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3/23/2007 ELEC A6 DC Machine1Lecture NotesELE-A6DC Machine
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3/23/2007 ELEC A6 DC Machine2DIRECT CURRENT (DC)MACHINES Fundamentals• Generator action: An emf (voltage) is induced in aconductor if it moves through a magnetic field.• Motor action: A force is induced in a conductor thathas a current going through it and placed in amagnetic field• Any DC machine can act either as a generator or asa motor.
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3/23/2007 ELEC A6 DC Machine3DC Generator Fundamentals( )ee B v= × ⋅=v B ll sin cosα βα - angle between the direction in which theconductor is moving and the flux is acting.β - smallest possible angle the conductor makeswith the direction of, the vector product, ( v × B)and for maximum induction, β = 0. Hence, e =Blv for most cases.( v × B) indicates the direction of the current flowin the conductor, or the polarity of the emf.e = induced voltage, v = velocity of the conductor, B =flux density and l is the length of the conductor
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3/23/2007 ELEC A6 DC Machine4Generated Voltage in a Loop(a coil of one turn)• For emf to be induced, theconductors must cut the fluxlines as they move.Otherwise, ( v × B) = 0.eloop = eab + ebc + ecd + edaeloop = Blv + 0 + Blv + 0eloop = 2 B l v
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3/23/2007 ELEC A6 DC Machine5Generated Voltage in a Loop(a coil of one turn)• Note: Induced voltages are always AC.
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3/23/2007 ELEC A6 DC Machine6Generated Voltage in a Loop(a coil of one turn)
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3/23/2007 ELEC A6 DC Machine7DC MACHINESReal DC machine Construction• Stator: Stationary part of the machine.The stator carries a field winding that isused to produce the required magneticfield by DC excitation. Often know asthe field.• Rotor: The rotor is the rotating part ofthe machine. The rotor carries adistributed winding, and is the windingwhere the emf is induced. Also knownas the armature.N SStator withwith polesBrushRotorField
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3/23/2007 ELEC A6 DC Machine8DIRECT CURRENT MACHINESDC machine Construction• The picture shows thestator of a large DCmachine with severalpoles.• The iron core is supportedby a cast iron frame.Dc motor stator constructionField poles
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3/23/2007 ELEC A6 DC Machine9DIRECT CURRENTMACHINESDC machine Construction• The rotor iron core is mounted on theshaft.• Coils are placed in the slots.• The end of the coils are bent andtied together to assure mechanicalstrength.• Note the commutator mounted onthe shaft. It consists of severalcopper segments, separated byinsulation.DC motor rotor constructionCommutatorPolesRotor windingFanBearingBrushes
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3/23/2007 ELEC A6 DC Machine10DC MACHINESGenerated EMF in a Real DC MachineWhereZ = total number of conductors, P = total number of polesa = P for lap winding, a = 2 for wave winding, φ = flux,ω = speed in rad/s and n = speed in rpm.φωφωπφφ mgG kaZPnknaZPE ====260
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3/23/2007 ELEC A6 DC Machine11DC Motor FundamentalsL is a vector in the direction of the flow of the current.direction indicates the direction of force.( )F B= ×l iF = induced force, B = flux density, I is the current passingin the conductor and l is the length of the conductor)( Bl ×
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3/23/2007 ELEC A6 DC Machine12DC Motor FundamentalsCounter EMF:When the motor is running, internally generatedemf, (EG = EC) opposes the applied voltage, thus:IV ERAT CA=−Where: VT = terminal voltage, Ec = counterEMF, RA is the armature resistance and IA isthe armature current
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3/23/2007 ELEC A6 DC Machine13The relationship between theinduced EMF and torque{ {E B vT B irthereforeETB vB irvirv rET iEi Tconductorconductorelectricpowermechanicalpower=== = ===llll,, ωωωTEI ZPaIT k IAm A= ==ω πφφ2Where: T is the torque,
