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Download Chapter 7 - 3 phase Induction Motor
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Review: Electromagnetism Magnetic field due to an electric current When a conductor carries an electric current, a magnetic field is produced around that conductor. Right Hand Rule Magnetic field of a solenoid If a coil is wound on a steel rod and connected to a supply, the steel becomes magnetized and behaves like a permanent magnet. Grip Rule if solenoid is gripped with the right hand, with fingers pointing in the direction of current, then the thumb outdtretched parallel to the axis of the solenoid points in the direction of the magnetic field inside the solenoid. Solenoid with a steel core Force on a conductor carrying current across a magnetic field F (force on conductor, N) = B (flux density, T) x L (length, m) x I (current through conductor, A) F BIL Left Hand Rule: First finger: Flux Second finger: Current Thumb: Mechanical Force Electromagnetic induction When a conductor cuts or is cut by a magnetic flux, an emf is generated in the conductor and a magnitude of the generated emf is proportional to the rate at which the conductor cuts or is cut by the magnetic flux. Ex : transformer Direction of induced e.m.f Two methods: a) Fleming’s right hand rule - If the first finger of the right hand be pointed in the direction of the magnetic flux and if the thumb be pointed in the direction of motion of the conductor relative to the magnetic field, then the second finger, held at right angles to both the thumb and the first finger, represents the direction of the emf. b) Lenz’s law Fleming’s right hand rule INDUCTION MACHINE The induction machine is the most rugged and the most widely used machine in industry. Like dc machine, the induction machine has a stator and a rotor mounted on bearings and separated from the stator by an air gap. However, in the induction machine both stator winding and rotor winding carry alternating currents. The induction machine can operate both as a motor and as generator As motors, they have many advantages. They are rugged, relatively inexpensive, and require very little maintenance. They range in size from a few watts to about 10,000 hp. The speed of an induction motor is nearly but not quite constant, dropping only a few percent in going from no load to full load. The main disadvantages of induction motors are: · The speed is not easily controlled. · The starting current may be five to eight times full-load current. · The power factor is low and lagging when the machine is lightly loaded INDUCTION MOTOR CONSTRUCTION Two different types of induction motor which can be placed in stator: a) squirrel cage rotor b) wound rotor Squirrel Cage rotor Wound rotor Types of rotor Squirrel cage rotor – consists of conducting bars embedded in slots in the rotor magnetic core, and these bars are short circuited at each end by conducting end rings. The rotor bars and the rings are shaped like squirrel cage. Wound rotor – carries three windings similar to the stator windings. The terminals of the rotor windings are connected to the insulated slip rings mounted on the rotor shaft. Carbon brushes bearing on these rings make the rotor terminals available to the user of the machine. For steady state operation, these terminals are short circuited. Squirrel Cage Rotor Rotor bars (slightly skewed) End ring Wound Rotor • Most motors use the squirrel-cage rotor because of the robust and maintenance-free construction. • However, large, older motors use a wound rotor with three phase windings placed in the rotor slots. • The windings are connected in a three-wire wye. • The ends of the windings are connected to three slip rings. • Resistors or power supplies are connected to the slip rings through brushes for reduction of starting current and speed control Induction Motor Components BASIC INDUCTION MOTOR CONCEPT A three phase set of voltages has been applied to the stator, and three phase set of stator currents is flowing. These produce a magnetic field Bs, which is rotating in a counterclockwise direction . The speed of the magnetic field’s rotation is nsync 120 f e P The rotating stator fields Bs induces