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Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET Two Pole Three Phase Synchronous Motors Fig.1 Two pole, three phase synchronous motor The stator windings of the synchronous machine are embedded in slots around the inside circumference of the stationary member. In a 2 pole machine each phase winding of the three phase stator winding is displaced 120 degrees with respect to each other as shown in the figure above. Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET The field or fd winding is wound on the rotating member. The as, bs, cs, and/d axes denote the positive direction of the flux produced by each of the windings. The as, bs, and cs windings are identical in that each winding has the same resistance and the same number of turns. When a machine has three identical stator windings arranged as shown in Fig. above, it is often referred to as a machine with symmetrical stator windings. It is assumed that each coil spans radians of the stator for a 2-pole machine. One side of the coil (coil side) is represented by a indicating that the assumed positive direction of current is down the length of the stator (into the paper). The indicates that the assumed positive direction of current is out of the paper. Each coil contains nc conductors. Therefore, in the case of the as winding, positive current flows in a conductor of coil a1 , then axially down the length of the stator and ' back the length of the stator and out at coil side a1 . This is repeated for nc conductors. ' The last conductor of the coil a1 - a1 is then placed in the appropriate slot so as to ' start coil a 2 - a 2 wherein the current flows down the stator via coil side a2 and then ' ' back through a 2 and so on until a 4 . The bs and cs windings are arranged similarly, and the last conductors of coil ' ' ' sides a 4 , b4 , and c 4 are connected together to form the wye-connected stator. Each coil consists of nc conductors, each of which makes up an individual single conductor coil within the main coil. Thus the number of turns of each winding is determined by the product of nc and the number of coils or the product of nc and the number of coil sides carrying current in the same direction. In the case of the/d winding, each coil (f1-f1’, for example) consists of nf conductors. The coil sides of each phase winding are considered to be distributed uniformly over 60° of the stator circumference. Another practical consideration is the so-called fractional-pitch winding. The windings shown in Fig above span radians for the 2-pole machine. This is referred to as full-pitch winding. In order to reduce voltage and current harmonics, the Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET windings are often wound so that they span slightly less than radians for a 2-pole machine. This is referred to as a fractional-pitch winding (Short Pitch Winding). For the purposes of deriving an expression for the air-gap MMF, it is convenient to employ the so-called developed diagram of the cross-sectional view of the machine shown in Fig. 1. The developed diagram is shown in Fig. 2. The length of the air gap between the stator and rotor is exaggerated in Figs. 1 and 2 for clarity. Fig 2 Development diagram for 2 pole 3 phase Synchronous motor The angular displacement along the stator circumference is denoted s and r , along the rotor circumference. The angular velocity of the rotor is r , and r is the angular displacement of the rotor. For a given angular displacement relative to the as axis we can relate s , r and r as s r r ...........................(A) Due to the high permeability of the stator and rotor steel the magnetic fields essentially exist only in the air gap and tend to be radial in direction due to the short length of the air gap relative to the inside stator diameter ( Fig 3). Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET Therefore, in the air gap the magnetic field intensity H and the flux density B have a component only in the ar direction, and the magnitude is a function of the angle s whereupon H r ,s , z H r s ar And Br s 0 H r s Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET With the assumed direction of the current /as, the magnetic field intensity H in the air gap due to the as winding is directed from the rotor to the stator for — from the stator to the rotor for < s < and 2 2 3 < s < 2 2 Ampere's law may now be used to determine the form of the air-gap MMF due to the as winding. In particular, Ampere's law states that H .dL i ....................................................(1) Where i is the net current enclosed within the closed path of integration. Let us consider the closed path of integration depicted in Fig. 3b. Applying Ampere's law around this closed path denoted as abcda, where the path bc is at s .we can 4 write and evaluate (1) r g 4 4 r 4 r 0 H dL H0dL 0 ...................(2) 4 r 0 g ( 0 ) H g H0g 0 0 4 4 Where r and r 0 the radius of the rotor at points b and a and g and g 0 4 4 are the corresponding air-gap lengths at points c and d respectively. The magneto motive force is defined as the line integral of H .Therefore, the terms on the left-hand side of (2) may be written as MMFs MMF + MMF(0) = 0.................................(3) 4 Where MMF(0) -H(0)g(0) Let us now consider the closed path aefda where path ef occurs at s 7 here 16 Ampere's law gives 7 MMF + MMF(0) = -nci as .................................