ELECTRICAL MACHINES I. Unit -I INTRODUCTION

Transcription

1 ELECTRICAL MACHINES I Unit -I INTRODUCTION

2 Hans Christian Oersted ( ) X 1822 In 1820 he showed that a current produces a magnetic field.

3 André-Marie Ampère ( ) French mathematics professor who only a week after learning of Oersted s discoveries in Sept demonstrated that parallel wires carrying currents attract and repel each other. attract A moving charge of 1 coulomb per second is a current of 1 ampere (amp). repel

4 Michael Faraday ( ) Self-taught English chemist and physicist discovered electromagnetic induction in 1831 by which a changing magnetic field induces an electric field. A capacitance of 1 coulomb per volt is called a farad (F) Faraday s electromagnetic induction ring

5 Joseph Henry ( ) American scientist, Princeton University professor, and first Secretary of the Smithsonian Institution. Discovered selfinduction Built the largest electromagnets of his day Unit of inductance, L, is the Henry

6 Magnetic Fields and Circuits A current i through a coil produces a magnetic flux, f, in webers, Wb. f B da i f BA + A v B = magnetic flux density in Wb/m2. N - B H H = magnetic field intensity in A/m. = magnetic permeability Ampere's Law: Hl Ni H dl i Magnetomotive force F Ni reluctance F Rf

7 Magnetic Flux Magnetic flux, f, in webers, Wb. i2 i1 + v1 - Current entering "dots" produce fluxes that add. + N1 N2 v2 - f11 flux in coil 1 produced by current in coil 1 f12 flux in coil 1 produced by current in coil 2 f21 flux in coil 2 produced by current in coil 1 f22 flux in coil 2 produced by current in coil 2 f1 total flux in coil 1 f11 f12 f2 total flux in coil 2 f21 f22

8 Faraday's Law i2 i1 + i v1 + N1 N2 - v2 - Total flux linking coil 1: 1 N1f1 d 1 df1 N1 Faraday's Law: induced voltage in coil 1 is v1 (t ) dt dt Sign of induced voltage v1 is such that the current i through an external resistor would be opposite to the current i1 that produces the flux f1. Example of Lenz's law Symbol L of inductance from Lenz

9 Mutual Inductance i2 i1 + + v1 N1 N2 - v2 - Faraday's Law v1 (t ) N1 df1 df df N1 11 N1 12 dt dt dt In linear range, flux is proportional to current di1 di2 v1 (t ) L11 L12 dt dt self-inductance mutual inductance

10 Mutual Inductance i2 i1 + + v1 N1 N2 - - di1 di2 v1 (t ) L11 L12 dt dt di1 di2 v2 (t ) L21 L22 dt dt L12 L21 M L2 L22 v1 (t ) L1 di1 di M 2 dt dt di1 di2 L2 v2 (t ) M dt dt Linear media Let v2 L1 L11

11 Core losses Hysteresis losses

12 Hysteresis losses

13 Hysteresis losses

14 Eddy current losses

15 Eddy current losses How do we reduce Eddy current losses SOLID LAMINATED

16 Eddy current losses

17 Eddy current losses in windings Can be a problem with thick wires - Low voltage machines - High speed machines

18 Force, torque and power Universal modeling of terminal characteristic of electro-magnetic devices based on energy balance

19 Induced EMF Induced emf could be classified into two types Dynamically induced EMF. Statically induced EMF.

20 Statically induced emf In statically induced emf, conductor is stationary with respect to the magnetic field. Transformer is an example of statically induced emf. Here the windings are stationary,magnetic field is moving around the conductor and produces the emf.

21 Statically induced emf The emf produced in a conductor due to the change in magnetic field is called statically induce emf.it could be classified into two 1)self induced emf and 2)mutual induced emf

22 Dynamically induced emf This is the EMF induced due to the motion of conductor in a magnetic field. Mathematically e = Blv volts e-induced emf B flux density of magnetic field in Tesla l = length of conductor in meters v- velocity of conductor in m/s

23 Dynamically induced emf If the conductor moves in an angle θ,the induced emf could be represented as e= Blvsinθ the direction of induced emf is given by flemmings right hand rule. Generator is an example of dynamically induced emf.