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3/23/2007 ELEC A6 DC Machine14Example 1Find the induced voltage fora road with a length of 1 m.B = 3 TV = 1 m/s0).(with90makespage)the(into3).(o=××==××=lBvlBvvBBvlBveind
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3/23/2007 ELEC A6 DC Machine15Example 2Example: Find theforce for a conductorwith a length of 1 m.B = 0.35 Ti = 2 ApagetheintoisdirectionThe7.0)35.0(*)2(*)1()( NiBlF ==×=
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3/23/2007 ELEC A6 DC Machine16Example 3VnaZPEG 8.17210*30*600*2*606*4*4860voltage.inducedthecalculatemWb,30ispoleperfluxtheandrpm600@rotatingisarmatureThe/slot.conductors4withslots48hasaramatureThewinding.WaveaasconnectedarmatureanhasmachineDCpole6A3=== −φ
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3/23/2007 ELEC A6 DC Machine17Example 4A six-pole DC machine has a flux per pole of 30 mWb.The armature has 536 conductors connected as a lapwinding. The DC machine runs at 1050 rpm and itdelivers a rated armature current of 225 A to a loadconnected to its terminals, calculate:A) Machine constant, KmB) Generated voltage, EGC) Conductor currentD) Electromagnetic torque.E) Power delivered by the machine.
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3/23/2007 ELEC A6 DC Machine18Example 4kW32.63).84.575225*03.0*31.85)5.376225)(:windinglapisitSince)V4.28196.109*03.0*31.85rad/s96.109601050*2602)31.856*2536*62)==================aGamamGmIETPemNIkTdAaIconductorIckEnbaZPkaωφφωππωππ
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3/23/2007 ELEC A6 DC Machine19DC MotorsEquivalent circuit.The equivalent circuit of DC Motors (and Generators) has twocomponents:• Armature circuit: it can be represented by a voltage source and aresistance connected in series (the armature resistance). The armaturewinding has a resistance, Ra.• The field circuit: It is represented by a winding that generates themagnetic field and a resistance connected in series. The field windinghas resistance Rf.
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3/23/2007 ELEC A6 DC Machine20Classification of DC Motors• Separately Excited and Shunt Motors- Field and armature windings are either connected separate or inparallel.• Series Motors- Field and armature windings are connected in series.• Compound Motors- Has both shunt and series field so it combines features of series andshunt motors.
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3/23/2007 ELEC A6 DC Machine21Shunt DC Motors– The armature and field windings are connected in parallel.– Constant speed operation.
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3/23/2007 ELEC A6 DC Machine22Shunt DC MotorsBy KVL around the outer loop:VT - IA RA - EC = 0 IVRFTF=nkIEmIInkEfCffgC==∝=φφφso,but
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3/23/2007 ELEC A6 DC Machine23Starting of Shunt DC Motors• At the starting of a DC motor, EC = 0, so:• To limit IA, a resistance is inserted in serieswith RA then removed after the developmentof ECaTARVI =
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3/23/2007 ELEC A6 DC Machine24Speed RegulationSpeed regulation is the percentage changein speed from no-load to full-load as afunction of the full load speed.SRn nnNL FLFL=−×100%
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3/23/2007 ELEC A6 DC Machine25Power Flow and Losses inDC MotorsPin = VtILI2R LossesCopper lossesCore Losses(Both cores)Vt = Terminal voltageIL= Line currentMechanicalLossesStray LossesPoutP(developed) = ECIa
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3/23/2007 ELEC A6 DC Machine26Example 1Q1) A 240 V, shunt DC motor takes anarmature current of 20 A when running at960 rpm. The armature resistance is 0.2 Ω.Determine the no load speed if the no loadarmature current is 1 A.
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3/23/2007 ELEC A6 DC Machine27Example 1rpm45.975960*2368.239.).(.).(.).(.).(.).(constantisIassumingV8.2392.0*1240load)(noV2362.0*20240load)(fullf==⇒=⇒=−==−=⇒+=lnnlFnlnnlFElnEEEERIVcccccaatloadfullload,no == F.l.n.l.
