase current as P = 3VI cos φ Q = 3VI sin φ (10) Load angle characteristic of a synchronous machine relates the active power with the load angle, δ. Note that, since the stator resistance loss and the rotational losses are neglected, input active power equals active power transferred from stator to rotor and it ultimately equals the output mechanical power. Substitution of (9) into (10) yields E sin δ VE P = 3VI cos φ = 3V =3 sin δ (11) Xs Xs The correlation between real (or output) power and load angle is called load angle characteristic of a synchronous motor. One observes from (11) that, for the given stator voltage and rotor current (that is, emf) power depends only on the sine of the load angle. This means that, higher the load is, higher the value of the load angle will be, and vice versa. In other words, rotor field in Fig. 6 will lag the total field more and more as the loading of the machine increases. The maximum power that a synchronous motor can deliver is met when load angle equals 90 degrees. Hence, for each rotor current setting (that is, for each value of the emf) there is a maximum power that the motor can deliver. The maximum power is VE Pmax = 3 (12) Xs If the power required by the load increases above the maximum power, load angle would go over 90 degrees and the so-called loss of synchronism would take place. Since increase of the load angle above 90 degrees leads to decrease of the motor power (compared to maximum power) rather than an increase, power demanded by the load cannot be met and the motor becomes unstable (it looses synchronism, that is, the speed starts decreasing). The stable operation of the motor is therefore possible only for load angle values between zero and 90 degrees. Torque of the motor can be directly obtained from the load angle characteristic by dividing the power with the mechanical synchronous angular speed of rotation: P E sin δ 1 1 VE Te = = 3V = 3P sin δ ωs Xs 2 πf P 2 πf X s (13) 1 VE Te max = 3 P 2 πf X s  E Levi, 2000 10
  11. ENGNG 2024 Electrical Engineering Hence the motor torque dependence on load angle is the same as for the power. The scaling factor is the constant, synchronous mechanical angular speed of rotation. Load angle characteristics of a synchronous motor are illustrated in Fig. 10, for a couple of values of the rotor current (i.e. emf). An increase in the rotor current causes an increase in the induced emf. Consequently, for the given load, the load angle decreases. An increase of the rotor current leads to an increase in the maximum power (maximum torque) that the motor can develop. Pmax1, Temax1 P, Te E1 Pmax2, Temax2 E2 Pmax3, Temax3 E3 STABLE UNSTABLE E1 > E2 > E3 0 90° 180° δ Fig. 10 - Load angle characteristic of a synchronous motor. Reactive power characteristic of the motor can be derived in exactly the same way as the active power characteristic. Consider the phasor diagram for under-excited operation, that is for convenience re-drawn in Fig. 11. The following two equations now correspond to (8): − X s I sin φ = E cosδ (14) X s I cos φ = E sin δ They are obtained by using projections of phasor in triangle ABC in Fig. 11. From (14) one expresses the reactive component of the stator current and then substitutes this expression into the equation for reactive power, (10). Thus Q = 3VI sin φ (10a) and further V − E cos δ I sin φ = Xs 2 V − E cos δ V − VE cos δ Q = 3VI sin φ = 3V =3 (15) Xs Xs 2 V VE cos δ Q=3 −3 Xs Xs The first term in this equation is the reactive power that the machine needs for its own magnetisation. The second term is the reactive power produced by the machine. It follows from (15) that reactive power is positive (i.e. absorbed) as long as V > Ecosδ. The reactive power is zero when V = Ecosδ and it becomes negative (i.e. the motor delivers reactive power to the grid) when V < Ecosδ. Reactive power characteristic is illustrated in Fig. 12.  E Levi, 2000 11
  12. ENGNG 2024 Electrical Engineering A jXsI V φ B C δ E φ I Fig. 11 - Phasor diagram of a synchronous motor for under-excited operation. It is easy to observe in Fig. 12 that the reactive power may be both positive and negative for the given value of the induced electro-motive force (i.e. rotor current). Whether the reactive power is delivered or absorbed will depend on the loading of the motor, since the load at the shaft determines how much the load angle is. The internal reactive power production of a synchronous motor is at maximum value when load angle is zero. This situation corresponds to the no-load operation, that is not normally met when the machine is used as a motor. However, this fact is made use of in so-called synchronous condensers that are exclusively used for reactive power production and delivery to the grid. A synchronous condenser is in essence a synchronous motor that operates at all times