a voltage in the rotor bars. eind (vxB) I The rotor voltage produces a rotor current flow, which lags behind the voltage because of the inductance of the rotor. The rotor current produces a rotor magnetic field BR lagging 900 behind itself, and BR interacts with Bnet to produce a counterclockwise torque in the machine ind kBR xBs THE CONCEPT OF ROTOR SLIP The voltage induced in a rotor depends on the speed of the rotor relative to the magnetic field. Slip speed is defined as the difference between synchronous speed and rotor speed nslip nsync - nm where nslip nsync nm = slip speed of the machine = speed of the magnetic fields = mechanical shaft speed of motor Slip is the relative speed expressed on a per unit or a percentage basis s n slip n sync x 100% s n sync - nm n sync x 100% CONTINUED… In term angular velocity (radians per second, rps) sync - m s x 100% sync If the rotor turns at synchronous speed, s = 0 while if the rotor is stationary/standstill, s = 1. n m (1 - s)n synx m (1 - s) synx THE ELECTRICAL FREQUENCY CONCEPT Like a transformer, the primary (stator) induces a voltage in the secondary (rotor) but unlike a transformer, the secondary frequency is not necessary the same as the primary frequency. If the rotor of a motor is locked, then the rotor will have same frequency as the stator. The rotor frequency can be expressed f r sf e P fr (n sync - nm ) 120 EXERCISE 1 A 208V, 10hp, 4 pole, 60Hz, Y connected induction motor has full load slip of 5%. Calculate, i) nsync (Ans:1800rpm) ii) nm (Ans: 1710rpm) iii) fr at the rated load (Ans: 3 Hz) iv) Shaft torque at the rated load (Ans: 41.7Nm) THE EQUIVALENT CIRCUIT OF AN INDUCTION MOTOR R1 = Per phase stator winding resistance jX1 = Per phase stator leakage reactance VP = Per phase terminal voltage RC = Core resistance jXM = Magnetizing reactance ER = Rotor voltage RR = Rotor resistance jXR = Per phase rotor leakage reactance THE ROTOR CIRCUIT MODEL The greater the relative motion between rotor and the stator magnetic fields, the greater the resulting rotor voltage and rotor frequency. The largest relative motion occurs when the rotor is stationary. (locked rotor or blocked rotor condition) jXR0 = blocked rotor rotor resistance ER IR RR jX R ER IR RR jsX R0 IR E R0 RR jX R0 s ER0 = Locked rotor voltage THE FINAL EQUIVALENT CIRCUIT In an ordinary transformer, the voltages, currents, and impedances on the secondary of the device can be referred to the primary side by means of the turn ratio of the transformer: VP VS' aVS IS IP I a ' S Z S' a 2 Z S The transformed rotor voltage becomes E1 ER' aeff ER0 The rotor current becomes I I2 R aeff Referred rotor resistance and reactance The rotor impedances becomes 2 RR Z 2 aeff jX R 0 s 2 R2 aeff RR 2 X 2 aeff X R0 THE FINAL EQUIVALENT CIRCUIT POWER FLOW DIAGRAM When the secondary windings in an induction motor (rotor) are shorted out, so no electrical output exists from normal induction motors. Instead, the output is mechanical. The relationship between input and output powers are shown below: POWER AND TORQUE IN INDUCTION MOTOR PSCL 3I 12 R1 Stator Copper Loss, Pcore 3E12 GC Core Losses, PAG Pin PSCL Pcore Air Gap Power, Rotor Copper Loss, PRCL 3I 22 R2 sPAG Developed Mech. Power, Output power, Pconv Pdev PAG PRCL Pout Pconv PF &W Pmisc Developed Torque, PAG R2 3I s 2 2 ind Pconv m PAG sync 1- s 3I R2 s 2 2 Separating the rotor copper losses and the power converted in Induction motor equivalent circuit The derivation of the induction motor induced-torque equation The induced torque in induction motor is: ind Pconv m ind PAG sync Air gap power: PAG I 22 R2 s Total Air gap power: PAG 3I 22 Air gap power is the power crossing the gap from the stator circuit to the rotor circuit. It is equal to the power absorbed in the resistance R2/s. R2 s Continued… If I2 can be determined, then the air gap power and the induced torque will be known. To solve circuit above, we use thevenin’s theorem: Continued… Z TH RTH jX TH X M X 1 R1 RTH XM R1 X1 X M X M X 1 2 X TH X 1 jX M ( R1 jX1 ) R1 j( X 1 X M ) THEVENIN’S EQUIVALENT CIRCUIT VTH I2 RTH PAG 3I 22 R 2 jX TH jX 2 s ind PAG sync 3VTH2 sync[( RTH R2 s 3VTH2 ( RTH R2 s R2 2 ) ( X TH X 2 ) 2 s R2 s R2 2 ) ( X TH X 2 ) 2 ] s