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3/23/2007 ELEC A6 DC Machine28Example 2Q2) A 120 V shunt motor has the followingparameters: Ra = 0.4 Ω, RF = 120 Ω androtational (core, mechanical and stray) losses are240 W. On full load, the line current is 19.5 A andthe motor runs at 1200 rpm, find:• The developed power• The output power, and• The output torque.
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3/23/2007 ELEC A6 DC Machine29Example 2N.m67.1412006021843)(watt18432402083(b)watt1.20835.18*6.112*V6.112)4.0(*)5.18(120A5.1815.191120120)(====−=−=====−=−==−=⇒==−=πωτ outoutdevelopedoutacdevelopedaaTcaffLaPclossesrotationalPPIEPRIVEIAIIIIa
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3/23/2007 ELEC A6 DC Machine30Separately Excited DC Motors– The armature winding supplies the load.– The field winding is supplied by a separate DC source whosevoltage is variable.– Good speed control.
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3/23/2007 ELEC A6 DC Machine31Separately Excited DC Motors
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3/23/2007 ELEC A6 DC Machine32Series DC Motors– The armature and field winding are connected in series.– High starting torque.
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3/23/2007 ELEC A6 DC Machine33Series DC MotorsBy KVL around the loop:VT - IA (RA + Rf) - EC = 0nkIEmIInkEACAAgC==∝=φφφso,but
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3/23/2007 ELEC A6 DC Machine34Example 3Q1) A DC series motor is operated at fullload from a 240 V supply at a speed of600 rpm. The EC is found to be 217.2V at a line current of 38 A, find:a) The armature resistance assuming theseries field resistance is 0.2 Ω.b) Find the no-load speed given that theno-load current is 1 A.
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3/23/2007 ELEC A6 DC Machine35Example 3rpm151,25V6.2396.0*1240:loadnoAt(b)6.0382.217240)(*)(2112212=⇒==−=Ω=−=+⇒++=nInInEEERRRRIEVaaacccfafaact
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3/23/2007 ELEC A6 DC Machine36Compound DC Motors
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3/23/2007 ELEC A6 DC Machine37Speed Control of DC MotorsSpeed can be controlled by varying:1) Armature circuit resistance using an externalresistance RA Ext.2) IF can be varied by using an externalresistance Radj in series with RF to control theflux, hence the speed.3) The applied voltage to the armature circuitresistance, if the motor is separately excited
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3/23/2007 ELEC A6 DC Machine38Torque developed by shuntmotor( )TKRkVRkTkVkTIRIkVRIEVmAmTAmmTmAAAmTAACT2forsolvingsobutsoφφωωφφωφφω−=+==+=+=
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3/23/2007 ELEC A6 DC Machine39Torque developed by shuntmotorIf VT and IF (hence )are constant, speed isdirectly proportional tothe torque
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3/23/2007 ELEC A6 DC Machine40Torque developed by seriesmotor
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3/23/2007 ELEC A6 DC Machine41Comparison betweendifferent DC Motors
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3/23/2007 ELEC A6 DC Machine42Comparison of DC MotorsShunt Motors: “Constant speed” motor (speed regulation is verygood). Adjustable speed, medium starting torque. (Start = 1.4 × TFL)Applications: centrifugal pump, machine tools, blowers fans,reciprocating pumps, etc.Series Motors: Variable speed motor which changes speeddrastically from one load condition to another. It has a highstarting torque.Applications: hoists, electric trains, conveyors, elevators, electriccars, etc.Compound motors: Variable speed motors. It has a high startingtorque and the no-load speed is controllable unlike in seriesmotors.Applications: Rolling mills, sudden temporary loads, heavymachine tools, punches, etc
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3/23/2007 ELEC A6 DC Machine43Home work problemsQ no.1
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3/23/2007 ELEC A6 DC Machine44Home work problemsQ no.2
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