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Chapter 3 Model of a Three-Phase Induction Motor

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Chapter 3 Model of a Three-Phase Induction Motor CHAPTER 3 MODEL OF A THREE-PHASE INDUCTION MOTOR 3.1 Introduction The induction machine is used in wide variety of applications as a means of converting electric power to mechanical power. Pump steel mill, hoist drives, household applications are few applications of induction machines. Induction motors are most commonly used as they offer better performance than other ac motors. In this chapter, the development of the model of a three-phase induction motor is examined starting with how the induction motor operates. The derivation of the dynamic equations, describing the motor is explained. The transformation theory, which simplifies the analysis of the induction motor, is discussed. The steady state equations for the induction motor are obtained. The basic principles of the operation of a three phase inverter is explained, following which the operation of a three phase inverter feeding a induction machine is explained with some simulation results. 3.2. Basic Principle Of Operation Of Three-Phase Induction Machine The operating principle of the induction motor can be briefly explained as, when balanced three phase voltages displaced in time from each other by angular intervals of 120o is applied to a stator having three phase windings displaced in space by 120o electrical, a rotating magnetic field is produced. This rotating magnetic field 45 has a uniform strength and rotates at the supply frequency, the rotor that was assumed to be standstill until then, has electromagnetic forces induced in it. As the rotor windings are short circuited, currents start circulating in them, producing a reaction. As known from Lenz’s law, the reaction is to counter the source of the rotor currents. These currents would become zero when the rotor starts rotating in the same direction as that of the rotating magnetic field, and with the same strength. Thus the rotor starts rotating trying to catch up with the rotating magnetic field. When the differential speed between these two become zero then the rotor currents will be zero, there will be no emf resulting in zero torque production. Depending on the shaft load the rotor will always settle at a speed ωr , which is less than the supply frequency ωe . This differential speed is called the slip speed ωso . The relation between, ωe and ωso is given as [13] ωso = ωe - ωr (3.1) If ωm is the mechanical rotor speed then P ω = ω . (3.2) r 2 m 3.3 Derivation Of Three-Phase Induction Machine Equations The winding arrangement of a two-pole, three-phase wye-connected induction machine is shown in Figure 3.1. The stator windings of which are identical, sinusoidally distributed in space with a phase displacement of 120o , with N s equivalent turns and resistance rs . 46 bs axis as' br axis ω ar' bs r cs br ar axis cr θr ωe ' as axis br' cr bs' ar cs' as cs axis cr axis Figure 3.1. Two-pole three-phase symmetrical induction machine. The rotor is assumed to symmetrical with three phase windings displaced in space by o an angle of 120 , with Nr effective turns and a resistance of rr . The voltage equations for the stator and the rotor are as given in Equations 3.3 to 3.8. For the stator: Vas = rs I as + pλas (3.3) Vbs = rs Ibs + pλbs (3.4) Vcs = rs I cs + pλcs (3.5) 47 where Vas, Vbs , and Vcs are the three phase balanced voltages which rotate at the supply frequency.For the rotor the flux linkages rotate at the speed of the rotor, which is ωr : Var = rr I ar + pλar (3.6) Vbr = rr Ibr + pλbr (3.7) Vcr = rr Icr + pλcr . (3.8) The above equations can be written in short as Vabcs =rsIabcs + pλabcs (3.9) Vabcr =rr Iabcr + pλabcr (3.10) where T (Vabcs ) = [Vas Vbs Vcs ] (3.11) T (Vabcr ) = [Var Vbr Vcr ] . (3.12) In the above two equations ‘s’ subscript denoted variables and parameters associated with the stator circuits and the subscript ‘r’ denotes variables and parameters associated with the rotor circuits. Both rs and rr are diagonal matrices each with equal nonzero elements. For a magnetically linear system, the flux linkages may be expressed as λabcs Ls Lsr iabcs = T (3.13) λabcr (Lsr ) Lr iabcr The winding inductances can be derived [16] and in particular 48 1 1 L + L − L − L ls m 2 m 2 m 1 1 Ls = − Lm Lls + Lm − Lm (3.14) 2 2 1 1 − Lm − Lm Lls + Lm 2 2 1 1 L + L − L − L lr m 2 m 2 m 1 1 Lr = − Lm Llr + Lm − Lm (3.15) 2 2 1 1 − Lm − Lm Llr + Lm 2 2 2π 2π cosθ cos(θ + ) cos(θ − ) r r 3 r 3 2π 2π Lsr = Lsr cos(θ r − ) cosθ r cos(θ r + ) . (3.16) 3 3 2π 2π cos(θ r + ) cos(θ r − ) cosθ r 3 3 In the above inductance equations, Lls and Lm are the leakage and magnetizing inductances of the stator windings; Llr and Lm are for the rotor windings. The inductance Lsr is the amplitude of the mutual inductances between stator and rotor windings. From the above inductance equations, it can be observed that the machine inductances are functions of the rotor speed, whereupon the coefficients of the differential equations which describe the behavior of these machines are time varying except when the rotor is at standstill. A change of variables is often used to reduce the complexity of these differential equations, which gives rise to the reference frame theory [16]. For the induction machine under balanced operating conditions the 49 synchronous reference frame of transformation is employed in which the reference frame rotates with the same frequency as that of the supply frequency ωe . The transformation matrix used for the synchronous reference frame transformation is 2π 2π cosθ cos(θ − ) cos(θ + ) s s 3 s 3 2 2π 2π K s (θ s )= sinθ s sin(θ s − ) sin(θ s + ) (3.17) 3 3 3 1 1 1 2 2 2 where θ = ω t +θ , θ being the initial angle of the reference frame. The inverse s ∫ s so so transformation for the above transformation is given as cosθ s sinθ s 1 −1 2π 2π (K s (θ s )) = sin(θ s − ) cos(θ s − ) 1 . (3.18) 3 3 2π 2π cos(θ + ) sin(θ + ) 1 s 3 s 3 The new variables after the transformation are related to the original variables as f qdo =(K s (θ s )) f abc (3.19) where f can be voltage current flux or any thing. For the equations in 3.9 and 3.10 after transforming them in to the synchronous reference frame these equations become Vqs =r s iqs + pλqs +ωeλds (3.20) Vds =r s ids + pλds −ωeλqs (3.21) Vos =r s ios + pλos (3.22) Vqr =r r iqr + pλqr + (ωe −ωr )λdr (3.23) 50 Vdr =r r idr + pλdr − (ωe −ωr )λqr (3.24) Vor =r s ior + pλor (3.25) where Vqs , Iqs ,λqs are the q-axis components, Vds ,Ids ,λds are the d-axis components, and Vos ,Ios ,λos belong to the 0- axis and usually represent the unbalances in the system. In case of balanced voltages the zero-axis currents, voltages and flux are zero under normal operating conditions. The flux linkage equations expressed in abc variables given in Equation 3.13 yields the flux linkage equations for a magnetically linear system −1 −1 λqd 0s K L (K ) K L (K )iqd 0s = s s s s sr s (3.26) λ −1 −1 i qd 0r K s Lsr (K s ) K s Lr (K s ) qd 0r where Ls , Lr ,and Lsr are as defined in Equations 3.14 to 3.16. Evaluating each term in the matrix of Equation 3.26 we get Lls + Lm 0 0 K L (K −1 ) = 0 L + L 0 (3.27) s s s ls m 0 0 Lls Llr + Lm 0 0 K L (K −1 ) = 0 L + L 0 (3.28) s r s lr m 0 0 Llr Lm 0 0 K L (K −1 ) = 0 L 0 (3.27) s sr s m 0 0 0 From above the flux linkages the expressions are modified as λqs = Lsiqs + Lmiqr (3.28) λds = Lsids + Lmidr (3.29) 51 λ0s = Llsi0s (3.28) λqr = Lr iqr + Lmiqs (3.30) λdr = Lr idr + Lmids (3.31) λ0r = Llr i0r (3.32) The expression for the electromagnetic torque in terms of the reference frame variables can be expressed as P −1 ∂ −1 Te =( )[(Ks ) iqdos ]T [Lsr ](Ks ) iqdor (3.33) 2 ∂θr The torque expression in Equation 3.33 can be expressed in terms of currents as 3 P T =( )( )L (i i −i i ) (3.34) e 2 2 m qs dr ds qr where Te is positive for motor action. Alternative expressions for the torque can be expressed in terms of flux linkages are given in Equations 3.35 and 3.36. 3 P T =( )( )(λ i −λ i ) (3.35) e 2 2 qr dr dr qr 3 P T =( )( )(λ i −λ i ) (3.36) e 2 2 ds qs qs ds The expression for the rotor speed is expressed in terms of torque as P pω = (T −T ) (3.37) r 2J e L Where Te ,TL are the electromagnetic and the load torque, respectively, depending on the load torque the motor settles at a speed, which is always less than the angular frequency with which it gets excited.
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