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. 54 Appendix A. Machine Data Output Power 7.5 KW Rated Voltage 340 V Winding Connection Delta Number of Poles 4 Rated Speed 1400 rpm Rated Frequency 50 Hz Stator resistance 2.52195 Ohm Stator Leakage Reactance 1.95145 Ohm Rotor Resistance 0.976292 Ohm Rotor Leakage Reactance 2.99451 Ohm Magnetizing Reactance 55.3431 Ohm Mechanical Shaft Torque 51.2636 N · m Estimated Rotor Inertia Moment 0.117393 Kgm2 Rated Current 19.2 A Moment of Inertia of the DC Motor 0.10958 Kgm2 Shaft Stiffness Constant 14320 Nm/rad Rotor Bar Shape Rectangular Number of Stator Slots 36 Outer Diameter of Stator 200 mm Inner Diameter of Stator 125 mm Slot Insulation Thickness 0.3 mm Air Gap 28 Inner Diameter of Rotor 30 mm Height of End Ring 13.2 mm Width of End Ring 4.4 mm Iron Core Length 170 mm Rotor Bar Length 239 mm Type of Rotor Cage Steel (cast copper) Appendix B. Estimated Rotor Circuit Parameter Resistance [mΩ] Inductance [µH] R1 L1 6.1150e−2 R2 0.656 L2 9.2940e−2 R3 0.321 L3 0.1896 R4 0.179 L4 0.3562 R5 1.338 L5 0.2596 Appendix C: Optimization Algorithm Illustration In order to realize an optimal height for each bar section as well as the optimal model impedance that gives a close correlation with the actual rotor impedance of the test machine, an algorithm that accomplishes such optimal division is developed. The total height of the bar is assumed to be a geometrical sum of the individual height of the section given by: hk = xi(di)k−1 (A1) where hk = height of each section, xi = fraction of the depth of the bar, k = number of the section whose depth is being computed, 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 optimization of the model, let us consider the rectangular bar shown in Fig. A1 below. Figure A1. Rectangular bar showing three unequal sections. Let us assume that h = 10 cm and user optimization index, di = 3.7. From (A1): h1 = xi(3.7)0 for k = 1 h1 = xi (A2) h2 = xi(3.7)1 for k = 2 (A3) h3 = xi(3.7)2 for 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 cm Note that if the UOI, di = 1.0, then: h1 = h2 = h3 = 3.333 cm 55

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