Unit 3


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Unit 3

  1. 1. 51 DC Machines UNIT 3 DC MACHINES Structure 3.1 Introduction Objectives 3.2 Principles of Electromechanical Energy Conversion 3.2.1 Basics of Electromagnetism 3.2.2 Generation of Electromotive Force (EMF) or Voltage 3.2.3 Torque Developed in DC Machine 3.3 Construction of DC Machine 3.3.1 Field System 3.3.2 Armature 3.3.3 Commutator 3.3.4 Armature Windings 3.4 Effect of Armature Current 3.4.1 Armature Reaction 3.4.2 Commutation 3.5 Characteristics of DC Generator 3.5.1 Separately Excited DC Generator 3.5.2 Self Excited DC Generators 3.6 DC Motor 3.6.1 Back EMF 3.6.2 Speed Regulation 3.6.3 Power Equation and Torque 3.6.4 Power Flow and Losses 3.6.5 Motor Current and Voltage Equations 3.7 Characteristics of DC Motors 3.7.1 DC Series Motor 3.7.2 DC Shunt Motor 3.7.3 DC Compound Motors 3.8 Starting, Speed Control and Application sof DC Motors 3.8.1 Starting of DC Motors 3.8.2 Speed Control 3.8.3 Applications 3.9 Summary 3.10 Answers to SAQs 3.1 INTRODUCTION Rotating electrical machines are electromechanical energy conversion devices. When these devices change electrical energy into mechanical then the device is said to function as a motor and when the device converts mechanical energy into electrical, it is said to act as a generator. In this unit, we will get an understanding of construction and working of DC machines. Further, we will discuss different excitation schemes used in DC machines. Objectives After studying this unit, you should be able to • identify and list the various components of energy loss,
  2. 2. 52 Electrical Technology • give an elementary description of DC armature winding and the commutator, • explain the various excitation schemes used in DC machines, • calculate induced emf and terminal voltage of DC generators, • know characteristics and applications of DC generators, • calculate torque and speed of DC motors, • describe characteristics, starting and speed control of DC motors, and • know applications of DC motors. 3.2 PRINCIPLES OF ELECTROMECHANICAL ENERGY CONVERSION 3.2.1 Basics of Electromagnetism In DC machines, there is an electromechanical energy conversion. In DC motor, the input is electrical energy and the output is mechanical energy. In a DC generator, it is vice- versa. An electromechanical conversion device interrelates electric and magnetic fields on one hand and mechanical force and motion on the other. For energy conversion, the principle of energy conservation is applicable. In DC machines, the input power is converted into output power and some heat loss. Thus, PInput = POutput + PLoss . . . (3.1) DC machine contains one or more than one magnetic circuit depending on the number of poles. Before discussing magnetic circuit of a DC machine, let us recall the basic laws of magnetic circuits. In any magnetic circuit, the magnetomotive force (MMF) depends on number of turns in coil N and the current I flowing through it, i.e. MMF = N . I (AT) . . . (3.2) The intensity of magnetic force H is given by H = N . I / l MMF l = = (AT/m) . . . (3.3) where l is length of magnetic path. This MMF sets a flux φ in the magnetic materials. There is a reluctance (S), which opposes the setting of flux. The reluctance is similar to resistance in electrical circuits and depends on the length l, area A and permeability µ of magnetic path, i.e. A l S µ = . . . (3.4) or A l S ro µµ = . . . (3.5) where µo is absolute permeability, and µr is relative permeability.
  3. 3. 53 DC MachinesThe relative permeability of a magnetic material can be obtained by plotting its B – H curve. The relative permeability is the ratio of flux density B and magnetic field intensity H of material. It can be determined by taking ratio of B and H in linear region of B – H curve. But actually the value of permeability depends on the value of flux density (B) as shown in Figure 3.1. It is known as permeability curve. B Permeability Figure 3.1 : Permeability Curve A DC machine has magnetic circuits which are not exactly in parallel and the flux passes through poles, air gap, armature teeth and magnetic core. The magnetic materials used for different parts have different permeability. This type of magnetic circuit is known as composite circuit. So, the total MMF required to setup the flux φ in this magnetic circuit is Total MMF p p g g t t c c y yH l H l H l H l H l= × + × + × + × + × . . . (3.6) where Hp and lp are field intensity and length of path, respectively, for the poles. Hg and lg are field intensity and length of path, respectively, for air gap. Similarly t, c and y refer to teeth, coils and yoke or magnetic core, respectively, of DC machine. 3.2.2 Generation of Electromotive Force (EMF) or Voltage EMF induced in rotating machines is dynamically induced emf. According to Faraday’s law of dynamically induced emf, if a conductor having length l is moving in a magnetic field of flux density B, with a velocity v m/s and θ as angle between B and direction of motions then induced emf e is given by θ= sinvlBe . . . (3.7) The direction of induced emf can be determined by Fleming’s right hand rule for emf. The induced emf can also be calculated by taking statically induced emf relations like Faraday’s law. The induced emf in a conductor is directly proportional to rate of change of flux linked with conductor dt d e φ = . . . (3.8) Consider a DC machine having the following : P = Number of poles, N = Speed of rotation of armature in RPM, Zt = Number of total conductors in armature,
  4. 4. 54 Electrical Technology φ = Flux per pole in Webers, and A = Number of parallel paths of armature conductors. The number of parallel paths depends on type of armature winding. For Wave wound armature A = 2 For Lap wound armature A = P Now, the flux linked with a conductor in one revolution Pd .φ=φ⇒ . Time taken in one revolution .ondssec 60 N dt == So emf induced per conductor in one revolution . 60 60 P P Nd e dt N φ φφ = = = . . . (3.9) The average emf induced in DC machine is always equal to emf across each parallel path. So average emf across each parallel path eav = e × No. of conductors in parallel path volts 60 A ZNP t × φ = . . . (3.10) For Wave wound armature volts 260 av tZNP e × φ = and . . . (3.11) for Lap wound armature volts 6060 av tt ZN P ZNP e φ =× φ = . . . (3.12) In case of generator the average emf is known as generated emf (eg) and in case of motor it is known as back emf (eb). 3.2.3 Torque Developed in DC Machine When a conductor carrying current is placed at right angles to an uniform magnetic field, a force is setup between the conductor and magnetic field, given by F = B l Ia . . . (3.13) here B is flux density, l is length of conductor and Ia is current carried by conductor. The direction of force can be obtained by Fleming’s Left hand rule. In DC motors, when the current carrying coil of armature is placed in uniform flux then a torque is developed. Average force on any conductor is given by F a av av I B l A   = ×     newtons . . . (3.14) where Bav is average flux density and A is number of parallel paths. For an armature of diameter D, average torque contributed by each conductor is armature radius 2 a av av av ID T F B l A    = × = ×        . . . (3.15)