(4) 16 Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET We can now start to plot the air-gap MMF due to the as winding. For the lack of a better guess let us assume that MMF (0) = 0. With this assumption, (3) and (4) tell us that the MMF is zero from s 0 to where our path of integration encircles the first ' coil side a1 .If we continue to perform line integrations starting and ending at a and each time progressing further in the s direction, we will obtain the plot shown in Figure 4. Fig 4.Plot of the air-gap MMF due to the as winding with the assumption that MMF (0) is zero. In any practical electric machine, the air-gap length is either (a) constant as in the case of a round-rotor machine or (b) A periodic function of displacement about the air gap as in the case of salient-pole machine. In particular, for a 2-pole machine g r g r substituting (A) in above equation w e get g s r g s r .........................(5) Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET Above equations is satisfied if the air-gap MMF has zero average value and, for a 2-pole machine, MMF (s ) MMF (s ).................(6) Hence, the air-gap MMF wave for the as winding, which is denoted as MMFas, is shown in figure 5. It is clear that the MMF due to the bs winding, MMFbs, is identical to MMFas but displaced to the left by 120°. MMFcs is also identical but displaced 240° to the left. Figure 5 Air-gap MMF due to as winding Actually most electric machines, especially large machines, are designed so that the stator windings produce a relatively good approximation of a sinusoidal air-gap MMF with respect to s so as to minimize the voltage and current harmonics. In order to establish a truly sinusoidal MMF waveform the winding must also be distributed sinusoidally A sinusoidally distributed as winding and a sinusoidal air-gap MMFas are depicted in Fig. 6. The distribution of the as winding may be written as N as N p sin s N as N p sin s 0 s ...............(7) s 2 Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET Where Np is the maximum turn or conductor density expressed in turns per radian Nr represents the number of turns of the equivalent sinusoidally distributed winding, then N s N p sin s ds 0 N p cos s 0 N p cos cos 0 N p 1 1 2 N p ........................................................(8) Where Ns is the average value of the sinusoidally distributed winding. Fig 6. MMF due to equivalent as winding. The MMF waveform of the equivalent as winding is (from Fig 5 and Fig 6) N MMFas s ias cos s .....................(9) 2 It follows that N 2 MMFbs s ibs cos s 2 3 ............(9a) Ns 2 MMFcs ics cos s 2 3 Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET Let us now express the total air-gap MMF produced by the stator currents. This can be obtained by adding the individual MMFs, given by MMFs Ns 2 2 ias cos s ibs cos s ics coss .......(10) 2 3 3 The above expressions is valid for 2 pole machine in general if P is the number of poles of the machine then the above equations can be modified as given below Ns P 2 MMFbs ibs cos s ........(11) P 3 2 N 2 P MMFcs s ics cos s P 3 2 MMFas Ns P ias cos s P 2 WINDING INDUCTANCES AND VOLTAGE EQUATIONS Once the concepts of the sinusoidally distributed winding and the sinusoidal air-gap MMF have been established, the next step is to determine the self- and mutual inductances of the machine windings. As in the previous section, it is advantageous to use the elementary 2-pole, 3-phase synchronous machine to develop these inductance relationships. Here the damper windings are not considered but this section can be extended with damper windings. SYNCHRONOUS MACHINE In a magnetically linear system the self-inductance of a winding is the ratio of the flux linked by a winding to the current flowing in the winding with all other winding currents zero. Mutual inductance is the ratio of flux linked by one winding due to current flowing in a second winding with all other winding currents zero including the winding for which the flux linkages are being determined. For the purpose of analysis the development diagram is again repeated as shown in the figure given below Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET For this analysis it is assumed that the air-gap length may be approximated as given below g r 1 ......................................(12) 1 2 cos 2r Where 1 2 is g min and 1 2 is g max When r 0 g r 1 1 1 1 g max 1 2 cos 2r 1 2 cos 2(0) 1 2 g min When r g r 2 1 1 2 cos 2r 1 1 2 cos 2 1 1 g min 1 2 g max 2 Which is evidnent from the above figure or 1 g s r .......................(13) s r r 1 2 cos 2s r MMF can be defined as Br 0 MMF .......................................(14) g Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET It is convenient to express the stator MMF in terms of MMF S we can write Br s , r 0 MMF s .......................................