EE 262 Synchronous Machines Lecture Notes PDF

Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES EE 262 – SYNCHRONOUS MACHINES COURSE CONTENT: Basic Theory. Construction, Phasor Diagram and Equivalent Circuit of Non-Salient Pole Machine. Machine Characteristics. Two-Axis Theory....

Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES EE 262 – SYNCHRONOUS MACHINES COURSE CONTENT: Basic Theory. Construction, Phasor Diagram and Equivalent Circuit of Non-Salient Pole Machine. Machine Characteristics. Two-Axis Theory. Phasor Diagram of Salient- Pole Machine. Assessment of Reactance. Determination of Voltage Regulation. Parallel Operation: Synchronizing, Effects of Changing Excitation and Mechanical Torque, Load Sharing of Two Machines, Hunting, Performance Equations, Circle Diagrams V-Curve of Synchronous Motor. Starting of Synchronous Motor and Its Industrial Control Circuit. Synchronous Induction Motor. Single-Phase Synchronous Generator. Prepared by E. K. Anto Page 1 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES EE 262 – SYNCHRONOUS MACHINES 1. THREE-PHASE SYNCHRONOUS MACHINES 1.1 Introduction Synchronous machines are a type of rotating machines where electromechanical energy conversion takes place when the change in flux is associated with mechanical motion. Generally in rotating machines, voltages are induced in windings or group of coils either by  rotating the windings mechanically through a magnetic field or  mechanically rotating a magnetic field past the windings or  designing the magnetic circuit so that the reluctance varies with rotation of the rotor. By any of these methods, the flux linking a specific coil is changed cyclically, and an induced voltage e   N  is generated in accordance with Lenz’s Law of d dt electromagnetic induction. The group of such coils so interconnected that their generated or induced voltages all make a positive contribution to the desired result is called an armature winding. The coils are wound on iron cores so as to concentrate the flux and thus ensure that the flux path through them is as effective as possible. Because the armature iron is subjected to a varying magnetic field, eddy currents will be induced in it. To minimize the eddy-current loss, the armature iron core is built up of thin laminations. The magnetic circuit is completed through the iron of the other machine member, and exciting coils, or field windings, are placed on that member to act as the primary sources of flux. Permanent magnets may be used in small machines. 1.2 Definition of Synchronous Machines A synchronous machine is an AC machine whose mechanical speed (or rotor speed) under steady-state conditions is proportional to the frequency of the voltage and current in its armature. The frequency in cycles per second (hertz or Hz) is the same as the speed of the rotor in revolutions per second, i.e., the electrical frequency is synchronized with the mechanical speed, and this is the reason for the designation synchronous machine. At synchronous speed, the rotating magnetic field created by the armature current travels at the same speed as the field created by the field current, and a steady torque results from the interaction of the two magnetic fields. For a P -pole machine, the synchronous speed N s (in rev/min.) is related to the frequency f of the armature induced voltages and currents and the number of poles 120 f P by the equation N s . P Prepared by E. K. Anto Page 2 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES The number of poles is chosen according to the desired speed. The speed is in turn determined by the driven equipment or prime mover attached to the shaft of the synchronous machine. Thus a 2-pole machine must revolve at 3000 rpm to produce a 50-Hz voltage. But a great many synchronous machines have more than 2 poles. It is worth noting that essentially, electric motors and generators are the less expensive the faster they run. With a few exceptions, synchronous machines are 3-phase machines because of the advantages of 3-phase systems for generation, transmission and heavy-power utilization. For the production of a set of three voltages phase-displaced by 120 electrical degrees in time, it is obvious that a minimum of 3 coils displaced 120 electrical degrees in space must be used. 1.3 Constructional Features of Synchronous Machines Unlike induction machines in which there are alternating currents in both stator and rotor windings, synchronous machines have alternating currents (AC) in the stator windings and DC currents in the rotor windings. The 3-phase synchronous machine has the following essential constructional parts:  stator comprising - stator frame or yoke - armature coils  rotor comprising - rotor core - rotor or field windings  slip-rings or collector rings  brushes and bearings Of these parts, the stator frame or yoke, the pole-cores, the rotor core and air gap between the poles and rotor core, form the magnetic circuit, whereas the rest form the electrical circuit. 1.4 The Stator: The stator unit is the stationary part of the machine and consists of the stator frame or yoke and the armature coils. The stator core contains the set of slots that carry 3- phase winding, and is laminated to minimize loss due to hysteresis and eddy- currents. The laminations are insulated from each other and have spaces between them for allowing cooling air to pass through. The windings have hollow passages through which cooling water is circulated. The stator frame or yoke serves the purposes of  providing mechanical support for the poles  protection for the machine and  carries the armature flux produced by the poles. In small machines, where weight is of little importance and cheapness is the main consideration, the yoke is made of cast iron. But for large machines, cast steel or rolled steel is usually employed to fabricate the stator frame. Prepared by E. K. Anto Page 3 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES 1.5 The Rotor: The rotor, which is the rotating part of the machine, is located on a shaft running on bearings, and is free to rotate between magnetic poles. The rotor core is cylindrical or drum-shaped, and is built of steel laminations with slots to house the field windings. Besides housing the field windings in slots and causing them to rotate to cut the magnetic flux of the magnetic fields, the rotor core also provides a magnetic path of low reluctance to the flux from the poles. The rotor or field windings are insulated from each other and placed in slots which are lined with tough insulating material. The rotor carries the dc windings or field windings. These field windings are excited by direct current conducted to them by means of carbon brushes bearing on slip rings or collector rings. 1.6 Damper Windings on Rotor In addition to the DC winding, the rotor carries the so-called damper windings (also called squirrel-cage winding). In the salient-pole (engine-driven) machines, the damper winding is embedded in the pole-shoes or pole faces and connected (short- circuited) at their end with brass or heavy copper rings. The damper windings are not usually required in cylindrical-rotor machines driven by reciprocating steam engines, water wheels or steam turbines. The damper winding serves the following functions: produces forces which dampen the oscillation of the rotor, thereby reducing hunting (momentary speed fluctuations) helps to start synchronous motors maintains balanced 3-phase voltage under unbalanced load conditions. improves parallel operation of salient-pole generators driven by internal- combustion engines. 1.7 The Slip-Rings (or Collector Rings): The slip-ring is made of copper segments, and has the same functions in the motor as in a generator. Its purpose is to facilitate the collection of current from the DC excitation source to the field windings. 1.8 The Brushes and Bearings: The purpose of brushes is to carry current from the external circuit to the commutator. They are usually made of blocks of carbon or graphite, and are rectangular in shape. The brushes should slide freely in their holder so as to follow any irregularity in the commutator. Because of their reliability, ball-bearings are frequently employed, though for heavy duties, roller-bearings are preferable. The ball and rollers are lubricated by hard oil for quieter operation and for reducing the wear of the bearings. Prepared by E. K. Anto Page 4 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES 1.9 Types of Synchronous Machines Synchronous machines may be classified broadly under:  Synchronous Generators (or Alternators)  Synchronous Motors (synchronous motors operating with load attached)  Synchronous Condensers (synchronous motors operating on no-load) 2. SYNCHRONOUS GENERATORS They are the main primary source of electrical energy we consume. They convert mechanical energy into electrical energy, in powers ranging up to 1500MW. In general, the greater the generated power, the higher the voltage rating. However the rated generated voltage seldom exceeds 25 kV because the increased slot insulation takes up valuable space at the expense of copper conductors. The stator winding (i.e., the armature winding which collects the generated voltage) is always star-connected and the neutral is connected to ground. The reasons for this arrangement are: 1. The highest effective voltage between a stator conductor and the grounded stator core is only 1 / 3 , that is, 57.7% of the line voltage. We can therefore reduce the amount of insulation in the slots, which in turn enables us to increase the cross-section of the conductor. A larger conductor permits us to increase the current and hence the power output of the machine. 