24 UNIT - II TRANSFORMERS

25 Introduction A transformer is a static machines. The word transformer comes form the word transform. Transformer is not an energy conversion device, but is a device that changes AC electrical power at one voltage level into AC electrical power at another voltage level through the action of magnetic field, without a change in frequency. It can be either to step-up or step down. Transmission System TX1 TX1 Generation Station 33/13.5k V 13.5/6.6kV Distribution s TX1 TX1 6.6kV/415 V Consumer

26 Transformer Construction Two types of iron-core construction: a) Core - type construction b) Shell - type construction Core - type construction

27 Transformer Construction Shell - type construction

28 Ideal Transformer An ideal transformer is a transformer which has no loses, i.e. it s winding has no ohmic resistance, no magnetic leakage, and therefore no I2 R and core loses. However, it is impossible to realize such a transformer in practice. Yet, the approximate characteristic of ideal transformer will be used in characterized the practical transformer. N1 : N2 I1 V1 I2 E1 E2 V2 V1 Primary Voltage V2 Secondary Voltage E1 Primary induced Voltage E2 secondary induced Voltage N1:N2 Transformer ratio

29 Transformer Equation Faraday s Law states that, If the flux passes through a coil of wire, a voltage will be induced in the turns of wire. This voltage is directly proportional to the rate of change in the flux with respect of time. d (t ) Vind Emf ind dt Lenz s Law If we have N turns of wire, d (t ) Vind Emf ind N dt MZS FKEE, UMP

30 Transformer Equation For an ac sources, Let V(t) = Vm sin t i(t) = im sin t Since the flux is a sinusoidal function; (t ) m sin t Then: Therefore: d m sin t dt N m cos t Vind Emf ind N Vind Emfind (max) N m 2 fn m Thus: Emf ind ( rms) N m 2 fn m 4.44 fn m 2 2MZS FKEE, UMP

31 Transformer Equation For an ideal transformer E fn1 m E fn 2 m (i) In the equilibrium condition, both the input power will be equaled to the output power, and this condition is said to ideal condition of a Input power output power transformer. V1 I1 cos V2 I 2 cos V1 I 2 V2 I1 E1 V1 and E2 V2 From the ideal transformer circuit, note that, MZS FKEE, UMP

32 Transformer Equation Therefore, E1 N1 I 2 a E2 N 2 I1 Where, a is the Voltage Transformation Ratio; which will determine whether the transformer is going to be step-up or step-down For a >1 E1 > E2 For a <1 E1 < E2 MZS FKEE, UMP

33 Transformer Rating Transformer rating is normally written in terms of Apparent Power. Apparent power is actually the product of its rated current and rated voltage. VA V1I1 V2 I 2 Where, I1 and I2 = rated current on primary and secondary winding. V1 and V2 = rated voltage on primary and secondary winding. Rated currents are actually the full load currents in transformer MZS FKEE, UMP

34 Example kVA single phase transformer has rated voltage of 144/240 V. Finds its full load current. Solution 1500 I1FL A I 2 FL 6A 240 MZS FKEE, UMP

35 Practical Transformer (Equivalent Circuit) I1 V1 R1 X1 Io I1 Ic Im RC Xm I2 N1: N2 R2 X2 Load V1 = primary supply voltage V2 = 2nd terminal (load) voltage E1 = primary winding voltage E2 = 2nd winding voltage I1 = primary supply current I2 = 2nd winding current I1 = primary winding current Io = no load current E1 E2 V1 = primary supply voltage V2 = 2nd terminal (load) voltage E1 = primary winding voltage E2 = 2nd winding voltage I1 = primary supply current I2 = 2nd winding current I1 = primary winding current Io = no load current V2

36 Single Phase Transformer (Referred to Primary) Actual Method R1 I1 X1 Io I2 Ic Im RC Xm R2 X2 N1: N2 I2 Load V1 N1 E1 V2 V2 OR V2 ' av2 N2 I2 I2 ' a ' E1 2 N1 R2 ' R2 OR R2 ' a 2 R2 N2 2 N X 2 ' 1 X 2 OR X 2 ' a 2 X 2 N2 E2 V2

37 Single Phase Transformer (Referred to Primary ) Approximate Method I1 X1 R1 R2 X2 Io Ic V1 RC I2 N1: N2 I2 Im Load Xm E1 E2 2 N R2 ' 1 R2 OR R2 ' a 2 R2 N2 2 N X 2 ' 1 X 2 OR X 2 ' a 2 X 2 N2 N ' E1 V2 1 V2 OR V2 ' av2 N2 I I2 ' 2 a V2