  5. 5. 55 DC MachinesTotal Torque due to all conductors, Nm 2 a av t t av ID T T Z Z B l A    = × = =        . . . (3.16) But f av P B Dl φ = π . . . (3.17) ∴ Nm. 2 1 aft IZ A P T φ            π = afa IKT φ= . . . (3.18) where       = A ZP K t a π2 3.3 CONSTRUCTION OF DC MACHINE Construction-wise both DC generator and DC motor are same. A DC machine has four essential parts : (a) Field System (b) Armature (c) Commutator (d) Carbon Brushes 3.3.1 Field System In DC machines, the field magnets provide uniform magnetic field surrounding the armature. Generally, electromagnets are preferred in DC machines in place of permanent magnets. The field magnet consists of four parts (a) Yoke or Frame (b) Pole Cores and Pole Shoes (c) Magnetic Coils The Yoke or Frame (a) Generally, cylindrical yoke is used in DC machines. (b) Material of yoke possess high permeability and it provides the path to the flux. (c) It is made of forged or cast steel or iron. Cast steel has very good magnetic properties. The yoke provides mechanical protection to the internal parts of a DC machine. Pole Cores and Pole Shoes Pole core has circular section and it carries the field magnets. The pole shoes are attached to pole core and act as support to the field coils. It also spreads out the flux uniformly over the armature periphery. Usually pole cores are made of iron and casted with yoke. The pole shoes are formed by laminated sheets of steel and bolted to the pole cores. Magnetic Coils
  6. 6. 56 Electrical Technology To set up the flux in magnetic circuit of DC machine the magnetomotive force or MMF is required. An electromagnet is formed by field or magnetic coils, which are supported by the pole core. The magnetic flux produced by MMF is developed in these magnetic coils. N S S N Slotted Armature Field Windings Yoke Typical Magnetic Flux Lines Pole Shoes Figure 3.2 : Sectional View of Field System In series field machines, the field coil is made of thick wire of copper with less turns. For a shunt field coil, many turns of fine wire are used. After proper winding of coils these are dipped in insulating varnish to provide mechanical strength and better insulating properties. 3.3.2 Armature Armature is the drum shaped rotating part of DC machine. The armature conductors are fixed at the upper surface of drum in slots. There is a small air gap between armature and pole shoes of field magnets to avoid any rubbing in DC machine. This air gap should be kept minimum. Usually armature is made of 0.3 mm to 0.6 mm thick laminated stampings of high grade steel to reduce hysteresis and eddy current losses. On the outer periphery the slots are formed by die cut or punch as shown in Figure 3.3. Also some air ducts are provided for proper ventilation. Key Way Tooth Slot Ventilating Ducts Figure 3.3 : Armature Lamination The armature conductors carry the current and are insulated using several layers of paper or mica insulation. Figure 3.4(a) depicts a slot containing two coil sides, each consisting of a single conductor. Two sides of one coil are housed approximately one pole pitch apart. One side occupies top layer and another side occupies the bottom layer of the respective slots. In a multiturn coil, each coil side consists of as many conductors as the number of turns in the coil.
  7. 7. 57 DC Machines Slot Wedge Insulation Conductors Active Length or Length of Armature Core Finish of CoilStart of Coil Overhang Sectional View Top Coil Side Bottom Coil Side (a) Coil Sides in Slot (b) Coil Viewed From Top Figure 3.4 : Arrangements of Coils in Slots 3.3.3 Commutator Commuator is an essential part of DC machines. It is placed between armature and the external circuit. The armature coils are connected with the commutator, which in turn gets connected to external circuit through carbon brushes sliding on commutator. Its serves the following purpose : (i) It completes electrical circuit by connecting the rotating armature coils and stationary electrical circuit. (ii) In generating action it works as a rectifier which converts the generated AC voltage into DC voltage. (iii) In motoring action it reverses the direction of DC current to maintain the torque in same direction. Commutator is made of wedge shaped segments of drop forged and hard drawn copper. A thin sheet of mica is used to separate or laminate the segments from each other. It has cylindrical shape and approximately same diameter as armature. The winding ends of armature are directly soldered to the commulator segments. Slotted Armature Segments Mica Shaft Figure 3.5 : The Commutator 3.3.4 Armature Windings Many different types of armature windings are employed in dc machines depending on requirements. The most common of these is the simplex lap, though the simplex wave is used in high voltage machines. (More elaborate windings, referred to as duplex and multiplex are also sometimes used). In this section, we will confine ourselves to a description of the simplex lap winding and refer in passing to the simplex wave.
  8. 8. 58 Electrical Technology Discussion of armature windings is greatly facilitated by using armature winding diagrams. Figure 3.6 is the armature winding diagram of a two-pole dc machine having 8 slots, 8 coils and 8 commutator segments operating in generator made. The diagram is a conventionalized representation of the cylindrical armature surface assuming that it is cut along an axial line and unrolled flat on to a plane. The 8 slots will then appear at equal distances from each other. Each slot has a top coil side and a bottom coil side. In the Figure 3.6, the top coil sides in the active length of the armature are represented by firm straight lines numbered 1 to 8 corresponding to the 8 slots. The bottom coil sides are indicated by dotted straight lines close to the firm lines. Since the machine is meant for two poles, there are 4 slots per pole. A coil whose top and bottom layers are 4 slot pitches is said to be a full-pitched coil. We will use full pitch coils. In this case, if the top layer of a coil is in slot number x, the bottom layer will be in slot number (x + 4). The triangular connections at the top are meant to indicate the interconnection of coil sides and belong to the overhang of the winding on the side away from the commutator. Thus, the firm line in slot 1 is connected to the broken line in slot 5 to represent coil number 1 etc. In an armature meant for four or more poles, top coil sides in slots 5, 6, 7 and 8 would in fact be connected to bottom coil sides in slots 9, 10, 11 and 12. Since our armature is for two poles and has only 8 slots, these coil sides would be connected to the bottom layers in slots 1, 2, 3 and 4. Thus, each side of a coil falls under separate poles. The segmented rectangular strip at the bottom represents the commutator. As seen from Figure 3.5. The representation in Figure 3.6 is the developed view of an equivalent commutator having the same diameter as the armature. In a lap winding, the ends of a coil are connected to adjacent commutator segments. Thus each commutator segment will be connected to top coil side of one coil and to the bottom coil side of a different coil. We will adopt the convention that the segment number is the same as that of the slot to whose top layer it is connected. In a lap-winding, the ends of coil 1 will be connected across segments 1 and 2, etc. The same type of diagram is used even for multiturn coils). Slot8 Velocity 7 8 1 2 3 4 5 6 7 B B S-pole Domain N-pole Domain Commutator Slot11 2 3 4 5 6 7 8 P Q Flux Directions Figure 3.6 : Winding Diagram 3.4 EFFECT OF ARMATURE CURRENT 3.4.1 Armature Reaction The distribution of flux under main poles is affected by magnetic field set up by the armature current; as a result the flux distribution or flux wave shape in air gap is distorted and demagnetised. This phenomenon is known as armature reaction.