2. When a generator is under load, the voltage per phase becomes distorted and the waveform is no longer sinusoidal. This distortion is mainly due to undesired third harmonic voltages. With star connection the harmonics do not appear between the lines. With delta connection, the harmonic voltages add up and produce large circulating current in the delta connected winding which causes additional I 2 R losses. 3. The neutral is available for protective gear. 2.1 Types of Synchronous Generators Synchronous generators are classified according to the type of rotors they use, which in turn depends on the speed of the prime mover. The two types of rotors used in generators are:  Salient-Pole Rotor Type  Cylindrical-Rotor (or Non-salient Pole) Type The constructional reasons for some synchronous generators having salient-pole rotor structures and other having cylindrical rotors can be appreciated with the aid of the 120 f synchronous speed equation N s . P 2.1.1 Salient-Pole Generators – (Low-speed Generators) Prepared by E. K. Anto Page 5 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES The power system in most countries, including Ghana, operates at a constant system frequency f of 50 Hz. And so for low-and medium-speed (engine-driven) rotors, a relatively large number of poles are required to produce the desired frequency. A salient- or projecting pole construction is thus characteristic of hydroelectric generators, because hydraulic and engine-driven turbines operate at relatively low speeds, with running speed between 50 and 300 rpm for hydraulic turbines and about 200 rpm for internal-combustion engines. The salient-pole construction with concentrated windings is thus best adapted mechanically to multi-polar slow- and medium-speed hydroelectric generators, hydraulic and engine-driven turbines and some synchronous motors. Such generators are characterized by their large diameters and short axial lengths. The poles and pole- shoes (which cover 2/3 of pole pitch) are laminated to reduce heating due to eddy currents. The field coils for small machines are wound with round wire, while rectangular copper strips wound on edge are used for large machines. 2.1.2 Cylindrical-Rotor Generators – (High-speed Generators) When the steam engine or gas turbine is operated at a high speed, it has a high efficiency, and for this reason, it is often used to drive generators at high speeds. But for high speeds, it is difficult to build a rotating field with projecting poles strong enough to withstand the centrifugal force. Projecting poles also cause excessive wind losses and make the generator noisy. To overcome these undesirable features, generators intended for high-speed steam- turbine or gas-turbine drive have their field structure made non-salient or cylindrical in form and small in diameter, with distributed field windings placed in slots and arranged so as to produce an approximately sinusoidal 2- or 4-pole field. Such high-speed turbine-driven generators are also called turbo generators, and are commonly 2- or 4-pole cylindrical- rotor machines running at speeds of 3000 rpm or 1500 rpm. In turbo generators, considerable heat is liberated in small spaces, and this heat must be carried away by air currents forced through passages in the heated parts. Hence forced ventilation is required in turbo generators. The turbo generator must be enclosed to control the direction of the air currents, as well as to reduce noise. 2.2 Principle of Operation of Synchronous Generator Rotating machines (be they AC or DC) operate on the same fundamental principles of electromagnetic induction. It may be recalled that they consist of an armature winding (i.e., a group of coils so interconnected that their generated or induced voltages all make a positive contribution to the desired result) and a field winding or exciting coils (to act as the primary sources of flux). The field winding (or rotor winding) is excited or energized by the so-called exciter system supplying direct current from a DC source that needs not exceed 250 volts. In most cases, the necessary exciting (or magnetising) current is obtained from a small DC shunt generator mounted on the shaft of the synchronous machine itself. Because Prepared by E. K. Anto Page 6 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES the field magnets are rotating, the direct current is supplied through two slip rings. As the exciting DC voltage is relatively small, the slip rings and brush gear are of light construction. Recently, brushless excitation systems have been developed in which a 3-phase AC exciter and a group of rectifiers supply DC to the machine. When the rotor is made to rotate by the turbine of the prime mover, the stator or armature conductors (being stationary) are cut by the rotated DC magnetic flux from the field coils on the rotor. Hence they have induced emf produced in them. Because the magnetic poles are alternately N and S , they induce an emf and hence current in the armature conductors, which first flow in one direction and then in the other. Hence an alternating emf is induced in the stator windings whose frequency depends on the number of N and S poles moving past a conductor in one second and whose direction is given by Fleming’s Right-Hand Rule. 2.3 Advantages of Stationary Armature and Revolving Field System In DC generators, a commutator assembly (consisting of split-rings, brushes, etc) is used to change the alternating emf induced in the coils to a unidirectional emf for the external circuit. Since the armature of a synchronous machine rather utilizes slip rings to supply alternating current to the external load and therefore has no need of a commutator, the armature needs not be the rotating member. In fact, it is more desirable to have a stationary armature with the field poles rotating inside it, as this structure has several advantages. The advantages of a having stationary armature and rotating field system include the following: 1. For polyphase power, a rotating armature would require three or more slip rings to deliver power to an external load. These slip rings, being exposed, are difficult to insulate, particularly for the high AC voltages (30 kV or more) synchronous generators are required to supply. 2. Because of the difficulty of insulation of the slip rings on a rotating armature, arc-overs and short circuits are apt to occur. 3. No slip rings are required in a stationary armature. The output current can be led directly from fixed terminals on the stator (or the armature windings) to the load circuit, without having to pass it through brush-contacts. With the doing away with slip rings, power and frictional losses due to contact resistance with slips are thus avoided. 