under no load conditions. The load angle, although not exactly zero since there are losses in the machine (that were neglected throughout here) is very small, so that the machine produces maximum possible amount of reactive power. Synchronous condensers are essentially reactive power generators and they are installed at various points in a power system to provide the necessary reactive power support. Generating mode of operation is discussed in the subsequent section. In principle, all the characteristics remain valid, but one has to take into account that the active power will now be produced and delivered to the grid, rather than absorbed. This fact does cause some inevitable differences,, especially in terms of the phasor diagrams. Q 3 V(V − E cosδ) /Xs 3V2 / Xs 0 90° 180° δ − 3VE cosδ /Xs Fig. 12 - Reactive power characteristic of a synchronous motor.  E Levi, 2000 12
  13. ENGNG 2024 Electrical Engineering EXAMPLES Example 1: Calculate the number of poles on the rotor of the synchronous motor whose frequency is 60 Hz and rated speed of rotation is 200 rpm. Solution: Since the frequency is 60 Hz and the synchronous speed is 200 rpm, then from (1) one calculates the number of pole pairs as P = 60 x f / ns, that is 60 x 60 / 200 = 18. The machine thus has 36 poles, 18 north and 18 south poles. Example 2: Calculate the synchronous speeds of two series of synchronous motors designed for 50 Hz and 60 Hz operation, respectively, if the number of pole pairs is from 1 to 10. Solution: Synchronous mechanical speed is given with (1), ns = 60 f / P . Hence f = 50 Hz P 1 2 3 4 5 6 7 8 9 10 ns 3000 1500 1000 750 600 500 428.5 375 333 300 f = 60 Hz P 1 2 3 4 5 6 7 8 9 10 ns 3600 1800 1200 900 720 600 514 450 400 360 Example 3: An induction motor drives a load and takes 350 kW from the mains at 0.707 power factor lagging. An over-excited synchronous motor is connected in parallel with the induction motor and it takes 150 kW from the grid. If the overall power factor of the two motors combined becomes 0.9 lagging, calculate the reactive power of the synchronous motor and its kVA rating. Solution: An induction motor always consumes reactive power. The synchronous motor in this example however produces reactive power. Active powers that the motors take are (1 stands for induction motor, 2 stands for the synchronous motor): P1 = 350 kW P2 = 150 kW so that the total power taken from the grid is P = P1 + P2 = 350 + 150 = 500 kW. Since the combined power factor of the two motors is 0.9 lagging, total reactive power that the two motors take from the grid is equal to Q = Q1 + Q2 = P tanφ = P tan (arccos0.9) = 500 tan (arccos 0.9) = 242.16 kVAr Since the power factor of the induction motor is known, one can determine further reactive power consumed by the induction motor as Q1 = P1 tan φ1 = 350 tan (arccos 0.707) = 350 kVAr Hence the synchronous motor delivers the reactive power of Q2 = Q - Q1 = 242.16 - 350 = - 107.95 kVAr  E Levi, 2000 13
  14. ENGNG 2024 Electrical Engineering where the negative sign indicates that the synchronous motor delivers reactive power. Power factor of the synchronous motor is cos φ2 = P2 / √(P22 + Q22) = 0.81 leading and the kVA rating is S2 = √(P22 + Q22) = 184.8 kVA Situation is illustrated in accompanying Figure. IM SM 350 kW 350 kVAr 150 kW 108 kVAr 500 kW 242 kVAr Example 4: A synchronous motor operates on mains of 6.3 kV phase to neutral voltage. Synchronous reactance is 14 Ohms and the stator resistance can be neglected. The machine operates under ideal no-load conditions. For induced electromotive forces equal to 6000 V, 6300 V and 7850 V construct the phasor diagram, and calculate stator current and reactive power. Solution: In this example the machine operates as a synchronous condenser with zero intake of active power. Hence it either delivers or consumes reactive power. With respect to rotating fields, this operation corresponds to the one analysed in conjunction with Fig. 5. As noted there, load angle is zero for such an operation. Hence the induced emf and the applied voltage are in all the three cases under consideration in phase. Since the voltage is 6300 V phase to neutral, the induced emf of 6000 V means that the machine consumes reactive power (E cosδ = E, E < V). Stator current is I = (V - E) / Xs = (6300 - 6000) / 14 = 21.4 A and the consumed reactive power is Q = 3 V I = 3 x 6300 x 21.4 = 404.46 kVAr In the second case the induced emf and the applied voltage are the same, 6300 V. There is consequently no current flow in stator and the reactive power is zero. The machine neither consumes nor delivers reactive power. In the third case induced emf is 7850 V. It is greater than applied voltage and the machine now delivers reactive power to the grid. Stator current is I = (V - E) / Xs = (6300 - 7850) / 14 = - 110.7 A and the delivered (negative sign) reactive power is Q = 3 V I = - 3 x 6300 x 110.7 = - 2.092 MVAr Corresponding phasor diagrams are given in Figure.  E Levi, 2000 14
