Lecture Notes M.Kaliamoorthy AP/EEE, PSNACET 3-Phase Induction Machines - Dynamic Modeling Using Reference Frame Theory Winding arrangement for a 2-pole, 3-phase, wye-connected symmetrical induction machine is shown in Fig.1. Stator windings are identical, sinusoidally distributed windings, displaced by120°, with Ns equivalent turns and resistance rs. Consider the case when rotor windings are also three identical sinusoidally distributed windings, displaced by120°, with Nr equivalent turns and resistance rr. Fig1. 2 pole Three Phase Induction Machine In abc reference frame, voltage equations can be written as Vabcs = rs iabcs + pλabcs Vabcr = rr iabcr + pλabcr ( f abcs )T = [ f as f bs f cs ], ( f abcr )T = [ f ar f br f cr ] s: denotes variables and parameters associated with the stator circuits and r: denotes variables and parameters associated with the rotor circuits. Lecture Notes M.Kaliamoorthy AP/EEE, PSNACET λabcs Ls λ = ( L ) T abcr sr Lsr iabcs Lr iabcr Where, Lls + Lms 1 L s = − Lms 2 −1L 2 ms 1 − Lms 2 Lls + Lms 1 − Lms 2 1 − Lms 2 1 − Lms , 2 Lls + Lms Llr + Lmr 1 L r = − Lmr 2 −1L 2 mr 1 − Lmr 2 Llr + Lmr 1 − Lmr 2 Llr + Lmr 1 − Lmr 2 1 − Lmr 2 Lls and Lms are, respectively, the leakage and magnetizing inductance of the stator windings. Llr and Lmr are, respectively, the leakage and magnetizing inductance of the rotor windings. 2π 2π cos(θ r + ) cos(θ r − ) cos θ r 3 3 2π 2π cos θ r cos(θ r + ) , L sr = L rs = Lms cos(θ r − ) 3 3 cos(θ + 2π ) cos(θ − 2π ) cos θ r r r 3 3 “Lsr” is the amplitude of the mutual inductances between stator and rotor windings. A majority of induction machines are not equipped with coil-wound rotor windings; instead, the current flows in copper or aluminum bars which are uniformly distributed in a common ring at each end of the rotor. This type of rotor is referred to as a squirrel-cage rotor. Rotor variables can be referred to the stator windings by appropriate turn’s ratio. 2 ′ iabcr Also, N N N N ′ = s Vabcr , λabcr ′ = s λabcr , Lms = s Lsr = r iabcr , Vabcr Ns Nr Nr Nr 2π 2π cos(θ r + ) cos(θ r − ) cos θ r 3 3 2π cos θ r cos(θ r + ) , [L′sr ] = N s [L sr ] = Lms cos(θ r − 2π ) 3 3 Nr cos(θ + 2π ) cos(θ − 2π ) cos θ r r r 3 3 2 N Lmr = r Lms , Ns 2 [L′r ] = N r [L r ] Ns Lecture Notes M.Kaliamoorthy AP/EEE, PSNACET Llr′ + Lmr [L′r ] = − 1 Lms 2 −1L 2 ms − 1 Lms 2 Llr + Lms − 1 Lms 2 Llr + Lms 1 − Lms 2 1 − Lms 2 Where, N Llr′ = s Nr 2 Llr Flux linkage may be expressed as λabcs L s λ ′ = ( L′ ) T abcr sr L′sr iabcs ′ L′r iabcr Voltage equations expressed in terms of machine variables referred to the stator windings may be written as Vabcs rs + pL s V ′ = p (L′ )T sr abcr pL′sr iabcs ′ rr′ + pL′r iabcr Where, N rr′ = s Nr 2 rr Energy stored in the coupling field may be written as 1 (iabcs )T (L s − L ls I )iabcs + 2 1 ′ + (iabcr ′ )T (L′r − L′lr I )iabcr ′ (iabcs )T (L sr )iabcr 2 Wc = W f = Where, I: identity matrix Voltage equations expressed in terms of machine variables referred to the stator windings may be written as P ∂Wc (i j , θ r ) Te (i j , θ r ) = 2 ∂θ r Since Ls and Lr are functions of θr, the above equation for the electromagnetic torque yields. Lecture Notes M.Kaliamoorthy AP/EEE, PSNACET p ∂ ′ Te = ( )(iabcs )T [L′sr ]iabcr 2 ∂θ r 1 1 1 1 1 1 ′ − ibr ′ − icr ′ ) + ibs (ibr ′ − iar ′ − icr ′ ) + ics (icr ′ − ibr ′ − iar ′ ) sin θ r ias (iar 2 2 2 2 2 2 P = − Lms 2 3 ′ ′ ′ ′ ′ ′ + 2 [ias (ibr − icr ) + ibs (icr − iar ) + ics (iar − ibr )]cos θ r The torque and rotor speed are related by 2 Te = J pωr + TL P Equations of Transformation for Rotor Circuit In the analysis of induction machines it is desirable to transform the variables associated with the symmetrical rotor windings to the arbitrary reference frame. cos β 2 ( f qd′ 0 r )T = f qr′ f