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  • 1. UNIT 7 ENERGY CONVERSION PRINCIPLES AND D.C. MACHINES Structure 7.1 lntroduction Objectives 7.2 Principles of Electromechanical Energy Conversion 7.2.1 Induced ElectromotiveForce and Voltage 7.2.2 Force on a Current Carrying Conductor 7.2.3 Back EMFand CounterTorque 7.3 Losses, Heating and Efficiency 7.3.1 ElectromagneticLosses 7.3.2 Mechanical Losses 7.3.3 Efficiency, Heating and Rating 7.4 Direct Current Generators 7.4.1 ConstructionalFeatures 7.4.2 Armature Windings,EMF and Counter Torque 7.4.7 ExcitationSchemesand MagnetisationCharacteristic 7.4.4 Load Characteristics 7.5 Direct Current Motors 7.5.1 Excitation Schemesfor DC Motors 7.5.2 Motoring Action and Speed Equation 7.5.3 Shunt Motors 7.5.4 Series Motors 7.5.5 CompoundMotors 7.6 Summary 7.7 Answers to SAQs 7.1 INTRODUCTION Rotating electrical machines are energy conversion devices which link an electrical system to a rotating mechanical system. When energy flows from the mechanical system to the electrical system, the device is said to function as a generator. When energy flows from the electrical system to the mechanical system, the device is said to function as a motor. This process of energy transfer is reversible, and a11electrical machine can be made to work either as a generator or as a motor. During such energy transfer, part of the energy is lost as heat energy, leading to an energy conversion efficiency which is always less than 100%. The electrical power system to which the machine is c o ~ e c t e dmay be an ac system at 50 Hz (60Hz in the USA) or a dc system. The three phase synchronous machine and the induction machine constitute the principal types of ac machines. DC machines are classified in terms of the schemes used for providing the magnetic field or excitationof the machines. In this unit we will confine ourselves to a study of dc machines. AC machines will be considered in Unit 8. Objectives After studying this unit, you should be aPle to use the Blv and Bli formulas for calculating emf and torque, identify and list the various components of energy loss, describe the constructional features of the heteropolar field syslem used in dc machines, give an elementary description of dc armature windings and the commutator, explain the various excilation schemes used in dc machines, calculate induced emf and terminal voltage of dc generators, and calculate torque and speed of dc motors.
  • 2. Ektrical Machiis Me~surioglmtrumene 7.2 PRINCIPLES OF ELECTROMECHANICAL ENERGY CONVERSION In an electromechanicalenergy conversiondevice,energy transfer between the electrical and mechanical systems takes place through the agency of the magnetic and electric fields. In the rotating electrical machines consideredin this unit, the energy stored in the electric field is negligible and conversion takes place mainly via the magnetic field.The principal phenomena to be considered in such machines are (i) the generation of electromotiveforce or voltage in the windings and (ii) the production of forces on current carrying conductors resulting in torque. 7.2.1 Induced ElectromotiveForce and Voltage Faraday's law of induction,as in the transformersdiscussed in Unit 6, plays an important role in understanding the phenomena governing the operation of rotating electrical machines. This law statesthat an electromotiveforce (emf)is induced in any coil whose magnetic flux linkages are changing with time. The voltageproduced satisfies the equation where N = number of turns of the coil; cp = flux linking the coil in webers; h = Ncp is the total flux linkagesof the coil and gq = rate of change of the flux. dt Eq. (7.1) gives the voltage drop in the directionof assumedpositive currentflow provided the convention is adopted that positive current in the coil setsup a positive flux cp and a positive flux linkage h Equivalently, the orientation of the currentin the loop must be so selected that a right handed screw,if turned in that direction,will advance in the directionof positive flux.In Eq. (7.1) it is assumedthat the sameflux cp links each one of theN turns. In the transformer, the rate of change of flux linkagewas entirely due to changingcurrents in the various coils, the coils themselves remaining stationary. In rotating machines, rotary motion is a very importantfactor in producing change of flux linkage.In fact using vector calculus, it can be shown that the induced emf can be split into two components viz., (i) a transformer emf and (ii) a motionalemf. When applied to a coil moving in a magnetic field which is itself changing with time, the transformeremf is the emf that would be produced if the coil remained stationary at its position at time t, while the flux continued to follow its own law of variationwith time. The motional emf, on the other hand, depends on the velocity of each part of the coil at time t, the flux density at each part being considered constant at its value at time t. If a conductor of length 1 is situated in a magnetic field with a uniform flux density of B teslas and the velocity of the conductor is v, the motional emf is givenby e =Blv volts, (7.2) provided the directionsof B, 1 and v are at right angles to each other.The conductor of length 1 meters is said to cut the magnetic flux densityB with a velocity of v metres per second. Eq. (7.2)is therefore often called theflux cutting rule. The directionin which e acts may be determinedby the so called Fleming's right-hand rulefor emf. In this rule, if the thumb, fore-fingerand middle fingerof the right hand are held mutually at right angles such that the fore-fingerpoints in the directionof magnetic field B and the thumb in the direction of motion (velocity),then the middle finger points in the direction of induced voltage rise. Example 7.1 (a) A coil of 100turns has a flux of 0.01 Wb linking it. If the flux is reversed in direction at uniform rate in 1ms (the coil being stationary),find the value of the induced voltage.
  • 3. (b) A conductor of length 20 cm is moved at a linear velocity of 10m/s at right Conversionprinciples angles to a magnetic field. The induced emf is 0.5 V. Find the flux density of *d D.CMachilles the field. Solution OL (a) Nett flux change =0.01 - (4.01) = 0.02 Wb This change occurs uniformly in 1 rns = 10-~s & - 0.02= 20&,s andTherfore - dt 0.001 e (b) B =-(from Eq. (7.2)) lv -- 0.5 = 0.25 tesla. 0.2 x 10 We will now use Faraday's law of induction to demonstrate the validity of the-fluxcutting rule by evaluating the induced voltage in a rectangular coil. (Many practical windings used in electricalmachines may be regarded as equivalent to rectangular coils. Consequently the results of this analysis will be of direct use later). Figure 7.1 shows a rectangular coil ABCD with its sides parallel to the X- and Y-axes of a right-handed rectangular coordinate system. The flux density B is assumed to be a function only of x, (independent of time t and the other space coordinates y and z), and to be directed in the positive z-direction. At time t, let the rectangularcoil be in the position shown and let the flux density at x =x,be B, Figure 7.1 : Motional Voltage In Moving Coil and atx =x,, B,. In the differentialtime dt, the entire coil will move towards the right through a distance (vdt) when v is the velocity of the movement. Due to'this motion, the flux linking the coil will increase by an amount B, lvdt due to the motion of conductor CD,and reduce by an amount B,lv dt due to the motion of conductor AB.The nett increase of flux linkage dq is therefore given by The induced voltage by Faraday's law is therefore Using the convention appropriate to Faraday's law, this is the induced voltage drop acting in the direction of the loop ABCD.(In applying Faraday's law, as stated earlier,the orientation of the loop must be so selected that a right handed screw, if turned in that direction,will advance-in the direction of,the positive flux linkage). By the flux cutting rule, no voltages will be produced in the conductors BC and DA as the motionhas no component perpendicular to these conductors. In conductorAB,using Fleming's right hand rule,-there is a voltage rise from A to B equal to B,lv and in conductor CD,a voltage fall from C to D given by B, lv. Therefore, round the loopABCD,the total , voltage fall is (B,-B,) lv, in agreement with Eq. (7.3a).
  • 4. Electrical ~ ~ h i n e s& If B is a function of time, Eq. (7.2) will no longer give the total induced emf. However, it Mem-g can be used to compute the component of emf we have described as the motional emf provided the'yalues used for B, and B, are their values B,(t) and B2(t)at time t when the coil is passing through the position shown. To get the correct values of induced emf we have to add the transformer emf which is given by Here x, and x, must be treated as constants (with time derivatives taken to be zero) having the values x,(t) and x2(t).These statements are clarified in the illustrative examples which follow. Example 7.2 PQ in Figure 7.2 is a representation of the rectangular coil of Figure 7.1, as seen in cross-section across XX, when located in a z-directed magnetic field of rectangular waveform and peak value Bo tesla. The coil of width 0.2 m and length 1= 0.1 m, is in motion at a constant velocity v= 20 m/s from left to right. The rectangular magnetic field is periodic and has a half wave-length of 0.2 m. At time t = 0, the coil PQ is in the position shown such that the flux through it has the maximum value of (0.2 x 0.1 x Bo ) Wb (a) Find the emf induced in the coil round the loopABCD using (i) the flux cutting rule; (ii) Faraday's law of induction. (b) What are the values of the motional emf and transformer emf in this case ? Explain. Figure 7.2 :Diagram for Example 7.2 Bz' v =20m/s Solution Pf, Assuming that P is at x = 0 when t = 0, at any time t conductor AB (see Figure 7.1) is at position xl = vt = 20 r and conductor CD is at x2 = (20t +0.2) length 1= 0.1 m ? Bo 4 2 m-,,Q (a) (i) During the time 0 < t < 0.01s, the voltage rise fromA to B by the flux cutting and right hand rules is B, x 0.1 x 20 = 2 Bo V. The emf rise from C to D is also B, x 0.1 x 20 = 2 BoV. Therefore, the emf rise round the loopABCD is a constant of value (2B0+2B0)= 4B0 V. V J C X --2m -- d o During the period .Ols < r < .02s,,theinduced emfs have the same magnitudes as before, but rise in the opposite direction. The emf rise round the loop ABCD is now a constant equal to (4B0) V. The induced emf thus has a rectangular waveform in time of peak value 4B0V, the period being T=(0.01 +0.01) = 0.02s. This corresponds to a frequencyf = (1/T) = 50 Hz. (ii) Flux linking the coil ABCD in the positive direction (as defined by the right hand corkscrew rule for traversal of the loop in the direction ABCD) is, during the period 0 < t < .Ols = i(0.2- vt) - vt)(O.lBo) = (0.2 -2 vt)(O.lB,) = (0.02- 4t) BoWb By Faraday's law, emf drop around the loopABCD This agrees with the result obtained in (a) (i). The result in a(i) for the next half cycle can be verified similarly. (b) Since the flux density distribution is not a fullction of time, there is no transformer emf. all emfs generated being purely motional.
