CHAPTER 16 Synchronous Generators 16.0 Introduction nree-phase synchronous generators. are the primary source of all the electrical energy we ‘consume. These machines are the largest energy con verters in the world. They c into electrical energy, in powers ranging up to 1500 MW. In this chapter we will study the construction and characteristics ofthese large, modern generators ‘They are based upon the elementary principles cov cred in Section 8.6, and the reader may wish (0 re= view this material before proceeding further: nvert mechanical energy 16.1 Commercial synchronous generators Commercial synchronous generators are built with either a stationary or a rotating de magnetic field. A stationaay-field synchtonous generator has the same outward appearance as a de generator. The salient poles create the de field, which is cut by a re- volving armature, The armature possesses a 3-phuse winding whose terminals are connected t three slip-rings mounted on the shall, A set of brushes, sliding on the slip-rings, enables the arma lure t@ be connected to an external 3-phase loa The armature is driven by a gasoline engine, or ‘some other source of motive power, As it rotates, & Z-phase voltage is induced, whose value depends ‘upon the speed of rotation and upon the de exciting ceurrent inthe stationary poles. The frequency of the voltage depends upon the speed and the number of Poles on the field. s Id generators are used when the power output is less than 5 kVA However, for greater outputs. it is cheaper. safer and more practical to employ a revolving de field. A revolving-field synchronous generator bas 3 Mationary armature called a stator. The 3-phase sta tor winding is directly connected to the load, with- cout going through large, unreliable slip-rings and brushes. A stationary stator also makes it easier 10 insulate the windings because they are not sub- jected to centrifugal forces. Fig. 16.1 isa schematic diagram of such led an erator, sometimes alternator. The field is excited by a de wenerator, usually mounted on the same shaft. Note that the brushes on the commutator have to be connected 10 another set of brushes riding on slip-rings to feed the de current / into the revolving field, 16.2 Number of poles The numberof poles on a synchronous generator de pends upon the speed of rotation and the frequencyNe ELC TCAL MACHES AD TanroReRS ee it G90 oe . x = ph rotor pilot erie AAS eee Si be an "Ee , St QB vain, 2/7 Pure 164 ‘Schematic diagram and cross-section view of atypical 500 MW synchronous generator and its 2400 KW de excitor. The dc exciting current J, (6000 A) flows through the commutator and two siiprings. The de contrl current J rom the pilot exciter permits variable field contra ofthe main excite, which, in turn, controls we wish to produce, Consider, for example, a stator ‘conductor that is successively swept by the N und S. poles of the rotor. Ifa positive voltage is induced when am N pole sweeps across the conductor, a si Har aegarive voltage 1S induced When the 3 pole speeds by. Thus. every time a complete pair of poles. crosses the conductor, the induced voltage through a complete excle. The same is true for e other conductor on the stator: we ean therefore de. luce that the alternator frequency is given by _m 120 (16.1) where J/= frequency of the induced voltage [Hz] umber of poles on the rotor speed of the rotor [r/mia} p Example 16-1 A laydaulie “a0 200 afin is 0 rected (0 a synchronous generator, If the induced age hay frequency of 60 H7, how many poles volt does the rotor have? Solusion From By. 16.1, we have p= 1209 = 120 x 60200 = 36 poles. oF 18 pairs of Nand S poles 16.3 Main features of the stator From an electrical standpoint. the stator of a syn Cchronous generator is identical to that of a 3-phase. induction motor (Section 13.17). I is composed of cylindrical laminated core containing a set of slots that carry «1 3-phase lap winding (Figs. 16.2, 16.3), ‘The winding is always connected in wye and the neutral is com wund. A wye conne: preferred to a delta connection because 1. The voltage per phase is only WA13 oF S8% of the voltage between the lines, This means that the highest voltage between a stator conductor and the grounded stator eore is only 58% ofthe line voltage. We can therefore reduce the amount of insulation in the slots whieh, in cura, ‘enables us to inerease the cross section of the conductors. A larger conductor permits us t0 in crease the current and, hence. the power output of the machineAW Figure 16.28 » SYNCHRONOUS GENERATORS 387 Stator of a 3-phase, 500 MVA, 0.95 power factor, 15 KV, 60 Hz, 200 rimin generator. Inernal diameter: 9250 mm: c’tfective axial langth of iron stacking: 2850 mm: 378 slots. (Courtesy of Marine Industrc) 2, When a synchronous generator is under lead. the voltage induced in ea and the wavetorm is distortion is mainly due to an undesired sid harmonic voltage whose trequeney is three times that of the fundamental frequency: With a wye ‘connection, the distorting line-to-neutral harmon jes do not appear between the lines because they each other. Consequently. the fine voltages remain sinusoidal under all lod becomes distorted Sinusoidal, The: ‘conditions, Unfortunately. when a delta connec monie voltages do not cancel aM tion is used th but ad up. B they proce ath ren which incre se the delta is closed 0 hharmonie circulating cur wsthe PR losses. The nominal line voltage ofa synchronou erator depends upon its KVA rat eneral, the greater the power rating, the higher the volta However. the nominal line-to-line vol seldom 5 KY because the increased slot insulation takes up valuable space atthe expense of the copper conductors‘The copper bars connecting successive stalor poles are designed to carry a current of 3200 A. The total output is 19250 A per phase rtesy of Marine Industrie) Figure 16.2¢ The stators bul up from toothed segments of high-quality silcon-iron steel laminations (0.5 mm thick), covered with fan insulating varnish. The siots are 22.3 mm wide and 169 rmvm deep. The salient poles of the rotar are composed of Inch Uomo (2 rn) on laraniatons. Thee larialiors are nat insulated because the de flux they cary does not vary The width of the poles from tip to tip is 600 mm and the air gap fongth is 33 mm. The 8 round holes in the face ofthe salient pole carry the bars of a squirrel-cage windingSl Figure 16.3 Slalor ofa 3-phase, 722 MVA, 3600 r/min, 19 kV, 60 Hz steam-turbine generator during the construction phase, “The windings are water-cooled. The stator wil eventually be completely enclosed in a metal housing (see back {ground}. The housing contains hydrogen under pressure to further improve the cooling (Courtesy of ABB)M0 ELECTRICAL MACHINES AND TRANSFORMERS 16.4 Main features of the rotor Synchronous generators are built with two types of rotors: salient-pole rotors cylindrical rotors. Salient-pole rotors are usually driven by low-speed hydraulic turbines, and eylindrieal rotors are driven by high-speed steam turbines. 1. Salient-pole rotors have to turn at Tow speods (hetween 50 and 300 rfmin) in order co extraet the maximum power from aa waterfall, Because the rotor is direetly coupled 10 the waterwhe (or 60 HY is required, a large number of poles are re {quired on the rotor. Low-speed rotors always poss sess large diameter 1 provid essary space Tor the poles. The salient poles are mounted on circular Steel frame which is fixed to a revolu~ 16.4), To ensure good cool: Most hydraulic turbines and because a frequency of 50 Hz Figure 16.4 This 36-pole rotor is being lowered into the stator shown in Fig, 16.2. The 2400 A de exciting current is supplied by 390, electronic rectifier, Other details are: mass: 600; moment of inertia: 4140 tm; air gap: 33 mm. (Courtesy of Marine Industrie) ing. the field coils are made of bare copper bars. with the turns insulated from each other by strips of mica (Fig, 16.5), The coils are connected in series. with adjacent poles having opposite polarities In addition to the de field winding a squirrel-cage winding, embedded in the po! (Fig, 16.6). Under normal conditions, this winding does not carry any current because the rotor turns: synchronous speed, However, when the load on the generator changes suddenly, the rotor speed begins to Muctuate, proxlucin ations ‘above and below synchronous speed. This induces a nthe squirtel- we often add voltag ‘winding, causing a large current to flow therein, The current reacts with the ‘magnetic field ofthe stator, producing forces which dampen the oscillation of the rotor. For this reason, the squirrel-cage winding is sometimes called a damper winding.Figure 16.5 Figure 16.6 This rotor winding for a 250 MVA salient-pole genera- Salient pole of a 250 MVA generator showing 12 slots tors made of 18 turns of bare copper bars having a to carry the squirel-cage winding, width of 89 mm and a thickness of 9 mm 7 Figure 16.72 FRotor of a 3-phase steam-turbine ganerator rated 1530 MVA, 1500 r/min, 27 KY. 50 Hz. The 40 longitudinal siots are being milled out ofthe sold steel mass. They will cary the de winding, Etlective axial magnetic length 7490 rm; diameter: 1800 mm, (Courtesy of Alis-Chalmers Power Systems inc., West lis, Wisconsin)Figure 16.7 ELECTRICAL MACHINES AND TRANSFORMERS Rotor wit its 4-pole de winding. Total mass: 204 t; moment of inertia: 85 tm; ar gap: 120 mm. The de exciting cur rent of 11.2 KA is supplied by a 600 V de brushless excites bolted to the end of the main shatt (Courtesy of Alis-Chalmers Power Systems inc., West Allis, Wisconsin) The damper windin anced 3-phase voltags Js to maintain bal- between the lines, even when the Tine currents are unequal due to uabal- anced load conditions, 2. Cylindrical rotors. I is well known that high speed steam turbines are smaller and more efficient irbines, The sume is true of high: speed synchronous ate the required frequency we eannot use less than 2 poles. and this Tixes the highest possible speed (On 2 60 Hz system itis 3600 e/min, The next lower speed is. 1800 r/min, corresponding to a 4-pole ma chine. Consequently, these stewm-rurbine genera tors possess either The rotor of a turbine-generator is @ long. solid than low-speed eneraiors, However, to for 4 poles, steel eylinder which contains a Series of ly slots millee out of the cylindrical mass (Fi Concentric eld coils, firmly wedged into the slots ined by high-strength end-rin create the N and $ poles. The high speed of rotation produces strong cen. Uifugal forces, which impose an upper limit on the diameter of the rotor. In the case of a rotor tuenin at 3600 r/min, the elastic limit of the steel requires the manufacturer to limit the diameter to a maxi mum of 1.2 m, On the other hand. 1 build the pow erful 1000 MVA to 1500 MVA generators the vol ume of the rotors has to be large. It follows that high-power, high-speed rotors have to be very long. 16.5 Field excitation and exciters The de field excitation of a large synchronous generator isan important part ofits overall design,SYNCHRONOUS GENERATORS 343 stonaey bls stator A | alternator Penson = ee T Fee bridge } oe Cr — , =F = 2 phase rotor ‘oe £4z2 Figure 168 Typical brushless exoter system, The reason is thatthe field must ensure not only a voltage may have to Five Yo Lge its momma val in Stable ac terminal voltage, but must also respond to sudden load changes in order to maintain sys- tom stability. Quickness of response is one of the important features of the field excitation, In order to attain it, two de generators are used: a main es citer and a pilor exciter Static exciters that involve no rotating paris at all are also employed. The main exciter feeds the exciting Current to the ld of the synchronous generator by way of brushes tnd slip-rings, Under normal conditions the exciter voltage lies between 125 Vand 600 V. I is regulated ‘manually or automatically by control signals that vary the current J, produced by the pilot exciter (Fig. 16.1), The power rating of the main exeiter depends ‘upon the capacity of the synchronous generator. Typically, 2 25 kW exciter is needed 10 excite a 1000 kVA alternator (2.5% of its rating) whereas a 2500 kW exciter suffices for an alternator of 500 MW (only 0.5% of its rating). ‘Under normal conditions the excitation is varied automatically. [responds