4. It is much easier to properly insulate a stationary armature (where winding space is less concentrated) for high voltage than it is to insulate a high-voltage rotating winding. The heavy insulation must have adequate mechanical strength to withstand the mechanical forces due to centrifugal force. Prepared by E. K. Anto Page 7 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES 3. PARAMETERS AND OPERATING CHARACTERISTICS OF SYNCHRONOUS GENERATORS 3.1 Rotating Magnetic Fields Polyphase AC Machines (e.g. Synchronous Generators To understand the theory of polyphase a-c machines, it is necessary to study the nature of the magnetic field produced by a polyphase winding. In particular, we shall consider the mmf patterns of a 3-phase winding such as those found on the stator of 3-phase induction and synchronous machines. In a 3-phase machine, the windings of the individual phases are displaced from each other by 120 electrical degrees in space around the airgap circumference. When the 3-phase windings are excited by a 3-phase power source, alternating currents flow in the windings. Under balanced 3-phase conditions, the instantaneous currents are given as: ia  I max cos t ib  I max cos(t  120 0 ) (1) ic  I max cos(t  240 0 ) where I max is the maximum value of the current and the time origin is arbitrarily taken as the instant when the phase-a current is a positive maximum. See Fig below ia ib ic t Fig: Instantaneous 3-Phase Currents The corresponding component mmf waves will also vary sinusoidally with time. Each component is a stationary, pulsating sinusoidal distribution of mmf around the airgap with its peak located along the magnetic axis of its phase and its amplitude proportional to the instantaneous phase current. In other words, a standing space wave is varying sinusoidally with time. Each component can be represented by an oscillating space vector drawn along the magnetic axis of its phase, with length proportional to the instantaneous phase current. The resultant mmf is the sum of the components from all three phases. Prepared by E. K. Anto Page 8 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES 3.2 Analytical Analysis of Rotating Mmf Wave To study the resultant field analytically, let the origin for angle  around the airgap periphery be placed at the axis of phase a. At any time t, all 3 phases contribute to the airgap mmf at any point . See Fig below. Axis of phase b a c X b X Axis of phase a b c X a Axis of phase c Fig: Simplified 2-Pole 3-Phase Winding The mmf contributions from the 3 phases are: Fa  Fapeak cos Fb  Fbpeak cos(  120 0 ) (2) Fc  Fcpeak cos(  240 0 ) The 120 electrical degrees displacements appear because the machine is so wound that the axes of the 3-phases are 120 electrical degrees apart in space. The resultant mmf at any point  is then F ( )  Fapeak cos  Fbpeak cos(  120 0 )  Fcpeak cos(  240 0 ) (3) But the mmf amplitudes vary with time in accordance with the current variations. Thus, with the time origin arbitrarily taken at the instant when the phase-a current is a positive maximum, the mmf amplitudes can be written as function of time t as: Fapeak  Fa max cos t Fbpeak  Fb max cos(t  120 0 ) (4) Fcpeak  Fc max cos(t  240 0 ) The quantities Fa max , Fb max and Fc max are respectively the time-maximum values of the amplitudes Fapeak , Fbpeak and Fcpeak. The 1200 displacements appear here Prepared by E. K. Anto Page 9 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES because the three currents are 1200 phase-displaced in time. Under balanced conditions, the currents in the 3-phases are also balanced and therefore of equal amplitude. The three amplitudes Fa max , Fb max and Fc max are also then equal, and the symbol Fmax may be used for all three. Equation (3) therefore becomes  F ( , t )  Fmax cos cos t  Fmax cos(  120 0 ) cos(t  120 0 )  (5)  F max cos(  240 ) cos(t  240 ) 0 0 Each of the three components on the right-hand side of Eqn (5) is a pulsating standing wave. In each term, the trigonometric function of  defines the space distribution as a stationary sinusoid, and the trigonometric function of t indicates that the amplitudes pulsate with time. The first of the terms expresses the phase-a component; the second and third terms express, respectively, the phase-b and phase-c components. By use of the trigonometric transformation 1 1 cos  cos   cos(   )  cos(   ) (6) 2 2 each of the components in Eqn (5) can be expressed as cosine functions of sum and difference angles. Thus using the trigonometric transformation, Eqn (5) is transformed into 1 1  F ( , t )   Fmax cos(  t )  Fmax cos(  t ) 2 2  1 1    Fmax cos(  t )  Fmax cos(  t  240 0 ) (7) 2 2  1 1    Fmax cos(  t )  Fmax cos(  t  480 0 ) 2 2  Now the three cosine terms involving the angles   t  ,   t  240 0 , and     t  480  are equal sinusoid displaced in phase by 120. Note that a lag angle of 0 0 4800 is equivalent to a lag angle of (4800 – 3600) = 1200. Their sum is therefore zero, and Eqn (13) reduces further to: F ( , t )  1.5 Fmax cos(  t ) (8) The resultant mmf wave described by Eqn (8) is a sinusoidal function of the space angle  , having a constant amplitude of 1.5 Fmax and rotating at angular velocity  or a space-phase angle t which is a linear function of time. Prepared by E. K. Anto Page 10 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES In general, it may be shown that a rotating field of constant amplitude will be produced by a q  phase winding excited by balanced q  phase currents when the respective phases are wound 2 / q electrical radians apart in space. The constant amplitude will be q / 2 times the maximum contribution of one phase, and the speed will be   2f electrical radians per second. 3.3 Generated Voltage in Rotating AC Machines (e.g. Alternators) The study of voltages induced in any of the armature windings of rotating machines resolves into study of the voltage induced in a single coil followed by addition of the individual coil voltages in the manner dictated by the specific interconnection of the coils forming the complete winding. The general nature of induced voltage has already been discussed. The focus in this section is the determination of the voltage magnitude by Faraday’s Law. Consider a single N -turn full-pitch coil, that is a coil that spans 180 electrical degrees or complete pole pitch. For simplicity, consider a 2-pole cylindrical rotor machine. The field winding on the rotor is excited by a d-c source, and assumed to produce a  sinusoidal space wave of flux density B  at the stator surface. The rotor is A spinning at constant angular velocity . When the rotor poles are in line with the magnetic axis of the stator (armature) coil, the flux linkage with the stator coil is N , where  is the airgap flux per pole. For the assumed sinusoidal flux-density wave, B  B max cos (9) where Bmax is the peak value at the rotor pole center and  is measured in electrical radians from the rotor pole axis. The airgap flux per pole is the integral of the flux density over the pole area. As the rotor turns, the flux linkage varies as the cosine of the angle  between the magnetic axes of the stator coil and rotor. By Faraday’s Law, the voltage induced in the stator coil is d d max e  N max sin t  N cos t (10) dt dt The minus sign associated with Faraday’s Law in Eqn (10) implies that while the flux linking the coil is decreasing, an emf will be induced in it in a direction to try to produce a current which would tend to prevent the flux linking it from decreasing. The first term N max sin t is called the rotational or speed voltage due to the d max relative motion of field and coil. The second term N cos t is the transformer dt voltage, and is only present when the amplitude of the flux density wave changes with time. Prepared by E. K. Anto Page 11 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES In the normal steady state operation of most rotating machines, the amplitude of the airgap flux wave is constant, and the induced voltage is simply the rotational voltage. e  N max sin t (11) In normal steady-state operation of AC machines, we are usually interested in the rms values of voltages and currents rather than their instantaneous values. And hence from Eqn (11), the maximum value of the induced instantaneous voltage is: E max  N max  2fN max (12) Its rms value is given as E max E rms   4.44 fN max (13) 2 These equations are identical in form to the corresponding emf equations for a transformer. Therefore relative motion of a coil (rotor field coil is this case) and a constant-amplitude spatial flux density wave (like the DC flux from the field windings on the rotor) produces the same voltage effect as does a time-varying flux in association with stationary coils in a transformer. Rotation, in effect, introduces the time element and transforms a space distribution of flux density into a time variation of voltage. For a distributed winding, the induced emf in Eqn (13) must be modified by the winding factor k w. E rms  4.44 k w fN (14) where N is the total series turns per phase. 3.4 Armature Reaction Flux in Synchronous Generators When the DC field winding on the rotor is energized, voltage is induced in the armature windings on the stator of the alternator. Under no-load condition, no current will flow through the armature windings. In that situation, the flux in the airgap is uniformly distributed and due only to that of the field windings. However, under loading situation when a synchronous generator supplies electrical power to a load, armature current flows. The armature current creates a component flux wave in the airgap which rotates at synchronous speed, as shown in Section 1.2.2. This armature flux reacts with the flux created by the field, and an electromagnetic torque results from the tendency of the two magnetic fields to align themselves. In a generator, this torque opposes rotation, and a counter mechanical torque must be applied from the prime mover in order to sustain rotation. The electromagnetic torque is the mechanism through which greater electrical power output calls for greater mechanical power input. 