  • 5. Example7.3 .. Energy Conversion Principles and D.C. Machines x In Figure 7.1 let the flux density be given by B, =B, cos 2nft cos ( 2n -) 2 corresponding to a stationarypulsating magnetic field whose wavelength in the x-direction is z and whose frequency isf Hz. (a) Determine the transformer emf at time t; (b)Determine the motional emf at time t; (c) Use Faraday's law to determine the induced emf and identify the componentscorresponding to (a) and (b). Assume that xl and x2are independent functionsof the time t. Solution Flux linkage of thetoil is a functionof xl, x2 and t and may be written as h (xl,xz,t) x2 17 x2 Xl = 1J B,,, cos 2nft.cos 2n dc = - Bmcos 2nfr (sin 2n -- sin 2n - ) (7.4) X I 2 2n z 2 (a) To determine-the transformer emf we must compute ( d m )while treating XI andx2 as constants.This merely means that e, is equal to the partial derivative of h with refetcnce to t. Thus, el =ah=-flr~, sin 2 m at (b) Using Blv and the right hand rule, the voltage rise fromA to B is Bllvl and that fromD to C is B~/V~,where vl = velocity of AB =(dxlldt)and v2 =velocity of DC = (d.x?ldt).The voltage drop round the loopABCDis (-Bllvl +B21v2)= E[B?(&$@ - Bl(dxlldt)].Therefore, (c) A right-handed screw which is rotated in the loop senseABCD will advance in the positive z-direction which is the assumed directionof positive flux. So,by Faraday's law, the voltagedrop round the loopABCDis given by d Vdt. Thus, ah It is readily verified that e,= -and at SAQ 1 (a) InExample7.3,assumethatB,=1.2T,~=0.4m,f=OHz,l=0.1m, x, = 20 t, x2= (20 t +0.2), (i.e., the coil is rigid with a span of 0.2m and a linear velocity of 20 m/s). Find (i) the transformer emf, (ii) the motional emf, and (iii) the total emf. (b) How will these change iff = 50 Hz? 7.2.2 Force on a Current Carrying Conductor Mechanicalforce is exerted on any conductor carryingcurrentin a magnetic field.;T'his force is given by F = B i Enewtons, (7.8)
  • 6. ElectricalMachines & where the force, directionof current flow and the direction of the magnetic field are at right MeasuringInstruments angles to each other, B is in teslas, 1 is in metres and i is in amperes. The direction in which the force acts on the conductor may be determined by the left-hand rule. This rule states that if the thumb, fore-fingerand middle fingerof the left hand are held perpendicular to each other, with the fore-fingerpointing in the direction of the magnetic field and the middle-finger pointing in the directionof current flow, the thumb indicates the directionof the force (or the motion) induced by it. SAQ 2 (a) (i)A linear conductor carrying a currentof 500 A is at right angles to a magnetic field of flux density 0.2tesla. Find the force on the conductor per metre length. (ii) If the current is directed along the positive x-direction and the flux density is directed along the positive y-direction, what is the direction of the force? (iii) If the current is directed along the positive y-direction and the flux density is along the positivex-direction,what is the direction of the force'? (b) The force on a conductor of length 50 cm in a uniformmagnetic field is 80 N. If the conductor-iscarrying a currentof 1200A, find B. 7.2.3 Back EMF and Counter Torque In a generator,the emf induced in the armature windings may be determinedusing Faraday's law (Eq. (7.1)) or, where applicable,the flux-cuttingrule (Eq. (7.2)).The voltage availableat the armature * terminals is this voltage less any voltage drop due to currentflow in the resistanceand reactanceof the armature. Even in a motor, because armature windings are in motion relative to the magnetic field, an emf will continueto be produced. This emf will equal in value the applied voltage less voltage dropsdue to the resistance and reactance in the winding and will act in a direction opposedto it. This induced emf 1stherefore referred to as the back emf. mechanical energy or vice versatakes place. In an electrical motor,the forces acting on current carrying conductorsin a magnetic field (Eq. (7.8)),result in a motoring torque which drives the mechanical load on the shaft of the machine. Even in a generator,because armature conductorscarry current and are situated in a magnetic field, a mechanical torque will be produced. However, such a torque will act in a directionopposite to the prime-mover torque driving the generator.Such a torque is therefore referred to as a counter torque. 7.3 LOSSES, HEATING AND EFFICIENCY Whenever a machine converts energy from one form to another,the useful energy output is always less than the energy input. The differencebetween input and output is the energylost internallyin the machine. Such energy losses resultin (i) an efficiencyof energyconversion / which is always less than 100%and (ii) an increasein the temperaturesof various parts of the machine. In staticenergy conversion devices such as transformersthere are only electromagneticlosses.In rotating electrical machines we alsohave mechanical losses corresponding to the energy required to overcome frictional and wind-resistanceforces. 7.3.1 Electromagnetic Losses Electromagneticlosses consistof ohmic losses in conductorscarrying current,electrical energy loss in the brush contactand magnetic flux associated losses in the iron parts of the machine. Whenever current flows through a coil, whether it is situated in the stationarypart of the machine (stator)or rotatingpart (rotor),ohmiclosses,often referred to as copper losses, occur.The power lost in this manner when a current of I amperes flows through a resistance of R ohms is I'R watts. This loss manifests itself as heat, causingthe conductor temperature to rise above the surroundingambienttemperature. The higher the current densityused in a conductor,the greater is the current and heat produced, and the higher the temperature rise. A P inrreaspd t e m w r a t ~ ~ r ~ cb a d t n deterinratinn nf t h ~plprtriral i n ~ l i l a t i n omaterialc
  • 7. surrounding thc conductors, highcr current densities require efficient cooling systems to PriI)(:iP1eS prcvent excessive temperature risc. and D.C. Machines Electric currents are suppl~edto rotating windings through stationary brushes which ride on slip-rings or. m the case of dc machines. a coillrnutator. The current density in the material ol thc brushes is usually kept low so that the 12Rlosses in them are negligible. However, in dc circuits, there is a voltagc drop across the contact surface between brush and slip-ring (or brush and commutator) which varies from 0.8 V to 1.3 V depending on the type of brush nlaterial, the applied pressure and the current through the brush. Ii the brush current 1sID anlperes and the brush contact voltage drop is V, volts, the electrical power loss in the brush is V,I, watts. Whenever there is a pulsating and lor rotating magnetic flux in iron, losses referred to as iron-lossesoccur as in Lhe case of transformers. Such fluxes induce ernfs and associated currents called eddy currents which result in 12Rlosses in the iron referred to as eddy-current losses. The use of laminations reduces eddy currents by forcing thc currents along longer paths of higher res~stailce,thereby reducing the eddy current loss. Another source of power loss in iron is due to the phenomenon of hysteresis. The associated hysteresis loss is a properly of the material and the heat treatment thc iron was subjected to. Both eddy current and hystcresis losses increase wilh increasing peak flux density in a non-linear manncr. They also increase with increasing frequency, thc hysteresis loss being proportional to the frequency of pulsationf; while the eddy current loss is proportional tof 2. 7.3.2 Mechanical Losses Mechanical losses are due lo bearing friction. brush friction and windage. Windage loss is caused by air-resistance to the motion of rotating parts and depends on the speed. I11order to reduce the temperature rise caused in the mach~neby the various power losses, a cooling system using a fan attached to the rotor, is commonly employed. Windage loss includes the power consumed by this fan as well. All mechanical losses depend upon the speed of the machine and on Ule turbulence associated with the moving parts. Consequently, these are greatly irllluenccd by thc dcsign of the bearings, brushes, slip-rings or commutator and the cooling systcm. In the absence oCprior information, tests have to be conducted on the machine itself to determine the value of these losses. 7.3.3 Efficiency, Heating and Rating of Machines The losses in an electrical machine generally increase substantially as the load increases. This is mainly due to increased current in the armature conductors leading to increased I 2~ losses. In a generator increased output power implies, in a straightforward manner, the increase of the armature current supplied by the 111acliine.Even in a motor, where increased load means a higher load torque on Ule shaft of thc machine, because of Eq. (7 8), the armature current must increase wilh load. Thus, as load increases, the copper loss progressively incrcascs, leading to greater heal production and consequent rise in temperature. However, the temperature must not exceed safe limits if the material used for insulating the conductors is not to deteriorate rapidly and lose its electrical insulating properties and mechanical strength. So, depending upon the type of insulat~ngmaterials used in a machine, the temperature rise of various parts of the machine have lo be restricted in value. This therefore imposes a restriction on the permissible value of the load current, and hence the output load of the machine. Temperature rise thus decides the armature current and power rating of the machine. The quality and thickness of the insulating materials used in various parts of the machine fix the voltage rating of the machine. FurUler discussion of these matters may be found in Section 9.3. SAQ 3 An electric generator is supplying an electrical power output of 100kW and is driven by a prime mover at a speed of 1000rpm. The electromagnetic losses arc found to be 3.5kW and the mechanical losses 1.5kW.Find (a) the power supplied to the shaft by the prime-mover; (b) the torque supplied by the prime-mover and (c) the efficiency.