  9. 9. 59 DC MachinesWhen the field is energized and there is no current in armature, the flux distribution is rectangular in space along armature periphery with axis of magnetisation through the centre of poles. When a armature carries current then there is some MMF due to number of turns in slots and it works as an electromagnet. The axis of magnetization differs for the various coils so the resulting axis of magnetization for the complete winding passes between active conductors and brushes are placed in interpolar region. In generating action, when the armature rotates in clockwise direction, the field produced by armature current acts opposite to the direction of rotation, so, in air gap flux is weakened under leading pole tips and strengthened under the trailing pole tips. This distorts the original rectangular flux wave shape leading to shift in resultant axis magnetisation and hence the brushes do not now fall on interpolar axis, thereby leading to poor commutation and speaking. The effect of armature MMF is summarized as The cross magnetising effect of armature reaction. Due to cross magnetizing effect DC machines have poor commutation. Compensating Winding In large capacity DC machines a compensating winding is used to neutralize effect of armature reaction. It maintains uniform flux distribution under pole faces. A copper winding is placed in the slots provided in pole shoes. It is connected in series with armature winding in such a way that the direction of current flowing through it is reverse of direction of armature current. This in effect neutralizes field produced by armature current and hence minimises cross-magnetising effect of armature reactions. The armature reaction is severe in large capacity and high speed machines. So compensating winding is used with inter poles arrangement. 3.4.2 Commutation In DC machines, the EMF induced in each coil of armature alternates with a frequency corresponding to product of number of poles and speed       = 120 PN f . To obtain the direct current in external circuit it is necessary to reverse periodically the connections of each armature coil with the external circuit. This reversal should take place at the instant when emf induced in coil is zero. For this purpose the commutator and carbon brushes are used. The reversal of current at this instant is affected by the following : Variable Contact Resistance The contact resistance between carbon brushes and commutator is not constant but largely depends on current density at contact surface. Also voltage drop across brush contact varies but this variation is very small compared to resistance variations. Current Density under the Brush When the brush thickness is less than one commutator segment, then reversal of current follows the simple linear law i.e. known as straight line commutation. A sparkles commutation is achieved by the following methods : (a) By using high resistance carbon brush. (b) By shifting carbon brushes in inter pole region. (c) By using inter poles or commutation poles. (d) By using equalizer rings.
  10. 10. 60 Electrical Technology 3.5 CHARACTERISTICS OF DC GENERATOR 3.5.1 Separately Excited DC Generator In separately excited DC generators, the field coils are excited by the external DC supply by using field regulating resistance. The open circuit characteristic or magnetizing characteristic is obtained by driving the machine at normal speed and varying the exciting current in steps from zero to maximum. Main circuit of machine is kept open and a voltmeter connected across the terminals. The excitation current is entirely independent of the load current in the armature. Equivalent circuit for mathematical calculations gives : Ia = I = Load current . . . (3.19) V = Eg – Ia Ra – Voltage drop across Carbon brush Power Input = Eg Ia . . . (3.20) Power Output = V . Ia . . . (3.21) E If I Ia RaEg FF FF + – V A Load Rheostat Figure 3.7 : Separately Excited DC Generator Magnetizing Characteristic Magnetizing characteristic or open circuit characteristic (OCC) is known as no load saturation curve which is obtained by method discussed above. The curve between the field current and armature terminal voltage is drawn when armature current is zero, i.e. at no load. Generally, it starts from zero but in self excited DC generator it has residual voltage and starts at slightly higher value. For DC generators, magnetizing curve is shown in Figure 3.8, its shape can be explained using theory of domain alignment. Self Excited Separately Excited If O a b c Eg Residual Voltage Figure 3.8 : Magnetic Characteristic Internal Curve
  11. 11. 61 DC MachinesInternal curve gives the relation between the emf actually generated in the armature and the armature current Ia. External Curve External curve also known as load curve is plotted between terminal voltage V and armature current Ia. For a separately excited DC generator these curves are shown in Figure 3.9. E Vg , Eg V Internal CurveExternal Curve Ideal (Generated emf) Ohmic Drops ( R )Ia a Armature Reaction Drop Ia Figure 3.9 : Internal and External Curves for Separately Excited DC Generators The internal curve is obtained by subtracting the drops due to armature reaction from generated emf (Eg). Also, it is a combination of external characteristic and Ia Ra drops. So, internal characteristic can be achieved by adding Ia Ra drops to external characteristic. Separately excited generators operate in stable condition with any excitation and generally used in laboratories. 3.5.2 Self Excited Generators Self excited machines have their field winding connected in series or parallel with the armature or main circuit. These are classified as : (a) Series wound (b) Shunt wound (c) Compound wound (i) Long shunt compound (ii) Short shunt compound To build up the voltage in self excited DC generators, following conditions should be met : (i) Presence of residual magnetism in pole curve, and (ii) Forward direction of rotation. When the armature of a self excited DC generator rotates at rated speed, the voltage across terminals increases up to rated voltage. Initially, due to residual magnetism there is a small flux which induces small residual voltage, which increases the field current. Finally, terminal voltage reaches the rated value. If the direction of rotation is reversed then induced emf due to residual magnetism demagnetizes the residual flux so there is no voltage built up in self excited generators.
  12. 12. 62 Electrical Technology E , E, Vg Eg Ia O Magnetic Characteristic And Armature Reaction Drop Drops Due to Armature and Field External Characteristic Stable Operating Point B’ B E V Figure 3.10 : Characteristics of Self Excited DC Generator External characteristics can be obtained by subtracting armature and field winding resistances drops from internal characteristic. External characteristic shows that first voltage increases with increase in load current, at max value of load current it decreases. This is known as dropping region. Under dropping region terminal voltage may fall to zero. Usually, this type of generators are used for welding purpose. Critical load resistance is that resistance of load above which a series wound generator fails to built up the voltage. The slope at external curve represent critical load resistance. Series Wound DC Generator In series wound DC generator, the field winding is series connected with main circuit and it delivers load current same as armature current. The equivalent circuit of DC series generator is shown in Figure 3.11. Ia Ra Eg V Rse Ise IL Figure 3.11 : Equivalent Circuit of DC Series Generator If series field resistance is Rse then Ia = Ise = IL V = Eg – Ia Ra – Ia Rse – voltage drop across carbon brush. . . . (3.22) V = Eg – Ia (Ra + Rse) – voltage drop due to carbon brush. . . . (3.23) Power Input = Eg . Ia . . . (3.24) Power Output = V . Ia . . . (3.25) Characteristics of DC Series Generator In a series wound dc generator the voltage cannot build up if the load terminals are open. To built up voltage first connect the load across output terminals of DC
  13. 13. 63 DC Machinesgenerator and rotate the armature in forward direction. Due to residual magnetism, there is small voltage across armature, which results in a small current in field winding. This small current developes a flux in field and voltage across armature increases due to increase in flux. This loop continues till the saturation. Now, voltage across the armature is rated. This given by magnetizing characteristic. Due to armature reaction there is some voltage drop, by subtracting it from magnetic characteristic, internal characteristic can be obtained. Shunt Wound Generator If field winding of a DC generator is connected across the main circuit then its known as DC shunt generator. The resistance of shunt field is very high as compared to armature resistance. The equivalent circuit is shown in Figure 3.12. IL Eg V Load Ia If Ra Rsh Figure 3.12 : Equivalent Circuit of DC Shunt Generator Rsh is the shunt field resistance then field current sh f R V I = . Here, Ia = IL + If Eg = V + Ia Ra + voltage drop due to carbon brush. Power Input = Eg . Ia Power Output = V . IL Voltage Built Up The emf induced by the armature winding due to its rotation through the residual magnetic flux sends a current in field winding. This current should be in such a direction as to tend to increase the magnetic flux. With proper connections and direction of rotation the magnetic field will be gradually strengthened and as a result the induced emf increases. This emf increases exciting current and so the voltage builds up until steady state conditions are obtained. Usually, shunt wound generators are connected with load after the rated voltage is built up. Generally the load resistance is less than field resistance otherwise the generator may fail to build up the rated voltage. On other land if field resistance is very high or more than critical field resistance then a dc generator fails to build up the voltage. The critical field resistance can be obtained by drawing the tangent at open circuit characteristic of DC shunt generator. Characteristics of DC Shunt Generator Magnetic characteristic is plotted between If and generated EMF. For a given magnetic flux, the emf generated varies in direct proportion to the speed of rotation. Then magnetization curve for various speeds plotted at the same scale gives different critical resistances at different speed. The shunt field resistance needs to be less than this critical value. Slopes OA, OB and OC give critical resistances at speeds N3, N2 and N1 respectively, here N3 > N2 > N1.