3.5 Distorting Effect of Armature Reaction Flux Prepared by E. K. Anto Page 12 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES The effect of the armature flux is to distort the main flux distribution in the airgap due to the field flux. The effect on the flux magnitude is important because both the generated voltage and torque per unit of armature current are influenced thereby. The armature mmf Fa combines with (or superimposes on) the main flux in the airgap due to the field mmf F f to give a resultant mmf Fr that is distorted. This net mmf distribution due to the combined magnetizing action of the field and armature fluxes magnetizes the machine to a resultant gap flux per pole  r , which in turn generates a voltage E g  4.44 k w fN r in the stator or armature winding. In the absence of magnetic saturation, Fa and F f can be considered to produce separate gap fluxes  a and  f , which superpose to give  r as resultant. The two flux components can then be considered to induce separate emfs E a and E f in the stator windings having the phasor sum E g. The net flux distribution in the airgap of an alternator depends on the amount of stator current and on the phase relation existing between the current and voltage; that is, the power factor of the load. 3.6 Armature Reaction Reactance in Synchronous Generators The net effect of the armature current is to distort or twist the main flux from the field windings and hence the generated voltage in the stator (armature windings). This effect is known as armature reaction. The mmf wave created by the armature current is called armature-reaction mmf. For salient-pole machines, the effect of the armature mmf is to be seen as creating flux sweeping across the pole-faces. Thus its path in the pole shoes crosses the path of the main-field flux. For this reason, armature reaction of this type is called cross- magnetizing armature reaction. It evidently causes a decrease in the resultant airgap flux density under one half of the pole and an increase under the other half. And so when the field winding is excited and the armature is connected to supply load, the resultant airgap flux distribution is the superposition of the flux distribution from the field windings and that from the armature windings. Because of saturation of iron, the flux density is decreased by a greater amount under one pole tip than it is increased under the other. Accordingly, the resultant flux per pole is lower than would be produced by the field winding alone, a consequence known as the demagnetizing effect of cross- magnetizing armature reaction. Since the demagnetizing effect of cross- magnetizing armature reaction is caused by saturation, its magnitude is a non-linear function of both the field current and the armature current. 3.6.1 Counteracting Armature Reaction The effect of cross-magnetizing armature reaction may be limited in the design and construction of the machine. The mmf of the main field should exert predominating Prepared by E. K. Anto Page 13 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES control on the airgap flux, so that the condition of weak field mmf and strong armature mmf may be avoided through the following measures: 1. Increasing the reluctance of the cross-flux path – essentially the armature teeth, pole shoes and the airgap by increasing the degree of saturation in the teeth and pole faces, by avoiding too small an airgap and by using chamfered pole face which increases the airgap at the pole tips. 2. The best but also the most expensive curative measure is compensate the armature mmf by means of a winding embedded in the pole faces called compensating or pole-face winding. The compensating winding is connected in series with the armature and in arranged in such a way as to supply a magnetizing action that is equal and opposite to that of the armature coils at all loads, thereby neutralizing the cross-magnetizing effect of the armature ampere-turns. Furthermore, the addition of the compensating winding improves the speed of response, because it reduces the armature circuit time-constant. The main disadvantage of commutating or pole-face winding is their expense. They are used therefore in machines designed for heavy loads or rapidly changing loads. 3.6.2 Armature Reaction Reactance As explained, the magnetomotive force (mmf) produced by the armature currents in the armature winding when the alternator is loaded is called armature reaction. The load current, flowing through the armature windings, builds up local flux which on cutting the winding generates a counter (or reactance) emf. This effect gives the armature a reactance that is numerically equal to 2fL , where L is the leakage inductance of the armature winding (similar to the leakage inductance in transformer windings). This leakage reactance arising from the armature reaction is called armature reaction reactance, since the flux which causes it is around the armature turns only and does not affect the field directly. This armature reaction flux is proportional to the armature currents, since the magnetic paths it covers is not normally saturated. 3.7 Phasor and Equivalent Circuit Diagrams of The Cylindrical-Rotor (Non- Salient Pole) Synchronous Generators An elementary picture of how a synchronous generator works has already been given in the previous sections. In this section, analytical methods of examining the study- state performance of polyphase synchronous generator will be considered. Initial Prepared by E. K. Anto Page 14 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES consideration will be given to non-salient pole (cylindrical-rotor) synchronous generators, with the effects of salient poles considered thereafter. 3.7.1 Phasor Diagrams of Cylindrical-Rotor Synchronous Generators It will be recalled from Eqn (8) that balanced polyphase in a symmetrical polyphase winding create an mmf wave whose space-fundamental component rotates at synchronous speed. Recall also that the mmf wave is directly opposite say, arbitrarily chosen phase a, at the instant when the phase a current has its maximum. When a synchronous generator is loaded, current flows in both armature and field windings, and they create mmf waves in the respective windings. The mmf wave created by the armature current is commonly called the armature-reaction mmf. The resultant magnetic field  r in the machine is the sum of the two components produced by the field current  f and the armature reaction  ar. Because the fields are sinusoid, and sinusoids can conveniently be added by phasor methods, the airgap flux and mmf conditions in a synchronous machine can thus be represented by phasor diagrams. The Fig below shows the space-phasor diagram for two situations of the armature current in phase with and lagging the excitation voltage in a synchronous generator.    T f T f r r  ar lag Eg  ar Ia Fig: Phasor Diagram of Fluxes in Cylindrical-Rotor Synchronous Generator Armature current in phase with excitation (unity power factor) Armature current lagging the excitation (lagging power factor) 3.7.2 Equivalent Circuit of Cylindrical-Rotor Synchronous Generator A very useful and simple equivalent representing the steady-sate behaviour of a cylindrical-rotor synchronous machine under balanced, polyphase conditions is obtained if the effect of the armature reaction flux is represented by an inductive reactance. For the start, let us consider an unsaturated cylindrical-rotor machine. Prepared by E. K. Anto Page 15 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES Although neglect of magnetic saturation may appear to be a drastic simplification, it can be shown that the results obtained can be modified so as to take saturation into account. As stated already, the resultant airgap flux in the machine can be considered as the phasor sum of the component fluxes created by the field and armature-reaction mmfs. From the viewpoint of the armature windings, these fluxes manifest themselves as generated emfs. The resultant airgap voltage E r can then be considered as the phasor sum of the excitation or generated voltage E g generated by the field flux and the voltage E ar generated by the armature-reaction flux. The component emfs E g and E ar are proportional to the field and armature currents respectively, and each lags the flux which generates it by 900. The armature-reaction flux  ar is in phase with the armature current I a , and consequently the armature- reaction emf E ar lags the armature current by 900. See Fig below. f  ar Eg r E ar Er Ia Fig: Phasor Diagram of Component Fluxes and Corresponding Induced Voltages The effect of armature reaction is simply that of an inductive reactance X ar (the armature-reaction or magnetizing reactance) accounting for the component voltage