dr′ f 0′r K r = sin β 3 ′ )T = [ f ar′ f br′ f cr′ ] ( f abcr 1 2 where, β=θ-θr from figure below ′ f qd′ 0 r = K r f abcr [ ] 2π 2π ) cos( β + ) 3 3 2π 2π sin( β − ) sin( β + ), 3 3 1 1 2 2 cos( β − Fig. 2. Axis of 2-pole, 3-phase Symmetrical machine. t θ r = ∫ ωr (t )dt +θ r (0) 0 Lecture Notes M.Kaliamoorthy AP/EEE, PSNACET cos β sin β 1 (K r )−1 = cos(β − 2π ) sin( β − 2π ) 1 3 3 2 2 π π cos( β + ) sin( β + ) 1 3 3 “r” subscript indicates the variable, parameters and transformation associated with rotating circuits. Voltage Equations in Arbitrary Reference Frame Variables For two-pole, 3-phase symmetrical induction, Vabcs = rs iabcs + pλabcs ′ = rr′iabcr ′ + pλabcr ′ Vabcr ′ λabcs = (Ls )iabcs + (Lsr′ )iabcr Vabcs = K sVqd 0 s , iabcs = K s iqd 0 s ′ = (Lsr′ ) iabcs + (Lr′ )iabcr ′ λabcr ′ = K rVqd′ 0 r , iabcr ′ = K r iqd′ 0 s Vabcr T Using the above transformation equations, we can transform the voltage equations to an arbitrary reference frame rotating at speed of ω. Vqd 0 s = rs iqd 0 s + ωλqds + pλqd 0 s ′ + pλqd ′ 0r Vqd′ 0 r = rr′iqd 0 r + (ω − ω r )λqdr [ where, (λqds )T = λds − λqs ] [ ′ )T = λdr′ 0 , (λqdr − λqr′ 0 ] Flux linkage equations in abc reference frame can be transformed to qd axes using Ks and Kr transformation matrices. λqd 0 s K s L s (K s )−1 K s L′sr (K r )−1 iqd 0 s = ′ 0 r ′ 0 r K r L′sr (K s )−1 K r L′r (K r )−1 iqd λqd Where K s L s (K s ) −1 K r L′r (K r ) −1 Lls + M = 0 0 Llr′ + M = 0 0 0 Lls + M 0 0 Llr′ + M 0 3 0 , M = Lms 2 Lls + M 0 3 0 , M = Lms 2 Llr′ + M 0 Lecture Notes M.Kaliamoorthy AP/EEE, PSNACET M 0 0 K s L′sr (K r ) = K r (L′sr ) (K s ) = 0 M 0 0 0 M Voltage equations written in expanded form can be expressed as −1 T −1 Vqs = rs iqs + ωλds + pλqs Vqr′ = rr′iqr′ + (ω − ωr )λdr′ + pλqr′ Vds = rs ids − ωλqs + pλds , Vdr′ = rr′idr′ − (ω − ωr )λqr′ + pλdr′ V0 s = rs i0 s + pλ0 s V0′r = rr′i0′ r + pλ0′r Flux linkage equations are λqs = Lls iqs + M (iqs + iqr′ ) λds = Lls ids + M (ids + idr′ ) λ0 s = Lls i0 s λqr′ = Llr′ iqr′ + M (iqs + iqr′ ) λdr′ = Llr′ idr′ + M (ids + idr′ ) λ0′r = Llr′ i0′ r Since machine and power system parameters are nearly always given in ohms or percent or per unit of a base impedance, it is convenient to express the voltage and flux linkage equations in terms of reactances rather than inductances. Let ϕ = λωb Then ω ϕ ds + pϕ qs ωb ω p Vds = rs ids − ϕ qs + ϕ ds , ωb ωb Vqs = rs iqs + V0 s = rs i0 s + p ωb ϕ0s (ω − ωr ) ϕ ′ p ϕ qr′ ωb ωb (ω − ωr ) ϕ ′ + p ϕ ′ Vdr′ = rr′idr′ − qr dr ωb ωb Vqr′ = rr′iqr′ + V0′r = rr′i0′ r + dr p ωb + ϕ0′ r And flux linkages become flux linkages per second with the units of volts. ϕ qs = X ls iqs + X m (iqs + iqr′ ) ϕ qr′ = X lr′ iqr′ + X m (iqs + iqr′ ) ϕ ds = X ls ids + X m (ids + idr′ ), ϕ 0 s = X ls i0 s ϕ dr′ = X lr′ idr′ + X m (ids + idr′ ) ϕ 0′ r = X lr′ i0′ r Lecture Notes M.Kaliamoorthy AP/EEE, PSNACET Equivalent circuits of a 3-phase, symmetrical induction machine with rotating q-d axis at speed of ω. Equivalent circuits of a 3-phase, symmetrical induction machine with rotating q-d axis at speed of ω. Equivalent circuits of a 3-phase, symmetrical induction machine with rotating q-d axis at speed of ω. Lecture Notes M.Kaliamoorthy AP/EEE, PSNACET Electromagnetic torque in terms of arbitrary reference frame variables may be obtained by substituting the equations of transformation in Te = P ′ (iabcs )T ∂ (Lsr′ )iabcr 2 ∂θ r [ ] T ∂ P (Lsr′ )(K r ) −1iqd′ 0r (K s ) −1 iqd 0 s 2 ∂θ r After some work, we will have the following: = 3 P ′ − ids iqr ′ ) Te = M (iqs idr 2 2 Where, Te is positive for motor action. Other expressions for the electromagnetic torque of an induction machine are 3 P Te = (λqr′ idr′ − λdr′ iqr′ ) 2 2 3 P Tem = (λds iqs − λqs ids ) 2 2 3 P 1 Te = (ϕ qr′ idr′ − ϕ dr′ iqr′ ) 2 2 ωb