Note for Electro Mechanical Energy Conversion 2 - EMEC2 By ajit sahu

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1 CHAPTER-I Induction Motor Construction : The induction motor mainly divided in to two parts. (1) Stator (2) Rotor In case of D. C. Motor basically it is divided into two main parts (i) Yoke (ii) Armature. Yoke is outer & stationary part, similarly the outer portion of the induction motor is known as stator. It is also stationary part of the induction motor. The stator of the induction motor is cylindrical in shape. The inner part of D. C. Motor i.e., armature is rotating in nature. Similarly the rotating part of the induction motor is known as rotor. The rotor lies inside the stator. It is cylindrical in shape. Rotor is divided into two types. (i) Squirrel cage Rotor (ii) Phase wound Rotor or Slip ring Rotor, Figure shows the disassembled view of an induction motor with squirrel cage rotor. (a) Stator (b) Rotor (c) bearing shields (d) Fan (e) Ventilation grill (f) terminal box. Fig 1.1 Similarly figure shows the disassembled view of a slip ring motor (a) stator (b) rotor (c) bearing shields (d) Fan (e) Ventilation grill (f) Terminal box (g) Slip ring (h) brushes & brush holder. Contd…

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2 Production of Rotating Magnetic Field : When 3 – phase stationary coils are fed with 3 – phase supply, a uniformly rotating magnetic flux of constant magnitude will produce. It will now be shown that when three – phase winding displaced in space by 1200, are fed by three phase currents, displaced in time by 1200, they produce a resultant magnetic flux, which rotates in space as if actual magnetic poles were being rotated mechanically. The principle of a 3 – phase, two pole stator having three identical windings placed 1200 space degree apart as shown in fig – 1.2. The flux due to three phase windings is shown in fig 1.3. Fig 1.2 Fig 1.3 Let the maximum value of flux due to any one of the three phases be fm. The resultant flux fr, at any instant is given by the vector sum of the individual fluxes f1, f2 and f3 due to three phases. Considering values of fr at four instants i.e. 1/6th time period apart corresponding to points marked 0, 1, 2 & 3. Contd…

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3 Proof : Case – 1 : Resultant flux at origin i.e. when q = 00 At that time f1 = 0, f2 = fm Sin < - 1200 = - 3 3 fm f3 = fm Sin < - 2400 = fm. 2 2 Fig 1.4 Resultant flux fr : As per law of parallerogram f 2r = f 22 + f 32 + 2 f2 × f3 × cos 600 2 2 æ 3 ö æ 3 ö 3 1 3 Þ f = çç f m ÷÷ + çç f m ÷÷ + 2. fm × fm 2 2 2 è 2 ø è 2 ø 2 r Þ f 2r = 3 2 3 3 f m + f 2m + f 2m 4 4 4 Þ f 2r = 9 2 fm 4 Þ fr = 3 fm 2 Þ fr = 1.5 fm Case – II : When q = 600 Therefore f1 = fm Sin < 600 = 3 fm 2 Contd…

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4 f2 = fm Sin < - 1200 + 600 = fm Sin < - 600 = - 3 fm 2 and f3 = fm Sin < - 2400 + 600 = fm Sin < - 180 = 0 case – III When q = 1200 f1 = fm Sin < 1200 = 3 fm 2 f2 = fm Sin < - 1200 + 1200 = fm Sin < 00 = 0 f3 = fm Sin < - 2400 + 1200 = fm Sin < - 1200 = - 3 fm 2 fr can be calculated as earlier Similarly fr = 1.5fm Case – IV When q = 1800 f1 = fm Sin < 1800 = 0 f2 = fm Sin < - 1200 + 1800 = fm Sin < 600 = f3 = fm Sin < - 2400 + 1800 = fm Sin < - 600 = 3 fm 2 - 3 fm 2 Similarly fr can be calculated as earlier fr = 1.5 fm Hence from the above four cases we can draw a conclusion that the resultant flux (fr) inside the stator winding at any time = 1.5 fm and the resultant flux (fr) rotates around the stator at syncronous speed. How the rotor rotates : The rotor lies inside the stator. There is an air gap in between the stator and rotor. The stator slots are provided with three Phase winding. When three phase stator windings are fed by a 3-phase supply then a rotating magnetic flux of constant magnitude will produce. This rotating flux passes through air gap and cuts the stationary conductors on the rotor . There is also a 3-phase rotor winding on the rotor. The stator and rotor windings act as Contd…