  • 8. Electrical Ma 2s & Memuring Jnstnrmenta SAQ 4 An electric motor is given ail electric supply of 100kW. It drives a mechanical load of constant torque at a specd of 1000rpm. If the electromagnetic losses are 3.5 kW and the mechanical losses are 1.5kW, find (a) the mechanical power output; (b) the output torque and (c) the efficiency. 7.4 DIRECT CURRENT GENERATORS Direct current (dc) is mainly produced these days using static electronic rectifiers which convert the current of an alternating current (ac) system to direct current without employing a rotating electrical machine. Electromechanical dc generators have thus become largely redundant. However, engine driven dc generators often find usage in situations where an ac power supply is not available. On the other hand, the dc motor is the machine of choice in many situations. An understanding of the principles of operation of the dc generator IS a great help in understanding the dc motor. Another motivating factor for the study of the dc generators is the fact that many industrial motors function as generators for brief periods during their operation. In this section we begin by briefly considering the constructional features of the field system, the armature, the commutator and the brush assembly of dc machines. We then briefly study the magnetisation characteristics and the load characteristics of dc generators provided with separate, shunt and compound excitation. 7.4.1 Constructional Features Commercial dc generators and motors are both built in the same way and possess four main components viz., (1) the field system, (2)the armature, (3) the commutator and (4)the brush assembly. Field System The field system is the means whereby the magnetic field is produced. The stationary part (stator) is essentially an electromagnet composed of a number of protruding poles bolted to the inside of the stator frame called the yoke. The yoke is usually made of solid cast steel, whereas the poles are made up of steel laminations. Field coils are mounted on the poles and carry the dc current (field current) required to produce the necessary magnetic field. The pole-end nearest the rotor flares out into what is termed the pole-shoe. The space between the pole-shoes and the rotor cylindrical iron surface is called the air-gap. The magnetic field crosses the air-gap essentially in the radial direction and constitutes the useful flux essential for electromechanical energy conversion. Figure 7.3 gives the sectional view of a four-pole heteropolardc machine. (In a practical machine the air gap is very small. However in Figure 7.3, it is shown to be quite large for the sake of clarity). The field system is described as being heteropolar because adjacent poles are of alternate polarity, the sequence of poles being North- South-North-South etc. Each pole carries a field winding, each winding having the same number of turns. The direction in which thc field current traverses each field winding fixes the polarity of the field. hl the figure each winding is shown in cross-section as a rectangular block though it actually contains many turns, the direction of current flow required to produce the polarities shown being as indicated. (N, S represent north, south magnetic poles respectively. By convention, flux leaves a north pole to enter the air-gap, whereas it enters a south pole from the air-gap. The @ sign indicates current entering perpendicularly into the plane of the paper from the side of the vicwer while 0 indicates current flow in the opposite dircction). The figure also shows typical flux lines passing through the yoke, a north pole, the air gap, the xmaturc, the air gap a second time and a south pole before closing on tl~emsclves.
  • 9. Energy ConversionPrinciples and D.C. Machines - TYPICAL MAGNETIC Figure 7.3 :Sectional V~ewof Field System / The Armature The main rotating part of the machine, called the armature, consists of an armature core of stacked laminated steel and the armature windings. The slotted armature core is indicated in both Figure 7.3 describing the field system, and Figure 7.6 describing the commutator. It consists of slotted iron (steel) laminations that are stacked to form a solid cylinder with axial slots in which the armature winding is housed as also i ventilating ducts for the passage of cooling air. Figure 7.4 represents a lamination suitable for a small size dc machine. r KEY WAY SLOT VENT1LATlNG DUCTS - Figure 7.4 :Armature Lamination The armature conductors carry the current which is delivered by the generator to a load. They are insulated from the rotor iron and from each other by several layers of paper or mica insulation and are held firmly in place by a slot wedge made of wood or other fibrous material. Figure 7.5 (a) shows a slot containing two coil sides, each coil side consisting of a single conductor. A single coil is housed in two slots, approximately one pole-pitch apart, one coil side occupying the top layer in a slot, and the other the bottom layer in its slot. This arrangement is used to facilitate the stacking of coils around the periphery, allowing the overhang of the coils to fit together neatly. If multiturn coils are used, each coil side will consist not of a single conductor as shown in Figure 7.5 (a), but as many conductors as the number of turns in the coil. The coil arrangement and the meaning of the terms overhang, active length of coil etc. should be clear from Figure 7.5(b). Since there are two coil sides in each slot, and each coil has two coil sides, the number of coils on an armature is equal to the number of slots in the armature.
  • 10. ElectricalMachines& MeasuringInstruments / INSULATION LENGTH OF ARMATLIRE CORE (a) Coil Sides in Slot (b) Coil Viewed from top. Figure 7.5 :Arrangement of Coils in Slots The Commutatorand Brushes The induced voltage in each coil of a dc machine is alternatingin nature.The commutator is the means whereby this voltage is rendered unidirectionalor dc. It is made up of a number of tapered copper segments insulated from each other by mica sheets and mounted on the shaftof the machine while being kept insulated therefrom. Figure 7.6 is a simplifiedrepresentationof a commutator in relation to its,shaftand the armaturecore. (Forclarity,the commutatorof Figure 7.6 is shown with an unrealistically small number of commutator segments). There are as many commutator segmentson the commutatoras the number of coils on the armature,the two ends of each coil being connected to two segments. (In the popular 'simplex' lap winding,discussed later in this section,the two ends of a coil are connectedto adjacentsegments). SHAFT SLOTTED RMATURE Figure 7.6 :'he Commutator
  • 11. Current is led intoand out of the armatureconductorsthrough brushes which j&on Energy Conversionprinciples the commutator segmentsand are made of carbon. If the armature currentis large, and D.C Machines currentmay be supplied through severalbrushes grouped together to form a brush-set.Twopole machineshave two fixedbrush-setsdiametricallyoppositeto each other, whereas, in machineshaving four or more poles, it is usual to employ as many brush-setsas there arepoles, spacedat equal intervals around the commutator. Adjacentbrush setshave positive andnegativepolarities alternately.Brushes having positivepolarity areconnectedtogether and the leads are brought out to the positive terminal, the other brushesbeing co~ectedto the negative terminal.Thesetwo terminals arereferred to as the armature terminals. When the armaturerotates, the commutatorrotates with it, and the commutator segmentsbrush past the stationarybrushes.The brushes are made ofcarbon because carbonhas good electricalconductivity and is soft enough not to wear out the commutatorrapidly. To improvethe conductivity,sometimesa small quantity of copper is added to carbon. In order to ensure goodcontact, the brushes are housed in brush holders and kept pressed against the commutator by means of adjustable springs,whose pressure can be manually adjusted.While toolow apressure can lead to imperfectcontact and sparking,excesspressurewill increasefrictionand lead to heat production and rapid wear of the brushes. - 7.4.2 Armature Windings, EMF and Counter Torque ArmatureWindings Many differenttypes of armaturewlndings are employedin dc machines depending on requirements. The most common of these is the simplexlap, though the simplex wave is used in high voltage machines. (More elaborate windings,referred to as duplex and multiplex are also sometimesused). In this section we will confine ourselves to a &scription of the simplexlap winding, and refer in passing to the simplex wave. Discussion of armature windings iskreatly facilitated by using armature winding diagrams. Figure 7.7 (a)is the armaturewinding diagramof a two-pole dc machinehaving 8slots, 8 coils and 8commutator segments.The diagramis aconventionalisedrepresentationof the cyl~ndricalarmature surfaceassuming that it is cut along an axial line and unrolled flaton to a plane. The eight slots will then appear at equal distances fromeach other. Each slot has a top coil side and abottom coil side as explainedearlier. In the figure, the top coil sides in the activelength of the armature are representedby firm straightlinesnumbered 1to 8 corresponding to the 8slots.The bottom coil sides are indicatedby dotted straightlines close to the firm lines. Sincethe machine is meantfor two poles, there are 4 slots per pole. A coil whose top and bortom layers are fourslot pitches apart is said to be a full-pitchedcoil. We will use full pitch coils. In this case, if the top layer of a co11is in slot number.x, the bottomlayer will be in slotnumber (x+4). The triangularconnections at the top are meant to indicate the interconnection of coil sides and belong to the overhangof the winding on the side away from the commutator. Thus,the firmline in slot 1is connectedto the broken line in slot 5 to represent coil no. 1,the firmline in 2 to the broken line in 6 to represent coil no. 2 etc. In an ! armaturemeant forfouror morepoles, top coil sidesin slots 5,6,7and 8 would in fact be connected to bottomcoil sides in slots 9, 10,11 and 12. Since our armature is for twopoles and has only 8 slots, these coil sides would be connectedto the bottom layers in slots 1,2,3 and4. In the diagramwe have indicatedthe half portion of the ovehang which is connected to the toplayerby afmline,and thehalf portion connectedto thebottomlayerby a dotted line. The segmentedrectangular strip at the bottom representsthe commutator.As seen from Figure 7.6, the actual commutatoris usually of much smallerdiameter than the armature. Therepresentationin Figure7.7 (a)may be regarded as the developed view of an equivalent commutatorhaving the same diameter as the armature. In a lap winding,the ends of a coil are connected to adjacentcommutatorsegments.Thus each commutatorsegment will be connected to the top coil sideof one ,oiland to the bottom coil sideof a differentcoil. We will adopt the conventionthat the segmentnumber is the same as that of the slotto whose top layer lt is connected.In a lap-winding,the ends of coil 1will be co~ectedacross segments 1and 2, the ends of coil 2 across segments2 and 3 etc., and finally the ends of coil 8across segments8 and 1.(Thoughthe diagramappears to represent only single turn coils,the same type of diagramis often used even for multiturncoils). Induced emf Voltages areinduced in the armaturebecause of its motion with reference to the magnetic fieldsetup by the fieldsystem.With referenceto the field poles, the positions of armaturk slotsand commutator'segmentsare continually changingfrom instantto instant.Consider