to the load chang ‘maintain a constant ae Tine voltage or to control the re active power delivered tothe electric utility system. A. serious disturbance on the system may produce a stu den voltage drop aevoss the terminals of the alterna tor, The exciter must then react very quickly to hep the se voltage from falling. For example, the exciter as Tittle as 308) 19-400 milliseconds. This represents very quick response, considering that the power the fer may be several thousand kilowatts 16.6 Brushless excitation Due to brush wear and carbon dust, we constantly have to clean, repair. and replace brushes. sips nal de excitation sys- and commutators on conve tems. Toeliminate the problem, brushless exeitation systems have been developed. Such a system con- sists oF a 3-phase stationary-field generator whose ac output is rectified hy a group of rectifiers, The de ‘output from the rectifiers is fed dirvetly into the Feld of the synchronous generator (Fig, 16.8. The urmature of the ae exciter ancl the rectifiers are mounted on the min shaft and turn together with the synchronous generutor, In comparing the excitation system of Fig, 16.8 with that of Fig. 16.1 we can see the} except that the Sphase rectifier replaces the commutator. slip- rings, and brushes, In other words, the commutator (owhich is really a mechanical wetitier) is replaced by an electronic rectifier, The result is that the brushes und slip-rings are no fon The de control current J, from the pilot e regulates the main exciter output /. sin the ease oF are Went:Mi ELECTRICAL, MACHINES AND TRANSFORMERS 4 conventional de exciter. The frequeney of the main exciter is zenerully We to thice times the syn= cchronous generator frequency (60 Hz). The in {erease in frequency is obtained by using more poles. fon the exeiter than on the synchronous generator 16.9 shows the rotating portion of a typical brushless exciter, Static exciters that involve n0 10. tating parts at al are also employed 16.7 Factors affecting the size of synchronous generators The prodigious amount of energy generated by el trical utility: compan scious about the efficiency of their ‘example, ifthe efficiency oF 1000 MW generating ation improves by only 1. itreprevents extra rev cenues of several thousand dollars per day, In this ne rd. the size of the tant because its efficiency automatically improves as the power ineteases, For example, if small has made them very con enerators, For encrator is particularly impor kilowatt synchronous zenerator has an efficiency of ‘50% a larger bul similar model having a capacity of 10 MW inevitably has an elficieney of about 90% ‘This improvement in efficieney with size is the « ators of 1000 MW and Lup possess efficiencies of the onder of 99% Another auvantage of large machines is that th power ouput per kilogram increases as the power increases. For example, ita AW 20 kg yielding 1000 W, nerator weighs aloMw Figure 16.9 This brushless exciter provides the de current for the rolor shown in Fig. 16.7. The exciter consists cof a 7000 KVA generator and two sets of diodes. Each set, cortesponding respectively tothe post live and negative terminals, is housed in the ei cular rings mounted on the shatt, as seen in the center ofthe photograph. The ac exciter is seen to the night The two round conductors protruding from the center of the shatt(loreground) lead the ‘exciting current to the 1530 MVA generator (Courtesy of Alls-Chalmers Power Systems ine. West ats, Wisconsin) Figure 16.10 Partial view of @ S-phase, salent-pole generator rated £87 MVA, 428 vimin, 50 Hz. Bath the rotor and stator are water-cooled, The high resistivity of pure water land the use of insulating plasic tubing enables the ‘water to be brought inta direct contact with the live parts of the machine. (Courtesy of ABB)generator of similar construction will weigh only 20.000 kg, thus yielding 00 Wikg. From a power Standpoint, large machines weigh relatively less than small machines: consequently, they are cheaper. Section 16.24 at the end of this chapter explains why the efficiency and output per kilogram increase with size Everything. therefore, favors the large machines. ase in size, we run inte seri- 1g problems. In effect. large machines in- horently produce high power losses per unit surface area (Wim?); consequently, they tend to overheat To prevent an unacceptable temperature rise, we ‘must design efficient cooling systems that become ‘ever more elaborate as the power increases. For ex- ample, a circulating cold-air system is adequate 10 coal synchronous generators whose rating is below 50 MW. but beiween 50 MW and 300 MW, we have tw resort fo hydrogen cooling. Very bi in the 1000 MW range have to be equipped with hollow. water-cooled conductors, Ultimately. a is reached where the increased cost of eooliny and this fixes exceeds the savings macle elsewher the upper Limit to size. To sum up. the evolution of big alternators has, ‘mainly been determined by the evolution of sophis- SYNCHRONOUS GENERATORS MS ticated cooling techniques (Figs, 16,10 and 16,11), Other technological breakthroughs. such as better materials, and novel windings have also played 3 ‘major part in modifying the design of early mt chines (Fig, 16.12). AAs regards speed, low-speed generators ane al ways bigger than high-speed machines. of equal power. Slow-speed bigness simplifies the cooling problem: a good air-cooling system, completed with ‘aheat exchanger, usually suffices. For example, the large, slow-speed 500 MVA, 200 r/min synchronous generators installed ina typical hydropower plant Ja smaller high-speed! ane air-cooled whereas the mu SO MVA. [800 r/min units instalfed in steam pint hhave to be hydrogen-cooled. 16.8 No-load saturation curve Fig, 16.130 shows a 2-pole synchronous generator operating at no-load. It is driven at constant speed by a turbine (not shown), The leads trom the have, wye-connected stator are brought out t0 terminals A. B,C. N. and a variable exciting current J, produces the fu in the air gap. Let us gradually increase the exciting while observing the ac voltage £, between terminal current Figure 16.11 The electrical energy needed on board the Concord aircraft is supplied by four S-phase generators rated 60 kVA, 200/118 V, 12.000 rimin, 400 H2. Each generator is driven by a hydraulic motor, which absoros a small portion of the enormous power developed by the turhoreactor engines, The hydraulic fluid streaming from the hydraulic motor is used to cool the generator and is than recycled. The generator isell weighs only 54.5 ka, (Courtesy of Air France)346 ELECTRICAL MACHINES AND TRANSFORMERS: Figure 16.12 ‘This rotating told generator was first installed in North America in 1888. I was used in a 1000-1amp street lighting system. The alternator was driven by an 1100 rin steam engine and had a rated output of 2000 V, $0 A at a tro- ‘quency of 110 Hz. It weighed 2320 kg, which represents 26 Wikg, A modern generator of equal speed and power ‘produces about 140 Wikg and occupies only one:tnird the floor space. A. say, and the neutral N. For small values of fy. the tage increases in direct proportion to the excit. ing current, However, asthe iron beginsto saturate, the voltage rises much less for the same inerease in J, Me we plot the curve of &,, versus f,, We obtain the no-load saturation curve of the synchronous generator. It is similar to that of a de generator (Seetion 4.13). Fig, 16.13h shows the actual no-load saturation curve of «36 MW, 3-phase generator having a nominal vohage of 12 kV (line to neutral). Up to bout 9 KY, the voltage inereases in proportion to the current, but then the iron begins (© saturate, Thus. an exciting curvent of 100.A produces an out put of 12 kV, but if the current is doubled, the volt- age rises only 10 15 kV. Fig. 16.13¢ isa schematic diagram of the gener ator showing the revolving rotor and the three phases on the stator. 16.9 Synchronous reactance— equivalent circuit of an ac generator Consider a 3-phase synchronous generator having terminals A, B, C feeding a balanced 3-phase load (Fig, 16.14), The xeneratoris driven by a turbine (not shown), and is excited by a de curtent f,. The ma chine and! its load are both connected in wye, yield ing the circuit of Fig. 16.15. Although neutrals Ny sand Np are not connected. they are at the samme po: tential because the load is balanced, ConsequentlyFigure 16.13 {. Generator operating at no-load, '. Norload saturation curve of a 36 MVA, 21 KV, S-phase generator. wwe cand connect them together (as indicated by the short dish ine) without affecting the behavior ofthe ‘oltages or curents in the circuit. The field caries an exciting current which pro- duces flux b. As the field revolves. the flux in- laces in the stator thre equal voltages, that are 120° out of phase Fig. 16.16) Each phase of the stator winding possesses a te- sistance R and a certain inductance L. Because this SYNCHRONOUS GENERATORS 347 i Gaertn \ Figure 16.13¢ Eloctric circuit representing the generator of Fig, 16 13a, is an alternating-current machine, the inductance ‘manifests itself as a reactance X.. given by X= Qafl where synchronous reactance, per phase [0] J-= generator frequency [Hz] L = apparent inductance of the stator wind: ing, per phase [HJ The synchronous reactance oa generator isan in- {eral impedance. just like its internal resistance R. The impedance is there, but it can neither he seen nor touched. The value of X, is typically 10 10 100 times greater than R: consequently, we can always. neglect the resistance. unless we are interested in efficiency oF heating effects. ‘We can simplily the schematic diagram of Fi 16.16 by showing only one phase of the stator In ef Jeet, the two other phases are identical, except that their respective voltages (and currents) are out of phase by 120°, Furthermore. if we neglect the tesis- tance of the windings. we obtain the very simple cir- cuit of Fig, 16.17. A synchronous generaior can there- fore be represented by an equivalent circuit composed ‘ofan induced voltage E, in series witha reactance X, Inthis circuit the exciting current J, produces the Fux cb which induces the imernal voltage E,, For aM8 ELECTRICAL MACHINES AND TRANSFORMERS Figure 16.14 Generator connected to a load Figure 16.15 Electic circuit representing the installation of Fig. 16.14 Figure 16.16 Voltages and impedances in a 3-phase generator and its connected load lows + Figure 16.17 Equivalent circuit of a 3-phase generator, showing oni one phase. given synchronous reactance, the volta terminals of the generator depends upon E,, and the load Z Note that £, and F are line-to-neutral volt ‘ages and Fis the line current 16.10 Determining the value of X, We can determine the unsaturated value of X. by the following open-circuit and short-circuit test During the open-circuit test the generators driven at rated speed ancl the exciting current is raised until the rated line-to-line voltage is attained. The corr. sponding exciting current J, and line-to-neutral volt age E,, are recorded. ‘The exeitation is then reduced to zero and the three stator terminals are short-circuited together, With the generator again running at rated speed. the exciting current is gradually raised to its original value fy The resulting short-circuit current 1. in the sti- tor windings is measured and X,is calculated by us- ing the expression ths (162) where X, = synchronous reaetanee, per phase [12)* E, = rated open-cireuit line-to-neuteal voltage vl This ave wf X, cones othe dean yess eaetance, Is widely used to deseo synconoas methine behavior1, = short-circuit current. per phase, using the same exeiting current Jy that was required 0 produce E, [A] ‘The synchronous reactance is not constant, but varies with the degree of saturation. When the iron is heavily saturated, the value of X, may be only hall its unsaturated value. Despite this broad range we usually take the unsaturated value for X, because it yields sufficient accuraey in most cases of interest rample 16-2 A 3ephase synchronous generator produces. an ‘open-circuit Fine voltage of 6928 V when the dee citing current is $0. The ae terminals are then short-circuited, smd the three fine currents are found to be 800 A, 44, Calculate the synchronous reactance per phase. b. Caleulate the terminal voltage if three 12.0 resistors are connected in wye aeross the ter- mminals, Solution a. The line-to-neutral induced voltag Ey = ENS BAY 69283 = 4000 xyo82 2088 e han he som 308.8 o 7 oo } BAT soos -o4 —— 2008, Te voltage = 63580 aaaarer Figure 16.18 a. See Example 16-2. . Actual ina voltages and currents. SYNCHRONOUS GENERATORS 349 When the terminals are short-circuited, the only impedance limiting the current flow is tht due to the synebronous reactance. Consequently. X= BM =50 000/800 The yynchronous reactance per pase is there- fore $0. b. The equivalent circuit per phase is shown Fig. 16.184, ‘The impedunce of the cireuit is Z-\R 4x 21 avin bs 130 ‘The current is EZ = AOS = 308. “The voltage across the lou resistor is 696 V E=IR= 38 x 12 The Hine voltage under foal is B= 3k V3 x 3096 = 6402 ‘The schematic diagram of Fig. 16,186 helps us vi- swalize what is happening in the actual circuit, 16.11 Base impedance, per-unit X, We recall that when using the per-unit system we first select a base voltage and a base power. In the cease of a synchronous generator. we use the rated ws the buse voltage Ey and line-to-neutral voltage the rated power per phase asthe base power.