DYNAMIC MODELS AND SIMULATION OF INDUCTION MACHINE WITH SKIN-EFFECT

O.I. Okoro

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  15. [15] I.R. Smith & S. Sriharan, Transient performance of the induction motor, Proc. IEE, 113 (7), 1966, 1173–1181.54Appendix A. Machine DataOutput Power 7.5 KWRated Voltage 340 VWinding Connection DeltaNumber of Poles 4Rated Speed 1400 rpmRated Frequency 50 HzStator resistance 2.52195 OhmStator Leakage Reactance 1.95145 OhmRotor Resistance 0.976292 OhmRotor Leakage Reactance 2.99451 OhmMagnetizing Reactance 55.3431 OhmMechanical Shaft Torque 51.2636 N · mEstimated Rotor Inertia Moment 0.117393 Kgm2Rated Current 19.2 AMoment of Inertia of the DC Motor 0.10958 Kgm2Shaft Stiffness Constant 14320 Nm/radRotor Bar Shape RectangularNumber of Stator Slots 36Outer Diameter of Stator 200 mmInner Diameter of Stator 125 mmSlot Insulation Thickness 0.3 mmAir Gap 28Inner Diameter of Rotor 30 mmHeight of End Ring 13.2 mmWidth of End Ring 4.4 mmIron Core Length 170 mmRotor Bar Length 239 mmType of Rotor Cage Steel (cast copper)Appendix B. Estimated Rotor Circuit ParameterResistance [mΩ] Inductance [µH]R1 L1 6.1150e−2R2 0.656 L2 9.2940e−2R3 0.321 L3 0.1896R4 0.179 L4 0.3562R5 1.338 L5 0.2596Appendix C: Optimization Algorithm IllustrationIn order to realize an optimal height for each bar sectionas well as the optimal model impedance that gives a closecorrelation with the actual rotor impedance of the testmachine, an algorithm that accomplishes such optimaldivision is developed. The total height of the bar isassumed to be a geometrical sum of the individual heightof the section given by:hk = xi(di)k−1(A1)where hk = height of each section, xi = fraction of the depthof the bar, k = number of the section whose depth is beingcomputed, and di = is the user optimization index (UOI),which is equal to or greater than one (di ≥ 1).To illustrate the algorithm used for the optimizationof the model, let us consider the rectangular bar shown inFig. A1 below.Figure A1. Rectangular bar showing three unequal sections.Let us assume that h = 10 cm and user optimizationindex, di = 3.7. From (A1):h1 = xi(3.7)0for k = 1h1 = xi (A2)h2 = xi(3.7)1for k = 2 (A3)h3 = xi(3.7)2for k = 3 (A4)But:h = h1 + h2 + h3 (A5)Solving (A2–A5), we have:xi = 0.5438, h1 = 0.5438 cm, h2 = 2.0120 cm,h3 = 7.444 cmNote that if the UOI, di = 1.0, then:h1 = h2 = h3 = 3.333 cm55

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