  • 12. Electrical Machines & the instant when slots 1to 4 are symmetricallybelow the N-pole, slots 5 to 8 then being Memaring ImtrumenLs under the S-pole. This situation is indicated in Figure 7.7(a) by identifying the areas of armature surfacewhich are acted upon by North polar flux and Southpolar flux respectively. The correspondingflux density B at various locations on the annature surface is sketched below in Figure 7.7 (b). It is against this stationary background of flux density that the armature conductors are in constantmotion. (a)Winding Diagram @) Induced Voltage in Conductors Figure 7.7 :Two Pole ArmatureWinding with 8 Slots In a single conductor, as given by Eq. (7.2), the magnitudeof the induced voltage is givenby e = Blv, where B is the flux density in teslas, I is the active armature length in metres and v is the peripheral velocity of the conductor in metres per second. For constant speed operation,since I and v are both constants, e is directly proportional to the value of B at the slot in which the conductor is located. For a full-pitchedcoil, the voltage induced in a single turn is equal to twice the induced voltage in a single-conductorand, for a multiturn coilhaving n, turns,the voltage induced per coil is (2n,) times that of a single conductor. So, to a different scale,Figure 7.7(b) alsogives the induced voltages in coils corresponding
  • 13. to the positionsof the slots containingtheir top layers. The induced voltages in coils 1to 8 Energy ConversionPrincipIes at the instant shown in Figure 7.7(a) are representedby el toe, in Figure 7.7(b). and D.C. Machines Using Fleming's right hand rule, we can determine the directions of induced voltage of the conductorsin the slots.These directions are indicatedin Figure 7.7(a),the assumed direction of movement being from left to right. N-pole flux is assumed to enter the plane of the paper (indicatedby @ @), while S-poleflux comes out of the paper (indicatedby 0 0). (On going round any coil we find that the emf directions coincide in these full- pitched coils,indicating that the induced voltages add). In a lap winding, segmentnumber x will be corlnected to the lead entering the top layer of coilx and to the lead leaving the bottom layer of the previous coil, namely coil number (x-1). The arrows in these leads are continuations of the arrows indicatinginduced voltage rise in the respective layers. An examination of Figure 7.7(a)indicatesthat two arrows converge on to a commutator segment only at segment no. 1and diverge from a segment only at segmentno. 5. We find there are two paths from segment no.5 to segment no.1. One path begins through the top layer of coil no. 5 and, after including coils 6,7 and 8,reaches segmentno.1 through the bottom layer of coil no. 8which is in slot 4. The total voltage rise from segmentno. 5 to segmentno. 1is equal to (e, +e, + e, +e,). The other path begins with the bottom layer of coil no. 4 in slot 8 and traverses coils 3,2 and 1in the reverse direction and reaches segment no. 1 through the top layer in slot 1.Sincethe coils are now being traversed in the opposite direction,the voltage rise from segment 5 to segment 1is the negative of the voltages shown in Figure.7.7(b) and is therefore equal to 4e, + e, + e3+ e,). From the figures it is clear that both voltages are equal in magnitudeand direction.If stationary brushes are located in the positions shown in the figure at B, and B,, the voltage between these brushes will be = (e, +e, +e, +e,)= -(el + e, +e, +e4).(The voltage dropbetween segments 1and 2 =el, that between 1 and 3 =el +e,, that between 1and 4 = el +e, + e3 ,that between 1 and 5 =el +e, + e, + e4,between 1and 6 = e, +e, +e3+e, +e, etc. Thus, the maximum voltage between segments occurs between segments 1and 5). As the conductorsmove with respect to the field,the induced voltages in individual conductorsand between segments will change. However, after a movement through I slot pitch (= 1/8revolution in our case), an identical pattern of voltages will be re-established with reference to the stationary poles and brushes. (All that happens is that each slot and commutator segment is replaced by a different slot and segment, their numbers being increased by unity). Because of this the voltage between segments 5 and 1will be a pulsating voltage, 1cycle of pulsation occurring in the time taken to move through 1slot pitch. Further, during this period the brushes, which have a definitewidth, will be bridging some adjacent segments thereby short-circuiting a few coils.For the brush width shownin Figure7.7(a) coils 1 and 5 arejust about to come out fromshort-circuitby the brushes and coils 8 and 4 are about to begin gettingshort-circuitedby the brushes. 1/8thof a revolution earlier,coils 1and 5 werejust about to get short-circuited.When coils are short-circuited, the voltage in those coils will not appear in the induced voltage between brushes. The brush voltage does not therefore include the voltage of the coils undergoing short-circuit. These points are illustratedquantitatively in Example7.4. I ! Example7.4 (a) Assuming that the maximum induced voltage in a coil is 10V, and using Figure 7.7(b) for making estimates,find the magnitude of the induced voltage between segments5 and 1(i) at the instant discussed in the text; (ii) 1/16th revolution earlier; (iii) 1/8threvolution eaflier. (b) For cases (a)(i), (ii) and (iii),find the voltagebetween brushes B1and B2 assuming that coils 1 and 5 are short-circuited by the brushes. (c) Which is the positive brush and which is the negative? Solution (a) (i) Induced voltage rise from segment 5 to segment 1 = (e5+e6+ e7+ e8)= (2.5+ 10+ 10+2.5) =25 volts. (ii) Position of slots is now to the left by 112slot pitch. Estimated induced voltage rise= (e< +eg/ +e i +eSf) =(0+10+10+ 10)=30volts (iii) Position of slots is now to the left by one slot pitch Estimated induced voltage rise = ( e y +e6" +e7" +e$) = (-2.5 +2.5 + 10+ lo) = 20 volts.
  • 14. (b) From the aboveexpressions we have to omit the terms correspondingto e5, e5', e5" as brush B2 will shortcircuit coil 5. So, voltagerise frombrush B2 to brush B1 is (i) (10+ 10+2.5) =22.5 volts (ii) (10+10+10)= 30 volts (iii) (2.5+10+10)=22.5 volts (c) Since there is alwaysa voltage rise fromB2 to Bl, BI is the positive terminal ,andB2 is the negative terminal. This could alsobe deduced fromFigure 7.7(a) by noting that BI rests on segment 1where both emf arrows convergebefore coming out together from the brush terminal. The opposite happens atbrush B2 and segment 5on which it rests. The annature winding discussedabovehas very few coilsper pole and very few commutator segments.It is because of this that the voltagefluctuationscalculatedin Example 7.4 arefairlylarge. In normal machineshaving very many more coils, the distance moved by the annatureduringone pulsationcycle will be very small. Further, the maximum voltagechangesoccur in conductorslocated in regions of low flux density where the voltageis small, and hence the voltagechangecaused is quite small. Also, coils short-circuitedby brushes (ascan be verified fromExample 7.4), are also in regions where the flux densityis close to zero. Because of these considerations,for allpractical purposes we may assume that the average voltage between brushes is equal to the averagevoltage produced by the coilsin any one parallel path. Thus, if the averageflux densityunder a pole is B,, the average induced voltage in a coil is ec= 2 nc lvB, volts, where nc = no. of turns per coil, 1 = active length of conductor, m v = -N = peripheral velocity,mls. 60 S If the total number of coilson the machine= S, the number of coils in one parallel path =, , where a =number of parallel paths between the positive and negative brush-sets.The induced emfbetween the annatureterminals is therefore Now the flux41per pole = (cylindricalactive surfacearea under apole) x (averageflux 7rDl density) =(7)B, + where P =number of poleson the machine. Substitutingin Equation (7.10) we get where Z = 2 nc S = total no. of conductorson the armature and 41 = fluxper pole inwebers. In our 2-pole machine, P= a= 2 and E --mt!volts 60 InducedVoltage when P 2 4 The simplex lap winding for 4 or more poles follows the samepattern as for two poles :The ends of any coil are connected to two adjacentcommutatorsegments.Conductorsin slots under a North pole have their induced voltages oppositely directed to those under a South pole. In a two-pole machine,at any instantthere isjust one commutator segmentat which the induced voltage arrowsin the connectingleads converge (diverge)from the segment.In a P-pole machine we will find that there are (Pl2)such segments.We locate (Pl2)positive brushes so as to be in contact with the segmentsat which the arrows converge, and (Pl2) negativebrushes in contact with the segmentsfrom which the arrowsdiverge. For lap windings,the directionof coil voltagesand the location of brushes can be further clarified by using a so called ring winding representation. Here each coi1,isshown as though it &wound on aring surroundingthe commutator.Figure 7.8(a) shows a ring winding
  • 15. representationof a two-pole, 8 slot lap-windingwhereas Figure 7.8(b)is for a four-pole 16 ~ ~ ~ ~ ~ ~ ~ slotlap-winding. and D.C Machines (a) 2-pole 8-slot Armature (b)4-pole l&slot Armature i L Figure 7.8 :Ring Windlog Representationsof Lap Winding In both thesefigures,from each negative brush twoparallel current paths divergeand at each positivebrush two parallel paths converge. So,in a P-pole lap winding, since the Pnumber of positivebrushes =number of negative brushes =- the number of parallel paths 2' in the armatureleadingfromthe +ve armature tenninalA+ to the -ve armature tenninalA- is a =P.The induced voltage in a lapwinding for any number of poles P is therefore given by -volts 60 Another fairly common armature winding is the simplex wave. Here, in a machine with P poles the number of coils connectedbetween adjacentsegments is (Pl2) (insteadof unity, as for the lapwinding).The wave-winding (which we willnot discussin this course),has only two parallel paths. Consequently,the induced voltage is givenby
  • 16. Electrical Machines& Meaquling Instruments CounterTorque If an electrical load,e.g. a resistance, is connected across the armature terminals, a current will flow through the resistance. Referring to Figure 7.8(a) or (b), if a resistance is connected across the armature terminals A+ andA- the d.c. generator will supply an armature current I, flowing from A+ toA-. Since the armaturehas a parallel paths, each parallel path will supply - ,and this is the current through the conductors on the (1.1armature. An examination of the winding diagrams, or even Figure 7.8 shows that this current is in the same direction as the induced emf. Thus, referring to Figure 7.7(a),the arrows indicating the directions of induced voltage rise also indicate the directions of current. We can compute the direction of the force on any conductor, whether it is under a N-pole or a S-pole using Fleming's left-hand rule. When we do so we find that all forces are in the same direction, viz. from right to left. But the direction of motion assumed was from left to right. Hence all these forces act in a direction opposing the motion. The magnitude of the force on any conductor is given by Eq. (7.8), viz., F= Bil. The .average force on any conductor is therefore Fa, = Ba [: )1 newtons. For an armature diameter D, the average torque contributed by each conductor is If there are Zconductors on the armature, since all conductor torques act in the same direction,nett developed torque But :. T nett torque produced = Z fi 1 I ( 2 ) n D l [an) = KT$ I,, where KT = [E) As we have just seen, this torque in a dc generator is a "counter-torque" as it acts in a direction opposing the motion and the driving torque of the prime-mover. Example7.5 A 10-pole dc generator has a simple wave-winding, 101 slots, 2 coils sides per slot and 3 turns per coil. (a) Is a full-pitch winding Fossible? If not, what is the best you can do? (b) If the machine generates 600 V and supplies 400 A current what is (i) the average voltage generated per turn (ii) the current flowing through a conductor? (c) If the armature is re-wound as a simple lap winding, for the same developed voltage per turn and current per conductor, what is (i) the voltage developed by the machine (ii) the current supplied? Solution 101 (a) Number of slot pitches perpole -= 10.1 is not an integer. So, a truly 10 full-pitched coil is not possible. The best would be to use a coil with sides 10 slot pitches apart.