* It fol- lows thatthe base impedance Z, is given by Be 4 (163) tn many por sts the bas parr is select 0 Fe ga errand ths ise vole ame vale Ze the tothe rated power of the Ath Hn-tine ols, Ths il tose pte350 ELECTRICAL MACHINES AND TRANSFORMERS where Zp, = base impednee (line-to-neutral) of erator [2] Ey, = base voltage (line-to-neutral) [V] Sip = base power per phase [VA] ‘The base impedance is used a for other impedances that the generator possesses. “Thus. the synchronous reactance may be expressed is a per-unit value of Z,. n general, X(pu) lies between (O.8and 2, depending upon the design of the machine usis of comparison Example 16-3 plieesi eee A30 MVA, 15 KV. 60 H7 ae generator has a synchro- nous reactance of 1.2 pu and at resistance oF 0,02 pu, Cateulate a. The base voltage. buse power and base imped lance of the generator >. The aetual value of the synchronous reactance ce. The actual winding resistance, per phase d. The total full-load copper losses Solution a. The base voltage is y= ENS = 18 0003 = 8660 V The base power is Sy = 30 MVA3 = 10 MVA, 0” VA (16.3) = 8607/10" =150 b. The synchronous reactance is X= Napuy x Zp 1.2Zq = 12x75 =90 . The resistance per phase is R= Ripuy X Zy = 002 Zy = 0150 0.02 x 75 Note that all impedance values are from Tine 1 neutral 1d, The per-unit copper losses at full-load are Pepa) Ripa) =F x 002 = 002 Pipu) Note thy equal to 1 The copper losses for all 3 phases are P= 0.02 Sy = 0.02 x 30 = 0.6MW = 600 kW 16.12 Short-circuit ratio Instead of expressing the synchronous reactance as a per-unit value of Zy. the short-circuit rusia is ‘Sometimes used, It isthe ratio of the field current fy needed to generate rated open-circuit armature vol age Ey to the Jha needed to produce rated current Jy, on a Sustained short-circuit. The short-circuit ratio (f/f) is exactly equal 10 the reciprocal of the per-unit value of X, as defined in Eq. 16.2. Thus. ifthe per-unit value of X is 1.2. the short-circuit ratio is 1.2 or 0.833, 1 Fullload the per-unit value of Js 16.13 Synchronous generator under load The behavior of a synchronous generator depends upon the type of load it has to supply. There are many types of loads. but they ean all be reduced to two basic eategor Figure 16.19 Equivalent circuit of a generator under load,1. Isolated loads. supplied by a sing! 2. The infinite bus We begin our study with isolated loads, leaving the discussion of the infinite bus to Section 16.16. Consider a 3-phase generator that supplies power toa load having a lagging power factor. Fig, 16,19 represents the equivalent circuit forone phase. In or- der to construct the phasor diagram for this ei swe list the following 1. Current fags bebind terminal voltage £ by an le 0 2. Cosine # generator power factor ofthe load. 3. Voltage E, across the synchronous reactance leads current / by 90°. ILis given by the expres sion E, = jIX. 4. Voltage £,, generated by the flux is equal to the phasor sum of E plus Ey and £, are voltages that exist inside Jhronous generator windings and cannot be mensured directly 6, Flux «bis that produced by the de exciting eur- rent Figure 16.20 Phasor diagram for a lagging power factor load, Figure 16.21 Phasor diagram for a leading power factor load. SYNCHRONOUS GENERATORS 351 ‘The resulting phasor 1620. Note that £, leads E by 8 degrees Furthermore, the interally-generated voltage By is greater than the terminal voltage, as we would expect. In some cases the lod is somewhat eapacitive, so that current / leads the terminal voltage by an angle 8. ‘What effect does this have on the phasor diagram? The answer is found in Fig. 16.21, The voltage E, across the synchronous reactance is still 0° ahead of the current, Furthermore isan equal to the phi sorsum of Eund £,. However. the terminal voltage is row greater than the induced voltage £,. which is very surprising result, In effect, the inductive reac- tance X, enters into partial resonance with the itive reactance of the load. Although it may appear we are getting something For nothing. the higher terminal voltage does not yield any more power If the load is entirely capacitive. a very high ter- minal voltage ean be produced with a sinall exeit- ing current. However, in later chapters. we wil see that such under-excitation is undesirable. fiagram is given in Fi Example 16-$ A386 MVA, 20.8 KV. 3-phuse alternator has a Syn cchronous reactance of 9 Q and a nominal current of KA. The no-load saturation curve giving the refa- tionship between £,, and f, is given in Fig, 16.13b, Ifthe excitation is adjusted so that the terminal volt- age remains fixed at 21 KY, calculate the exeiting current required and draw the phasor diagram for the following conditions: 1a, No-load b. Resistive load of 36 MW cc. Capacitive load of 12 Mvar Soluion ‘We shall immediately simplify the circuit to show only one phase. The line for all cases is fixed at E= 2083 = 12kV c is no voltage drop ince: consequently. the syn IkV352 ELECTRICAL MACHINES AND TRANSFORMERS, Fb Gaara Figure 16.228 PPhasor diagram at no-load. ‘The exciting current is 1, = 100 (see Fig. 16.130) The phasor diagram is given in Fig. 16.2 With a resistive load of 36 MW: bs. The power per phase is P= 363 = 12MW The full-oad line eurrent is, = 12 x 10712. 000 = 1000. The current isin phase with the ‘The voltage across X, is = JIX, = j1000 x 9 = 9 RV 290° 1 PIE al voltage is. 90° ahead of J. F,, generated by [is equal to the phasor stim af Find #, Rofhering te the pha. ‘Sor diagram, its value is given by ‘The required exciting current is 1, = 200A (see Fig. 16.136) ‘The phasor diagram is given in Fig. 16.22b. With a capacitive load of 12 Mya ce. The reactive power per phase is be fo sxy| wy to, 1 i 1 T = 1a kv Figure 16.220 PPhasor diagram with unity power factor load. Q= 128 = 4 Mvar ‘The line current is 1 = QIE = 4x 10°12 000 =33A The voltage across Xi By = IK, = 333 <9 = 34V290° As before E, leads / by 90° (Fig. 16.22e) aa z tak anv ow Figure 16.226 Phasor diagram with a capacitive load ‘The voltage £, generated by J is equal to the phasor sum of E and E E,= E+ B= 12+ (“9 = 9kV ‘The corresponding exciting current is 1 = TOA (see Fig. 16.135) Note that £ jn less than the terminal volt age E. ‘The phasor diagram for this capacitive load is given in Fig. 16.22¢, 16.14 Regulation curves ‘When a single synchronous generator feeds variable Toxsd, we are interested in knowing how the terminal voltage £ changes as a function of the load current ‘The relationship between E and fis called the regu son curve, Regulation curves are plotted with the field excitation fixed and for a given load power fucto. ig. 16.23 shows the regulation curves for the 36 MVA, 21 KV, 3phase generator discussed in Example 16-4. They are given for loads having unity power factor, 0.9 power factor lagging. and 0.9 power factor leading, respectively. These curves ‘were derived using the method of Example 1-1,0 leading ao soo 780—*1000«— es Load curent | ——+ Figure 16.23 Regulation curves of @ synchronous generator at three different load power factors. except that £,, was kept fixed instead of E. In each ‘of the three cases, the value oF E, was set so that the tuting point for all the curves was the rated line- neutral terminal voltage (12 KV at rated line cur sent (1000 A). ‘The change in voltage between no-load and full- load is expressed as a percent of the rated terminal voltage. The percent regulation is given by the equation Eu Fi % regulation = "100 Ey Eq, = no-s Ey voltage [V] ated voltage [VI Example 16-5 — Calculate the percent regulation corresponding. to the unity power factor curve in Fig. 16.23, Solution ‘The rated line-to-neutral volta Ey = 12AV at full-load is ‘The no-load terminal voltage is Egy = ISKV. SYNCHRONOUS GENERATORS. 358 ‘The perweat regulation is We note that the percent regulation of a synchro: nous generators much greater than that of ade ‘erat, The reason isthe Ligh inpedsice of the yt Cchronous reactance. 16.15 Synchronization of a generator We olten has parallel fo supply a common load, For example, as the power requirements of « large utility: system build up during the day. generators are successively connected to the system to provide the extra power Later, when the power demand falls, selected g cerators are temporarily disconnected from the sys ‘tem until power again builds up the Following day Synchronous generators are therefore regularly b ing connected and disconnected fron large power _grid in response to customer demand. Sueh 3 grid is, said to be an infinite bus because it contains so ‘many generators essentially connected in parallel ‘that neither the voltage nor the frequency of the grid ccan be altered. Before connecting a generator to an infinite bus. (or in parallel with another generator), it must be synchronized. A generator is said to be synchro nized when it meets all the following conditions: 1. The generator frequency is equal to the system, frequency, ‘The voltage 10 connect Wo oF more generators in snerator voltage is equal to the system 3. The generator voltage isin phase with the sys- tem voltage 4. The phase sequence of the generator isthe same as that OF the system, ‘Tosynehronize an alt tor. we proceed as Follows:S84 ELECTRICAL MACHINES AND TRANSFORMERS | Adjust the speed regukitor ofthe turbine so that 3, Observe the phase angle between E, and E by the acy is close to the system means of a syeloscope (Fig, 16.24). This iaseu Trequeney, ment has a pointer that continually indicates the phase angle between the two voltages. covering the entire range from zer0 to 360 d Although the degrees are not shown, the dial has. zero marker to indicate when the voltages are in phase. In practice, when we synchronize an alter ator. the pointer rotates slowly as it tracks the phase voltages. If the generator frequency is lightly ystem frequency. the pointer ro 2. Adjust the excitation so that the generator volt ‘between the alternator and syste tates clockwise tendeny to lea he system fequeny. Fe Conversely. i the generator frequency is lig A rncunaseont Tow the pve tte countrlockwise The We 7 bine speed regulators ine-uned acondingys0 thatthe poimter barely erp aronthe dil A hal check is made See hat the allem v isl equal the system voltage. Then atthe Figure 16.24 Synchroscope 4. The line circuit breaker is closed, connecting (Courtesy of Lab-Voln the generator to the syst Figure 16.25 This loatng oil derrick provides its own energy needs. Four diesel-drivan generators rated 1200 KYA, 440 V, 900 ‘min, 60 Hz supply all the electrical energy. Athough ac power is generated and distributed, all the motors on board are thyistor-cantralled de motors, (Courtesy of Siemens)In modern generating stations, synchronization is usually done automatically 16.16 Synchronous generator on an infinite bus We seldom have to connect only two generators in parallel except in isolated locations (Fig. 16.25). ‘As mentioned previously, itis much more common to connect a generator to a large power system (infinite bus) that already hay many alternators connected to it An infinite bus is a system so powerful that it imposes its own voltage and frequency upon any apparatus. connected 10 its terminals. Once con nected 10 a large system (infinite bus). 