The frequency f 2 of the induced voltage and current

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Unformatted text preview: frequency f 2 of the induced voltage and current in the rotor circuit will correspond to this slip rpm, because this is the relative speed between the rotating field and the rotor winding. Thus, from Equation (4.9): f2 = p 120 (ns − n ) = p 120 sns = sf1 (4.11) Three-Phase Induction Machine 227 This rotor circuit frequency f 2 is also called slip frequency. The voltage induced in the rotor circuit at slip s is: E2 s = 4.44 f 2 N 2φ p KW 2 = 4.44 sf1 N 2φ p KW 2 = sE2 (4.12) Where E2 is the induced voltage in the rotor circuit at standstill, that is, at the stator frequency f1 . The induced currents in the three-phase rotor windings also produce a rotating field. Its speed (rpm) n2 with respect to rotor is: n2 = 120 f 2 120 sf1 = = sns p p (4.13) Because the rotor itself is rotating at n rpm, the induced rotor field rotates in the air gap at speed n + n2 = (1 - s)ns + sns = ns rpm. Therefore, both the stator field and the induced rotor field rotate in the air gap at the same synchronous speed ns. The stator magnetic field and the rotor magnetic field are therefore stationary with respect to each other. The interaction between these two fields can be considered to produce the torque. As the magnetic fields tend to align, the stator magnetic field can be visualized as dragging the rotor magnetic field. 228 Chapter Four Example 4.1 A 3-phase, 460 V, 100 hp, 60 Hz, four-pole induction machine delivers rated output power at a slip of 0.05. Determine the: (a) Synchronous speed and motor speed. (b) Speed of the rotating air gap field. (c) Frequency of the rotor circuit. (d) Slip rpm. (e) Speed of the rotor field relative to the (i) rotor structure. (ii) Stator structure. (iii) Stator rotating field. (f) Rotor induced voltage at the operating speed, if the stator-to-rotor turns ratio is 1 : 0.5. Solution: (a) n s = 120 f p = 120 * 60 = 1800 rpm , 4 n = (1 − s )ns = (1 − 0.05) * 1800 = 1710 rpm (b) 1800 rpm (same as synchronous speed) (c) f2 = sf1 = 0.05 x 60 = 3 Hz. (d) slip rpm = s ns = 0.05 * 1800 = 90 rpm (e) (i) 90 rpm (ii) 1800 rpm (iii) 0 rpm (f) Assume that the induced voltage in the stator winding is the same as the applied voltage. Now, E2 s = sE2 = s N2 460 E1 = 0.05 * 0.5 * = 6.64V / Phase N1 3 229 Three-Phase Induction Machine 4.6 Equivalent Circuit of the Induction Motor In this section we develop the equivalent circuit from basic principles. We then analyze the characteristics of a low-power and high-power motor and observe their basic differences. Finally, we develop the equivalent circuit of an asynchronous generator and determine its properties under load. A 3-phase wound-rotor induction motor is very similar in construction to a 3-phase transformer. Thus, the motor has 3 identical primary windings and 3 identical secondary windings one set for each phase. On account of the perfect symmetry, we can consider a single primary winding and a single secondary winding in analyzing the behavior of the motor. When the motor is at standstill, it acts exactly like a conventional transformer, and so its equivalent circuit (Fig.4.6) is the same as that of a transformer, previously developed. Fig.4.6 Equivalent circuit of a wound-rotor induction motor at standstill. 230 Chapter Four In the case of a conventional 3-phase transformer, we would be justified in removing the magnetizing branch composed of jX m and Rm because the exciting current I o is negligible compared to the load current I P . However, in a motor this is no longer true: I o may be as high as 40 percent of I P because of the air gap. Consequently, we cannot eliminate the magnetizing branch. The stator and rotor winding can be represented as shown in Fig.4.7 (a) and (b), where, V1 = per-phase terminal voltage. R1 = per-phase stator winding resistance. L1 = per-phase stator leakage inductance. E1 = per-phase induced voltage in the stator winding Lm = per-phase stator magnetizing inductance Rc = per-phase stator core loss resistance. E2 = Per-phase induced voltage in rotor at standstill (i.e., at stator frequency f,) R2 = Per-phase rotor circuit resistance L2 = Per-phase rotor leakage inductance Note that there is no difference in form between this equivalent circuit and that of the transformer primary winding. The difference Three-Phase Induction Machine 231 lies only in the magnitude of the parameters. For example, the excitation current I o is considerably larger in the induction machine because of the air gap. In induction machines it is as high a: 30 to 50 percent of the rated current, depending on the motor size, whereas it is only 1 to 5 percent in transformers. Moreover, the leakage reactance X 1 is larger because of the air gap and also because the stator and rotor windings are distributed along the periphery of the air gap rather than concentrated on a core, as in the transformer. 232 Chapter Four Fig.4.7 Development of the induction machine equivalent circuit Note that the rotor circuit frequency and current are: f 2 and I 2 . I2 = sE2 R2 + jsX 2 (4.14) T...
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This document was uploaded on 03/12/2014 for the course ENGINEERIN electrical at University of Manchester.

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