  14. 14. 64 Electrical Technology Eg If O N3 N2 Slope B C N1 A Figure 3.13 : DC Shunt Generator Magnetic Characteristic at Different Speeds Ideally, the terminal voltage of any dc shunt generator should be constant. But due to armature reaction, load voltage decreases with increasing load current. It further decreases due to Ia Ra drops. Example 3.1 The emf induced in the armature of a 450 kW, 250 volt shunt generator is 258.8 volt, when the field current is 20.0 amp and the generator is supplying power to a load at rated terminal voltage. The armature circuit resistance is 0.005 ohm. Determine (i) load current, (ii) power generated, (iii) power output, and (iv) electrical efficiency. Neglect brush contact drop. Solution Let the load current be I amp Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop or 258.8 = 250 + (I + 20) × 0.005 or amp1740 005.0 7.8 ==I (i) Load current = 1740 amp (ii) Power generated = E . Ia = 258.8 (1740 + 20) watts = 455.488 kW (iii) Power output V . I = 250 × 1740 watts = 435 kW (iv) Electrical efficiency = %5.95100 488.455 435 =× SAQ 1 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts. Its armature has 150 single turn coils and has a resistance of 0.01 ohm. The air gap flux per pole is 0.02 weber. Shunt field resistance is 50 ohm. Calculate the speed at which it is being driven. Neglect brush contact drop.
  15. 15. 65 DC Machines DC Compound Generator In compound generators the field winding is connected in series as well as parallel to the armature circuit. In long shunt generators, the shunt winding is connected across armature and series field but in short stunt it covers only armature winding. If the flux produced by series and shunt windings is additive, the dc generator is known as cumulative compound generator and if fluxes are deductive then it is known as differential compound generator. For a long shunt generator as shown in Figure 3.14, Ia = IL + If . . . (3.26) sh f R V I = . . . (3.27) )( seaag RRIEV +−= − voltage drop due to carbon brush . . . (3.28) Input Power = Eg . Ia . . . (3.29) Output Power = V . IL . . . (3.30) IL Eg V Load Ia If Ra Rsh Rse Figure 3.14 : Long Shunt DC Compound Generator In a short shunt generator as shown in Figure 3.15, Ise = IL Ia = IL + If . . . (3.31) sh sese f R RIV I + = . . . (3.32) V = Eg – Ia Ra – IL Rse – voltage drop due to carbon brush . . . (3.33) Power Input = Eg Ia . . . (3.34)
  16. 16. 66 Electrical Technology Power Output = V . IL . . . (3.35) IL Eg V Load If Rsh Ia Rse Ise Rb Figure 3.15 : Short Shunt DC Compound Generator The degree of compounding depends on flux shared by series or shunt windings. SAQ 2 (a) A long shunt compound generator delivers a load current of 50 A at 500 V and has armature, series-field and shunt field resistances of 0.05 Ω, 0.03 Ω and 250 Ω respectively. Calculate the generated emf and the armature current. Allow 1.0 volt per brush for contact drop. Also draw the circuit diagram of this arrangement. (b) A 4 pole, lap wound, 11.5 kW, 230 volts, dc shunt generator has armature resistance of 0.2 ohm and field resistance of 100 ohms. Calculate (i) emf generated if the brush contact drop is one volt per brush, and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage. Characteristics of Compound Generators Usually the series winding is arranged to assist the shunt winding and the terminal voltage variation with increasing load is determined by the relative strengths of two windings. Under suitable conditions terminal voltage may remain constant from no load to full load. By using relatively strong series windings the terminals voltage may increase with load. Its known as over compounded generator. If the shunt winding is strong then terminal voltage decreases with rise in load. It is under compound generator. If two fluxes are subtracting then the external curve goes down sharply, as shown in Figure 3.16.
  17. 17. 67 DC Machines O Differential Compounded Under Compounded Flat Compounded Over Compounded V IL Figure 3.16 : Characteristics of Compound Generators 3.6 DC MOTOR DC motor converts electrical energy into mechanical energy. When a current carrying conductor is placed in a magnetic field, a force acts on the conductor and conductor moves in the direction of force. When the DC machine is connected to DC supply a current passes through the armature winding. When conductors of armature winding carry outward current under north and incoming current under south pole then those conductors experience a force in clockwise direction according to fleming’s left hand rule. Due to this force, conductors move in clockwise direction. The direction of current is reversed by commutator, which causes the moving conductor coming under different pole to carry reverse current. This causes the force on the conductor to be again in the same direction as flux and current both change direction simultaneously. Thus armature conductors always experience force in same direction. 3.6.1 Back EMF When the armature of a DC motor rotates, an emf is induced in armature conductors known as back emf which opposes the applied voltage. volts 60 A PNZ Eb φ = . . . (3.36) If Ra is armature resistance then aab RIEV += . . . (3.37) here Ia is armature current and V is applied voltage so a b a R EV I − = . . . (3.38) Back emf makes a dc motor self regulating. When speed is low then back emf will less and armature current will be large. 3.6.2 Speed Regulation The back emf in armature is A NZP Eb 60 φ = . . . (3.39)
  18. 18. 68 Electrical Technology so ZP AE N b φ = 60 . . . (3.40) since P, Z and A are constant for a machine then bE N ∝ φ or a aV I R N − ∝ φ . . . (3.41) For a shunt motor φ is constant then orb a aN E N V I R∝ ∝ − . . . (3.42) For a series motor φ is proportional to Ia so a b I E N α . . . (3.43) percentage speed regulation 100× − = f fo N NN . . . (3.44) here No is speed at no load in RPM and Nf is speed at full load in RPM. Example 3.2 A shunt generator delivers 50 kW at 250 volts and 400 rpm. The armature and field resistances are 0.02 and 50 ohms respectively. Calculate the speed of the machine running as a shunt motor and taking 50 kW input at 250 volts. Allow one volt per brush for contact drop. Solution As Generator : Line current amp200 250 1050 3 = × =I Shunt field current amp5 50 250 ==shI Armature current amp2055200 =+=+= sha III Armature drop volts1.402.0205 =×== aa RI Induced emf dropcontactbrushdroparmature250 ++=gE 250 4.1 2 1gE = + + × [There are two brushes] = 256.1 volts As Motor : Armature current amp1955200 =−=−= sha III Armature drop volts9.302.0195 =×== aa RI Back emf dropcontactbrushdroparmature250 −−=bE
