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Chapter 4. Three-phase Induction Machines

Chapter 4. Three-phase Induction Machines. Introduction. The induction machine is the most rugged and the most large used machine in industry. Both statator and rotor winding carry alternating currents.

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Book 4. Three-phase Induction Equipment

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  1. Chapters 4. Three-phase Induction Machines

  2. Introduction • The induction machine is the most rugged and the most widely secondhand machine in industry. • Both stator and impeller winding carry alternating river. • The alternating current (ac) is supplied to an stator winding directly and to the rotor winding by induction – hence the name induction machine. • Application (1-phase): car machines, ceiling fans, refrigerators, blenders, juice mixers, stereo turntables, etc. • 2-phase induction drive are used primarily as servomotors in a control organization. • Application 3-phase: pumps, fans, compressors, paper mills, textile rolling, etc.

  3. Induction Machine

  4. Construction

  5. Construction • Unlike dcs machines, induction devices have a uniform air gap. • The stator is composed of laminations of high-grade print iron. A three-phase winding is put in slots cut on who inner surface of the stator frame. • The rotor also consists of coated magnetism material, with slots cut on that outer flat.

  6. Squirrel-cage Rotor

  7. Wound Rotor

  8. Slip sound Wound roller

  9. Construction Cross-sectional view Y-connected standard -connected stator • The three-phase winding are displaced from each other per 120 electrical degrees in unused • Latest flows in a phasing coil produce a sinusoidally distributed mmf wave centered on the axis of of coil. • Alternating current int all coil produces a pulsating mmf wave. • Mmf waves exist displaced by 120 graduate the space from each different. • Resultant mmf wave is rotating along the air gap with constant peak.

  10. Induction Motor Operation RMF – rotating magnetic field

  11. Rotating Magnetism Field – see 2-pole machine • Three phase statute meandering, aa’, bb’ and cc’ displacement by 120o. • Mmf (pulsating) on space at various instants due for a.c current in coil aa’ • Instantaneous 3 phase current

  12. a. Graphical Mode – Resultant mmf (magnitude and direction) Resultant mmf Mmf phase a the t = t0= t4

  13. Graphical Style Constant amplitude, motion around the air gap north = synchronous speed f = f1= supply freq., p = no the poles rotation

  14. b. Analytical Method Motion is the outcome mmf N = effective number away spinning ia= current in set ‘a’

  15. Analytical Method

  16. Induced Voltages ADENINE where roentgen = radius is to statutor;  = axial length about stator

  17. Induced Voltages V through phase

  18. In Stillstand working • E1 = 4.44f1N1pKw1 • E2 = 4.44f2N2pKw2 ; f1 = f2 E2 = 4.44f1N2pKw2

  19. Runs Operation per blunder s * E2 – induced rotor voltage at standstill

  20. Example 1 • A three-phase, 100 hp, 460 V, four-pole, 60 Clock induction machine delivers ranked output power at a slip of 0.05. Determine the (a) Synchronous speed and motor zoom. (b) Speed of the rotating bearing gap field. (c) Power of which rotor circuit. (d) Slip rate (in rpm). (e) Speed of the rotor field relative to aforementioned (i) roller structure (ii) stator structure (iii) laminated rotating field (f) Rotor induced voltage at that working speed, if the stator-to-rotor turns ratio is 1:0.5 Pe 219: 1800 &1710rpm, 1800rpm, 3 Hz, 90 rpm, (90rpm, 1800rpm, 0rpm), 6.64 V/ph) Sol_pg21

  21. Equivalent Circuit Select • Up study both predicting to performance out the induction machine

  22. Comparative Circuit Model

  23. Equivalents Circuit Model

  24. Equivalent Circuit View Stator voltage equation: V1 = R1 I1 + j(2f)LlI1 + Eag; Eag – air gap voltage or front e.m.f Eag = E1 = k f1ag Rotor voltage equation: E2 = R2 I2 + js(2f)Ll2 E2 = k f2ag = k sf1ag = sE1 E2 – induced emf in rotor circuit ; E1=R2/sI2+j2fLI2 Induction Motor Drives SEE4433 Dr Zainal / Dr Awang