2 synch: ‘nous generator becomes part of a network compris ing hundreds of other generators that deliver power to thousands of loads. It is impossible, therefore. 10 specily the nature of the load (large or small, resis- tive oF capacitive) connected to the terminals of this particular generator. What, the power the machine delivers? To answer this qt tion, we must remember that both the value and the frequency of the terminal voltage across the gener- ator are fixed. Consequently, we can vary only two ‘machine parameters: determines the 1, The exciting e 2. The mechanical torque exerted by the turbine rent I, Let us see how a change in these parameters a fects the performance of the machine. 16.17 Infinite bus—effect of varying the exciting current Immediately after we synchronize a generator and connect it 10 an infinite bus, the induced voltage F,, is equal , and in phase with, the terminal voltag E ofthe system (Fig, 16.26). There is no difference ‘of potential aeross the synchronous reactance and, ly, the load current / is zero. Although is connected to the system, it delivery no power: its said to float on the line SYNCHRONOUS GENERATORS 355 If we now inereave the exciting current, the volt- age E, will inerease and the synchronous reactance 3 \ill experience a difference of potential E, given by E.=B-E A current will therefore circulate in the circuit given by Because the synchronous seis inductive the current lags 90° behind E, (Fig. 16.26). The ‘current is therefore 90° behind £, which means that the generator sees the system as iit were an induc e. Consequ ‘ayynchronous generator, it supplies 1 to the infinite bus. The reactive power increases as we raise the de exciting current. Contrary 19 what \we might expect. iis impossible 10 make a genera- tor deliver active power by raising its excitation, Let us now decrease the exciting current so that E, becomes smaller than E. Asa resull, phasor E, = E,,~ E becomes negative and therefore points to the left (Fig. 16.26e). As always, curnent = EI, lags 90° behind £,. However. this puts 190° ahead of E, which means thatthe alternator sees the system as if it were a capacitor, Consequ ‘excite an alternator. it draws reactive power trom the system. This reactive power produces part of the ‘magnetic field required by the machine: the remain der is supplied by exciting current I, tive react ly. when we over-encite ive power 16.18 Infinite bus—effect of varying the mechanical torque Let us return to the situation with the synchronous generator floating on the line. £, and F being equal ‘and in phase. If we open the steam valve of the 1ur- bine driving the generator, the immediate result is an increase in mechanical torque (Fig. 16.274). The rotor will accelerate and, consequently. E,, will t= tain its maximum value 3 fttle sooner than before. Phasor £;, will slip ahead of phasor F, leading it by 4 phase angle 8 Although both voltages have the same value, the phase angle produces a differenceinfrite bus Figure 16.26a ‘Generator floating on an infinite bus. Figure 16.260 ‘Overexcited generator on an infinite bus. Figure 16.26¢ Under-excted generator on an infinite bus. urbine + L-G save [+ infioite bus bara by £ f avi ow 00 av ah Tv I Figure 16.27 a Turbine driving the generator. . Phasor diagram showing the torque angle 8 ® 356of potential £, = E, ~ Eacross the synchronous re actance (Fig. 16.27) A current F will flow (again lagging 90° behind ,), but this time itis almost in phase with E. I fol: lows that the generator feeds active power into the system. Under the driving force ofthe turbine, the t0- tor wi the angle 8 will con- tinue to diverge, and the electrical power delivered to the system will gradually build up. However, as soon ay the electrical power delivered to the system is {equal to the mechanical power supplied by the tur bine. the rotor will cease to accelerate. The generator Will again run at synchronous speed, and the torque angle 8 between E,, and E will remain constant. nis important to understand that a difference of potential is created when two equal voltages are out ‘of phase, Thus, in Fig. 16.27, a potential difference of 4 KV exisis between £, and £, although beth voltages have a value of 12 kV. continue to accelerate, 16.19 Physical interpretation of alternator behavior ‘The phasor diagram of Fig, 16.27b shows that when, the phase angle between E, and £ increases, the value of E, inereases and, hence, the value of Fin- creases. But a larger current means that the tive SYNCHRONOUS GENERATORS 357 power delivered by the generator also increases. To ‘understand the physical meaning of the diagram. let us examine the currents, fluxes, poles inside the machine. ‘Whenever 3-phase currents flow in the stator of erator, they produce a rotating magnetic field identical to that in an induction motor. Ina synchro. hous generator this field rotates at the same speed and in the same direction as the rotor. Furthermor i has the same number of poles. The respective fields produced by the rotor and stator are, therefore, stationary with respect to cach other. Depending on the relative position of the stator poles on the on hand and the rotor poles on the other hand, powerful Forces of attraction and repulsion may be set up be tween them, When the generator oats on the line the stator current J is zero and so no forees are de veloped. ‘The only flux is that ereated by the rotor, and it induces the voltage Ifa mechanical torque is applied to the gener- ator (by admitting more steam to the turbine). the rotor aecelerates and gradually advances by a angle «., compared to its original po- sition (Fig. 16.28b). Stator currents immediately begin to flow, owing to the electrical phase angle 8 between induced voltage £,, and terminal volt- age E, The stator currents create a revolving ind position of the ‘mechanica Figure 16.28a ‘The N poles of the rotor are lined up withthe S poles ofthe stator, Figure 16.28 ‘The N poles of the rotor are ahead af the S poles of the stator.ASN ELECTRICAL. MACHINES AND TRANSFORMERS. xl a corresponding set of N and § poles, Forces of attraction and repulsion are developed be: ‘oyeen the stator poles and rotor poles, and these magnetic forces produce a torque that opposes the mechanical torque exerted by the turbine, ‘When the electromagnetic torque is equal 10 the mechanical torque. the mechanical angle will no. longer increase but will remain at a constant value @, ‘There is a direct relationship between the 1 chanical angle ec and the torque angle 8. given by (16.4) = pa where ' = torque angle between the terminal voltage E and the excitation voltage al degrees| number of poles on the generator 7 «c= mechanival angle between the centers Cf the stator and rotor poles [mechar ical degrees} Example 16-6 — ‘The rotor poles of an S-pole synchronous generator shift by 10 mechanical degrees from no-load 10 full Fou a, Calculate the torque angle between E,, and the terminal voltage & at full-load E oor Eq. is leading? Solution a, The torque angle is: = pad =8X WOR = 40° tor delivers active power, B, ol- ways leads & 16.20 Active power delivered by the generator We can prove (see Section 16.23) thatthe active power delivered by a synchronous generator is sivem by the equation 6s) active power, per phase [W|] E,, = induced voltage, per phase [V] E = terminal voltage, per phase [V] X\ = synehronous reactance per phase [12] 8 = torque angle between E,, and E[°| This equation ean be used under all load eandi- tions, ineluding the ease when the generator is eon nected to an infinite bus To understand its meaning, suppose a generator is connected to an infinite bus having a voltage E. Furthermore. suppose thatthe de excitation of the generator is kept constant so that &,, is constant The term E,EIX, is then fixed, and the active power which the alternator delivers to the bus will vary directly with sin 8, the sine of the torque a ele. Thus. as we admit more steam, 8 will increase and so, 100, wil the active power outpat. The rel tionship between the two is shown eraphieally in Fig. 16.29. Note that between zero and 30° the powcr increases aha Heatly wit the wrque angle. Rated power is typically attained ut an an ale of 30° " rete om aD — angles grees Figure 16.29, Graph showing the relationship between the active ower delivered by a synchronous generator and the tarque angle.However, there is an upper limit to the active power the generator can deliver. This limit is reached when 6 is 90°, The peak power output is then Pry ~ E,EIX, HE we try 10 exceed this limit (such as by admitting more Steam to the turbine), the rotor will accelerate andl lose synchronism with the infinite bus. The rotor will wen faster than the rotating field ofthe stator, and large, pulsating eur- rents will low in the stator, In practice. this condi tion is never reached because the cireuit breakers {rip as soon as synchronism is lost, We then have 10 resynchronize the generator before it can again de- liver power to the grid. Example 16-7, —___ A236 MVA, 21 KV, 1800 r/min, 3-phase generator connected 10 a power grid has 2 synchronous reac~ tance of 9.0 per phase. Ifthe exciting voltage is 12 KV (line-to-neutral), and the system voltage is 17.3 KV (line-to-line), calculate the Follow a, The active power which the machine delivers when the torque angle 8 is 30° (electrical) 'b, The peak power that the generator ean delive belore it falls our of step (loses synchronism) Solution a. We have E, = 12k E> ITRVIN3 = 1OKV b= 30" ‘The active power delivered to the power grid is P= (BEIX) sin (12 1099) x 05 6.67 MW The total power delivered by all three phases is (3 x 667) = 20MW b. The maximum power, per phase. is attained when 8 = 90" P= (E,EIX) Sin 90 = (12 x 1019) x1 SYNCHRONOUS GENERATORS 389 = 1aMW ‘The peak power output of the alternator is, therefore x 13.3) = 40MW 16.21 Control of active power When a synchronous generator is connected (0 3 system, its speed is kept constant by an extremely sensitive governor. This device can detect speed changes as small as 0.01%. An automatic control system sensitive to such snmill speed changes in mediately modifies the valve (or gate) opening of the turbine so as to maintain a constant speed and ‘constant power output On a big uility network, the power delivered by ‘each generator depends upon a program established in advance between the various generating stations. The station operators communicate with each other to modify the power delivered by each station so that the generation and transinission of energy is one as clficiently as possible, In-more elaborate systems. the entire network is under the contra of a computer, In addition, individual overspeed detectors are always ready 10 respond to a karge speed cha particularly ifa generator for one reason or another. should suddenly become disconnected from the system, Because the steam valves are still wide: ‘open. the generator will rapidly accelerate and may attain a speed 50 percent above normal in t0 5 se ‘nds. The centrifugal forces at synchronous speed are already close to the Limit the materials can with stand. so any excess speed can quickly dangerous Situation. Consequently, steam valves ‘must immediately be closed off during such eme gencies. Atthe same time, the pressure build-up in the steam boilers must be relieved and the fuel bbumers must be shut att: ea very 16.22 Transient reactance A synchronous nected (0 3 system is subject fo unpredictable load changes that some- times occur very quickly. In such eases the simple wenerior e36). RLECTRICAL MACHINES AND TRANSFORMERS yoru fe iat ate Figure 16.30 Variation of generator reactance following a short- cut. equivalent circuit shown in Fig. 16.17 does not re- Tlect the behavior of the machine, This circuit is ‘only valid under steady-state conditions or when the load changes gradually, For sudden load current changes, the synchro ouy reactance X, must be replied by another reac- tance X” whose walue varies as a function of time, ig, 16.30 shows how X’ varies when a generator is euited. Prior to the short-circuit, y cance is simply