  • 17. (b) Number of parallel paths in simple wave = 2 Average number of turns per parallel path = 101 x 3 2 = 151.5 Therefore average voltage per turn = -'0° -- 3.96 volts 151.5 la 400 Currentper conductor = - = - a 2 (c) For a lap winding there are 10parallel paths. Average number of turns per path = 101 x 3 10 = 30.3 Therefore average voltage developed = 30.3 x 3.96 = 120V Current supplied by 10parallel paths = 10x 200 = 2000 A The armature current that the lap winding can supply is thus 2000 A. Energy ConversionPrinciples and D.C. Machines 7.4.3 Excitation Schemes and Magnetization Characteristic From an electriccircuit point of view, a dc machine is characlerisedby ils annature and field windings. The armature is connected into an electrical system through the two terminalsA+ andA- shown in Figure7.8. The magnetic field is set up by one or more coils on each pole. A seriesconnectionof one coil fromeach pole, having two end terminals to whch external connectionsmay be made, constitutesa field winding. Field windings are of two kinds,one being a shunt winding meant tobe connectedacrosseither the armature voltage or an equivalentseparate voltage. The other is aseries winding, meant to be connected in serieswith the armature and designedto carry the armalurecurrent. The shunt field has comparatively many turns, a high resistanceand is meant to carry a maximum current which is only a fraction (about 10%or less) of the full-load armature current. The series field,because it has to carry the full armature current, is of a few turns and low resistance. DC generatorsuse severaldifferent schemes of energising the field windings, resulting in different excitation schemes.A diagrammaticrepresentation of these schemesis given in Figure 7.9. These diagramsemploy the accepted method of representing the annature and its brush systemby a circlebetween two small rectangles.The field windings are represented as single coils, a shunt type field winding being shown with more turns than a series type winding.In all these figures,the terminals through which the armature gets connected across an electricalload or into an electrical system are indicatedby the polarity signs +,- appearing across the load voltage V,. Figure (a)denotes a schemeof separate excitation in which the field winding, in serieswith a variable resistance for adjusting the fieldcurrent If is connectedacross a separate field voltage V/.In (b),the field excitationis not obtained from a separatevoltage source,but is connected across the armature terminals and the voltage generated there. This is thereforecalled a self-excited shunt excitation scheme. Figures (c)and (d) are also self- excited schemes,but use in addition to a shunt field winding, a series field winding also. They are referred to as cornpound excitation schemes. In (c),the shunt field winding comes directlyacross the armature terminals,the seriesfield winding carrying only the load current.In (d), the shunt field winding is connected, not across the armature,but across the load voltage V,, and the series field carries the armaturecurrent la,which is different fromI,. Figures (c) and (d) are respectively referred to as short-shunt and long shunt. PositivecurrentsI, and I, (or I d in the shunt and serieswindings may set up fieldswhich reinforceeach other (representedby firm arrows in the same directionfor both $ and $,) or oppose each other (representedby a firm arrow for $f and the dotted arrow for @, ,the arrows being in oppositedirections). If the fieldsreinforce each other, the compoundingis said to be cumulative.If they oppose each other,the compounding is differential.
  • 18. Electrical Machines & Messwing htnunents (a)Separate excitation (b)Shunt excitation if-- a A. #f -+ - A 1 -s 0 6-- VL 8 I I (c)Short-shuntcompound - (d)Long-shuntcompound Figure 7.9 :Excitation schemes for D.C. Generators The "magnetizationcqaracteristic" correspondingto the shuntfield winding alone is the relationshipbetween the voltageE, induced in the armature at a constantspeedand the field current$ From Eq. (7.1I), this induced voltage is givenby PZ 27t= K N N $ , where KN = -= ---KT a60 60 (7.15) If the speedis held constant, this voltage is directly proportional to the flux per pole. Now, as the fieldcurrent is increased, the flux $ will increase approximatelylinearly for low values of currents,and increaseat a slowerrate for higher values due to the phenomenonof
  • 19. magnetic saturation.The magnetisationcharacteristicat a speed N will be as shown in FigGre 7.10. Since$, the flux per pole is proportional to Eawhen Nis constant, to a Energy ConversionPrinciples and D.C Machines ,---SPEED =kNc - - ---SPEEDSPEED,= Nk N R % / - LKNEE Figure7.10 :Magnetisation Characteristics different scale, any one of the curves in Figure 7.10 can also be regarded as relating $ to I/. At a speedkN, the voltage correspondingto each value of I, will be k times the value for speedN, as shown by the dotted curves in the figure. The curved portion of the characteristicis called the "knee" of the curve. The rated voltage is generally selectedto be the voltage E,, just above the knee, for the case when N is equal to the rated speed of the machine. The flux $ per pole is due to the magnetomotiveforce set up by the field windings. If the shuntfield is acting alone, on each pole the magnetomotiveforce is N, If where N,is the number of turns on a pole. If a series winding is also operating, the current through the coil being I,, , field turns N,, , an additional magnetomotiveforce N,, I,, would be in operation. The nett magnetomotive force is then (N,If+ N,, I,,), the plus sign corresponding to cumulative action and the minus sign to differential action.This magnetomotiveforce is Ulerefore equivalentto a current If' flowing through the shunt field winding alone, where N I + N,, I,, + [ , )={+*(%)i] Using the value of If' as calculated above, magnetisationcharacteristics such as those of Figure 7.10 can be utilised for determining the induced voltage in the armature of compoundwound generators. Self-Excitation On the applicatibnof a voltage Vf across the field circuit, the current Ifflows through the field winding and a flux $ will come into being. So, the process of voltage generationin a separately excited generator is easy to understand. In the self-excited schemes, this is far from clear. This is because, to start with, when the generator is at rest, nolvoltages exist anywhere and so the field current Ifcan only be zero. If we now bring thk armatureupto a speedN it is not at all clear if there will be any Ifat all, and if it exists, what its value will.be. Actually if a shuntgenerator is being rotated at a speed N, with the switch Sin Figure 7.9(b) kept open, and the switch is closed at t = 0, the generator can "self-excite" provided there is some residual magnetism in the irorr Consider the magnetisationcharacteristic shown in Figure 7.11,where we have assumed that there is residual magnetism,i.e, even with I,= 0, a residual voltage E,is induced in the armature at speed N. Let the current Ifbe a function of time I,(t)*. *. In this discussion we shall make a departure from convention and use capital letter&for the time varying quantities If (t ) and Ea(t)to facilitate reference to the magnetisation characteristic relating steady stateE, to If. I 53
  • 20. ElectricalMachines& Measuring lastmrnents CURVES FOR WHICH V = If RT 0 * If Then, the total flux linking the field winding will also be a function of time given by h,(t). The current Ifflowing through the armature will produce a flux linkage h, (t) with the armature. Let the total resistance in the circuit formed by the armature, the field winding and the variable resistance Rfbe Rj-. The flux linkages hf and ha are both stationary in space and so whereas a speed voltage E,(t), related to If(I) by the magnetisation characteristic, is (lh d 1, produced in the armature, the additional emfs are only the transformer emfs A and - (it (lt We may regaid the speed voltage as balancing the resistance drop in the circuit and the transformer emfs according to the equation dh, . d h In practice, -is negligible compared to and we may write dt dt Referring to Figure 7.11, if the field current is glven by OA, at time t, Ulen AC = E,(t), AB d h d h = IfRTand so BC corresponds to --f .So, at time t = 0,-2= E, and so hf must increase. dr dl This is possible only if the field current ifresponiihle for producing $increases. So both current Ifand flux hf increase to new values as t increases. As long as the magnctisation characteristic lies above Ule V = RTlfcurve, both Ifand hf will continue to increase, the rate d h of increase --f corresponding lo the amount by which E, exceeds RTIf at that instant. This fit rate of increase becomes zero when If=OP, and so this is the final value reached by If. The armature voltage, after a transient period, has therefore built up from the value .Fr to the value give11by PQ,. If Rf and therefore RTis increased, the slope of the V = If RTcurve 'keeps increasing and the voltage to which the armature builds up will move to the left along - the magnetisation characteristic. Beyond a certain value R, (called the critical resistance) the built up voltage will abruptly fall to a small value as at Q,. 1; there is no residual magnetism in the field circuit, this build up process can not begin. Also, even if there is residual magnetism, but it is in such a direction that any field current produced by it tends to decrease the original flux, there is reduction i11Uie voltage arid, therefore the voltage build up process car1not take place. Further, as we saw earlier, if thc resistance in Ule field circuit exceeds a critic211value, Ule build up process call only lead to a very slight change in the voltage from the residual voltage. So, we must ensure that thcre is residual flux. Next we must ensure that the resistance in the field circuit is less than the critical value. If despite these two conditions being met, the armature voltage does not build up, this must be because the polarity of Uie residual enlf is in the wrong dircclion. This can
  • 21. be rectified either by reversing the direction of current flow in the fieldby interchangingthe E n e w anversionPrinciples field terminalsor by reversing the directionof rotation. and D.C. Machines Example 7.6 When separatelyexcited, the opencircuit characteristicof a shuntgenerator at 1500 rpm is given by: If(A) 0 0.2 0.3 0.4 0.5 0.6 0.7 0.8 E,(V) 6 96 142 180 215 240 258 264 The correspondinggraph is given in Figure 7.12 (a) If the shunt field resistance is 100ohms, what should be the value of external resistance added if the machine is to generate 200 V at (i) 1500rpm; and (ii) 1200rpm? (b) What is the critical resistance at (i) 1500rpm (ii) 1200rpm? Solution Refer to the graph shown in Figure 7.12 (a) (i) Field current required fo generate 200 V at 1500rpm =0.45 A 200 ResistanceRTrequired is therefore - = 444.4 R 0.45 Therefore externalresistance to be added = (444.4 - 100) = 344.4 R Figure 7.12:Graphfor Example7.6 (Speed= 1500 RPM)
  • 22. Eleetrid Machines& (ii) 200 V at 1200rpm is equivalent to -1500 x 200 = 250 V at 1500rpm. The Me~slvinglmtrmneats 1200 correspondingfield current is therefore ( as seen from the graph) 0.65 A. 200 ResistanceRT required to yield this current = - 0.65 Therefore external resistance to be added = (307.7 - 100) = 207.7 i2. PR 150 (b) (i) Considering the tangent to the graph, the critical resistanceRT =-= - PQ 0.3 = 5 0 0 n Since the graph corresponds to a speed of 1500rpm, this is the critical resistance at 1500rpm. 1200 (ii) If the graph were redrawn for 1200rpm, all ordinates would reduce to -= 1500 0.8 times their values at 1500rpm. Therefore critical resistancecorresponding to 1200rpm = 150 x 0.8 0.3 7.4.4 Load Characteristics By adjustingthe fieldcurrent Ifin the separatelyexcited generator, or the value of the variable resistanceRfin the case of the self-excited generator. the voltage VLcan be adjusted to a desired value (rated value usually) on "no load, i.e., with I,, = 0. If now the load resistance is decreased the load current will increase.When this happens, the load voltage VLwill, in general, undergo a change because of armature reaction and armatureresistance. Armature Reaction When a current I, flows in the armature, a magnetic field is set up which interacts with the magnetic field produced by the current in the field windings. l'his interaction is referred to as armature reaction.When armaturecurrent flows in a generator, the direction of current flow is the same as that of the induced voltage rise in the armatureconductors. Hence, referring to Figure 7.7(a), the arrows shown represent the directions of current flow also. We find that in all the conductors under the S-polecurrents flow from top to bottom, whereas they flow from bottom to top under the N-pole. The action of these currentb is to set up a South polar flux in the interpolarregbn with axis between slots 4 and 5 and a North polar flux in the interpolarregion between slots 8 and 1.The effect will be to add to the magnetic field produced by the field windings in the,right half of the polcs and subtract frorgthis flux in the left half of the poles. If the iron did not show any saturation effects, Ihe flux added would be exactly equal to the flux subtxacted,leaving the flux produced per pole by the field windings unaffected.However, because of saturation, the increase in flux in the right half of the poles tends to be less than the decrease, leading to a nett reduction in the flux per pole, the effect being more pronounced at higher values of flux per pole. Figure 7,.13indicatesthe manner in which different values of armature current affect the flux per pgle. In the figure we have plotted the flux per pole vs If ', where If' is the equivalent field cdrrent as given by Eq. (7.16). 1; Figure 7.13 :Effect of Armature Kea~T~onor1I-lux per Pole