  19. 19. 69 DC Machinesor volts1.244129.3250 =×−−=bE Since field current is constant 21 and NENE bg ∝∝ where N1 and N2 are respectively the speeds of the machine as a generator and as a motor. rpmor 257.381 1.256 4001.2441 2 2 1 = × === g b b g E NE N N N E E SAQ 3 A shunt wound motor runs at 500 rpm from a 200 volt supply. Its armature resistance is 0.5 ohm and the shunt field resistance is 100 ohms and takes 32 amperes line current from the supply. What resistance must be reduced to 300 rpm, the armature and field currents remaining the same? Neglect brush drop. 3.6.3 Power Equation and Torque The voltage equation for motor is aab RIEV += . . . (3.45) By multiplying Ia in this equation by Ia we get aaaba RIIEIV 2 += . . . (3.46) It is a power equation, V Ia is input power, Eb I is power developed in armature and Ia 2 Ra represents power losses in armature. So mechanical power developed by the motor is aaaabm RIIVIEP 2 −== . . . (3.47) or Pm = input power – losses Differentiate the Eq. (3.47) with respect to Ia we get 2m a a a dP V I R dI = − . . . (3.48) For max power output 0m a dP dI = . or aa RIV 2= but baa EVRI −= . . . (3.49) so )(2 bEVV −=
  20. 20. 70 Electrical Technology or 2 V Eb = . . . (3.50) In any motor mechanical power developed in armature is maximum when back emf is half of applied voltage. 3.6.4 Power Flow and Losses Electrical Losses in Core/Iron Parts In the iron part of machine some electrical losses occur in the form of hysteresis and eddy current losses. Hysteresis losses occur due to magnetic reversals caused by the rotating armature. Hysteresis losses are directly proportional to the number of magnetic reversal per second. hysteresis loss Pn = n (Bmax)x f V watts these losses occur in armature core and teeth of the dc machine. To reduce the hysteresis loss armature core is made of silicon steel. When armature core rotates in magnetic fields of poles which induce emf in armature core and yoke. Due to this induced emf eddy currents circulate in armature core, the eddy current losses mainly depend on thickness of material. Pe = K Bmax f 2 V t2 watt . . . (3.51) To minimize eddy current losses the armature core is made of laminated stampings. Hysteresis and eddy current losses are known as core losses and are about 20% to 30% of full load losses. Mechanical Losses Due to friction of bearings, air friction or windage some losses occur in dc machines. These are known as mechanical losses. The brush friction losses are quite large. These losses are about 10% to 20% of full load losses. Losses and Efficiency In electrical machines, the efficiency is always less then one. It means that the output is less than the input. For any machine, efficiency Input Output = . . . (3.52) In electrical machine input power is sum of output power and power loss i.e. Power (Input) = Power (Output) + Power loss So, efficiency PowerInput PowerOutput = PowerInput lossPowerPowerInput − = PowerInput lossPower 1 −= . . . (3.53) The various machine losses may be classified as electrical losses and mechanical losses.
  21. 21. 71 DC MachinesElectrical Losses In DC machines electrical losses occur in several parts of machine. The maximum electrical losses occur due to I2 R losses because a large current flows through various machine windings. In addition to I2 R losses there is brush contact loss at the contacts between the brushes and commutator. These losses are known as copper losses and amount to 40% to 60% of the full load losses. The power input of any DC machine is distributed as shown in Figure 3.17 below. POWER INPUT Total Losses Useful Output Mechanical LossesIron LossesCopper Losses Armature losses Field Winding Losses Eddy Current Loss Hysterisis Loss Friction Loss Windage Loss Figure 3.17 : Power Input Distribution in DC Machine Power Flow Analysis In DC machines the sequence of energy conversion is necessary in analysis of the operation and characteristics of the machine. Power Flow in DC Generator DC Generator Mechanical Power Converted to Electrical Power Output PowerMechanical Power Input from Turbine, Engine etc. Armature Copper Loss Mechanical Power Loss and Hystirisis and Eddy Current Loss Power Lost in Shunt Field Figure 3.18 : Power Flow in DC Generator Power Flow for DC Motor DC Motor Electrical Power Input Power Converted In Mechanical Power Loss in Shunt Field Winding Copper Losses in Armature Output Core and Mechanical Losses Figure 3.19 : Power Flow for DC Motor Torque In a DC motor, output power is converted to torque. If at a wheel of radius r metre, a force F acts on circumference then
  22. 22. 72 Electrical Technology Torque T = F . r . . . (3.54) work done per revolution = F . 2π r joules work done per second = F . 2π r . n . . . (3.55) here, 60 N n = and N is speed of rotation in RPM. r F Figure 3.20 Now power developed, P = work done per second or P = F r. 2π n or P = T . ω . . . (3.56) where, T = Torque, and ω = angular velocity. here ω = 2π n 60 π2 N = . . . (3.57) Power developed in armature is Eb . Ia So, Eb Ia = T . ω . . . (3.58) or ω = ab IE T . . . (3.59) or 60 2 60 N I A NZP T a πφ = or A I PZI A ZP T a a φ= π φ = 159.0 2 . . . . (3.60) or N IE T ab 55.9= . . . (3.61) Since Z, P and A are constant so T ∝ φ Ia . . . (3.62)
  23. 23. 73 DC MachinesFor DC shunt motor φ is constant so T α Ia and for dc series motor φ α Ia, so T α Ia 2 . Example 3.3 A 125 V, dc shunt motor at its rated conditions develops 1 kW at 1800 rpm. Its line current is 10.67 A. The motor has a field resistance of 110 ohms and armature circuit resistance of 1.23 ohms. If the motor torque is increased by 20% determine (i) its probable new speed, and (ii) line current. Solution Shunt field current amp136.1 110 125 === sh sh R V I Under rated conditions amp534.9136.167.101 =−=−= sha III Torque ∝ φ Ia For a shunt motor, φ is constant ∴ T ∝ Ia . . . (3.63) When the motor torque is increased by 20%, the new value of torque becomes 1.2 T and let the new value of armature current be Ia2. We have, 1.2 T ∝ Ia2 . . . (3.64) Dividing Eq. (ii) by Eq. (i), we have amp44.112.1534.9or 534.9 2.1 2 2 =×== a a I I T T Under rated conditions, 23.1534.912511 ×−=−= aab RIVE or volts27.1131 =bE But back emf ∝ speed [φ being constant] ∴ 11b NE ∝ . . . (3.65) with increased torque 23.144.1112522 ×−=−= aab RIVE or 93.1102 =bE Let the new speed be N2, then Ebe ∝ N2 . . . (3.66) Dividing Eqs. (3.66) by Eq. (3.65), we have 2 2 110 93 110.93 1800 1762.8 113 27 1800 113.27 or rpm N. N . × = = = New speed = 1762.8 rpm and New line current = Ia2 + Ish
  24. 24. 74 Electrical Technology = 11.44 + 1.136 = 12.576 amp Example 3.4 The flux per pole of a 4 pole, 220 volt dc series motor is 0.025 weber for a load current of 52 amperes. The armature is wave wound with 500 conductors. Calculate : (i) the gross torque, (ii) the speed, (iii) the output torque, and (iv) the efficiency. The corresponding iron, friction and windage losses total 860 watt. The armature and field resistances are 0.21 ohm and 0.15 ohm respectively. Solution (a) Gross torque T = 0.159 φ Z IA       A P or       ×××= 2 4 52500025.0159.0T = 206.7 N-m (b) Back emf ( )b a a seE V I R R= − + volts28.201)15.021.0(52220 =+−=bE But A PNZ Eb ×φ= 60 or 0.025 500 4 201.28 60 2 N× × = × or N = 483.072 rpm (c) Torque lost in friction windage etc. 860 2 60 fwT N = π 17 072.4832 60860 = ×π × = N-m ∴ Output torque = T – 17 = 206.7 – 17 = 189.7 N-m (d) Efficiency = a ab IV IE 860 Input Output − = %97.83 52220 8605228.201 = × −× = SAQ 4 A dc series motor develops 30 kW and takes a current of 80 amp when running at 1200 rpm. Find the starting torque if at starting armature current is 120 Amperes. Magnetic circuit remains unsaturated.