X, However, at the instant of short-circuit, the reactance immedi= ately falls toa much lower value X’,, Tt then in- creases gradually until itis again equal to X, after a time interval T. The duration of the interval depends, upon the size of the generator, For machines below 100 KYA it only lasts a traction of a second, but for ‘machines in the 1000 MVA range it may fast as long as 10 seconds ‘The reactance Xy isealled the trutsient reactance ‘of the alternator. H may be as low as 15 percent ofthe synchronous reactance. Consequently. the initial shori-circuit current is much higher than that corre the synchronous reactance X, This hasa direct bearing on the eapacity of the eireut breakers atthe generator output, In effect, because they must interrupt short-circuit in three 10 six cscs, it follows that they have to inkerrupt a very current, On the other hand, simplifies the voltage regulation problem when the load on the generator increases rapidly. Firs the in ternal voltage drop due to X’y is smaller than it would be if the synchronous rouctanes N. wore set ing. Second, ” stays at a value far below X. fora sufficiently long time t quickly raise the exciting current f,. Raising the excitation increases Ey. which helps to stabilize the terminal vole. Example 16-8 A250 MVA, -phase steam-turbine gener- aor has a synchronous reactance of 1.6 pu and & Uwansient reaetance X', of 0.23 pu. It delivers ity rated output at a power Factor of 100%. A short- circuit suddenly occurs on the line, close to the generating station. Cateutate a, The induced voltaze B,, prior to the short-cireuit bb, The initial value of the short-circuit current cc. The final value of the short-circuit current if the circuit breakers should fal 10 0p. Solution ‘The base impedance of the gene 2y> ES = 25 0007/250 10") 250 ‘The synchronous reactance is (pw) Zp 6x8 40 The rated line-to4 EH cutral voltage per phase is MakV The rated load current per phase isFigure 16.31 See Example 16-8 I= SNE = 250 x 10°1.73 x 25 000) = SIT4A ‘The internal voltage drop E, is = IX= 9774 x4 3.1 kV “The current is in phase with & because the power factor ofthe load is unity. Thus. reer: ring to the phasor diagram (Fig, 16.31), £, is E,=V = Vid + 23 = 27.2kV b. The transient reaetance is Xy = X pw Zy = 0.23 x25 = 05750 ‘The initial short-cireuit current is Lo EX 27.200575 4TSKA which is 8.2 times rated current . I the short-circuit is sustained and the excita tion is unchanged, the current will eventually level off at a steady-state value: SYNCHRONOUS GENERATORS 36t : rua * |5.78 Kal BBkA Figure 16.32 CChange in current when a short-circuit occurs across the terminals of a generator. See Example 16-8. I= BUX = 274 = 68 kA which is only 1.2 times rated current Fig. 16.32 shows the generator curn sand during the short-circuit, We ass terval T of 5 seconds, Note that in practice the eit cuit breakers would vertainly trip within 0.1 afer the short-circuit occurs. Consequently. they have 10 current of about 47 KA. K prior 0. yea time ine interrupt 16.23 Power transfer between two sources ‘The cireuit of Fig. 16.33a is particutarly important because itis encountered in the study of generators, synchronous motors, and transmission lines. In such circuits we are often interested in the active power transmitted from a source A to a source B or Vive versa, The magnitude of voltages Ey and Es, as well as the phase angle between them. are quite ar bitrary. Applying Kirehhoti"s voltage law to this circuit, we obtain the equation Ey = By 4 jx If we assume that / lass behind Es by an arbitia angle # and Ey leads Ey by an angle 8, we obra the phasor diagram shown (Fig. 16.335). Phasor 1X Teads / by 90°, The active power absorbed by Bis P> Esl cos (16.6)302 ELECTRICAL MACHINES AND TRANSFORMERS or . 8 @ x © * Figure 16.33 Power flow between two vollage sources, From the sine law for triangles, we have Whin B= Khio Ey/sin (90 + 8) = E\leos 8 Consequently, feos 8 = E sin BI (16.7) Substituting (16.7) in Bq. 16.6, we find EE pa PE in (16.8) x where sansmitted [W] voltage of source 1 |V] voltage of source 2 [VI phase angle hetw ‘The active power P received by B is equal to that delivered by A, because the reactance consumes 0 active power. The magnitude of P is determined by the phase angle between £ and Ey: the angle # be- tween E, and / does not have to be specified, Theactive power always flows from the leading 10 the lagging voltage. In Fig. 16.33. itis obvious that E, leaus Ey: hence power flows from left to right Example 16-9 Referring to Fig, 16.338, source A generates a volt. age Ey = 20 KV 2 5° and source B generates a volt age Es = 15 RV Z 42°, The transmission fine co them has an inductive reactance of 14 0, Calculate the active power that flows over the fine sand specify which source is actually a load. Solution The phave angle between the two sources is 42° — 5° = 37°. The voltage of source B leads that of Source A because its phase angle is more positive. Consequer Iy. power Hows from B 0.4 and so. is tactually & load. The active power is given by: BE n= FE ns sa) oAV kV 200 150 aga = 19 1 - 20 sn Note that. strange as it may seem, power flows, from the source having the lower voltage (15 KV) t0 the one having the higher voltage (20 kV). 16.24 Efficiency, power, and size of electrical machines The physical size of an electrical machine hasa pro- found effet upon its efficiency. power output, rela- tive cost, and temperature rise. The following analysis reveals why these characteristics are inti- mately related, [Let us consider a simall ae geuerawor having the following characteristics: power output raw rated voltage 120-V.3 phase raed current ABA rated speed 1800 r/minefficiency 3% input torque 727m mosnent of inertia 0.0075 kgm external diameter 0.180.m external length Os m mass 20ke power ourpatimass 50 Wikg Using this information, we can calculate the losses oF the machine P. x 100 awe eg. 3.6 B 100 input power P, = 1.37 kW logses = 1.37 KW — LOKW = 0.37 kW ‘The losses comprise the 17R losses in the windings, the hysteresis and eddy-current losses in the iron and the windage and friction losses. Let us increase the size of the mal a way that ils Finear dimensions are raised in ex- actly the same proportion. while keeping the same materials throughout, Thus, ifa particular type of iron lamination was used in the stator, the same type is used in the larger machine. The same type oF insulation is also used, thereby duplicating and ‘magnifying everything, including the bearings, ‘nuts and bolts In this larger generator we will keep the sume current densities (A/a) ay inthe original machine We will also maintain the same Mux densities (1es- las) in the various parts of the magnetic circuit (core, air gap, stator teeth, ee.) Asa result, the FR losses per em" and the iron losses perenr’ will be everywhere the original machine, It follows that the copper losses and iron losses will inerease in proportion wits vole lume, Let's assume that the windage and friction losses also increase the sume way. We further assume that the number of slots, eon. ‘ductors and interconnections remain the same as. befowe und that the speed of rotation (1800 min) is left unchanged, ne in such nes inthe SYNCHRONOUS GENERATORS 363 Under these conditions, we ean predict the prop: erties of the generator as its size is increased. For example, suppose that all the linear dimen- sions ate sripled, The volume will therefore in- crease by a factor of 3° = 27. Consequently, the ‘mass will increase by a factor of 27 and so. 00, will the losses, The mass of the bigger machine will therefore be 27 X 20 kg = 540 kg. The losses will rise to 27 % 0.37 kW = JOKW ‘The slots are 3 times wider and 3 times deeper. As a result, the cross section of the conductors is 9 times greater which means they can carry 9 times ‘more current. The larger machine can therefore de- liver a current of 9 X48 A= 43.2.4, AAs regards the generated voltage per conductor, it is determined by equation (2.25) = Bl: We recall that B isthe flux density. is the length ofthe condue- torand vis the speed at which the flux cuts across it ‘The flux density in the larger machine is the same as before, However, the length has tripled Furthermore. the peripheral speed ¥ has increased 3 limes because the diameter ofthe rotor has tripled. As a result the voltage generated per conductor also in- creases by a factor of 9, Because the langer generator has the same number of conductors ts before and be ‘cause they are connected the same way. the genera- {or will produce a voltage of 9 % 120 V = 1080 V. ‘Thus, by tripling the linear dimensions. the volt- ‘age and current both increase by a factor of 9. This ‘means that the power output inereases 9 9 = 81 times, The power output of the new generator is therefore 81 1 AW = 81 KW. ‘The power input needed! to drive the ae genera tor is P, = 81 KW + losses = 81 kW + 10 KW = ST KW. The new efficiency is therefore: SEW 100 ork * = 089 89% ‘The efficiency has increased from 73% to 89% Which is a dramatic improvement, The reason is that the power output has inereased 81 times, whil30s ELECTRICAL MACHINES AND TRANSFORMERS losses increased only 27 times. Consequently. the ef ciency ofthe machine was bound to increase with size. inal machine produced an output of 50 Wikg. The larger machine has a mass of S40 kg and produces 81 kW. Consequently. it produces 81 KW/S4O ky = 150 Wokg which is 3 times greater than before. ‘The larger generator is therefore relatively Tighter ane cheaper than the snraller machine, AS proof. if eighty-one I kW generators were used to prosluce 81 kW, their combined mass would be 81 X 20 ke ~ 1620 kg, This generating center would obviously be more costly and take up more floor space than the single 81 kW generator As another matter of interest, we recall that the ‘moment of inertia J of a rotor is proportional to its sand the square of its radius (see Table 3A) Hence. when linear dimensions are tripled. J will we by a faetor of J = mir? = 77 x ¥ = 3 = 243. The moment of inertia of the larger machine is therelove 243 0.0075 ky-m? = L8 ky: ‘The characteristics of the larger generator are summarized below. They are in striking contrast to the original | kW machine, power output Br kw rated voltage 1080 V, 3 phase rated current BIA rated speed 1800 r/min elticieney 89% input torgu 483 Nom moment ofinentia 1.8 kgm? external diameter O.s4.m external length 0.45 m mss 540 kg power outpuiimass 150 Wa The one big problem is temperature rise. When linear dimensions are tripled, the heat-dissipating surface area of the machine inereases 9 times but the loyses inerwase 27 times. Hence, the power dis- Sipated per square meter inereases by a factor of 3. Consequently, unless better cooling means are used, the langer machine is howd to be hotter: To prevent damage to the insulating materials. the tem- perature rise has t0 be limited t a maximum of ns about 200°C. Consequently, the evoling of large machines isa very important matter. In conclusion, ihe general principles covered here reganling physical size, power output, efficiency, temperature rise and so forth, apply’ to all machines, including ae and de motors and transformer. Questions and Problems Practical level VoL What are the advan of using a stutionury large synchronous generators? Why is the stato always connected in wye? 16-2 State the main differences between st turbine generators and salient-pole gener ators. For a given power output. whieh of these machines is the larger? 16-3. In analyzing a hydropower site it is Found that the turbines should turn at close &9 350 r/min, Ifthe directly-coupled gener tors must generate a frequency of 60 Hz, calculate the following: The number of poles on the ror >. The exact turbine speed 16-4 An isolated 3-phave generator proces 3 ‘no-load line voltage oF 13.2 kV. Ifa load hhaving a lagging power factor of 0.8 is cconnevted to the machine, must the exei- tation be increased or decreased in order {© maintain the same line voltage? 16-5 What conditions must be met before a gener ‘ator ca be connected 0 a 3-phase system? 