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182 Chapter 7 Analysis and Simulation of Ipm Generator 182 CHAPTER 7 ANALYSIS AND SIMULATION OF IPM GENERATOR DRIVING AN INDUCTION MACHINE 7.1 Introduction In this chapter, the analysis and simulation results of the IPM generator driving a three phases, one horsepower induction machine will be presented. A schematic of the topology can be seen in Figure 7.1. Shunt capacitors placed at the terminals of the IPM are used to boost the terminal voltage and to supply reactive power. For the two machines used in the experiment, the use of the capacitors was not optional because it was found that the induction machine would not even start without the presence of the capacitors. In the first part of the chapter, the dynamic and steady state equations used to model the induction machine will be presented. Next, the value of the parameters of the induction machine (and the methods used to determine the parameters) will be given. The comparison between the steady state experiment and calculated results will then be compared. Finally, simulated waveforms will be presented which show how the system behaves for three different types of load. 183 7.2 Derivation of Dynamic and Steady State Equations of a Squirrel Cage Induction Machine The d q stator (after the stator resistance and core loss (see Figure 7.2)) and rotor voltage equations of the induction machine for the voltage can be written as V qss = ω λ ds + p λ qs V dss = - ω λ qs + p λ ds (7.1) V qr = rr I qr + (ω - ωr )λ dr + p λ qr V dr = rr I dr - (ω - ωr )λ qr + p λ dr , where λ qs = Llsm I qsm + Lmm ( I qsm + I qr ) λ ds = Llsm I dsm + Lmm ( I dsm + I dr ) (7.2) λ qr = Llr I qr + Lmm ( I qsm + I qr ) λ dr = Llr I dr + Lmm ( I dsm + I dr ) . It is advantageous to have the equations entirely in terms of flux linkage and thereby eliminate the current terms. Letting Lss = Llsm + Lmm (7.3) Lrr = Llr + Lmm , 184 ωr Load Figure 7.1 Schematic diagram of IPM machine feeding an induction machine Figure 7.2. DQ representation of induction machine with capacitive compensation 185 the flux linkage equations can be written as ⎡λ qs⎤ ⎡ Lss 0 Lmm 0⎤ ⎡I qsm⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢λ ds⎥ ⎢ 0 Lss Lmm 0⎥ ⎢I dsm⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ . (7.4) ⎢λ qr⎥ ⎢Lmm 0 Lrr 0⎥ ⎢ I qr⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢λ dr⎦⎥ ⎣ 0 Lmm 0 Lrr⎦ ⎣⎢ I dr⎦⎥ Solving for the currents in terms of the flux linkages gives ⎡I qsm⎤ ⎡ Lrr 0 - Lmm 0⎤ ⎡λ qs⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢I dsm⎥ 1 ⎢ 0 Lrr 0 - Lmm⎥ ⎢λ ds⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ , (7.5) D ⎢ I qr⎥ ⎢- Lmm 0 Lss 0⎥ ⎢λ qr⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ I dr⎦⎥ ⎣ 0 - Lmm 0 Lss⎦ ⎣⎢λ dr⎦⎥ where 2 D = Lss Lrr - Lmm . (7.6) Substituting the values of currents in terms of flux linkages into the voltage equations gives ⎡ p ω 0 0⎤ ⎢ ⎥ ⎡V qss⎤ ⎢ -ω p 0 0⎥ ⎡λ qs⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢V dss⎥ ⎢ ⎥ ⎢λ ds⎥ - r mm 0 r ss (ω - ωr ) ⎢ ⎥ = ⎢ r L r L + p ⎥ ⎢ ⎥ . (7.7) ⎢ V qr⎥ ⎢ D D ⎥ ⎢λ qr⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ V dr⎦⎥ ⎢ ⎥ ⎣⎢λ dr⎦⎥ 0 - r mm -(ω - ωr ) r ss ⎢ r L r L + p⎥ ⎣ D D ⎦ 186 It is also worth noting that the voltages Vqr and Vdr are equal to zero for the squirrel cage rotor used in this thesis (since the rotor windings are shorted together and can not support a voltage). Now, the q axis terminal voltage of the stator may be found by recognizing that V qs - V qss = I qsmm , (7.8) r sm and V qss I qsmm - = I qsm . (7.9) Rc Substituting Equation (7.8) into (7.9) and solving for Vqss gives Rc r sm V qs V qss = ( - I qsm ) . (7.10) Rc + r sm r sm Substituting the value for Iqsm found in Equation (7.5) into (7.10) yields Rc r sm V qs 1 V qss = ( - ( Lrr λ qs - Lmm λ qr )) . (7.11) Rc + r sm r sm D Equating Equation (7.11) with the solution for Vqss found in (7.7) and solving for Vqs gives r sm Lrr Rc + r sm Rc + r sm r sm Lmm V qs = λ qs + p λ qs ( ) + ( ) ω λ ds - λ qr . (7.12) D Rc Rc D 187 The d axis terminal voltage of the stator may also be found by recognizing that V ds - V dss = I dsmm , (7.13) r sm and V dss I dsmm - = I dsm . (7.14) Rc Substituting (7.11) into (7.14) and solving for Vdss gives Rc