  25. 25. 75 DC Machines 3.6.5 Motor Current and Voltage Equations Separately Excited DC Motor Eb = V – Ia Ra . . . (3.67) Ia = IL, f f E I R = Power drawn from supply P = V IL . . . (3.68) Mechanical power developed Pm = Eb Ia = Ia (V – Ia Ra) . . . (3.69) = V Ia – Ia 2 Ra 2 L L aV I I R= − = Powers drawn from supply – I2 R losses in armature . . . (3.70) V IL + – Ia Eb Ra A E AA Battery F FF If Rf Figure 3.21 : Separately Excited DC Motor Series Wound DC Motor )( seaab RRIVE +−= . . . (3.71) Power drawn from supply aIVP =⇒ . . . (3.72) seaaaam RIRIIVP 22 −−= . . . (3.73) or ( )( )m a a a seP I V I R R= − + . . . (3.74) bam EIP = . . . (3.75)
  26. 26. 76 Electrical Technology Ra RSE A IL Eb Ia AA F FF + – Figure 3.22 : Series Wound DC Motor Shunt Wound DC Motor shaL III += . . . (3.76) sh sh R V I = . . . (3.77) aab RIVE −= . . . (3.78) LIVP .= aashLm RIIVIVP 2 −−= . . . . (3.79) 2 2 ( )L sh a a a a aV I I I R V I I R= − − = − )( aaa RIVI −= . . . (3.80) bam EIP = . . . (3.81) IL Eb V Ia Ra A AA F FF Rsh Ish e Figure 3.23 : Shunt Wound DC Motor Direction of rotation of DC motor can be reversed by reversing the direction of current either in armature or in field windings. If current through both windings will reverse, then direction of rotation will remain same. 3.7 CHARACTERISTICS OF DC MOTORS To determine performance and stability of a dc motor following are drawn :
  27. 27. 77 DC Machines(a) Torque – Armature current : Electrical characteristics (b) Torque – Speed : Mechanical characteristics (c) Speed – Armature current : Speed characteristics 3.7.1 DC Series Motor For series motors the flux is directly proportional to armature current. From speed equation, we get ( ) µ b a a seE V I R R N − + = φ φ . . . (3.82) aT I∝ φ 2 aT I∝ ( )a se a V N R R I ∝ − + ( )a se V N R R k T ∝ − + (in saturated region) and aT I∝ 2 ( )a a seV I R R N k − + ∝ 2 ( )a seV T R R N k − + ∝ (in saturated region) A curve between armature current and speed will be rectangular hyperbola before saturation. Series DC motor have variable speed but on no load speed is dangerously high. To start a series DC motor, first we must put a mechanical loading on it. N N T Ia T Torque (a) (b) Figure 3.24 : DC Series Motor Characteristics (a) Torque-current Characteristics of Armature Current Speed Characteristics, and (b) Speed-torque Characteristics It is suitable for gear drive because gear provides some load on account of frictional resistances. Before saturation, flux (φ) is proportional to Ia, so torque (T) is proportional to Ia 2 but after saturation flux is constant so torque is directly proportional to armature current. From torque current characteristic (Figure 3.24(a)) we see that the series motor develops high starting torque for heavy loads. So these are used for electric traction. From speed torque characteristic (Figure 3.24(b)) we see that the speed are sharply falls with increase in torque. So another application of DC series is in fans whose speed falls with load. 3.7.2 DC Shunt Motor DC shunt motors are constant speed motors, there is only slight variation in speed from no load to full load. These are suitable for constant speed requirement. Torque-armature current characteristic is a straight line according to its speed torque characteristic, shunt motor is used for medium starting torque i.e. in pumps, fans, conveyors, shapers, milling etc. Here since φ is constant
  28. 28. 78 Electrical Technology b a aN E V I R∝ = − a aT I I∝ φ α 3 aN V R R T∝ − N T N T Ia No Load Speed Speed Torque (a) (b) Figure 3.25 : DC Shunt Motor Characteristics (a) Torque-current Characteristics of Armature Current Speed Characteristics, and (b) Speed-torque Characteristics 3.7.3 DC Compound Motors Differential compound motors are rarely used. Cumulative compound motors have combined characteristic of series and shunt motors. Shunt field provides constant speed and series field provides high starting torque. They are used in – rolling mills, shearing machines, lifts, mine hoist, etc. N T Ia Speed Differential Commulative Shunt Torque Commulative Differential Figure 3.26 : DC Compound Motor Characteristics 3.8 STARTING, SPEED CONTROL AND APPLICATIONS OF DC MOTORS 3.8.1 Starting of DC Motors When the armature of a dc motor is at standstill, the back emf is zero as N = 0.       φ = A NZP Eb 60 ∵ . If a dc motor is connected directly to the supply then a heavy current will flow through armature conductors       − = a b a R EV I∵ because Ra is very small (i.e. less than 1 Ω). This heavy current may damage the armature windings. Thus, for the protection of the motor against the flow of large current during starting period starters are used. Starter provides high resistance at the time of starting and removes this additional resistance when motor attains its normal speed. Starters are :
  29. 29. 79 DC Machines(a) Two point starter – used for DC series motors. (b) Three point starter – used for DC shunt and compound motors. (c) Four point starter – for DC shunt and compound motors when speed variation is achieved by flux control method. Here we will discuss three-point starter only. Three Point Starter It consists of series starting resistance in several sections, connected on brass studs. A spiral spring is placed on starter arm to bring this arm to OFF position when holding coil is de-energised. Under normal running condition the arm is held at ON position by holding coil which is magnetized by field current. Operation To start motor, the starter arm is moved gradually from OFF position to ON position. At position A, total starting resistance is inserted in series with armature and field is directly connected across supply through Brass arc and holding coil, hereby magnetising the holding coil. On moving starter arm clockwise further the resistance decreases in steps. At ON position, starting resistance is totally out of circuit. The soft iron on starter arm latches with the magnetised holding coil and hence keeps the starter arm at ON position. + – 220 VDC A AA F FF L F A A C D Overload Release Spring Starter Arm Soft Iron OFF Brass Arc ON Starting Resistance Holding Coil or No Volt Release Figure 3.27 : Three-point Starter No Volt Release No volt release coil is an electromagnet which is series connected with the field winding. In the abnormal position (i.e. supply failure and field winding open) the holding coil is demagnetized and starter arm goes back to OFF position due to spring force. Over Load Release Coil To provide overload protection to motors overload release is used. This coil is connected in series with armature winding when motor draws heavy currents
  30. 30. 80 Electrical Technology then coil magnetises it to such an extent that it pulls lever and closes contact L of overload release coil. This short circuits the No volt release coil. The No volt release coil is demagnetized and releases the starting arm which goes back to OFF position. 3.8.2 Speed Control Speed of any dc motor is directly proportional to back emf and inversely proportional to magnetic flux. i.e. φ ∝ bE N ∵ aab RIVE −= so φ − ∝ aa RIV N . . . (3.83) Thus the speed of any DC motor can be controlled by adjusting the following : (a) Applied voltage across armature, V. (b) Voltage drop across armature, Ia Ra. (c) Flux per pole. Speed Control of DC Shunt Motor Field or Flux Control Method V + – I Ia F AA A FF Eb If Figure 3.28 : Field Control for DC Shunt Motor Speed variation is accomplished by means of a variable resistance inserted in series with shunt field. By controlling field current the speed of DC shunt motor is controlled. This method is independent of load on the motor. Drawbacks of this system are : (i) Creeping speeds cannot be obtained. (ii) Top speeds are only obtained at reduced torque due to very weak field. (iii) At higher speeds very weak field increases armature current required for developing the desired torque. SAQ 5 The speed of a 500 V shunt motor is raised from 700 rpm to 1000 rpm by field weakening, the total torque remaining unchanged. The armature and field resistance are 0.8 Ω and 750 Ω respectively, and the line current at lower speed is 12 amps. Calculate the additional shunt field resistance required, assuming the magnetic circuit to be unsaturated and neglecting all losses.