  15. ENGNG 2024 Electrical Engineering V = 6300 V E= jXs I 7.85 kV V jXs I V E = 6.3 kV V E = 6 kV I I Example 5: Construct the phasor diagram of a synchronous motor when it operates in under- excited and over-excited mode and give the equivalent circuit. Solution: jXs jXsI I + V jXsI V E V E δ E φ δ φ I I lagging power factor leading power factor Equivalent circuit Phasor diagram for motoring operation in under- excited and overexcited modes Example 6: A synchronous motor is connected to a 3980 V, 3-phase line and rotor current is such that the induced emf is 1790 V (line to neutral). The synchronous reactance is 22 Ohms and the load angle between the voltage and the emf is 30 degrees. Determine the stator current and the power factor angle. Solution: Since the given voltage is line to line, the corresponding phase to neutral voltage is 3980 / √3 = 2300 V. As phase to neutral voltage is greater than induced emf, the motor operates in under-excited mode. The solution is then most easily found using the phasor diagram given in Figure and equations A jXsI V φ B C δ E φ I  E Levi, 2000 15
  16. ENGNG 2024 Electrical Engineering − X s I sin φ = E cosδ X s I cos φ = E sin δ 2300 − 22 ⋅ I ⋅ sin φ = 1790 cos 30 = 1550.185 22 ⋅ I ⋅ cos φ = 1790 sin 30 = 895 Hence I = 40.682 / cos φ 22 ⋅ ( 40.682 / cos φ ) sin φ = 749.815 tan φ = 0.83777 φ = 40 o cos φ = 0.766 lagging I = 40.682 / cos φ = 53.1 A Example 7: A 150 kW, 1200 rpm, 460 V, 3-phase synchronous motor has a synchronous reactance of 0.8 Ohms and the induced emf is 300 V phase to neutral. Determine the power - load angle characteristic, torque - load angle characteristic and maximum power and torque. Solution: Phase to neutral voltage is 460/√3 = 266 V. Synchronous speed of the machine is 1200 rpm. Hence, from the table in Example 2, it follows that the machine is a 60 Hz one and that the number of pole pairs is 3. Thus the power vs. load angle and the torque vs. load angle characteristics are VE 266 ⋅ 300 P=3 sin δ = 3 sin δ = 299.25 sin δ [kW] Xs 0.8 P VE 3 Te = 3 sin δ = 3 299,250 sin δ = 2381.4 sin δ [Nm] 2πf Xs 2π 60 Corresponding maximum power and maximum torque are obtained, by setting load angle to 90 degrees, as 299.25 kW and 2381.4 Nm. Example 8: A 3 MW, 6.6 kV, 60 Hz, 200 rpm synchronous motor operates at full load at a leading power factor of 0.8. If the synchronous reactance is 11 Ohms, determine: the apparent power that the motor develops on stator terminals, the stator current, the induced emf, and the load angle. Solution: As the motor operates at full load it delivers 3 MW on the shaft. All the losses are neglected, so that the input active power is 3 MW as well. The apparent power is then S = 3 VI = P / cosφ = 3 / 0.8 = 3.75 MVA As the line to line voltage is 6.6 kV, the corresponding phase voltage is 3815 V. The stator current is then found from the apparent power, S = 3 VI = 3.75 MVA I = S /(3V) = 328 A  E Levi, 2000 16
  17. ENGNG 2024 Electrical Engineering The motor operates as over-excited (leading power factor). Hence the phasor diagram is the one shown in Figure. C B φ A jXs I D V E φ δ I O It is described with: + X s I sin φ = E cos δ X s I cos φ = E sin δ so that the unknowns are the induced emf and the load angle. Thus ( −1 ) 3815 + 11⋅ 328 ⋅ sin cos 0.8 = E cosδ 11 ⋅ 328 ⋅ 0.8 = E sin δ E sin δ = 2886.4 E cosδ = 5979.8 tan δ = 2886.4 / 5979.8 = 0.4827 δ = 25.77 o E = 5979.8 / cos 25.77 = 6640 [V] 4. GENERATING MODE OF OPERATION In generating, active power is produced from input mechanical power and delivered to the grid. On the other hand, reactive power may again be delivered to the grid (over-excited mode) or it may be consumed (under-excited mode). Positive current direction in the equivalent circuit of Fig. 7 for motoring was from the grid to the machine. Hence both the real power and the reactive power are positive when consumed. This positive direction for the generator stator current is inconvenient, since produced real power would come with a negative sign. It is therefore customary to change the positive stator current direction for generation, so that the current is positive when it flows from the machine to the grid. This means that the active and reactive power are now positive when delivered to the grid (reactive power becomes negative when consumed, i.e. for under-excited mode). The corresponding equivalent circuit is shown in Fig. 13 and it differs from the one of Fig. 7 only in the positive direction for stator current flow. Hence the phasor equations that describes a synchronous generator now takes the form  E Levi, 2000 17
