basics of a electric motor

Views:

Category: Entertainment

Presentation Description

No description available.

Presentation Transcript

Slide 1:

LP11 1 Basics of a Electric Motor

Slide 2:

LP11 2 A Two Pole DC Motor

Slide 3:

LP11 3 A Four Pole DC Motor

Slide 4:

LP11 4 Operating Principle of a DC Machine

Slide 5:

LP11 5 FORE FINGER = MAGNETIC FIELD MIDDLE FINGER= CURRENT THUMB = MOTION FORCE = B IAl Fleming’s Left Hand Rule Or Motor Rule

Slide 6:

LP11 6 FORE FINGER = MAGNETIC FIELD 900 900 900 MIDDLE FINGER = INDUCED VOLTAGE THUMB = MOTION VOLTAGE = B l u Fleming’s Right Hand Rule Or Generator Rule

Slide 7:

LP11 7 Action of a Commutator

Slide 8:

LP11 8 Armature of a DC Motor

Slide 9:

LP11 9 Generated Voltage in a DC Machine

Slide 10:

LP11 10 Summary of a DC Machine Basically consists of An electromagnetic or permanent magnetic structure called field which is static An Armature which rotates The Field produces a magnetic medium The Armature produces voltage and torque under the action of the magnetic field

Slide 11:

LP11 11 Voltage and Torque developed in a DC Machine Induced EMF, EA = Km (volts) Developed Torque, Tdev = KIA (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

Slide 12:

LP11 12 Interaction of Prime-mover DC Generator and Load Prime-mover (Turbine) DC Generator Load IA Tdev m EA + - VL + - Tpm EA is Generated voltage VL is Load voltage Tpm is the Torque generated by Prime Mover Tdev is the opposing generator torque

Slide 13:

LP11 13 Interaction of the DC Motor and Mechanical Load DC Motor Mechanical Load (Pump, Compressor) Tload m EA + - Tdev EA is Back EMF VT is Applied voltage Tdev is the Torque developed by DC Motor Tload is the opposing load torque IA VT + - -

Slide 14:

LP11 14 Power Developed in a DC Machine Input mechanical power to dc generator = Tdev m= KIAm =EA IA = Output electric power to load Input electrical power to dc motor = EA IA= K m IA = Tdev m = Output mechanical power to load Neglecting Losses,

Slide 15:

LP11 15 Equivalence of motor and generator In every generator there is a motor (Tdev opposes Tpm) In every motor there is a generator (EA opposes VT)

Slide 16:

LP11 16 Field Coil Armature RA Vf Separately Excited DC Machine + -

Slide 17:

LP11 17 Shunt Field Coil Armature RA Shunt Excited DC Machine

Slide 18:

LP11 18 Series Field Coil Armature RA Series Excited DC Machine

Slide 19:

LP11 19 Shunt Field Coil Armature RA Compound Excited DC Machine Series Field Coil If the shunt and series field aid each other it is called a cumulatively excited machine If the shunt and series field oppose each other it is called a differentially excited machine

Slide 20:

LP11 20 DC Machine-Example I A 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 same IA. Solution on Greenboard

Slide 21:

LP11 21 Field Coil Armature RA Vf Separately Excited DC Motor Torque-speed Characteristics + - m Tdev

Slide 22:

LP11 22 Series Field Coil Armature RA Series Excited DC Motor Torque-Speed Characteristics m Tdev

Slide 23:

LP11 23 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 applicable for speeds below rated or base speed. m VT Tdev1 Tdev2 Tdev3 Tdev1<Tdev2< Tdev3

Slide 24:

LP11 24 Speed Control of Separately Excited DC Motor By Controlling(reducing) Field Current If or  and keeping VT at rated value. This method of speed control is applicable for speeds below rated speed. m Tdev1 Tdev2 Tdev3 Tdev1<Tdev2< Tdev3 

Slide 25:

LP11 25 A separately excited dc motor with negligible armature resistance operates 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 II Solution on Greenboard 