  31. 31. 81 DC Machines Armature Control Method Speed variation of shunt motors by armature control requires applied voltage to change without altering the field current. So speed is adjusted by armature control method using : (a) Armature resistance control. (b) Shunted armature control. (c) Terminal voltage control impressed upon armature circuit. Armature Resistance Control – F FF + V Ia Ish A Eb AA Figure 3.29 : Armature Resistance Control When a resistance is connected in series with armature it increases armature drop so voltage applied across armature is reduced. Thus, the speed is reduced, which is directly proportional to voltage drop, without altering shunt field current. In this method, large amount of power Ia 2 R (R is resistance in series with armature) is wasted so this method is not used for long operation. Its used for short period in printing machines, hoists, cranes, fans and blowers. Shunted Armature Method Eb – F FF + V Ish Ia AA A Series Resistance Divider I Figure 3.30 : Shunted Armature Control Speed variation in armature resistance control method also depends on load current. This double dependence makes speed sensibly constant on rapidly changing loads. For more stable operation a divertor is used across armature. It is generally applied in low power rating machines. Armature Voltage Control Speed variation is obtained by using variable supply voltage which is provided by the adjustable electronic rectifier. This gives large speed range with good speed regulation and efficiency.
  32. 32. 82 Electrical Technology SAQ 6 The load torque and the loss torque of a dc shunt motor are constant and independent of speed. When the motor is connected directly to the 440 V supply and the speed is steady, the armature current is 20 A. The resistance of the armature is 1.5 Ω. Find the resistance of a starter to limit the armature starting current to 40 A. The starter contact is to be held on the first active stud until the motor acquires a steady speed. The contact is then to be moved to the second stud and the initial current is to be limited to 40 A. Find the resistance between the first and second studs of the starter. Assume that the field flux is constant. Ward Leonard Method Speed control by means of an adjustable voltage generator connected across armature terminals of the motor is called Ward Leonard system. M2 M1 G2 G1 R Y B Rheostat RS R A S B Figure 3.31 : Ward Leonard Method M1 is work motor powered by generator G1 which is driven by a synchronous or Induction motor M2 which gives constant speed drive to generator G1, and exciter G2 mounted on the same shaft as the generator. The field current for the work motor M1 and the generator G1 is obtained from exciter E. The Ward Leonard set starts the driving motor. Variable voltage across the terminals of the generator or across motor is obtained by varying the exciting current of the generator G1 by means of shunt regulator R. Switch S, at positions A and B, is used to reverse direction of voltage of G1 and hence rotation of M1. By controlling the field current of G1 through R, the armature voltage of M1 and hence its speed can be controlled. Advantage (a) Very fine speed control over whole range from zero to normal speed in both directions. (b) Uniform acceleration can be obtained.
  33. 33. 83 DC Machines(c) Speed regulation is good. Disadvantage (a) Arrangement is costly due to three extra machines. (b) Low overall efficiency of the system. Speed Control of DC Series Motor Armature Control Armature Resistance Control Speed variation is obtained by varying armature voltage drops. The maximum range of speed control will be available on the load. This is most economical for constant torque drives. Control Resistance+ – F FF Eb AA A + – V Figure 3.32 : DC Series Motor : Armature Resistance Control Shunted Armature Control This method gives slow speeds at light loads. There is speed control both by lowering armature voltage and flux variation. Armature voltage is varied by series rheostat R1 and flux is varied by rheostat R2. Eb + – F FF AA A V I R1 R2 Figure 3.33 : DC Series Motor : Shunted Armature Control Armature Voltage Control Speed varies by varying supply voltage using adjustable electronic rectifiers. Field Control Method Field Divertor Flux can be reduced by shunting a portion of field winding. This method gives speeds above normal speed. This method is economical and provides speed control in range not exceeding 2 : 1.
  34. 34. 84 Electrical Technology + – F FF AA A V Diverter Eb Figure 3.34 : Field Divertor Method Tapped Field Control This is another method of increasing speed by reduced flux. It is obtained by reducing number of turns of field winding. This method is generally used in traction. F FF AA A Diverter – – I Eb Figure 3.35 : Tapped Field Control Paralleling Field Coil Method By regrouping the field coils in series or parallel the variable speed is obtained. It is used in electric traction. 3.8.3 Applications DC Series Motors (a) Electric Traction • High starting torque and reduced torque at high speeds. • Large tractive effort, so a number of motors in series. • High η, large       Weight Power ratio ∴ Motors robust in construction. (b) Hoists, cranes, excavations, electric vehicles, streetcars, battery – powered portable tools, automotive starter motors : all because of high starting torque. (c) Drive fan load • Torque requirement increases with the square of speed. (d) Battery Operated Vehicles • Cars and other battery – powered vehicles have traction characteristics. • Speed control can be carried out by thyristers. DC Shunt Motors (a) Constant speed applications.
  35. 35. 85 DC Machines(b) Applications where a wide range of speed control is employed e.g. in lathes, in paper industry etc. (c) As a separately – excited motor when field winding is disconnected from armature and connected to an external voltage source. DC Compound Motors (a) Rolling Mills : To improve characteristic and have higher starting torque for the lower rolling motor of the twin-drive. Cummulative compound motors are better suited than shunt motors. In conjunction with flywheel, they can take sudden temporary loads and are ideal for rolling mills and coal-cutting machines. (b) Punching Press. (c) Milling Machine. (d) Traction Motors : Only where supply voltage is likely to vary considerably. (e) Hoisting and Lowering of Loads (along with regenerative braking). (f) to (v) were for cumulatively compounded motor. (g) Differential compound motors find only rare application as in research and experimental work. Example 3.5 A DC generator has the following magnetization characteristics at 800 rpm. Generated emf (V) 27.5 53 75 88 95 106 112 Field Current (A) 1 2 3 4 5 6 7 If the machine is shunt excited, determine the induced emf for a field circuit resistance of 19 ohm. Calculate the load current for a terminal voltage of 76 volts. The armature resistance in 0.1 ohm. Neglect the armature reaction. Solution The magnetization characteristic at 800 rpm is plotted in Figure 3.36, the field resistance line OA for 19 Ω is drawn as follow : Take any current say 3 amp, multiply it by 19 Ω, we get 57 V. Locate point B (3 amp, 57 volts) and draw a line joining the origin (0, 0), and the point B. This gives 19 Ω, resistance line. It cuts the induced emf corresponding to this point C is 100 volts. Hence the generator will develop 100 volts corresponding to field circuit resistance of 19 Ω.