  18. ENGNG 2024 Electrical Engineering E = V + jX s I (16) This change of the positive direction of the current has some important consequences on the corresponding phasor diagrams and definitions related to the power factor of the machine. The issue will be clarified shortly, upon considerations of the rotating fields in the machine during generation. jXs I + V E Fig. 13 - Equivalent circuit of a synchronous generator. In motoring, as shown in Fig. 6, rotor field lags the total field by the load angle. Hence induced emf, that corresponds to the rotor field, lags in the phasor diagram stator voltage (that corresponds to the total field) by the load angle. That is, the emf phasor appears to the right with respect to the stator voltage phasor. When the machine generates, the load angle again appears between the total field and the rotor filed. However, as illustrated in Fig. 14, rotor field now leads the total field by the load angle. This means that in the corresponding phasor diagrams induced emf phasor will lead the stator voltage phasor by the load angle. In other words, for generation, the emf phasor will appear to the left of the stator voltage phasor. Fs Fr Ftotal δ ω Phase a axis Fig. 14 - Rotating fields in generating operation in over-excited mode. The other important difference with regard to the phasor diagram is related to the meaning of the lagging and leading power factor. For motoring one has the standard definition related to the positive current flow into the machine: power factor is lagging when the machine consumes reactive power (under-excited operation), while it is leading when the machine delivers reactive power (over-excited mode). For generation however, due to the change of the positive stator current direction, the situation reverses: over-excited mode that means delivery of the reactive power to the grid is now described with the lagging power factor, while under-  E Levi, 2000 18
  19. ENGNG 2024 Electrical Engineering excited mode that means consumption of the reactive power corresponds to the leading power factor. The phasor diagrams for the generator operation in under-excited and overexcited mode are shown in Fig. 15. Summarising, if it is said for a generator that it operates with lagging power factor, this means that it is in the over-excited mode and delivers the reactive power to the grid. If a generator operates with leading power factor, it is under-excited and it consumes reactive power from the grid. jXsI E jXsI V V E δ δ φ I φ I leading power factor lagging power factor Fig. 15 - Phasor diagram for generating operation in under-excited and overexcited modes. As far as the load angle characteristic of a synchronous generator is concerned, nothing changes. Active power delivered to the grid for generation has been made positive by changing the direction of the stator current. Hence the equations for power vs. load angle, torque vs. load angle and maximum power and torque remain to be given with (11),(12) and (13). That is, VE P = 3VI cos φ = 3 sin δ (11) Xs VE Pmax = 3 (12) Xs 1 VE Te = 3 P sin δ 2 πf X s (13) 1 VE Te max = 3 P 2 πf X s Load angle characteristic is therefore the same as in Fig. 10 and is repeated for convenience in Fig. 16. Stable operating region is again for load angles between zero and 90 degrees. Remember however that the load angle represents now the angle by which the induced emf phasor leads the stator voltage phasor, since the rotor field leads the total field by the same angle (while the angle is lagging for motoring). As far as the reactive power characteristic is shown, reactive power is now positive when delivered to the grid. Hence the terms related to V and Ecosδ in (15) interchange the position and the equation is 2 VE cosδ V Q=3 −3 (17) Xs Xs The condition that describes transition from under-excited mode to over-excited mode  E Levi, 2000 19
  20. ENGNG 2024 Electrical Engineering however remains the same. This means that the reactive power is zero when V = Ecosδ and it becomes positive (rather than negative, as for motor) when the generator delivers reactive power to the grid when V < Ecosδ. Reactive power characteristic is illustrated in Fig. 17. Pmax1, Temax1 P, Te E1 Pmax2, Temax2 E2 Pmax3, Temax3 E3 STABLE UNSTABLE E1 > E2 > E3 0 90° 180° δ Fig. 16 - Load angle characteristic of a synchronous generator. Q 3 V(E cosδ − V)/Xs 3VE cosδ /Xs 0 90° 180° δ - 3V2 / Xs Fig. 17 - Reactive power characteristic of a synchronous generator. EXAMPLES Example 9: A synchronous generator operates on mains of 6.3 kV phase to neutral voltage. Synchronous reactance is 14 Ohms and the stator resistance can be neglected. The machine operates under ideal no-load conditions. For induced electromotive forces equal to 6000 V, 6300 V and 7850 V construct the phasor diagram, and calculate stator current and reactive power.  E Levi, 2000 20
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