  • 23. ResistanceDrop ELergy ConversionPrinciples The induced voltage produced by the motion of the armaturein the magnetic field, being and D.C. Machines proportional to the flux per pole at constant speed,will be influencedby armaturereaction. However, this voltage will not be available as the voltage VLacross the load because of the voltage dropproduced by the armaturecurrent flowing through the resistanceof the armature and the brush contact (See Seclion 7.3.1). If the induced armature voltage is E,, for generatorswith separate or shunt excitalion VL = E, - I, Ra - 2 V, , (7.18 a) where R, = resistance of the armature winding and leads, VB = Brush contact voltage drop (- 0.8 to 1.3V, often taken to be 1 V) I, = armature current The term 2VBoccurs because in any closed path through the load two brushes will appear in series.For the compoundwound machines there is an additionalresistance drop I$,,, in the resistance R ,of the series winding. Hence, for long-shuntcompound generators, For short-shuntconnections I, R,, in the above equation should be replaced by I, R,,. Load Characteristics In a separatelyexcited generatorbeing driven at the rated speed by a prime mover, let a field current If,produce the rated voltage Eoon no load. Then, as load current increases from0, as seen from Figure 7.9 (a),I, =I, also increases.Becauseof armaturereaction, there is an increasedreduction in the flux per pole and so the induced voltage falls as given by curve (a) in Figure 7.14. (In this graph, I, is the variable on the x-axis while voltage is the variableon the y-axis). If we neglect the brush contact voltage drop there is a resistance drop I,R,. This has been plotted at the bottom of the figure. From Eq. (7.18a), the terminal voltage V, =E, - I,R, and will be given by curve (b). Curve (a) is referred to as the "internal characteristic"and curve (b) as the "external characteristic" of the separately excited generator. In the shuntgenerator(Figure 7.9(b)) for the same rated speed, the samefield current If,will induce an armature voltageEo.However, since I, =If +I,, I, = If for I, = 0. Due to current V, will be slightly less thanE, because of armatureresistance drop If Ra evtiSn at no load. So, for VL= Eo,the field current will have to be adjusted to a slightlyhigher value by varying Rf.Then, as the load current increases, there will be reductions in voltage due to ,, both armaturereaction and armatureresistance drop. However, since the field current depends on VLfor a given field branch resistance as seen from Figure 7.9 (b), unlike the fieldcurrent in the separatelyexcited machine, Ifdoes not remain constantat its initial value but falls with the voltage. Therefore the external characteristicof the shunt generator will be as seen in curve (c),and liesbelow the external characteristicof the separatelyexcited generator.The addition of I$, to curve (c) yields curve (a'), which is called the internal characteristicof the shunt generator. Id2) Id11 (a1 (a'1 (b1 Figure 7.14 :Load Characteristicsof D.C. Generators
  • 24. EloaridMachines& . In the case of the sh~rt-shuntcompound generator (Figure 7.9(c)), from Eq. (7.18 b), the Memuring Imtrumed effective field current will differ from the actual field current Ifand will be given by I; = [I j k ),'+' for cumulative and '-' for differential operation. As a consequence, the internal and external characteristicsof the cumulatively compounded generator will lie above that of the shunt generators, while these will lie below in the case of the differentialcompound generator. Curves(d,) and (d, ) are typical external characterestics for the case of cumulativecompounding.In curve (dl), the terminal voltage at full load is equal to that at no load. and the generatoris said to be level compounded. If with cumulativecompounding,the external characteristiclies above (dl), (as for curve (d,)), the generator is said to be over compounded.If the curve lies below curve (d,), the generator is said to be under compounded.Curve (e)shows the external characteristicof a differential generator.The characteristicsfor the long-shuntconnection are very similar, marginal differences appearing because the current though the seriesfield winding is (If+Id instead of merely Ic Example7.7 A 25 kW, 250 V shunt generator is driven at a speed of 1000rpm. The armaturehas 4 poles, and employs 100coils having 2 turns each, in a simple lap winding. Neglecting brush voltage drop and armaturereaction, and taking the resistanceof the armature as 0.03ohms and that of the fieldcircuit as 250ohms, find the induced emf and the flux per pole at full load, rated voltage. Solution Refer to Figure 7.9 (b). 25000 - l00ALoad current at full load =-- 250 Field current If=250250 = 1A I, = 101A Ea= V, +I, R,= 250 +101x 0.05 = 255.05 V 2,Total no of conductors = 100 x 2 x 2=400 SAQ 5 Acompound generator delivers full load power of 50k W at 250 V to a load. The resistancesof armature,shunt field and series field are respectively R,= 0.012Q, Rf= 62.5 !2and R, =0.006 !2. Neglect armaturereaction. Find the induced emf in the armature if the contact drop at a brush = 1 V. The generator uses a short shunt connection. 7.5 DIRECT CURRENT MOTORS DC motors are extremely versatilemachines used for driving a variety of devices such as pumps, fans, hoists and vehicles. In a dc generator the speed is fixed by the prime-mover and field conditions are adjusted in order to provide the desired terminalvoltage at a particular load current. In a dc motor, on the other hand, the need is to match the speed torque characteristicof the driven device (which constitutes the load on the motor) by adjustment of the field conditionsand, if required, the armature voltage. Electric motors may be classified according to their speed-torquecharacteristics as constant speed,
  • 25. adjustable speed and variable speed motors. A constantspeed motor will maintain a nearly Energy Goversion Principles constant speed at all loads. An adjustablespeed motor is one in which adjustmentscan be and D.C Marhies made to vary the no-load speed of operationover wide-limits, the adjusted speed remaining fairly constantat all loads. A variable speed motor has an inherentcapacity to vary its speed with the load, usually, by reducing speed as the torque increases,so that excessivepower is not drawnby the motor at high torques.In this section we will learn the characteristicsof shunt,compound and seriesexcited dc motors and their speed torquecharacteristics. 7.5.1 Excitation Schemes for DC Motors As stated in Section7.4.1,both dc generators and dc motors are built in the same way and have the sameconstructional features. Further,dc motors employ the same excitation schemes as dc generators. In fact the connection diagramsfor separately excited,shunt excited and compoundexcited dc motors can be readily obtainedfrom those shown in Figure 7.9 by replacing the load voltage V, by supply voltage V,. If the shunt field winding in either Figure 7.9(c) or (d)is disconnected,we will be left with a series excited generator. Series generators are quite uncommon. However, seriesexcited dc motors are widely used in a variety of applications. Figure 7.15 (a), (b), (c) and (d) show dc motor connectionsfor (a) separatelyexcited,(b)shunt,(c) series and (d) short-shunt compound motors. In these figures the supply voltage V, is shown as a dc battery. - (a) SeparatelyExcited Motor - (b) Shunt Motor -(c) Series Motor
  • 26. (d) Short-shuntCompoundMotor Figure 7.15 :Excitation Schemes For DC Motors 7.5.2 Motoring Action and SpeedEquation The transition fromgenerating to motoring action canbe readily understood by considering the separatelyexciteddc machine of Figure 7.15 (a).Imagine that the fieldexcitation is held constantand that the shaftis supplied at all times with such a torque that the speed of the motor is alsoheld constant in both magnitude and direction. Under these circumstances. neglecting armaturereaction, the induced emf Eawill alsoremain constantand will be given by Eq. (7.11).viz The annature will generate electricalpower if the armature current I, flowsout of the positive terminal of the armature.For the polaritiesshown in the figure for E, and Vs,this can occur only if Vsis smallerthan E,. The armaturecurrent,(neglectingV, or incorporating its effectin R,) will be, from Eq. (7.18a) The electricalpower generated in the armature is then Eh,, that absorbed by the supply is VsIuthe difference I: I?, being dissipated as heat in the armatureresistance.There must be an input of mechanicalpower at the shaftfor conversion into the electricalpower generated, implying that the shaft torque must act like a prime-mover torque in the same direction as that of rotation. The armaturecurrent,interacting with the magneticfieldresults in a counter-torquesuch that the nett torque on the shaftis zero, thus ensuringrotation at a constantspeed. If Vsis gradually increased,the armaturecurrent will fallbecoming zero when Vs=Ea indicating neither generationnor absorptionof power in the armature.Under these circumstancesthe counter-torque is zero and the shafttorque will be small,being required to overcome merely the frictional torque. Vs=E,, therefore marks the transitionfrom generatingtomotoring action, for,if V, is increased further,the armaturecurrentwill reversedirection,flowingout of the positive terminal of Vsintothe positive terminal of E, implying that the armatureis now absorbingelectricalpower fromthe supply voltage Vs.As the directionof the magneticflux is unchanged,the torque produced by armature current will reverse.This torqueis thereforeno longer a counter-torquebut a motoring torque as it acts in the samedirectionas that of rotation. Again, for the rotor to run at constantspeed, the nett shafttorque must be zero implying that the torque on the shaft must now constitute a braking or load torque. Motoring action requires that Vsis greater than E,. The direction of flow of I, shown in Figure 7.15 (a) and in all other schemes of Figure 7.15, is consistent wiq motoring action. SpeedEquation : The supply voltage and the magneticflux per pole, together with the armature current, determine the speedof a dc motor. For motoring aetion,since Vsis greater than Ea,the equation corresponding to Eq. (7.18 a) for generators, for the currentreferenceshownin Figure 7.15becomes
  • 27. for separateor shunt excitation, and E,= Vs-IaRa-IsRse-2 VB for seriesand short-shuntcompound excitation. From Eq. (7.1I), Hence,the speed can be calculated as EnergyConversionPrinaples and D.C Machines (7.20 b) PZ where KN=- 60a Example 7.8 A separately excited dc motor is running in the clockwise direction .What is the effect of (a) reversing the supply Vfalone ;(b) reversing the supply VSalone; (c) reversingboth band VS? Solution I The torque reverses in direction if either the direction of the armature current or that I of the magnetic flux reverses. So, in cases (a) and (b), the direction of the developedtorque, and hence that of rotation,reverses,becoming counterclockwise. In case (c), since the directions of both the armaturecurrent and the field reverse together, the direction of the torque, andhence that of rotation,remains unchanged. SAQ 6 What is the effect of reversing the polarity of the supply voltageon the direction of rotation in the case of the shunt, series and compoundmotors shown in Figure 7.15? Comment. SAQ 7 A dc sh nt machine generates an armature voltage of 200 V on no load at 1000rpm, the fiedcurrent being 2 A. Ths machine is next operated as a shunt motor on a 200 V supply.If the armature resistance is 0.5 a,find the speed of the motor if the supplycurrent is (i) 5 A (ii) 50 A (Neglectarmature reaction and'brush,voltage drop).