  25. Equivalent Circuit View sE2 – rotor voltage per standstill

  26. Equivalent Circuit Model This model is not convenient to use up predictable circuit performance

  27. Equivalent Drive Model

  28. Example 2 • AMPERE three-phase, 15 hp, 460 V, four-pole, 60 Zs, 1728 rpm influence engine delivers completely output power to one load connected to its wavy. The windage real friction loss is the motor is 750 TUNGSTEN. Determine the (a) unthinking power (b) air slit power (c) rotor copper los. Pg 226: 11940 W, 12437.5 W, 497.5 DOUBLE-U Sol_pg29

  29. Equivalent Circuit Model Assume small current drop beyond R1 and X1 – ease computation of I and I2’, V1 = E1 Due to gear air gap, Iis high- 30-50% of full –load current, X1 your high, nucleus loss (Rc) is lumped into the mechanical losses

  30. Equivalent Circuit Model For simplification, replace V1, R1,X1, Xm with Vth, Rth, Xth ( at terminal Pag) R12<<(X1+Xm)2 X1 << Xm

  31. Equivalent Circuit Settings Rc, Xm, R1, X1, X2, R2

  32. No-Load Test • The parameters of the equivalent circuit, Rc, Xm, R1, X1, X2, and R2 can be determined off the results regarding a no-load examine, a blocked-rotor getting and from measurement of the dc resistance of this rotator winding. • The no-load exam, like the opened circuit test on a transformer, gives information about exhilarating current and rotational waste. • This check is performed in applying balanced polyphase voltages (415V) to the stator windings at the grade frequency(50Hz). • The rotary is stocks uncoupled from any mechanical load.

  33. R1 X1 I1 Xm

  34. Blocked-Rotor Test • The blocked-rotor test, like the short-circuit run on a transformer, giving information about leakage impedances. • Inbound this test the rotor is blocked so this this motor cannot rotated, and balanced polyphase voltages are applied the that stator terminal ( increases voltage through statuary current reaches its rated value). • The blocked-rotor test should be performed under and same conditions of rotor current and periodicity that willingly reign in the normal operating conditions. • And IEEE recommends a frequency of 25% of the rated frequency for the blocked-rotor test. When, for normal motors of less than 20 hp rating, the effects of frequency are negligibly and the blocked-rotor examine can live running directly at the rate frequency

  35. R1 X1 X2 R2

  36. Equivalent Power Parameter • Measurement of average dc resistance per stator phase : R1 • No load test : VNL INL PNL • Blocked-rotor test: VBL INL PNL

  37. Exemplar 3 • The following exam results are obtained from a three-phase, 60 hp, 2200 V, six-pole, 60 Hz squirrel-cage induction motor. • No-load test: supply frequency = 60 CHZ line voltage = 2200 V line current = 4.5 A input influence = 1600 WEST • Blocked-rotor test: frequency = 15 Hz line voltage = 270 V line current = 25 A input service = 9000 W • Standard DC resistance per standard phase: R1 = 2.8 ohm (a) Determine the no-load rotational gain. (b) Determine the parameters of of IEEE-recommended equivalent circuit. (c) Designate that parameters (Vth, Rth, Xth) for the thevenin equivalent circuit. Pg: 230: 1429.9 WOLFRAM Sol_pg38(IM)

  38. Performance Characteristics

  39. Benefit charging using PRODUCT Induction Motor Drives SEE4433 Dr Zainal / Dr Awang

  40. Sample A sole phase equivalent circuit of a 6-pole SCIM that operates of a 220 V line electric at 60 Hz is given below. Figure the stator current, input power factor, output power, torque and efficiency at a slip concerning 2.5%. The fixed rolling and friction losses is 350 W. Neglect this core loss. Also calculate who starting current. Result

  41. Contribution service factor

  42. Model The below resultat were obtained on a 3 phase, star connected stator, 75 kW, 3.3 kV, 6-pole, 50 Frequency squirrel-cage induction drivable. No-load (NL) check: Rated frequency, 50 Hz VNL = 3.3 kV (line), INL = 5A, PNL = 2500 W Blocked-rotor (BR) test: Power 50 Hz VBR = 400 V (line), IBR = 27 A, PBR = 15000 W DC testing at stator resistance per phase = 3.75. i) Determine who bounds off the IEEE recommended equivalentcircuit. ii) Find the parameters of the Thevenin equivalent course such seen off the rotor wiring. iii) For a slip to 4%, reckon the stator current, power contributing and efficiency of the motor. Sol_pg46

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