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LP11 1 Basics of a Electric Motor

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  • LP11*Basics of a Electric Motor

    LP11

  • LP11*A Two Pole DC Motor

    LP11

  • LP11*A Four Pole DC Motor

    LP11

  • LP11*Operating Principle of a DC Machine

    LP11

  • LP11*FORE FINGER = MAGNETIC FIELDMIDDLE FINGER= CURRENTTHUMB = MOTIONFORCE = B IAlFlemings Left Hand Rule Or Motor Rule

    LP11

  • LP11*FORE FINGER = MAGNETIC FIELD900900900MIDDLE FINGER = INDUCEDVOLTAGETHUMB = MOTIONVOLTAGE = B l uFlemings Right Hand Rule OrGenerator Rule

    LP11

  • LP11*Action of a Commutator

    LP11

  • LP11*Armature of a DC Motor

    LP11

  • LP11*Generated Voltage in a DC Machine

    LP11

  • LP11*Summary of a DC MachineBasically consists of

    An electromagnetic or permanent magnetic structure called field which is staticAn Armature which rotates

    The Field produces a magnetic mediumThe Armature produces voltage and torque under the action of the magnetic field

    LP11

  • LP11*Voltage and Torque developed in a DC MachineInduced EMF, EA = Km (volts) Developed Torque, Tdev = KIA (Newton-meter or Nm)

    where m is the speed of the armature in rad/sec., is the flux per pole in weber (Wb) IA is the Armature current K is the machine constant

    LP11

  • LP11*Interaction of Prime-mover DC Generator and LoadPrime-mover(Turbine)DC GeneratorLoadIATdevmEA+-VL+-TpmEA is Generated voltageVL is Load voltageTpm is the Torque generated by Prime MoverTdev is the opposing generator torque

    LP11

  • LP11*Interaction of the DC Motor and Mechanical LoadDC MotorMechanicalLoad(Pump,Compressor)TloadmEA+-TdevEA is Back EMFVT is Applied voltageTdev is the Torque developed by DC MotorTload is the opposing load torque

    IAVT+--

    LP11

  • LP11*Power Developed in a DC MachineInput mechanical power to dc generator

    = Tdev m= KIAm =EA IA

    = Output electric power to loadInput electrical power to dc motor

    = EA IA= K m IA = Tdev m

    = Output mechanical power to loadNeglecting Losses,

    LP11

  • LP11*Equivalence of motor and generatorIn every generator there is a motor (Tdev opposes Tpm)

    In every motor there is a generator (EA opposes VT)

    LP11

  • LP11*Field CoilArmatureRAVfSeparately Excited DC Machine+-

    LP11

  • LP11*Shunt Field CoilArmatureRAShunt Excited DC Machine

    LP11

  • LP11*Series Field CoilArmatureRASeries Excited DC Machine

    LP11

  • LP11*Shunt Field CoilArmatureRACompound Excited DC MachineSeries Field CoilIf the shunt and series field aid each other it is called a cumulativelyexcited machineIf the shunt and series field oppose each other it is called a differentiallyexcited machine

    LP11

  • LP11*DC Machine-Example IA dc motor has Ra =2 , IA=5 A, EA = 220V, m = 1200 rpm. Determine i) voltage applied to the armature, developed torque,developed power . ii) Repeat with m = 1500 rpm. Assume sameIA.Solution on Greenboard

    LP11

  • LP11*Field CoilArmatureRAVfSeparately Excited DC MotorTorque-speed Characteristics+-mTdev

    LP11

  • LP11*Series Field CoilArmatureRASeries Excited DC MotorTorque-Speed CharacteristicsmTdev

    LP11

  • LP11*Speed Control of Separately Excited DC Motor(2)By Controlling Terminal Voltage VT and keeping If or constant at rated value .This method of speed control is applicablefor speeds below rated or base speed.mVTTdev1Tdev2Tdev3Tdev1
  • LP11*Speed Control of Separately Excited DC MotorBy Controlling(reducing) Field Current If or and keeping VT at rated value. This method of speed control is applicable for speeds below rated speed.mTdev1Tdev2Tdev3Tdev1
  • LP11*A separately excited dc motor with negligible armature resistanceoperates at 1800 rpm under no-load with VT =240V(rated voltage). The rated speed of the motor is 1750 rpm.i) Determine VT if the motor has to operate at 1200 rpm under no-load.ii) Determine (flux/pole) if the motor has to operate at 2400 rpm under no-load; given that K = 400/.iii) Determine the rated flux per pole of the machine.DC Machine-Example IISolution on Greenboard

    LP11

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