16-6 Caleulate the number of poles on the gener torin Fig. 16.12 using the information given. Calculate the number of poles on th ceralt generator shown in Fig, 16.1 16-7 16-8 A 3-phave generator turning at 1200 r/min generates @ no-load voltage of 9 KY. 60) Hy. How will he terminal voltage be al fected ifthe following loads are connected to its terminals? a Resistive oad Bb. Inductive lod © Capwitive load169 In Problem 16-8, ifthe field current is kept constant, calculate the no-load volt~ age and frequency if the speed is 1000 e/min b. Simin, Intermediate level 16-10 tot told lols lol 16-15 16-16 ‘What is meant by the synchronous reac tance of a 3:phase generator? Draw the equivalent circuit of a generator and ex plain the meaning of all the parameters. State the advantages of brushless excita tion systems over conventional systems Using a schematic circuit diagram, show how the rotor in Fig. 16.7 is excited. Referring to Fig. 16.13, calculate the ex- ng current needed to generate a no: load fine voltage of a. 42kV BRT A 3ephaye generator possesses a yynehro- nous reactance of 6 $1 and the excitation voltage E, is 3 kV per phase (ref. Fig. 16.19). Calculate the fine-to-neutral volt tage E for a resistive load of § Q and draw the phasor diagram, 28, In Problem 16-13. draw the curve of E ver- sus / forthe following resistive loads: in ity. 24, 12.6, 3.0 ohms, i. Caleulate the ative power P per phase in cach ase, «e. Draw the curve of £ versus P For what value of load resistance isthe power output Referring to Fig. 16.2, calculate the length ‘of one pole-pitch measured along the in- ternal circumference of the stator ‘The 3-phase generator shown in Fig. 16.16 has the following characteristics: E, = 2440 X= 440 R=179. load impedance Z = 175.0 (resistive) SYNCHRONOUS GENERATORS WS Catewtate 1. The total resistance of the phase «6 The total reactance of the circuit, per phase 4. The line curent €. The line-to-neutral voltage eros the load 1. The line voltage aerons the load 18 The power of the turine driving the alter hh The phase angle between E, and the vol age across the load A 3.phase generator rated 3000 KA, 20 KV, 900 r/min, 60 Hz delivers power 10 22400 KVA, 16 kV load having a lagging power factor of 0.8. If the synchronous re- actance is 1002, calculate the value of E,, per phase. 16-17 16-18 ‘The generator in Fig. 16.2 has a synchro nous reactance of 0.4 £2, per phase. It is connected to an infinite bus having a line voltage of 14 KY, and the excitation volt- ic adjusted 10 1 14 pu Catealate erator de 1 The torque angle 8 when the g livers 420 MW bb. The mechanical displacement angle o «6. The linear pole shift (measured along the inside stator cireuference) corresponding to this displacement angle {i} 16-19 test taken on the S00 MVA alternator of Fig, 16.2 yields the following results: 1. Open-circut ine voge is 15 kV for a de ‘exciting current of 1400 A. 2. Using the same de current, withthe arma- ture short-cireited the resulting ae line eur rent is 21 000 A. Catewlare ‘8, The base impedance of the generator, per phase ‘The valve ofthe synehionous reactance The perunit value of X, I The short-circuit ratio aoe366 Advane 16.20 16-23 16-24 PLECTRICAL MACHINES AND TRANSFORMERS ved level The synchronous generator in Fig. 16.2 hasan efficiency of 98.4% when it deliv- cers an output of S00 MW. Knowing that the de exciting current is 2400 A at a de ‘voltage of 300 V, calculate the Following: a, The total losses in the machine D. The copper losses in the rotor The torque developed by the turbine 4. The average difference in temperature be tween the coo! incoming airand warm out oing ai, ithe airflow is 280 Referring to Fig, 16.4, each coil on the ro- torhas 21.5 tums, and carries a de current ‘of 500 A. Knowing that the air gap length is 1.3 inches, calculate the Mux density in the air gap at no-load. Neglect the mmf re- {quired forthe iron portion of the magnetic citeuit. (See Section 2.17), g. 16.17. the following in- ven about a generator: I2kV B= 14kV 20 , leads E by 30° CCaleuate the total aetive power output of i Dy, Draw the phusor diagram for one phase «6. Cleat the power factor of the load ‘The steam-turbine generator shown in Fig 16.3 has. a synchronous reactance of 1.3 pu The excitation voltage E, is adjusted 10 1.2 pat and the machine is connected tan infi- nite bus of 19 KV. IFthe torque angle & is 20° eaeulate the following: a, The active power output bi. The line easrent (6 Dray the phasor ukagram, cor ome pase In Problem 16-23, ealeulate the active power output of the generator ifthe steam valves are closed. Does the alternator re- ceive or deliver reactive power and how ‘much? 16.25 ‘The generator in Problem 16-20 is driven by a hydraulic turbine whose moment of inertia is 54 x 10? Ih, The rotor has a J of 4.14 x 10° kg’ a Ite line circuit beakers sakleny tip. ca culate the sped ofthe generating unit (tur bine an alternator) secon! ater. as ing that the wicket zates remain wie open b, By how many mechanival deszees do the poles advance (vith espe positon aunt hhow many electrical degrees? A 400 Hz alternator has a 2-hour rating of TSKVA, 1200 rimin, 3-phase, 450 V. 80 percent power luctor (Fig, 16.34a). The stator possesses 180 slots and has an inter- nal diameter of 22 inches and an axial length of 9.5 in, The rotor is designated for a Field current of 31 A.at 115 V. 16-26 Cateutate ‘a, The number of poles on the rotor bi. The number of coils on the stator 6 The number of coil por phase group on the stator 4, The lensth of one pole pateh, measured slong the cireumference ofthe stator fe. The resistance of the de winding on the ro tor and the power needed to excite i Industrial application 16-27 A338 RVA. 480 V. 3-phase, 60 Hz diesel driven emergency alternator is designed 10 ‘operate ata power factor of 80 percent, The following additional information is given: Efficiency: 83.4% Weight: 730 Ib WA" (moment of inertia) : 15.7 1b. Insulation: class B Caleutate 8. The minimum horsepower rating of the icsel engine to dive the generator b. The maximum allowable temperature of the ‘windings, using the resistance methoA220 MVA, 500 r/min, 13.8 kV, 50 Hy, 0.9 power factor, water-turbine synchro- nous generator, manufactured by Siemens, has the following properties: Insulation class: F Moment of inertia: 525 tm” Tonal mass of stator: 158 1 mmettie ton) Total mass of rotor: 270 t SYNCHRONOUS GENERATORS 367 Efficiency at full-load. unity power Factor 98.95% insaturated synchronous reactance: 1.27 pu Transient uectance: 0.37 pu Runaway speed in generator mode: 890 r/min Static excitation is used and the excitation current is 2980 A under an excitation volt- age of 258 V. Figure 16.34a Rotor and stator of a 75 kVA, 1200 «min, 3-phase, 450\V, 400 He alternator for shipboard use. The alternator is. riven by @ 100 hp, 1209 rimin synchronous motor. Figure 16.34b Stator and rotor ofthe 100 hp, 1200 r/min, 60 Hz synchronous motor. The stator is mounted on a bedplate that also serves as a base forthe alternator. The rotor is equipped with a squirel cage winding to permt starting as an in- duction motor. (Courtesy of Electro-Mécanik)Sok ELECTRICAL MACHINES AND TRANSFORMERS ‘The generator is also designed to oper- 16-29 ae as a motor, driving the turbine as a pump. Under these conditions, the motor develops an output of 145 MW, Both the stator and rotor are water- ‘cooled by passing the water through the hollow current-carrying conductors, The el So tha ivity is less than 5 S/em, The pure water ows through the stator at a rate of 8.9 liters per second and through the rotor at 5.9 Titers per second, Given the above information, make the following calculations: a. The unity power factor and at 0.9 lagging power factor bb The rated reactive power output, in Mvar . The short circuit ratio 4. The value ofthe line-0-ne nous reactance, per phase the total losses of the generator at Fall load and unity power factor its con active power oatpat, in MW at 1630 ul synch: In industry application Problem 16-28, calculate the following: 4a. The horsepower rating of the generstor when it rans asa pump moto 1, The kinetic energy of the rotor when it uns at rte spec. 6. The kinetic energy of the rotor when it sehes its maimum allowable runaway speed 44. The time wo each the ranawsay speed inthe ‘erator is delivering its rate! load and asi ing thatthe water continues to flow unchecked. though the turbine (gates wide open) In Problem 16-28 calculate the power di sipated in the rotor windings and the ower loss per pole. Knowing the rate of water flow and that the inlet eemperature is 26°C, calculate the temperature of the ‘water flowing out of the rotor windings. ‘What is the minimum resistivity (Qn) of the circulating water?CHAPT ER 17 Synchronous Motors 17.0 Introduction he synchronous generators described in the pre vious chapter can operate either as generators ‘oF ay motors, When operating as motors (by con necting them to a 3:phase source), they are called svaichronous morors, As the name implies, synchro- nous motors run in synchronism with the revolving field, The speed of rotation is therefore tied 0 the Frequency of the source. Because the frequency is fixed, the motor speed stays constant. irrespective of the load oF voltage of the 3-phase line. However. synchronous motors are used not so much because they rum at constant speed but because they possess fother unique electrical properties. We will study wpe. these features in this Most synchronous motors are rated. between 150 KW (200 hp) and 15 MW (20 000 hp) and turn at speeds ranging from 1500 1800 /min, Consequently these machines are mainly used in heavy industry igure 17.1 Three-phase, unity power factor synchro: nous motor rated 3000 hyp (2200 KW), 327 rimin, 4000 V, 60 He driving a compressor Used in a pumping station on the Trans- Canada pipeline. Brushless excitation is provided by a 21 kW, 250 V altematorract tier, which Is mounted on the shaft bonwaen the bearing pedestal and the main rotor (Courtesy of General Electric)310 ELECTRICAL. MACHINES AND TRANSFORMERS 1. 17-1). Atthe other end of the power spectrum, ‘we find tiny single-phase synchronous motors used in control devices und electric clocks. They are dis cussed in Chapter 18, 17.1 Construction ‘Synchronous motors are identical in construction 10 salient-pole ac generators. The stator is composed of a slotted magnetic core. which carries a 3-phase Tap winding. Consequently, the winding is also identical to that of a 3-phase induction motor ‘The roror has aset of salient poles that are excited by a de current (Fig, 17.2), The exciting coils are ‘connected in series to two slip-rings. and the de cur = Figure 17.3 rent is fed into the winding from an external exciter Slots arc also punched out along the circumference of the salient poles. They carry ing similar to that ina 3-phase induction motor. This damper winding serves to start the motor. Modern synchronous motors often employ brush: less excitation, similar to that used in synchronous wonerators. Refers tively small hase generator, called exciter and a 3-phase recti- The a squirrel-cage wind: o Fig. 17.3. are Fier are mennted at one end of the motor sh ide current f, from the rectifier is fed directly into the salient-pole windings, without going through brushes and slip-rings, The current ean be varied by controlling the small exciting current f, that flows in the stationary field winding of the excites, Fig. 17.4 Figure 17.2 Rotor ofa 50 He to 16 23 Hz frequency corwerter used to power an electric railway. ‘The 4-pole rotor atthe leitis associated with a single-phase alternator rated 7000 KVA, 16 2/3 He, PF 85%, The rotor on the ght tora GYUU KVA, SU Mz, YO%e PE ‘synehronaus motor which dives the sin- le-phase alternator. Both rotors are ‘equipped with squitrel-cage windings. (Courtesy of ABB) Sphawe input to tator Diagram showing the main components of a brushless excite fora synchronous motor. tis similar to that of a syn- Cchronous generator.shows how the exci ‘mounted in 3000 kW synchronous motor: The rotor and stator always have the same mum: ber of poles, As in the case of an induetion motor: teeter, and salient poles are the number of poles determines the synchronous speed of the motor: a _ n= 1204 any where = motor speed {r/min} J = frequency of the source [Hz] p= number of poles. SYNCHRONOUS MOTORS Example 17-1 Calculate the number oF salient poles on the rotor of the synchronous motor showa