  36. 36. 86 Electrical Technology 1 2 3 4 5 6 7 10 20 27 53 52 75 88 95 100 106 112 Field current (A)Generatede.m.f.(V) A B E V If Ia I Ra Rsh Rse + – Figure 3.36 Given terminal voltage = 76 volts Field current amp4 19 76 === a f R V I Induced emf E corresponding to If = 4 amp from the given OCC is 88 volts. ∴ Voltage drop in the armature Ia Ra = E − V or volts127688 =−=aa RI ∴ amp120 1.0 1212 === a a R I ∴ Load current I = Ia − If = 120 – 4 = 116 amp. 3.9 SUMMARY You learnt about principles of electromechanical energy conversion and calculations of emf and force or torque experienced by current carrying conductor. After a brief introduction to the constructional features of dc machine in Section 3.3, you learnt the calculations of emf for generators, torque and speed regulation for motor. After study of characteristics of generator and motors you were able to decide the applications of DC machines. You also learnt the starting and different speed control schemes for DC motors which are used for industrial purpose in Section 3.8. 3.10 ANSWERS TO SAQs SAQ 1 Load current amp55.204 11 2250 220 1045 3 == × == V W I
  37. 37. 87 DC Machines Field current amp4.4 50 220 === sh sh R V I Armature current amp95.2084.455.204 =+=+= sha III Armature drop volts0895.201.095.208 =×=ss RI Induced emf volts0895.2220895.2220 =+=+= aa RIVE But A PNZ E × φ = 60 or 60 222.0895 60 2 0.02 (150 2) 4 E A N Z P × × × × = = φ × × × 1110.45 rpm= SAQ 2 (a) The circuit diagram for the given system is shown in Figure 3.37. 1 2 3 4 5 6 7 10 20 30 40 50 60 70 80 90 100 110 120 Field current (A) Generatede.m.f.(V) A B E V If Ia Ra Rf + – Figure 3.37 Current in the shunt field amp2 250 500 ==shI Armature current sha III +== or amp52250 =+=aI Voltage drop in the armature and series resistance )( seaa RRI += = 52 (0.05 +0.03) = 4.16 volts Generated emf voltageTerminal=E )( seaa RRI ++ + brush drop
  38. 38. 88 Electrical Technology 500 4.16 2E = + + = 506.16 volts (b) Load current amp50 230 10005.11 = × == V P I Field current amp3.2 100 230 === sh sh R V I ∴ Armature current sha III += or amp3.523.250 =+=aI Armature drop 52.3 0.2 10.46 va aI R = × = Brush drop volts221 =×= Generated emf = 230 + 10.46 + 2 = 242.46 volts But A PNZ E × φ = 60 For lap wound generator A = P ∴ NZ ENZ E 60 60 or =φ φ = or 60 242.46 0.012123 1200 1000 webers/pole × φ = = × SAQ 3 Line current I = 32 amp Field current amp2 100 200 === sh sh R V I Armature current 32 2 30 ampa shI I I= − = − = Back emf 1 200 30 0.5 185 voltsb a aE V I R= − = − × = We know that back emf ∝ a flux per pole × speed. It is given that field current is same, therefore, flux per pole remains constant Back emf ∝ speed ∴ 1 1bE N∝ . . . (i) Let R ohm be the external resistance to be added in the armature circuit to reduce the speed to N2 = 300 rmp. At this speed let the back emf be Eb2. ∴ 22b NE ∝ . . . (ii) Dividing Eq. (ii) by Eq. (i), we have voltsor 111 500 300 185 2 2 == b b E E Since armature current remains same ∴ )(2 RRIVE aab +−= or 111 = 200 – 30 (0.5 + R) or R = 2.467 Ω
  39. 39. 89 DC Machines∴ Required resistance to be added in the armature circuit = 2.467 Ω. SAQ 4 Output of dc series motor = watts3 1030 ×=ab IE Output is also equal to 2 1200 2 60 60 NT T π = π × × ∴ 3 1030 60 1200 2 ×=××π T or π = 750 T N-m For a series motor with unsaturated field : 2 aT Iα 2 aT k I= 2 )80( 750 k= π . . . (i) At starting Ia = 120 amp. Hence torque at starting, Ts = k (120)2 . . . (ii) Dividing Eq. (ii) by Eq. (i) we get, 2 80 120 750       = π sT π ×      = 750 80 120 2 sT π ×      = 750 2 3 2 sT π ×= 750 4 9 sT 15.537=sT N-m. SAQ 5 In the first case : amp 3 2 750 500 1 ==fI amp 3 34 3 2 121 =−=aI 1 34 500 0.8 3 b a aE V I R= − = − × or volts93.490 3 2.27 5001 =−=bE
  40. 40. 90 Electrical Technology 500 V Ia IL Ish + – 750 Ω 0.8 Ω 0.03 Ω Eb Let the new shunt field resistance be R. Hence in the second case, 2 500 f V I R R = = Let the new armature current be Ia2 ∴ 2 2 2500 0.8b a a aE V I R I= − = − × . . . (i) As torque is the same in two cases ∴ 22112211 or aaaa IIIIT φ=φφ∝φ∝ . . . (ii) Since saturation is neglected, hence flux is proportional to field current ∴ R II ff 500 3 2 2211 and ∝∝φ∝∝φ Substituting the values in Eq. (ii) ∴ 2I 500 3 34 3 2 a R ×=× or 3 2 101.15 50033 342 − ×= ×× × = R R Ia Substituting in Eq. (i), we have REb 8.00151.05002 ×−= R01208.0500 −= Using the relation 2 1 1 2 1 2 φ φ ×= b b E E N N , we get R R 500 3 2 93.490 01208.0500 700 1000 × − = or 5003 2 93.490 )01208.0500( 7 10 RR ×× − = or R (500 – 0.01208 R) = 525996.42 or 0.01208 R2 – 500 R + 525996.42 = 0 Solving this equation, we get R = 1080.18 Ω Hence additional shunt field resistance required = 1080.18 – 750
  41. 41. 91 DC Machines= 330.18 Ω SAQ 6 Let the total resistance required (including armature resistance) in the armature circuit to limit armature current to 40 A be R Ω== 11 40 440 R [as Eb = 0 at starting, i.e. N = 0] External resistance to be inserted in the armature circuit Ω=−=− 5.95.111aRR With this resistance present in the armature circuit the back emf developed under steady speed. volts2201120440440 =×−=−= RIE ab [Since T α Ia for shunt motor and it remains constant due to constant load torque, Ia at steady state remains same, i.e. 20 A]. Let the resistance present between the first and the second stud be r Ω. When this resistance is cut out by moving the arm to the second stud, the resistance remaining in the armature circuit = (R – r) ∴ 5.5 40 220440 )(40 or)given(amp = − =−= − − rR rR EV b ∴ r = R – 5.5 = 11 – 5.5 = 5.5 Ω