generated by the space-fundamental flux created by armature reaction. Part of the excitation or generated voltage E g is used to overcome the voltage E ar generated by the armature-reaction flux and the rest appears as the resultant airgap voltage E r. Thus, E g  E r  E ar (15)  E r  jI a X ar The resultant airgap voltage E r differs from the terminal voltage V by the armature resistance Ra and leakage-reactance X a drops. See Fig (a) below. The armature leakage reactance accounts for the voltages induced by the component fluxes which are not included in the airgap voltage E r. Prepared by E. K. Anto Page 16 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES These fluxes include not only leakage across the armature slots and around the coil ends, but also those associated with the space-harmonic field effects created by the departure from a sinusoid necessarily present in actual armature mmf wave. Finally, the equivalent circuit for an unsaturated non-salient-pole (cylindrical-rotor) machine under balanced polyphase conditions reduces to the form shown in Fig (b) below, in which the machine is represented on a per-phase basis by its excitation voltage E g in series with a simple impedance. X ar Xa Xs Ra Ra Eg Er V  Eg V Fig: Equivalent Circuits of Cylindrical-Rotor (Non-Salient Pole) Synchronous Machines 3.7.3 Synchronous Reactance This simple impedance is called synchronous impedance. Its reactance X s is called the synchronous reactance. In terms of the magnetizing and leakage reactances, X s  X ar  X a (16)  Z s  Ra  jX s  Ra  j ( X a  X ar ) The synchronous reactance takes into account all the flux produced by balanced polyphase armature currents, while the excitation or generated voltage takes into account the flux produced by the field current. In an unsaturated cylindrical-rotor (non- salient pole) machine at constant frequency, the synchronous reactance is a constant. Furthermore, the excitation voltage is proportional to the field current and equals the voltage which would appear at the terminals if the armature were open- circuited, the speed and field current being held constant. The value of X s is 10 to 100 times greater than Ra. Therefore we can neglect the resistance, unless we are interested in efficiency or heating. 4. STEADY-STATE PERFORMANCE CHARACTERISTICS OF THE SYNCHRONOUS MACHINE Two basic sets of characteristic curves for a synchronous machine are involved in the inclusion of saturation effects and in the determination of the appropriate machine constants. The tests performed to obtain these characteristics are the open-circuit Prepared by E. K. Anto Page 17 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES and short-circuit tests. These tests yield the open-circuit and short-circuit characteristics, which are necessary for determining the performance of synchronous machines. Except for a few remarks on the degree of validity of certain assumptions, the discussions apply to both non-salient-pole (cylindrical-rotor) and salient-pole machines. 4.1 Open-Circuit Characteristics Like the magnetization curve for a d-c machine, the open-circuit characteristic of a synchronous machine is a curve of the armature terminal voltage on open circuit (or no-load) as a function of the field excitation, when the machine is running at synchronous speed. Essentially, the open-circuit characteristic represents the relation between the space-fundamental component of the airgap flux and the mmf on the magnetic circuit when the field winding constitutes the only mmf source. The open-circuit characteristic is determined experimentally by driving the machine mechanically at rated synchronous speed as an unloaded generator, that is, with armature terminals on open circuit, and measuring the open-circuit terminal voltages for a series of field current values. The OCC is shown in the Fig below. Open-circuit Voltage If Fig: Open-Circuit Characteristic (OCC) of a Synchronous Machine Since the generated voltage is proportional to the flux and therefore the flux density, the open-circuit characteristic (OCC) is, to some scale, just like the B  H magnetization or hysteresis curve. The OCC may be plotted in per-unit terms, where unity voltage is the rated voltage and unity field current is the excitation corresponding to rated voltage on the airgap line. The straight line tangent to the lower portion of the magnetization curve is called the airgap line. It indicates very closely the mmf required to overcome the reluctance of the airgap. If it were not for the effects of saturation, the airgap line and OCC would coincide, so that the departure of the curve from the airgap line is an indication of the degree of saturation present. In a normal machine, the ratio at rated voltage of the total mmf to that required by the airgap alone usually is between 1.10 and 1.25. Prepared by E. K. Anto Page 18 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES 4.2 Short-Circuit Characteristics The short-circuit characteristic (SCC) is obtained experimentally by short-circuiting the armature (i.e. stator) windings of a synchronous machine being driven as a generator at synchronous speed through low-resistance ammeters. The excitation or field current is gradually adjusted from its zero value until the armature current has reached a maximum safe value, about 1.5 to 2 times rated or full-load current. Corresponding values of short-circuit armature current and field current are measured, and the SCC is drawn as shown in Fig below. If Fig: Short-Circuit Characteristic (SCC) of a Synchronous Machine The SCC curve is linear because the armature reaction mmf Far is almost as large as the field excitation, so that the resultant excitation and flux are small and therefore the iron is not saturated. The phasor relation between the excitation (or generated) voltage E g and the steady- state armature current I a under polyphase short-circuit conditions is: E g  I a ( R a  jX s ) (17)  I sc Z s Because the resistance Ra is much smaller than the synchronous reactance X s , the armature current I a lags the excitation voltage E g by very nearly 900. Consequently, the armature-reaction mmf wave is very nearly in line with the axis of the field poles and in opposition to the field mmf. The resultant mmf, obtained from the phasor sum of the field and armature-reaction mmfs, creates the resultant airgap flux wave which generates the airgap voltage E r equal to the voltage consumed in armature resistance Ra and leakage reactance X a. As an equation, E r  I a ( R a  jX a ) (18) Prepared by E. K. Anto Page 19 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES In most synchronous machines, the armature resistance is negligible, and the leakage reactance is about 0.15 p.u., that is, at rated armature current, the leakage- reactance voltage drop I a X a is about 0.15 p.u. And so from Eqn (18), the airgap voltage E r at rated armature current on short-circuit is about 0.15 p.u., that is to say, the resultant airgap flux is only about 0.15 of its normal-voltage value. Consequently, the machine is operating in an unsaturated condition. The short-circuit armature current therefore is directly proportional to the field current over the range from zero to well above rated armature current. From Eqn (17), for the same field current, Eg Zs  (19) I sc for same field current If This impedance is called unsaturated synchronous impedance. At low excitations, the OCC is linear and the unsaturated synchronous impedance is constant. But at high excitations, the OCC is nonlinear and the unsaturated synchronous impedance is not constant. The unsaturated synchronous impedance can be found from the open- circuit and short-circuit data. See Fig below. ag line Short-circuit Armature Current Open-circuit Voltage c Rated voltage a OCC SCC Rated Armature Current o f f  o Field Excitation If Fig: Typical Open-Circuit and Short-Circuit Characteristics of Synchronous Machine 4.3 Determination of the armature resistance Ra The armature resistance Ra per phase can be measured directly by voltmeter and ammeter method or by using Wheatstone bridge. However, under working conditions, the effective value of Ra is increased due to skin effect. The value of Ra so obtained Prepared by E. K. Anto Page 20 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES by direct measurement is increased by 60 % or so to allow for this effect. Generally, an effective value of 1.6 times the DC value Rdc is taken. Example 1 The following test results were obtained from a 3-phase, 6000 kVA, 66 kV star- connected, 2-pole, 50 Hz turbo alternator. With a field current of 125 A, the open circuit voltage is 8000 V at rated speed. With the same excitation and rated speed, the short-circuit current was 800 A. If at the rated full-load, the resistance drop is 3%, determine the following: i. synchronous impedance Z s ii. armature resistance Ra iii. synchronous reactance X s Solution 1 i. The synchronous impedance is calculated using the equation E oc 8000 / 3 Zs  =  5.77  I sc for same field current I f 800 ii. The voltage per phase is E oc  8000 / 3  3810 V S 3 ph 6000 Full load current, I    525 A V L 3 6.6  3 Resistance drop, IRa  3 %  E oc  0.03  3810  114.3 V Hence Ra  114.3 / 525  0.218  iii. Synchronous reactance , X s  Z s 2  Ra 2  5.77 2  0.218 2  5.74  Example 2 A three-phase 50 Hz star-connected 2000 kVA, 2300 V alternator gives a short-circuit current of 600 A for a certain field excitation. With the same excitation, the open circuit voltage was 900 V. The resistance measured between a pair of terminal was 0.12 Ω. Determine the following: Prepared by E. K. Anto Page 21 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES i. synchronous impedance Z s ii. armature resistance Ra iii. synchronous reactance X s Solution 2 i. The synchronous impedance is calculated using the equation E oc 900 / 3 Zs  =  0.866  I sc for same field current I f 600 ii. Resistance between the pair of terminals is 0.12 Ω. It is the resistance between two phases connected in series. Thus measured resistance per phase 0.12/2 = 0.06 Ω. Hence effective armature resistance, Ra  1.6  Rmeasured  1.6  0.06  0.096  iii. Synchronous reactance , X s  Z s 2  Ra 2  0.866 2  0.096 2  0.86  4.4 Output Power Delivered By Cylindrical-Rotor Synchronous Generator Consider a simple circuit of a cylindrical-rotor synchronous generator of excitation voltage E g supplying power to a bus of voltage Vbus through an impedance Z such that current I flows. The phasor diagram is also shown below. Eg jIX Z  R  jX R I X  Eg Vbus  Vbus IR I Fig: (a) Circuit Diagram (b) Phasor Diagram The power P2 delivered to the load or bus end Vbus is P2  Vbus I cos  (20) Prepared by E. K. Anto Page 22 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES where  is the phase angle of the current I with respect to Vbus. The phasor current is E g  Vbus I (21) Z If the phasor voltages and impedance are expressed in polar forms, E g   Vbus 0 Eg Vbus I  (   z )  ( z ) (22) Z z Z Z where  is the phase angle (called power angle) by which E g leads Vbus , Z is the magnitude of the impedance. The real part of the phasor Eqn (22) is the component of I in phase with Vbus. Hence  Eg V  I cos   Re al part  (   z )  bus ( z )  Z Z  (23) Eg V  cos(   z )  bus cos( z ) Z Z We note that cos( z )  cos( z )  R / Z. Substituting Eqn (23) into (20), we obtain, P2  Vbus I cos   Eg V   Vbus   cos(   z )  bus cos( z )  Z Z  E g Vbus V 2 R (24)  cos(   z )  bus2 Z Z E g Vbus Vbus 2 R  sin(   z )  Z Z2 R where  z  tan 1  90 0   z (25) X and is usually a small angle. Similarly, the power P1 at the source end E g of the impedance can be expressed as E g Vbus Vbus 2 R P1  sin(   z )  (26) Z Z2 Prepared by E. K. Anto Page 23 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES If, as is frequently the case, the resistance is assumed negligible, then E g Vbus P1  P2  Preal  sin  (27) X 4.5 Maximum Power Output From Cylindrical Rotor Alternator There is a maximum output that an alternator is capable of delivering for given values of terminal voltage, frequency and excitation. For a cylindrical rotor, if the resistance is negligible (in which case the IRa drop is neglected), and the synchronous reactance X s and voltages V and E g are constant (of course E g is fixed by excitation), then the maximum power per phase occurs at   90 0. Then from Eqns (26) and (27), the maximum power per phase is given as: E gV Pmax  if Ra is neglected Xs  V E g  V cos   if Ra is considered (28) Zs Ra where cos   Zs Example 3 A 3-phase 11 kV 5 MVA star-connected alternator has a synchronous impedance of ( 1  j12)  per phase. Its excitation is such that the generated line emf is 12 kV. If the alternator is connected to an infinite busbar, determine the maximum output at the given excitation. Solution 3 12000 The generated phase voltage is E g   6928 V 3 11000 The terminal voltage per phase is V   6351V 3 Since the armature resistance is NOT negligible, we need to calculate the internal R 1 angle cos   a   0.083 Zs 1  12 2 2 Hence the maximum power output per phase is Prepared by E. K. Anto Page 24 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES Pmax  V E g  V cos   Zs 6351  6928  6351  0.083 12  12 2  3375.88 kW  Total Pmax  3  3375.88  10127.64 kW 4.6 Phasor and Equivalent Diagrams of Salient-Pole Rotor Synchronous Generator (Two-Axis or Two-Reactance Theory to Account for Saliency) The effect of salient poles is taken into account in the so-called two-axis or two- reactance theory, by resolving the armature current or the sinusoidal armature flux into two perpendicular components, one in time quadrature (say  aq ) and the other in time phase (say  ad ) with excitation voltage E g taken as the reference or d-axis. Each of the fluxes  ad and  aq may then be thought of as inducing its own voltage E d and E q respectively. 4.6.1 Phasor Diagram of Salient-Pole Synchronous Generator This diagram is drawn for an unsaturated salient-pole generator operating at a lagging power factor. f r Iq  aq Eg  ad  ar Id Ia Fig: Phasor Diagram of Salient-Pole Synchronous Generator (Lagging pf) The d-component I d of the armature current, in time quadrature with the excitation voltage, produces its component fundamental armature-reaction flux  ard along the axes of the field poles. On the other hand, the q-component I q of the armature current, in phase with the excitation voltage, produces its component fundamental armature-reaction flux  arq in space quadrature with the field poles. Prepared by E. K. Anto Page 25 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES A direct-axis quantity is one whose magnetic effect is centered on the axes of the field. Direct-axis mmfs act on the main magnetic circuit. A quadrature-axis quantity is one whose magnetic effect is centered on the interpolar space. For an unsaturated machine, the armature-reaction flux  ar is thus the phasor sum of the components  ard and  arq. The resultant flux  r is then the phasor sum of the armature-reaction flux  ar and the main field flux  f. Thus  r   ar   f , where  ar   ard   arq   ard   arq   f 4.6.2 Equivalent Circuit of Salient-Pole Synchronous Machines With each of the armature component currents I d and I q , there is associated a component synchronous reactance voltage drop jI d X d and jIqX q respectively, where the reactances X d and X q are, respectively, called the direct- and quadrature axis synchronous rectances. Principally, the synchronous reactance accounts for the inductive effects of all the fundamental-frequency-generating fluxes created by the armature currents, including both armature leakage and armature-reaction fluxes. Thus the inductive effects of the direct- and quadrature axis armature-reaction flux waves  ard and  arq can be accounted for by the equivalent d-axis and q-axis magnetizing reactances X ard and X arq respectively, similar to the magnetizing armature-reaction reactance X ar of the non-salient-pole (cylindrical-rotor) theory. Expressed mathematically, the direct- and quadrature-axis synchronous reactances X d and X q are given as X d  X a  X ard (29) X q  X a  X arq where X ard = d-axis armature-reaction reactance (due to I d or  ard ) X arq = q-axis armature-reaction reactance (due to I q or  arq ) X a = armature leakage reactance and is assumed to be the same for both d-axis and q-axis currents. Prepared by E. K. Anto Page 26 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES X d = d-axis synchronous reactance X q = q-axis synchronous reactance Note: The synchronous reactance expression of Eqn (29), deduced from the salient- pole two-reactance theory, compares with that of Eqn (16), deduced from the cylindrical-rotor theory. The generated voltage E g equals the terminal voltage V on no-load. If the armature resistance is introduced, then the excitation voltage must equal the terminal voltage V plus the armature resistance drop I a Ra and the component synchronous-reactance drops due to the d- and q-axis armature currents. Thus E g  V  I a Ra  jI d X d  jI q X q (30) See Fig below for the phasor representation of Eqn (30) Iq   jI q X q V Ia R Id Ia a jI d X d Fig: Phasor Diagram of Salient-Pole Synchronous Generator The q-axis synchronous reactance X q is less than the d-axis synchronous reluctance X d because of the greater reluctance of the airgap in the quadrature axis. Usually, X q is between 0.6 X d and 0.7 X d. Note that a small salient-pole effect is present in turbo-alternators, even though they are cylindrical-rotor machines, because of the effect of the rotor slots on the quadrature-axis reluctance. 