r sm V ds V dss = ( - I dsm ) . (7.15) Rc + r sm r sm Substituting the value for Idsm found in Equation (7.5) into (7.15) yields Rc r sm V ds 1 V dss = ( - ( Lrr λ ds - Lmm λ dr )) . Rc + r sm r sm D Equating (7.16) with the solution for Vdss found in (7.5) and solving for Vds gives Rc + r sm r sm Rc + r sm r sm V ds = - ( ) ω λ qs + Lrr λ ds + ( ) p λ ds - Lmm λ dr , (7.16) Rc D Rc D and the state equations may be written as 188 ⎡rsm Lrr Rc + rsm Rc + rsm rsm 0⎤ + p ( ) ω ( ) -Lmm ⎢ D D ⎥ ⎢ Rc Rc ⎥ ⎢ ⎥ ⎡Vqs⎤ ⎡λqs⎤ ⎢ Rc + rsm rsm Lrr Rc + rsm 0 rsm Lmm ⎥ ⎢ ⎥ -ω ( ) +p( ) - λdr ⎢ ⎥ ⎢ D D ⎥ ⎢Vds⎥ ⎢ Rc Rc ⎥ ⎢λds⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ . (7.18) ⎢Vqr⎥ ⎢λqr⎥ ⎢ -rr Lmm 0 rr Lss (ω - ωr )⎥ ⎢ ⎥ ⎢ +p ⎥ ⎢ ⎥ D D ⎣⎢Vdr⎦⎥ ⎢ ⎥ ⎣⎢λdr⎦⎥ ⎢ ⎥ ⎢ 0 -rr Lmm -(ω - ωr ) rr Lss ⎥ +p ⎣⎢ D D ⎦⎥ The equations for the shunt capacitors in the dq plane are given as (- I qs - I qsmm ) - ω C V ds = p C V qs (7.19) (- I ds - I dsmm ) + ω C V qs = p C V ds . At steady state, the rate of change of the states is zero, and Equations (7.18) and (7.19) may be reduced to r sm Lrr Rc +r s r sm Lmm V qs = λ qs + ω( ) λ ds - λ qr D Rc D Rc +r s r sm Lrr r sm Lmm V ds = -ω ( ) λ qs + λ ds - λ dr Rc D D - rr Lmm rr Lss 0 = λ qs + λ qr + (ω - ωr ) λ dr D D (7.20) - rr Lmm rr Lss 0 = λ ds - (ω - ωr ) λ qr + λ dr D D V qs 0 = (- I qs - I qsm - ) - ω C V ds Rc V ds 0 = (- I ds - I dsm - ) + ω C V qs . Rc 189 The equations given in Equation (7.20), along with the two voltage equations of the IPM machine, i.e. V qs = I qs r s + λ ds ω (7.21) V ds = I ds r s - λ qs ω , are the equations necessary to analyze the IPM feeding the induction motor (IPM-IM) when shunt capacitive compensation is present. Due to the increased number of states created by the induction machine, no attempt at finding a closed form solution for the IPM-IM scheme was undertaken. Matlab’s Fsolve function files were used to solve the system numerically. The empirical relationships between the q and d axis inductances and the magnet flux as functions of the stator current of the IPM machine ( given in Chapter 2) must also be included. In addition, as will be shown in the next section, equations for the mutual inductance and the core loss of the induction machine will also need to be included into the set of nonlinear equations to be solved numerically. 190 7.3 Induction Motor Parameter Determination The determination of the parameters of the induction motor involves three different tests. They are the dc test, the blocked rotor test, and the no load test. The per phase, equivalent circuit of the induction motor is shown in Figure 7.3(a), where X1 =ω Llsm , X2 =ω Llr , and Xm =ω Lm. The dc test involves applying a dc voltage across any two of the three terminals of the induction machine. The voltage applied divided by the current flowing through the two windings gives the sum of the stator resistance for each winding. Therefore, on a per phase basis 1 V dc r sm = . (7.22) 2 I dc The short circuit test (also known as the blocked rotor test) involves measuring the power, voltage and current flowing into the induction machine when the rotor is prevented from spinning. For this case, the slip is equal to one and the equivalent circuit can be drawn as shown in Figure 7.3(b). During stall, the impedance rr + jX2 is much smaller than the core loss resistance and, therefore, Rc can be neglected and the equivalent circuit can be redrawn as shown in Figure 7.3(c). The equivalent short circuit impedance can be written as ( rr + j X 2 )(j X m ) Z sc = + r sm + j X 1 , (7.23) ( r2 + j ( X 2 + X m ) 191 rsm X I1 X1 2 rr V1 R Xm c s (a) rsm X Isc X1 2 Vsc Xm Rc rr (b) rsm X Isc X1 2 Vsc Xm rr (c) (d) Figure 7.3. Schematic diagram of (a) per phase equivalent circuit, (b) short circuit equivalent circuit, (c) simplified short circuit equivalent circuit, (d) no load equivalent circuit for induction machine 192 which reduces to 2 - rr X 2 X m + rr X m ( X 2 + X m ) X 2 X m ( X 2 + X m ) + rr X m = + + j ( + ) . (7.24) Z sc 2 2 r sm X 1 2 2 r r + ( X 2 + X m ) rr + ( X 2 + X m ) The real and imaginary parts may be written separately as 2 rr X m = + Rsc r sm 2 2 rr + ( X 2 + X m ) (7.25) 2 2 2 rr X m + X 2 X m + X 2 X m = + .
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