(15) g s r The air-gap flux density due to current in the as winding ias with all other currents zero may be obtained by substituting (9) and (12) in to (15) we get MMF s N Br s , r 0 0 s ias cos s 1 2 cos 2s r (16) g s r 2 Similarly, the flux density with all currents zero except ibs is Br s , r 0 MMF s N 2 0 s ibs coss 1 2 cos 2s r (17) g s r 2 3 Similarly, the flux density with all currents zero except ics is Br s , r 0 MMF s N 2 0 s ics coss 1 2 cos 2s r (18) g s r 2 3 In the case of salient-pole synchronous machines the field winding is generally uniformly distributed and the poles are shaped to approximate a sinusoidal distribution of air-gap flux due to current flowing in the field winding. In the case of round-rotor synchronous machines the field winding is arranged to approximate more closely a sinusoidal distribution. We will assume that the damper windings (additional rotor windings) may also be approximated by sinusoidally distributed windings. Thus, the air-gap MMF due to current i fd flowing in the fd winding may be expressed as, MMF fd Nf 2 i fd sin r .....................(19) Hence the air-gap flux density due to i fd with all other currents zero may be expressed as Br r 0 Nf MMF s 0 i fd sin r 1 2 cos 2r (20) g s r 2 Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET Let us consider the flux linkages of a single turn of a stator winding which spans radians and which is located at an angle s . In this case the flux is determined by performing a surface integral over the open surface of the single turn. In particular s , r s B , rl d ..................(21) r r s Where is the flux linking a single turn oriented s from the as axis l is the axial length of the air gap of the machine r is the radius to the mean of the air gap is the dummy vari able of integratio n Computation of the flux linkages of an entire winding involves a double integral. As an example, let us determine the total flux linkages of the as winding due to current flowing only in the as winding. Here as Llsias N as s s , r d s L lsias N as s Br , r rl .d .ds ......(22) s Where Lls is the stator leakage inductance due primarily to leakage flux at the end turns. Substituting (7ii) in (22) instead (with Np replaced by Ns ) and (16) for Br , r in 2 the above equation we get 2 N as Llsias s sin s 2 s s 0 Ns ias cos 1 2 cos 2 r .rl .d .ds ........(22) 2 N Llsias s 0 rl 1 2 cos 2 r ias ....................................................(23) 2 2 2 We know that Lasas as ias therefore dividing (23) by ias we get, N L asas Lls s 0 rl 1 2 cos 2 r ....................................................(24) 2 2 2 Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET The mutual inductance between the as and bs windings may be determined by first computing the flux linking the as winding due to current flowing only in the bs winding. Thus s as N as s Br , r .rl.d .ds .......(25) s Substituti ng (7) and (17) in the above expression we get 2 N as s sin s 2 s 0 s Ns 2 ibs cos 1 2 cos 2 r .rl .d .ds .......(26) 2 3 Therefore, the mutual inductance between the as and bs windings is obtained by evaluating (26) and dividing it by ibs we get, N Lasbs s 0 rl 1 2 cos 2 r ........(27) 3 2 2 2 Therefore, the mutual inductance between the as and bs windings is obtained by evaluating (20) expressed in terms of s r in to (25) Thus as 2 N 2s sin s s s 0 Nf 2 i fd sin r 1 2 cos 2 r .rl .d .ds ..................(28) Evaluating and dividing by i fd yeilds N N Lasfd s f 0 rl 1 2 sin r .....................................................................(29) 2 2 2 The self-inductance of the field winding may be obtained by first evaluating the flux linking the fd winding with all currents equal to zero except ifd. Thus, with the fd winding considered as sinusoidally distributed and the air-gap flux density expressed by (20) Thus 3 2 fd Llfd i fd 2 Nf 2 cos r r r 0 Nf 2 i fd sin 1 2 cos 2 .rl .d .dr .............(30) From which 2 Nf 0 rl 1 2 ....................................................................(31) L fdfd Llfd 2 2 Where L lfd is the leakage inductane of the field winding Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET The remaining self- and mutual inductances may be calculated using the same procedure. We can express these inductances compactly by defining Let 2 N LA s 0 rl1............................(32) 2 N LB s 0 rl 2 ...........................(33) 2 2 N N f 0 rl 1 2 .........(34) Lsfd s 2 2 2 2 Lmfd Nf 2 2 0 rl 1 2 ..............(35) 2 The machine inductances may now be expressed as Lasas Lls LA LB cos 2 r ............................(36) 2 Lbsbs Lls LA LB cos 2 r .................(37) 3 2 Lcsc s Lls LA LB cos 2 r .................(38) 3 L fdfd Llfd Lmfd .............................................(39) 1 Lasbs LA LB cos 2 r .....................(40) 2 3 1 Lascs LA LB cos 2 r .....................(41) 2 3 1 Lbscs LA LB cos 2 r .....................(42) 2 Lasfd Lsfd sin r .............................................(43) 2 Lbsfd Lsfd sin r 3 2 Lcsfd Lsfd sin r 3 .................................(44) .................................(45) The voltage equations for the elementary synchronous machine is given by Lecture Notes M.Kaliamoorthy A.P/EEE, PSNACET das ..................................(46) dt d Vbs rs ias bs ..................................(47) dt d Vcs rs ias cs ..................................(48) dt d fd V fd rfd i fd ...............................(49) dt Vas rs ias Where rs is the resistance of the stator winding and rf is the resistance of the field winding. The flux linkages are expressed as as Lasasias Lasbsibs Lascsics Lasfdi fd ........(50) bs Lbsasias Lbsbsibs Lbscsics Lbsfdi fd ........(51) cs Lcsasias Lcsbsibs Lcsc s ics Lcsfd i fd ........(52) ds Ldsasias Ldsbsibs Ldscsics Ldsfdi fd .......(53)