4.6.3 Relations Among Component Voltages In Phasor Diagram For Salient- Pole Synchronous Generators Operating at Lagging Power factor In using or drawing the phasor diagram of the salient-pole synchronous generator in the Fig above, the armature current must be resolved into its d-axis and q-axis components. This resolution assumes that the phase angle (    ) of the armature Prepared by E. K. Anto Page 27 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES current with respect to the excitation voltage is known. Often, however, the phase angle (power-factor angle)  at the machine terminals is explicitly known, whilst the power angle  (i.e. angle between the excitation voltage E g and terminal voltage V ) is not known and must be calculated. The phasor diagram above is repeated by the solid-line phasors in the Fig below.  Eg a  Iq o E  a Eg c   jI q X q V Id Ia Ia R b a b  a o b jI d X d Fig: Relations Among Component Voltages in Phasor Diagram of Salient-Pole Synchronous Generator Operating at Lagging Power Factor The study of the diagram shows that the dashed phasor o a  , perpendicular to the armature current I a , equals jI a X q. This result follows geometrically from the fact that triangles o a b  and oab are similar, because their corresponding sides are perpendicular. Thus using the properties of similar triangles, oa  ba   oa ba (31) ba  jI q X q  oa   oa  I a  jI a X q ba Iq Similarly, other relations are obtained. oa   jI a X q ob  jI d X q a c  jI d ( X d  X q ) ba   bc  jI q X q ob  jI d X d oa   jI a X d The phasor sum V  I a Ra  jI a X q then locates the angular position of the excitation voltage E g and therefore the d- and q-axes. Physically, this must be so, because all the field excitation in a normal machine is in the direct axis. One use of these relations in determining the excitation requirements for specified operating conditions at the terminals of a salient-pole machine is illustrated in the following example. Prepared by E. K. Anto Page 28 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES Example 4: The reactances X d and X q of a salient-pole synchronous generator are 1.00 and 0.60 p.u., respectively. The armature resistance is negligible. Compute the excitation voltage when the generator delivers rated kVA at 0.80 pf lagging current, and rated terminal voltage. Solution 4: The power factor angle  is known but the power angle  is unknown. And so first of all, the phase (    ) of the excitation voltage E g must be found, so that the armature current I a can be resolved into its d-axis and q-axis components.  E g a Eg E c a Iq oa   jI a X q a c  jI d ( X d  X q ) o V  o a   jI a X d Ia Id Fig: Salient-Pole Generator Phasor Diagram for Example 1 The armature current is given as: I a  0.80  j 0.60  1.00  36.9 0 p.u. (32) With the terminal voltage as the reference phasor, V  10 0  1  j 0 , and the phasor sum E  in the phasor diagram is: E   V  jI a X q  (1  j 0)  j (0.80  j 0.60)(0.60)  1.36  j 0.48  1.4419.4 0 p.u. The power angle  = 19.40 and so the phase angle between the excitation voltage E g and the armature current I a is (    ) = (36.90 + 19.40) = 56.30. The armature current can now be resolved into its d- and q-axis components. Their magnitudes are Prepared by E. K. Anto Page 29 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES I d  I a sin(   )  1.00 sin 56.30  0.832 (33) I q  I a cos(   )  1.00 cos 56.30  0.555 As phasors, I d  I d (90 0  19.4 0 )  0.832  70.6 0 (34) I q  I q (19.4 0 )  0.55519.4 0 We can now find the excitation voltage E g by simply adding numerically the length a c  I d ( X d  X q ) to the magnitude of E . Thus the magnitude of the excitation voltage is the algebraic sum Eg  E  Id (X d  X q )  1.44  (0.832)(1.00  0.60) (35)  1.77 p.u. As a phasor, E g  1.7719.4 0 p.u. 4.6.4 Output Power Delivered by Salient-Pole Synchronous Generators The discussion will be limited to a simple system shown in the schematic diagram below comprising a salient-pole machine SM connected to an infinite bus (e.g. a large power system network) of voltage Vnet through a series impedance (e.g. a transmission line) of reactance X ser per phase. Resistance is neglected. V Vnet Iq Eg X ser Eg jI q X q Xd Id Ia V jI d X d  Xq r se jI q X ser a X jI jI d X ser Fig: Salient-Pole Synchronous Machine and Series Impedance Prepared by E. K. Anto Page 30 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES (a) Single-Line Diagram (b) Phasor Diagram The dashed phasors show the external reactance drop resolved into components due to I d and I q. The effect of the external impedance is merely to add its reactance to the reactances of the machine; i.e., the total values of reactance interposed between the excitation voltage E g and the infinite bus voltage Vnet are X dT  X d  X ser (36) X qT  X q  X ser (37) If the infinite bus voltage Vnet is resolved into components Vnet sin  and Vnet cos  in phase with I d and I q respectively, then the active power P delivered to the bus per phase is P  I d Vnet sin   I qVnet cos  (38) Also, from the Fig (b) above, E g  Vnet cos   I d X ser  I d X d  I d ( X ser  X d )  I d X dT E g  Vnet cos  Hence I d  (39) X dT Similarly, Vnet sin   I q X q  I q X ser  I q ( X ser  X d )  I q X qT Vnet sin  Hence I q  (40) X qT Substituting Eqns (39) and (40) into Eqn (38) gives E g Vnet X dT  X qT P sin   Vnet 2 sin 2 (41) X dT 2 X dT X qT NB: Eqn (41) applies to all possible combinations of a synchronous machine and an external system, so long as the resistance is negligible. This power-angle characteristic is shown in the Fig below. Prepared by E. K. Anto Page 31 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES P Resultant P E g Vnet sin  X dT  90   180 0 90 180 V 2 net ( X dT  X qT ) sin 2 2 X dT X qT Fig: Power-Angle Characteristic of a Salient-Pole Synchronous Machine Showing Fundamental Component Due to Field Excitation and Second Harmonic Component Due to Reluctance Torque The first term of Eqn (41) is the same as the expression obtained for a cylindrical-rotor machine. This term is merely an extension of the concept to include the effects of series reactance. The second term introduces the effects of salient poles. It represents the fact that the airgap flux wave creates torque tending to align the field poles in the position of minimum reluctance. This term is the power corresponding to the reluctance torque. Note that the reluctance is independent of field excitation. Note also that if X d  X q as in a uniform-airgap machine, there is no preferential direction of magnetization, the reluctance torque is zero, and Eqn (41) reduces to the power-angle equation for a cylindrical rotor machine whose synchronous reactance is X d. Because of the reluctance torque, a salient-pole machine is stiffer than one with a cylindrical rotor, i.e., for equal voltages and equal values of X d , a salient-pole machine develops a given torque at a smaller value of  , and the maximum torque which can be developed is somewhat greater. For a normally excited machine, the effect of salient poles usually amounts to a few percent. Only at small excitations does the reluctance torque become important. Except at small excitations or when exceptionally accurate results are required, a salient-pole machine usually can be treated by simple cylindrical-rotor theory. Example 5: A 1500 kVA, star-connected, 2300 V, three-phase salient pole synchronous generator has reactances X d =1.95 ohms and X q =1.40 ohms per phase. All losses may be neglected. Find the excitation voltage for operation at rated kVA and power factor of 0.85 lagging. Solution 5: The terminal phase voltage is given as Prepared by E. K. Anto Page 32 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES 2300 V   1328 V 3 The rated armature current per phase is given as (1500  10 3 ) / 3 S ph Ia    377 A V 1328 Using the terminal voltage as reference V  13280 0  (1328  j 0) V The per phase armature current is then I a  377( cos 1 0.85 0 )  377  31.8 0  (320  j199) A The phasor sum E  is given as E   V  jI a X q  (1328  j 0)  j (320  j199)(1.40)  1607  j 448  166815.6 0 V Hence the power angle  = 15.60, and so the phase angle  between the excitation voltage E g and the per phase armature current I a is  = (    ), when  is lagging, = (15.60 + 31.80) = 47.40, The armature current can now be resolved into its d- and q-axis components. Their magnitudes are I d  I a sin(   )  377 sin 47.4 0  278 A I q  I a cos(   )  377 cos 47.4 0  255 A As phasors, the d- and q-axis components of the armature currents are Prepared by E. K. Anto Page 33 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES I d  I d (90 0  15.6 0 )  278  74.4 0 I q  I q (15.6 0 )  25515.6 0 We can now find the excitation voltage E g by simply adding numerically I d ( X d  X q ) to the magnitude of E . Thus the magnitude of the excitation voltage is the algebraic sum Eg  E  I d ( X d  X q )  1668  (278)(1.95  1.40)  1821V