  • 28. Electdcd Machims& Mmuring Instrmoene 7.5.3 Shunt Motors Speed-torqueCharacteristics Consider a dc shunt motor running at no load on rated voltage with the field current held constant.Under steady conditionsthe motor will run at a constant speed, the electricallyproduced driving torque beingjust sufficientto balance the no load windage and friction torque acting on the rotor. As seen from Eq. (7.14), for a constant flux the torque is proportional to the armature current. So, the no load armature currentof tlie motor will be quite small, being ideally equal to zero. On a mechanical load being suddenly applied to the shaft, since the driving torque due to armature current is quite small, there will be a braking torque on the shaft causing the motor to decelerate and slow down. As the motor slows, the counter-emf (Eq. (7.15)),being proportional to speed at constant flux,will reduce. This in lum from Eq. (7.20 a),results in an increased armature current and increaseddriving torque. Finally, a new steady statewill be reached at a lower speed with ahigher armaturecurrent such that the nett torque on the rotor is once again zero. The speed-torquecharacteristicof the shunt motor thus has a droopingcharacteristicas shown in Figure 7.16. As seen earlier,armature reaction leads to a furtherdecrease of the flux per pole because of tnagnetic saturation.Calculationof the speed on load, neglecting armature reaction, has already been illustrated in SAQ 7: Example 7.9 furtherillustratesthe calculationof torque, again neglecting armaturereaction. The reduction with load of the flux per pole caused by armature reaction will lead to increased speed and armature current for a particularload. Example 7.10(b)requires calculationswhich take this flux reduction into account. CURVE SPEED NEGLECTING I ARM-REACTION I I FULL LOAD TORQUE I/ TORQUE Figure 7.16 :Speed torque curveof a shunt motor Example7.9 A 20 kW,250 V shunt motor has an armature resistanceof 0.3 C2 and shunt field circuit resistanceof 100a.At no load and rated voltage the speed is 1500rpm, the armature currentbeing 5 A. If the full load line currentof the supplyis 90 A, find (i) the full-load speed and (ii) the developed torque. Neglect armature reaction and brush contact drop. Solution Refer to Figure 7.15(b) 250 Field currentIf=-= 2.5 A 100 At no load, Ea= 250 -0.3 x 5 = 248.5 V = KNQOX1500 ThereforeKN,Qo= -z:i -- 0.1657 At full-load,armature currentI, = (IS-If) = (90 - 2.5) = 87.5 A ThereforeEa= 250- 0.3 x 87.5 = 250 - 26.25 = 223.75 V =KNQo NL where NL= speed on load. E, -(i) ThereforeNL=- - ---223'75 -- 1350.3 rpm KNQo 0.1657
  • 29. 27c (ii) Also, as shown in the text, KN=-KT 60 60 Therefore = -KN@o = x 0.1657 = 1.5823. 27c 27c Therefore full load torque TL= 1.5823I, = 1.5823x 87.5 = 138.45Nm Starting In order to ensure low losses and good efficiency,the voltage drop at full load due to armatureresistanceand brush contact is designed to be just a few percent of the rated voltage.When the motor is at standstill,no counter-emf will be generated in the armature. Consequently if full voltage is applied to the armature of a stationary motor, very heavy armaturecurrents will flow,causing damage to the commutator and the armature winding. To prevent this, a starter is required to be used in order to ensurethat the armature current does not exceed about twice full load current during the starting process. Essentially,the starter introduces a resistance in series with the armature such that the initial current is restricted. Because of this starting current the motor accelerates from zero speed,and a counter-emf appears reducing the armature current. In a commercialstarter, the series resistance is cut out in steps and finally eliminated altogether when the machine comes up to full speed. A more detailed description of a commercialstarter is presented in Unit 9, Section 9.5. Stopping When a large dc motor is coupledto a heavy inertial load, if the supply is merely switched off, the motor may take an hour or more to come to a stop. Such a long period for stopping is often unacceptable and it becomes necessary to apply a braking torque to ensure a quick stoppage.Electrical methods of achieving this include (a) dvnamic braking and (b)plugging, in both of which the field is kept energised and the armature is made to carry a current in a directionopposite to normal so that a braking torque is produced. A brief discussion of these methods is also presented in Unit 9. Section9.5. Speed Control The speed of a dc motor is given by Eq. (7.21),repeated below : Energy Conversion Principles and D.C. Machines Speed control can be effectedby (i) armature control, wherein the flux is held constantand E, is varied, (ii) field control, where the armature voltage is held constant and the field current is varied or (iii) by a combinationof the two. In armaturecontrol, one practical way of changing E, is to alter the armature supply voltage Vs. In the Ward-Leonardmethod of control, illustrated in a rudimentary form in Figure 7.17, Vs is varied by connecting the annature of the motor M to be controlled to the armature of a variable voltage dc generator G. The field current of the generator can be varied from zero to a maximum value in either direction.By this means the generator voltage can be made to vary from zero to maximum with any polarity. Even if the field current of the motor is kept fixed, the variable nature of the generator voltage makes it possibleto run the motor from zerospeed to a maximum value in either direction. Varying the field current gives an additional method of control. Figure 7.17 :Ward-Leonard Method
  • 30. ElectricalMachines& Memuring ~ ~ e a C s In field control,the flux $ is varied by controllingthe field current.Referring to Figure7.15 (b),this is quite simply effectedby changing the resistanceof the field rheostatRfshown in the figure. If the field current is abruptly reduced by a small amount, the induced emf will alsoreduce abruptly by a small amount.However, the Vs - VB- Ea armaturecurrent given by I, = will increase by a much higher R, percentageresulting in increased torque.This will cause the motor to speedup, increasing the emf and reducing the armature current and acceleratingtorque,a higher value of steady speedbeing finally attained. Applications For a fix@ field current, the shunt motor is essentiallya constant speed motor whose speed drops by 5 to 15% fromno load to full load. The no load speed can be adjusted by varying the field current, the amount of adjustmentpossible dependingon the manufacturer's design specification. Because of its good speed regulation,the shunt motor is used for applicationsrequiring approximatelyconstantspeed servicesuch as in lineshafts, milling machines,individual drivesfor lathes, centrifugalpumps, conveyors,fans etc. Example 7.10 The magnetisationcharacteristicfor the shunt winding of a dc motor is the same aq that given in Figure 7.12 of Example7.6. When delivering full load, the shunt motor draws an armaturecurrent of 20 A from a 230 V supply,the field current being held constantat 0.5 A. (a) Find the speed and developed torque at full load, given that the armature resistance is 0.75 R.Neglect armature reaction. (b) Find the actual speed and developedtorque if armaturereaction reduces the flux per pole'by 7%. Solution Refer to Figure 7.15(b). (a) From the graph,the emf induced at 1500rpm by a field currentof 0.5 A =215 V. On full-load,the induced emf in the armature, E, = (230-20 ?< 0.75) =215 V. ~ e n c ethe speed is 1500rprnneglecting armature reaction effect. 215 KN$o=-a 60 215 1500 ,therefore KT oo=- x -2n 1500 60 215 x20=27.4NmThereforetorque =KTOola=- x - 271 1500 (b) Let the flux per pole, taking armaturereactioninto account be (o. Then, $0 =0.93 4,. The new value of KN (b - 215 - x 0.93 '1500 - 1612.9rpmThe new speed = --0.93 The new torque =0.93 x 27.4 = 25.5 Nm. 7.5.4 SeriesMotors In a series motor, the flux per pole $is establishedby the armaturecurrent which flows through it. If there wereno magneticsaturation,this flux would be directly proportional to armature currentand could be represented as 0=KI, .The torque in this case would be given by T=KAIa = K~KI:,and would be parabolic in shape. However,because of saturation,as I, increases0will increaseat a slower rate. On a constant voltage'supplyV, to the motor, E, will decrease from V, for zero armaturecurrentto slightly less than V, at full load. If we treatE, as approximately constant, the speed N given by
  • 31. will yield a speed-armatdecurrent characteristic which is ahyperbola. Figure 7.18 gives Energ w v e * M n d r k the torque-currentand speed-currentcurves of a seriesmotor. In the figure,dotted curves and D.c Machines are theoreticalcurves which neglect the effectof saturationwhile the firm lines give the actual curves. Figure 7.19 illustratesthe speed-torquecurve of the series motor. When a series motor is operated from a constant supply voltage, its speed at light loads is very much larger than at full load, the speed droppingrapidly as the torque increases. In fact,on no load the speed will be dangerouslyhigh (theoreticallyinfinite),and hence series motors shouldnot be allowed to function onno-load. Figure 7.18 :N, Tvs I, of a D.C. Series Motor I C RATED 7TORQUE Figure 7.19 :Torque-speedCurve of a L).C. Series Motor The speed-torquecharacteristicof a seriesmotor is excellentfor traction,as the large value,' ' of armaturecurrent when the annatureis at standstillprovides a large starting torque and quick acceleration.Also,under steady operating conditionsincreased torque is met by decreasedspeed. In electrictraction this means that the motor will automatically slow down when climbingup a gradient,and run at good speed on level ground. Seriesmotors are also used in cranesand hoists,where light loads will be lifted quickly and heavy lopds slowly. Series motors are alsoused in devices which must run at high speedson light loads. The startingmotorsused in automobilesare also series motors.