in Fig. 17.4 Solution The motor operates at 60 Hz and runs at 200 s/mins consequently n= 120 fp 200 = (120 x 60y4p p= 36 poles The rotor possesses 18 north poles and 18 south poles Figure 17.4 ‘Synchronous motor rated 4000 hp (3000 KW), 200 rimin, 6.9 KV. 60 He, 80% power factor designed to drive an ore crusher. The brushless exciter(allerna- torirectfer) is mounted on the overhung ‘shatt and ig rated 50 KW, 250 V. (Courtesy of General Electric) Figure 17.4b Close-up ofthe 50 kW exciter. showing the armature winging and 5 of the 6 diodes used to recy the ac current (Courtesy of General Eiectric)372 ELECTRICAL MACHINES AND TRANSFORMERS 17.2 Starting a synchronous motor A synchronous motor cannot start by itself; cons quently, the rotor is usually equipped with a yquirrel- ceage winding so that it ean start up as an induction ‘motor, When the stator is connected to the 3-phase fine, the motor accelerates until it reaches a speed slightly below synchronous speed, The de excitation iy suppressed during this starting period. While the rotor accelerates, the rotating flux ere tated by the stator sweeps across the slower movin salient poles. Because the coils on the rotor possess a relatively large number of turns, x high voltage is duced in the rotor winding when it turns at low speeds. This voltage appears between the slip-rings nd it decreases as the rotor accelerates, eventually becoming negligible when the rotor approaches synchronous speed. To limit the voltage, and to im- prove the starting torque, we either short-circuit the slip-rings or connect them to an auxiliary resistor during the starting period. If the power capacity of the supply Tine is Tim- ited, we sometimes have to apply reduced voltage {o the stator. Asin the ease of induction motors, we Use either autotransformers oF series reactors to limit the starting current (see Chapter 20). Very large synchronous motors (20 MW and more) are sometimes brought up to speed by an auxiliary mo~ tor, called a pony motor Finally. in some big instal- lations the motor may be brought up to speed by a variable-frequency electronie sour. 17.3 Pull-in torque As soon as the motor is running at close t0 53 cchronous speed. the rotor is excited with de current ‘This produces alternate N and S poles around the circumference of the rotor (Fig. 17.5). Ifthe poles. at this moment happen to be facing poles of eppo- site polarity on the stator. a strong magnetic attrac on is set up between them, The mutual attction Jocks the rotor and stator poles together, and the ro- tor is literally yanked into step with the revolving field, The torque developed at this moment is ap- propriately called the pul-in torque. miateee—ob gain Figure 17.5. ‘The poles ofthe rotor are attracted tothe opposite poles on the stator. Al no-load the axes of the poles coincide. ‘The pull-in torque of @ synchronous motor is, powerful, but the de current must be applied at the right momeat, For example. if it should happen that the emerging N.S poles ofthe rotor are opposite the N, S poles of the stator, the resulting magnetic re- pulsion produces a violent mechanical shock, The ‘motor will immediately slow down and the circuit breakers will wip. In practice, starters for synchro ‘nous motors are designed to detect the precise mo: ‘ment when excitation should be applied. The motor then pulls automatically and smoothly into step with the revolving field Once the motor tums at synchronous speed, no, voltage is induced in the squirrel-cage winding and so it carries no current, Consequently, the behavior of 2 synchronous motor is entirely different from that of aan induction motor. Basically, a synchronous motor rolatesbecause of the magnetic attraction between the poles ofthe rotor and the opposite poles ofthe stator. To reverse the direction of rotation, we simply interchange any two lines connected to the stator. 17.4 Motor under load— general description ‘When a synchronous motor runs at no-load, the ro: tor poles are directly opposite the stator poles and their axes coincide (Fig. 17.5). However, if we ap- ply a mechanical load, the rotor poles fall slightlyauisofN poe Figure 17.6 “The rotor poles are displaced with respect to the axes ‘of the stator poles when the motor delivers mechani- cal power, behind the stator poles, but the rotor continues to tum at synchronous speed. The mechanical angle «between the poles increases progressively as we increase the load (Fig. 17.6). Nevertheless, the magnetic attraction keeps the rotor locked to the revolving field, and the motor develops an ever ‘more powerful torque as the angle increases But there is a Fimit, If the mechanical load ex- ‘ceeds the pull-out torque of the motor, the rotor poles suddenly pull away from the stator poles and the motor comes 10 a halt, A motor that pulls out of sep creates a major disturbance on the line, and the citeuit breakers immediately trip, This protects the motor because both the squirrel-cage and stator windings overheat rapidly when the machine ceases to run at synchronous speed. The pull-out torque depends upon the magneto- motive force developed by the rotor and the stator poles, The mmf ofthe rotor poles depends upon the de excitation J,, while that of the stator depends upon the ac current Mowing in the windings. The pull-out torque is usually 1.5 t0 2.5 times the nom inal full-load torque. The mechanical angle « between the rotor and stator poles has a direct bearing on the stator current AAs the angle increases, the current increases, This is tobe expected because a larger angle corresponds to SYNCHRONOUS MOTORS 373 aa bigger mechanical load, and the increased power ccan only come from the 3-phase ae source 17.5 Motor under load— simple calculations We can get a better understanding of the operation of a synchronous motor by referring tw the equiva: lent circuit shown in Fig. 17.7a, It represents one phase of a wye-connected motor. It is identical 10 the equivalent or, because both machines are built the sam ux eb created by the rotor induces a vol the stator. This flux depends on the de exci rent 4. Consequently, £, varies with th As already mentioned, the rotor and stator poles are lined up at no-load, Under these conditions, in- duced voltage &, isin phase with the line-to-neutral voltage E (Fig. 17.7b). If in addition, we adjust the excitation so that E,, = £, the motor “floats” on the line and the line current J is practically zero. la ef- fect, the only current needed is to supply the small swindage and friction losses in the motor, and so itis negligible ‘What happens if we apply a mechanical load 10 the shaft? The motor will begin to slow down, causing the rotor poles to Fall behind the stator poles by an angle a, Due to this mec shift, Ey litte tater than before. Thus, referring to Fig. 17.7e, Eis now 8 electrical degrees behind £, The mechanical dis placement a produces an electrical phase shift 5 between E, and E. ‘The phase shift produces a difference of poten. tial E, across the synehronous reactance X, given by iches its maximum value Consequently, a current / must Mow in the circuit, ‘given by JK, from which = jE IK, HE ~ BUX,AM ELECTRICAL MACHINES AND TRANSFORMERS Figure 17.74 Equivalent crcut of a synchronous motor, showing fone phase, EE, Figure 17.7b ‘Motor at no-load, with, adjusted to equal E 5= pala Figure 17.76 ‘Motor under load has the same value a in Fig, 17-70, butitlags behind E. The current lags 90° behind E, because X, is in- ductive, The phasor diagram under load is shoven in 17-Je, Because /is nearly in phase with E, the motor absorbs active power, This power is entirely transformed into mechanical power, except for the relatively small copper and iron losses inthe stator. In practice, the excitation voltage B, is adjusted, to be greater or less than the supply voltage &. Tts value depends upon the power output of the motor and the desired power factor. Example 17-2¢ 'AS(\) hp. 720 ehnin synchronous motor connected to 4 3980 V, 3-phase line generates an excitation volt- age E,,of 1790 V (line-to-neutral) when the de excit- ing current is 25 A. The synchronous reactance is 22 {and the torque angle between E,, and E is 30°, Calewlare a, The value of Ey b. The ac line current ce. The power factor of the motor 4. The approximate horsepower developed by the motor . The approximate torque developed at the shat Solution ‘This problem can best be solved by using vector no- tation, a The vole £ ineto-netr applied to the motor has value B= IVI = 39809 OV Let us select Eas the reference phasor. whose angle with respeot to the horizontal axis is as sumed to be zero, Thus, E 3002.0" I follows that E, is given by the phasor 7902 —30° ‘The equivalent circuit per phase is given in Fig. 178, Moving clockwise around the circuit and ap- Kirchhoft's voltage law we can write 30° 2300 (cos 0° + j sin 0°) 1790 (cos =30° + j sin 30° 2300 ~ 1550 + 895 = 750 + j 895 = 11682.50°Figure 17.88 Equivalent circuit ofa synchronous motor connected toa source E. ‘Thus, phasor £, has a value of 1168 V and it leads phasor £ by 50° . The line current Fis given by pars Ey 1168 2 50° ward 532-40" ‘Thus, phasor J has a value of 53 A and it lags 40° behind phasor E. cc. The power factor of the motor is given by the cosine of the angle between the line-to-neutral voltage E across the motor terminals and the current J, Hence, power Factor = cos = cos 40° = 0.766, 01 76.6% ‘The power factor is lagging because the current lags behind the voltage. ‘The complete phasor diagram is shown in Fig. 17.8b, 4d, ‘Total active power input to the stator: P,= 3X Fixh cos 3.6 2300 ¥. $3 ¥ eon 10 = 280 142 W = 280.1kW Neglecting the FR losses and iron losses in the stator, the electrical power transmitted across the airgap to the rotor is 280.1 KW. Approximate horsepower developed: P= 280.1 x 109746 = 375 hp SYNCHRONOUS MOTORS 375 e 2300 fo sh 1 53a D vay Figure 17.80 ‘See Example 17-2. fe. Approximate torque 955 XP _ 9.55 x 280.1 x 10° n 70 15Nem Example 17-2b ‘The motor in Example 17-2a has a stator resistance of 0.64 Q per phase and possesses the following osses: PR losses in the rotor: Stator core loss: ‘Windage and friction loss Catewlate a, The actual horsepower developed. b. The actual torque developed at the shaft cc. The efficiency of the motor Solution a4, Power imput to the Stators 280.1 kW Stator PR losses = 3X 53? X 0.64.2 = S4kW Total stator losses = 5.4 + 3.3 = 8.7 kW Power transmitted fo the rotor = 280.1 ~ 87 = 14 kW “The power at the shaft is the power to the rotor ‘minus the windage and friction losses. The rotor376 ELECTRICAL MACHINES AND TRANSFORMERS FR losses are supplied by an external de source and so they do not affect the mechanical power, Power available at the shat P= ITA 1S = 269.9kW = 200910" ae gy ieee te ‘This power is very close to the approximate ‘value calculated in Example 17-24, '. The corresponding torque is pa DSS XP _ 955 X 260.9 10° Sn 0. = 3580 Nem c. Tonal lowes = $4433 +324 15 = BARW ‘Total power input = 280.1 + 3.2 = 283.3 KW “Total power output ~ 269.9 KW Eificiency = 269,9/283,3 = 0.9527 95.3 % Note that the stator resistance of 0.64 is very small compared to the reactance of 22. Q. Consequently. the true phasor diagram is very close tthe phasor diagram of Fig, 17.86, 17.6 Power and torque ‘When « synchronous motor operates under load, it draws active power from the Tine. ‘The power is iziven by the same equation we previously used for the synchronous generator in Chapter 16: P= (EEIX) sin (16.5) Asin the ease of a absorbed by the motor depends upon the supply voltage E, the excitation voltage E,, and the phase angle 6 between them. If we neglect the relatively small 7° and ion losses in the stator, all the power is transmitted across the air gap 10 the rotor. This is analogous to the power P, transmitted across the air gap of an induction motor (S 13.13), However, in a synchronous motor, the 10> tor FR losses are entizely supplied by the de source, Consequently. all the power transmitted nerator, the active power tio stcross the air gap is available in the form of me nical power, The mechanical power developed by a synchronous motor is therefore expressed by the equation sind 7.2) where P = mechanival power of the motor, per phase [WI ., = line-to-neutral voltage induced by 