As a phasor, E g  182115.6 0 V (with reference to the terminal voltage). Example 6: A 36 MVA, 21 kV, 1800 rev/min alternator has a synchronous reactance of 5 ohm per phase. If the excitation voltage is 12 kV (line-to-neutral) and the output voltage is 17.3 kV (line-to-line), calculate the power delivered by the machine when the torque angle is 30o. Solution 6: Power delivered per phase is VE g 12 (17.3 / 3 ) P sin   sin 30 o  12 MW Xs 5 Therefore the total power delivered by all the three phases = 3 x 12 = 36 MW 4.7 Voltage Regulation and Load Excitation It is a known fact that with a change in load, there is a change in terminal voltage of an alternator. The magnitude of this change depends not only on the load but also on the load power factor. The voltage regulation of a synchronous generator is defined as the ratio of the change (actually a rise) in voltage when full load at rated voltage and a given power factor is removed from the machine, to the rated terminal voltage, the field excitation and speed remaining constant. Prepared by E. K. Anto Page 34 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES Thus if V is the rated voltage, and the terminal voltage becomes E0 when full-load is thrown off, then the per-unit voltage regulation is given as E0  V VR  (42) V Notes on Eqn (42). i. ( E0  V ) is the arithmetic difference and not the vector difference. ii. For a leading load power factor, the terminal voltage will fall on removing the full-load. Hence regulation is negative in that case. iii. The rise in voltage when full-load is thrown off is not the same as the fall in voltage when full-load is applied. 4.7.1 Determination of Voltage Regulation The voltage regulation of a generator depends on the armature resistance and synchronous reactance and power factor. Small Machines: In the case of small machines, the voltage regulation is found by direct loading. The procedure is as follows: The alternator is driven at synchronous speed and the terminal voltage is adjusted to its rated value V. The load is varied until the wattmeter and ammeter (connected for the purpose) indicate the rated values at desired power factor. Then the entire load is thrown off while the speed and field excitation are kept constant. The open-circuit or no-load voltage E0 is read. The voltage regulation can E V thus be determined from the formula VR  0. V Large Machines: For large machines, the cost of finding the voltage regulation by direct loading is prohibitive. Hence other indirect methods are used as described below. It must be pointed out that all these methods differ chiefly in the way the no-load voltage E0 is found in each case. The indirect methods employed for determining the voltage regulation and load excitation of large machines include the following: Prepared by E. K. Anto Page 35 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES i. Synchronous Impedance or EMF Method (due to Behn Eschenberg) ii. Ampere-Turn or MMF Method (due to Rothert) iii. Zero Power Factor or Potier Method (as name implies, due to Potier) All these methods require the determination of a. armature (or stator) resistance Ra b. open-circuit/no-load characteristic (OCC) and c. short-circuit characteristic SCC. The value of the armature resistance per phase can be measured directly by the voltmeter and ammeter method or by using the Wheatstone Bridge method. However under working conditions, the effective value of Ra is increased due to the so-called skin effect, and so the value of Ra so obtained is increased by 60% to account for the skin effect. Generally, a value 1.6 times the d-c value is taken. 4.7.2 Synchronous Impedance or EMF Method: The general data required in this method are the OCC, SCC and the synchronous impedance which is determined in the following procedure. I. The O/C test is performed by running the machine on no-load and noting the values of induced voltage and filed excitation. The OCC is then plotted from data obtained. It is just like the B-H curve. II. Similarly, the SCC is drawn from the data given by the S/C test. It is a straight line passing through the origin. E0 IX s O V I A IRa B E1 D O.C. Voltage S.C. Current E0 IX s IX s V C O  V sin  I V cos  A IRa B I1 O I E0 If  IX s V IRa Fig: Synchronous Impedance or EMF Method of Determining Voltage Regulation Both these curves are drawn on a common field-current base as shown below in Fig (a) and its vector diagram in Fig (b). Prepared by E. K. Anto Page 36 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES III. Consider a field current I f. The O/C voltage corresponding to this field current is, say, E1 , and I1 is the corresponding short-circuit current obtained on the SCC. When the winding is shorted-circuited, the terminal voltage is zero. Hence it may be assumed that the whole of this voltage E1 is used to circulate the armature short-circuit current I1 against the synchronous impedance Z s. Thus E1  I 1 Z s  E1 (open  circuit ) Zs  (43) I 1 ( short  circuit ) IV. Since the armature resistance Ra can be found as described earlier, the synchronous reactance is obtained as X s  ( Z s2  R 2a ) V. Knowing Ra and X s , the vector diagram of Fig (b) can be drawn for any load and any power factor. From it, the no-load generated voltage is determined as E0  OB 2  (OA  AB)  BD 2  2  ( BC  CD) 2  or E0  V cos   IRa 2  V sin   IX s 2 (44)  sign : lagging pf - sign : leading pf   0 : unity pf E V Hence the voltage regulation is obtained as VR  0. The internal voltage V E g behind synchronous impedance is taken as the no-load voltage. The method is referred to as the saturated synchronous impedance method. Practical experience has shown and confirmed that reasonable good results are obtained by this method. From the vector diagram above, it is clear that for the same field excitation, the terminal voltage is decreased from its no-load value E0 to V (for a lagging power factor). This is because of the following: 1. drop in voltage, IRa , due to armature resistance 2. drop in voltage, IX a , due to armature leakage reactance 3. drop in voltage, IX ar , due to armature reaction The drop in voltage due to armature reaction has been accounted for by assuming the presence of the fictitious reactance X ar in the armature winding. The value of the armature reaction reactance X ar is such that IX ar represents the voltage drop due to armature reaction. Hence the vector difference between the no-load voltage E0 and Prepared by E. K. Anto Page 37 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES terminal voltage V is equal to the drop IZ s across the synchronous impedance Z s. Note that IZ s represents the total drop in the alternator under load. Example 7 A 3300 V , 100 kVA 3-phase star-connected synchronous generator has an effective armature resistance of 0.5  / ph. A field current of 5 A was necessary to produce a rated current on short-circuit and a voltage of 1000 V (line) on open-circuit. Determine the full-load voltage regulation for the following types of loads: a) unity power factor b) 0.80 leading power factor c) 0.71 lagging power factor Solution 7 For a star-connected winding, the rated phase current equals the line current. S 100  10 3 I  IL    17.5 A 3  VL 3  3300 The synchronous impedance is given as E oc 1000 / 3 Zs    33  I sc sameI f 17.5 The synchronous reactance is then calculated as X s  Z s 2  Ra 2  (33) 2  (0.5) 2  32.99  E V The voltage regulation is VR  , where V is the terminal voltage and E is the V generated voltage, and is given as E  V cos   IRa 2  V sin   IX s 2. The terminal voltage per phase is V  V L / 3  3300 / 3  1905V a) Unity power factor cos   1.0 , sin   0.0 The generated voltage is Prepared by E. K. Anto Page 38 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES E V  IRa 2  IX s 2  (1905)  (17.5  0.5)2  (17.5  32.99)2  1999 V Hence the percentage voltage regulation is  E V   1999  1905  VR     100     100  4.93 %  V   1905  b) 0.8 leading power factor cos   0.8 , sin   0.6 The generated voltage is E V cos   IRa 2  V sin   IX s 2  (1905  0.8)  (17.5  0.5)2  (1905  0.6)  (17.5  32.99)2  1634 V Hence the percentage voltage regulation is  E V   1634  1905  VR     100     100   14.24 %  V   1905  b) 0.71 lagging power factor cos   0.71 , sin   0.71 The generated voltage is E V cos   IRa 2  V sin   IX s 2  (1905  0.71)  (17.5  0.5)2  (1905  0.71)  (17.5  32.99)2  2362 V Hence the percentage voltage regulation is  E V   2362  1905  VR     100     100  23.97 %  V   1905  In the Table below, find a summary of the values for the generated voltage, terminal voltage and voltage regulation for the different loads and power factors. Prepared by E. K. Anto Page 39 Dept. of Electrical & Electronic Engineering, KNUST Lecture Notes for Course EE 262 – SYNCHRONOUS MACHINES Resistive load Capacitive load Inductive load (unity pf) (leading pf)