  • 32. Elect-ticdMachines ,% SAQ 8Memuring Imirumen'q A series motor has a magnetisation characteristicat 1500rpm given by E,(V) 6 96 142 180 216 240 258 264 The resistance of the annature is 0.5 Q and of the series field also 0.5 Q. On a supply of 230 V, the motor is found to draw an armature current of 20 A at full-load. Find the speed and developed torque (a) at full-load; (b) when the arm& currentis 10A. 7.7.5 Compound Motors Compound motors may be connectedin long shunt or short shunt as in generators.(The long shunt connection is not shown explicitlyin Figure 7.15). As in generators,a compound motor is said to be cumulativelycompounded if the magnetomotive forces of the series and shunt windings aid each other, the compounding being diflerentinl when they oppose each other. Since the flux per pole 41increases with load in a cumulativelycompounded motor, the speed will fall more rapidly with load than with shunt excitation alone. Like a series motor, though to a lesser extent,high starting currentswill enable the motor to gencrate higher startingtorques than the shunt motor, and increasing torque will make the speed drop more rapidly. A differentiallycompounded motor, as the flux per polc reduces with increasingarmature current, can provide a rising speed-torque characteristic.If a small number of series turns are used, it is possible to ensure that the no load and full load speeds are equal,producing a motor which has a more nearly constant speed than the shunt motor. However differcntial motors are rarely used becauseof the risk of dangerouslyhigh speeds in the event of an accidential overload in which thc scricsfield magnetomotiveforce becomes large enough to cancel the shunt field resulting in vkry small flux per pole. For comparison,speed-torquecharacteristicsof compound motors are shown along with those of series and shunt motors in Figure 7.20. SPEE I I C TORQUE Figure 7.20 :Speed-torqueCurves of I1.C. Motors RON A compound motor with a strong series field has a widely variable speed and, when a large torque demand of short durationoccurs, it decreases its speed much more than a shunt motor and thus allows its kineticenergy tohelp carry the load. This effect cm be increased by increasingthe inertia of the motor by adding a flywheel.Such motors are common in punch presses where the motor slows down slightlyduring its power stroke, thc kinetic energy stored in the flywheel doing most of the work. Between punching operations, the motor brings the flywheelback to normal speed. Compound motors are also used wherever slightly variable speed is an advantageas in rolling mills. Sometimesthey are also used in
  • 33. hoists and elevators where quick starting(provided by the series field) is followedby a E n e w ConversionPrinciples fairly long period where a fairly constant speed is required,during which time the series and D.C. Machines field can be removed from operationby short-circuiting it. 7.6 SUMMARY The extensiveuse of electrical generatorsand motors was emphasized in Section7.1, which introduced the Unit. In Section 7.2 you were introduced to Faraday's law of induction and the Blv rule for computing emf in generators and back-emf in motors. You next learnt to use the Rlirule to compute forces on conductorsas also torque in motors and counter-torque in generators. In Section 7.3, you learnt aboutcopper losses,iron losses and mechanical losses in electrical niachiiies and their importantrole in causing temperature rise and fixing the efficiency and rating of electrical machines. After a brief introduction to the constructional features of dc machines, you leamt'in Scction7.4 how to calculate the emf and torque in armatureshaving simple lap or wave-windings.The Sectionalso considered shunt,series and compound schemesof excitationof generators,magnetization characteristicsand load characteristicstaking armature resistance and armature reaction into account. Section 7.5 dealt with motoring action, the speed equation of dc motors and their torque-speedcharacteristics. 7.7 ANSWERS TO SAOs SAQ 1 At any frequencyi B, = 1.2cos 2nft cos 5m, for B, = 1.2teslas, and z = 0.4 m. (a) Substituting the values given in SAQ 1into Eq.(7.4). x(t, xI, x2)= XO e 4 x 1.2 {sin5m2- sin 5m1) - -- 0'024 {sin5 m2-sin 5m1 n ah (i) e, = - at = 0 ax 3 ax dx2 (ii) e,,, = -- + - . - dxl - dx2 = 20 ds axl . dt ax2 dt . Here -- - / dr dt I , O.OM x 5n {cos5n (201+0.2)- cos 5n (201) x 20Therforeem=- 'It i =-4.8 cos 100nt. The sameresult could alternatively be obtainedusing Blv and the right hand rule. (iii) Sincethere is no transformer emf,em= the total emf. O.OM cos 1O O ~ {sin 5m2- sin 5ml1.(b) Here A. (t, XI,x2)=- n (1) e,= -a' = -4'0M x 1OOn sin 1OOnt { sin (100nt +n ) -sin 100nt) at = -2.4 sin 100ntx (-2 sin 100trt) = 4.8 sin21OOnt a3L dx1 a3L h 2 (ii)em = - - + - -axl dt ax2 dt --- OaoM cos L O O ~t { 5n cos 57~x2-5n cos5ml x 20 n = 2.4 cos100nt{cos(100nt+n )- cos 1OOnt) =-4.8 cos2100nt
  • 34. Electrical Machines & Mesclwing btromeots (iii) Therefore total emf = e, +en,=-4.8{cos2100nt- sin2100nt) =- 4.8 cos200nt. Alternatively,sincexl and x2 are functions of t, we could express h entirely as a function of t alone: h(t)=-0024 coslOOnt { sin(100nt+n) - sin 1Mht n --- cos100 nt .sin100 nr = - - n 0.024 sin 200 nt n Therefore = - x 200n cos2OOnt =-4.8 cos 200nt, as before. dt n SAQ 2 (a) (i) F=0.2 x 500 x 1= 100N (ii) Using Fleming's left-hand rule, Fis in the positive z-direction. (iii)Using the left-handrule,Fis in the negative z-direction. (b) From the Bil = Fformula, FB = - = 80 il 1200 x 0.5 = 0.133 tesla SAQ 3 Total losses = 3.5 +1.5= 5kW. (c) SAQ 5 Power supplied by the prime mover = Output power +losses = 105kW If prime-mover torque is T Nm, and w, is the speed of the rotor in radians per second,power outputof prime-mover = w,T watts. -27Ch' 2 n x 1ooo=104.72rad/sIf N = speed in rpm, w, =--- - 60 60 Therefore 105,000= 104.72T Output 100 - 0.952 Efficiency = - --- Input 105 Mechanical power output = { 100-(3.5 + 1.5))= 95 kW w, = 104.72rads Therefore T =-95'000 -- 907.2 Nm 104.72 u ut Efficiency = = $=0.95 Input Figure for A~lswerto SAQ 5
  • 35. I Refer to the figure given. L 509000-200 ALoad current IL=-- 250 Energy (&version ~ r i n c i ~ l e s and D.C Machines Terminal voltage at the armature= VL+I,& = 250 +200x 0.006 = 251.2 V I Current through shunt field If =-251'2 --4.02 A I 62.5 Therefore armature currentla= 200 +4.02 = 204.02 A I Armature resistance drop I S , =204.02 x 0.012 = 2.45 V Brush voltage drop through 2 brushes in series=2 V Therefore induced armature emf = 251.2+2.45 +2 = 255.65 V SAQ 6 Reversingthe polarity of the supply voltage reverses the directions of both the armature and field currentsin all three cases.Consequently,since the directionsof both armature current and the magnetic flux reverse together, the direction of the developed torque remainsunchanged. If the directionof rotation is to be reversed,additionalswitching arrangementsmust be provided to ensure that, on reversing either the armature or the field current,the other continues to flow in its original direction. SAQ 7 When operated as a generator,I, = IL+If and E, = VL+I,&. On no-load, I, =If = 2 A; E, = 200 +2 x 0.5 = 201 V . (i) As a motor, Ea= Vs-I,Ra, la= (Is-If) = 5 -2 = 3 A Therefore Ea= 200 - 3 x 0.5 = 198.5 V . Since the field current both as a generator and a motor is the same,Ea is pdoportional to speed. 198.5 x1000=987.6rpmSo, speed when Is = 5 A is - 201 (ii) ForIs=50A,Ia=50-2=48A Therefore Ea = (200-48 x 0.5)= 176 V Therefore speed = x 1000= 875.6 rpm 201 SAQ 8 (a) When the armature current is 20 A, the induced emf Ea in the armature is given by E, = 1230- 20 x (0.5 +0.5))= 210 V At 1500rpm, we find from the magnetization characteristicthat 20 A induces an emf of 216 V. For the same current to produce an induced emf of 210 V, the speed of the motor should be From Eqs. (7.12)and (7.14),we obtain T = -Ea 'x 60 = 27.5 Nm. 27th' (b) The induced emf in the armature is now (230- 10 x 1) = 220 V . At 1500 rpm, 10 A induces an emf of 120 V. 220 Therefore speed of motor = - x 1500 = 2750 rpm. 120 Torque T = 220x lo x 60 = 7.6 Nm. 2n x 2750