1, [V1 re of the source [V] = fine-to-neutral volt synchronous reactance per phase [0] torque angle between E, and E [electrical deg: This equation shows that the mechanical power ineteases with the torque angle. and ity maximum value is reached when 8 is 90°. The poles of the rotor are then midway between the N and S poles of the stator, The peak power P, (per-phase) is given by 7 (150/600) = 600 V ‘The frequency is also proportional to the speed, and so f> 60 (15/60) S Hz ‘The synchronous reactance is proportional 10 insane comeateny Figur 1.208 160 2e00v Motor turning at 600 vimin (Example 17-6) 600 ein600 150 ri Figur 1.208 Motor turning at 180 rin (Example 17-6), = 16 x (15/00) = 4.0 Referring to Fig. 17.22b the new impedance per phase at 150 r/min is Z-\OP +P =40 “The current phase is T= EZ = 6004 = 150 Thuis the short chan; rmin to 150 r/min, The power dissipated in the 3 phases is therefore the same as before: reuit current remains un- ed as the motor decelerates from 600 P= 13.5kW The kinetic energy at 600 e/min is By = 548 x 10 Te G8) = 548 x 10-* x 275 x 600° = 542.5 kl 4. The kinetic Eqs = 548 x 10-8 x 275 x 150" =339k ray at 150 r/min is ‘e. The loss in kinetic energy in decelerating from (600 r/min to 10 r/min is W=6-& = 5425 — 33.9 = 508.6 KI This energy is lost as heat in the armature re tance, The time for the speed 10 drop from 600 min 10 150 eimin is given by SYNCHRONOUS MOTORS 385 P= Wh Gay 13.5 = 508.6i7 whence 1 = 37.7. Note that the motor would stop much sooner if external resistors were connected tor termi oss the Ste 17.14 The synchronous motor versus the induction motor We have already seen that induction motors have excellent properties for speeds above 600 r/min, But at lower speeds they become heavy. costly. and have relatively low power factors and efficiencies, Synchronous motors are particukarly autrctive for low-speed drives because the power factor ean altways be adjusted to I, and the efficiency is high, Although moze complex to build, their weight and cost are ofien less than those of induction motors of cequall power and speed. This is particularly true for speeds below 3000 r/min, A synchronous motor ean improve the power fae: tor of a plant while cumying its nated load. Furthermore, its starting torque eam be made consid cably greater than that of an induetion motor. The reason is that the resistance ofthe squirtel-cage wind- ing can be high without alTecting the speed or eff ciency at synchronous speed. Figure 17.23 compares the properties of a squirrel-eage induction motor and asynchronous motor having the sume nominal r The biggest difference is in the starting torque, High-power electronic converters. generating Jhronous very low frequencies enable us to fun motors at ullraclow speeds, ‘Thus. h the 10 MW range drive crushers, rotary kilns, and variable-speed ball mills, 17.15 Synchronous capacitor Asynchronous capacitor is essentially asynchronous, motor running at no-load. Its only purpose isto ab. sorb or deliver maetive power on a 3-phase system in order to stabilize the voltage (see Chapter 25), The ‘machine aets as an enormous 3-phase capacitor (or386 ELECTRICAL MACHINES AND TRANSFORMERS x PS eee peels eee ee » & ; » 2 9939 409080 Too % eet eaad Figure 17.23 Comparison between the ficiency (a) and starting torque (b) ofa squitel-cage induction motor and a ssynehzonous motor, both rated at 4000 hp, 1800, rimin, 6.9 KV, 60 He. Inductor) whose reactive power ean be varied by changing the de exci Most synchronous capacitors have ratings that range from 20 Myar to 200 Mar and many are hydrogen-cooled (Fig. 17.24). They are started up like synchronous motors. However, if the system cannot furnish the requited starting power, a pony motor is used to bring them up to synchronous speed. For example, in one installation, a 160 Mvar Figure 17.24a “Three-phase, 16 kV, 900 rimin synchronous capacitor rated ~200 Mvar (supplying reactive power) to +300 ‘Mvar (absorbing reactive power). Ils used to regulate the voltage of @ 735 kV transmission line, Other char- acteristics: mass of rotor: 143 t; rotor diameter: 2670 rim; axal length of stator iron: 3200 mm; air gap length: 39.7 mm Figure 17.24b ‘Synchronous capacitor enclosed ints stee! housing containing hydrogen under pressure (300 kPa, or about 44 loin’) (Courtesy of Hydro-Québec)synchronous capacitor is started and brought up 10 speed by means of a 1270 kW wound-rotor motor, Example 17-7 7 A synchronous capacitor is rated at 160 Mvar, 16 kV. 1200 rimin, 60 Hz. It has a synchronous re- actance of 0.8 pu and is connected to a 16 kV Tine. Calculate the value of E, so that the machine 1a. Absorbs 160 Mvar b. Delivers 120 Myar Solution a. The nominal impedance of the machine is ‘The synchronous reactance per phase is X, = Xe(pu) Z, = 08 1.6 128.0 ‘The line current for a reactive load of 160 Mvar SUV3 Bg 160 x 10°21.73 16000) STS ‘The drop across the synchronous reactance is E.= IX. = S780 % 1.28 7400 V The line-to-neutral voltage is BING = 16 000/173 9250V Selecting Fas the reference phasor. we have E= 9250 ‘The current Flags 90° behind E because the machine is absorbing reactive power conse- quently 1= 57802 -90" SYNCHRONOUS MOTORS 387 4295.8 5850 9250 14800 Figure 17.25b Over-excited synchronous capacitor delivers reactive power (Example 17-7). From Fig, 17.25a we ean write HE + jIX, + E,=0 hence: Ey = B= jIX, = 925020" ~ S780 x 1.282(90° — 90°) = 185020 Note that the excitation voltage (1850 V) is sch less han the fine voltage (9250 V), b. The load current when the machine is dativer- ing 120 Moar is .= OUT) = 120 10°.73 * 16.000) = 454 This time / leads £ by 90° and so 14335290" From Fig. 17.25b we ean write E,= B= JIX, = 925020" — 4335 1.287180" (9250 + $550)20 = 14 80020388 ELECTRICAL MACHINES AND TRANSFORMERS 5780.8 Figure 17.254 Under-excitad synchronous capacitor absorbs reac- tive power (Example 17-7), ‘The excitation voltage (14 800 V) is now con siderably greater than the fine voltage (925 Questions and Problems Prcctictl level 17-1 Compare the construction of a synehro- ‘nous generator, a synchronous motor, and fa squirrel-cage induetion motor. Explain how a synchronous motor starts up. ‘When should the de excitation be applied? Why does the speed of a synchronous motor remain constant even under variable load? Name some of the advantages of a syn- cchronous motor compared to a squirrel cage induction motor. m5 What is meant by a synefyonous capacitor and what is it used for? 1, What is meant by an under-esvited sy chronons motor? b, Ir we overexcitea synchronous motor, does its mechanical power output increase? Agynchronous motor draws 2000 kVA at a power factor of 90% leading. Calculate the approximate power developed by the 176 m7 motor [hp] knowing it has an efficiency of 95%. 17-8 Asynchronous motor driving a pump ‘operates at a power Factor of 100% ‘What happens if the de excitation is in- creased? 17-9 A 3-phase, 225 r/min synchronous motor connected t0 a4 KV, 60 Hz line draws a current of 320 A and absorbs 2000 kW. Cateulate 4. The apparent power supplied the mator b. The power factor The reactive power absorbed 44. The number of poles on the rotor A synchronous motor draws 150 A from a 3-phuse line, IF the exciting current is raised, the current deops to 140.4. Was the motor over- or under-excited before the excitation was changed? 17-10 Intermediate level IT-I1 a, Calculate the approximate full-load current ‘of the 3000 hp mevor ia Fig. 17.1, i thas an efficiency of 97%, 17-12 Referring to Fig. 17.2, at what speed must the rotor turn to generate the indicated fre~ quencies? 17-13 A 3:phase synchronous motor rated 800 hp, 2.4 KV, 60 Hz operates at unity power factor, The line voltage suddenly drops 10 1.8 kV, but the exciting current remains unchanged. Explain how the Following quantities are affected: 2, Motor speed and mechanical power output D. Torque angle 8 €. Position ofthe rotor poles 4. Power factor ©. Stator current A.synchronous motor has the followin, parameters, per phase (Fig. 17,74): E = 24KV:E, = 3kV 20 900.4, Wt x,Draw the phasor diagram and determine: ‘a. Torgue angle 8 by Active power, per phase «Power factor ofthe motor 4. Reactive power absorbed (or delivered), per phase 17-15 a, In Problem 17-14 calculate the Hine current and the new torque sigh 8 i the mechani cal load is suddenly eemoved. De. Calculate the new reactive power absorbed (or delivered) by the motor, per phase 4.500 hp synchronous motor drives a compressor and its excitation is adjusted so that the power factor is unity. IV the ex- citation is increased without making any ‘other change, what isthe effect upon the following: a. The active power absorbed by the motor bb. The line cure «6. The reactive power absorbed (or delivered) bythe motor 4. The torque angle 1716 Advanced level 17-17 The 4000 hp, 6.9 KY motor shown in Fig 174 possesses a synchronous reactance of 10.2, per phase. The stator is connected in wye, and the motor operates at full-load (4000 hp) with a leading power factor of (0.89. Ifthe efficiency is 97%, calculate the following: a, The apparent power bb The line current «. The value of E,. per phase 4. The mechanical displaceme From theie no-load position , The worl reactive power supplied to the clecricl system 6 The approximate maximum power the motor ein develop. ssithout palling oat of step (hp In Problem 17-17 we wish to adjust the power factor to unity ‘oF the poles i718 Cateutare a. The exciting volge E, required. per phase i. The new torque angle 119 SYNCHRONOUS MOTORS 389 A3-phase, unity power factor synchro- hous motor rated 400 hp, 2300 V, 450 min, 80 A, 60 Hz, drives a compressor. ‘The stator has a synchronous reactance of 0.88 pu, and the excitation , is adjusted 10 1.2 pu. Catewlate 17.20 ‘a, The value of X.and of E, bb. The pull out torgue [ft 1bt) «6 The line current when the motor is about to pull out of synchronism ‘The synchronous capacitor in Fig, 17.28 possesses a synchronous reactance of 0.6 © per phase. The resistance per phase is 0.007 Q. If the machine coasts toa stop, it will run for about 3 h. In or- der to shorten the stopping time, the sta- tor is connected to three large 0.6 0 braking resistors connected in wye. The de excitation is fixed at 250 A so that the initial line voltage across the resistors is ‘one-tenth of its rated value, oF 1600 V, at 900 min, per phase Cateulate ‘a, The total braking power and braking torque 900 g/min be. The braking power and braking torque at 450 min The average braking torque between 900 rien and 480 einin ‘4. The time for the speed ofall rary 900 chm 1 450 ein, knowing that the moment of ria of the rotor is 7 % HOP Industrial application 1721 ‘A 500 hp, 3-phase, 2200 V, unity power factor synchronous motor has a rated cur- rent of 103.A. It can deliver its rated out- pput so long as the air inlet temperature 40°C ov less. The manufacturer states that ‘the output of the motor must be decreased by I percent for each degree Celsius above 40°C. If the air inlet temperature is 46°C. calculate the maximum allowable ‘motor current.ELECTRICAL MACHINES AND TRANSFORMERS An 8800 KW. 6.0 kV, 1500 r/min, 3-phase, 50 Hz, 0.9 power factor synchronous mo- tor manufactured by Siemens has the fol- lowing properties: 1. Rated curren: 2. Rated tog 3 Pall-cut torque: 4. Locked-rotor eucrent: Excitation voltage: 160.V Excitation currents 387A Fui-ioad ettciency. excluding exenation system losses: 97.8% 8. Moment of inertia of rotor 9. Temperature rise of cooking water: Ie Flow of cooling water: Maximum permis inertia: 1370 kgm? Mass of ro1or 6.10 111 = metric ton) Mass ofstator: 7.501 Mass ofenclosure: 3.971 962 S60kNm 15 pu 4.9 pu 5 6. S20ke 2c 465 Limin 2 13 MW Usi lowing: he above information, calculate the fol- 8, The total ass of the motor inching its enclosure. in metric tons “The flow of cooling water in gallons (U.S.) per eninute “The maximum total moment of inertia (in Ibs), which the motor ean pull nto sya hronismt 4d. The total losses oF the motor at Ful-ead fe The sal etic oF te nn all load 6. The resetive power delivered by the motor a fll-load If the iron losses are equal to the stator cop: per losses, caleulate the approximate resis tance between two terminals ofthe Stator 1h, Caleulate the resistance of the Field circuit b.