38 Example Problem 1. A 10 kva single phase transformer 2000/440V has primary resistance and reactance of 5.5 and 12 respectively, while the resistance and reactance of secondary winding is 0.2 and 0.45 respectively. Calculate: i. The parameter referred to high voltage side and draw the equivalent circuit ii. The approximate value of secondary voltage at full load of 0.8 lagging power factor, when primary supply is 2000V

39 Example 1 (Cont) Solution R1=5.5, X1=j12 R2=0.2, X2=j0.45 i) Refer to H.V side (primary) I1 =(4.55)2 R2 (0.2) = 4.14, X2 =j(4.55)20.45 = j9.32 Therefore, R01=R1+R2 = = 9.64 X01=X1+X2 =j12 + j9. 32 = j21.32 R V1 X av2

40 Example 1 (Cont) Solution ii) Secondary voltage p.f = 0.8 Cos = 0.8 =36.87o Full load, From eqn I FL VA 5A 2000V V1 0o ( R01 jx 01)( I1 o ) av o (9.64 j 21.32)( o ) (4.55)V2 V o

41 Transformer Losses Generally, there are two types of losses; i. ii. i. Iron losses :- occur in core parameters Copper losses :- occur in winding resistance Iron Losses Piron Pc ( I c) 2 Rc Popen circuit ii Copper Losses Pcopper Pcu ( I 1) 2 R1 ( I 2) 2 R2 Pshort circuit or if referred, Pcu ( I 1) 2 R01 ( I 2) 2 R02

42 Transformer Efficiency To check the performance of the device, by comparing the output with respect to the input. The higher the efficiency, the better the system. Efficiency, Output Power 100% Input Power Pout 100% Pout Plosses V2 I 2 cos 100% V2 I 2 cos Pc Pcu Where Pcu = Psc Pc = Poc ( fullload) (load n ) VA cos 100% VA cos Pc Pcu nva cos 100% 2 nva cos Pc n Pcu Where, if ½ load, hence n = ½, ¼ load, n= ¼, 90% of full load, n =0.9

43 Voltage Regulation The purpose of voltage basically to determine the voltage drop between no load. Voltage Regulation can based on 3 methods: a) Basic Defination b) Short circuit Test c) Equivalent Circuit regulation is percentage of load and full be determine

44 Voltage Regulation (Basic Definition) In this method, all parameter are being referred to primary or secondary side. Can be represented in either Down voltage Regulation VNL VFL V.R 100% VNL Up Voltage Regulation VNL VFL V.R 100% VFL

45 Tap Changer A transformer tap is a connection point along a transformer winding that allows a certain number of turns to be selected. By this means, a transformer with a variable turns ratio is produced, enabling voltage regulation of the output. The tap selection is made via a tap changer mechanism.

46 Three Phase Transformers 3 single phase transformers connected together 1.Star/Delta winding arrangements 2. Easy to replace failed units Common core device 1. Lighter and cheaper than 3 individual units 2. 6 rather than 12 external connections 3. Whole transformer must be replaced if single winding fails. For both cases analysis procedure identical!

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51 UNIT-III Electromechanical Energy Conversion

52 Electromechanical Energy Conversion The electromechanical energy conversion theory allows the representation of the electromagnetic force or torque in terms of device variables, such as the currents and the displacement of the mechanical systems. An electromechanical system consists of an electric system, a mechanical system, and a means whereby the electric and mechanical systems can interact. 2

53 Electromechanical Energy Conversion Consider the block diagram depicted below. Coupling Field Electric System WE Energy supplied by an electric source = We + Energy transferred to the coupling field by the electric system WeL + Energy losses of the electric system. Basically, I2R Mechanic System WeS Energy stored in the electric o magnetic field 3

54 Electromechanical Energy Conversion W M The energy= transferred to thew coupling m field can+be represented by WmL Energy supplied by a mechanical source Energy transferred to the coupling field from the mechanical system WF = Total energy transferred to the coupling field WF We Energy losses of the mechanical system + Energy transferred to the coupling field by the electric system = Wf Energy stored in the electric system + WmS Energy stored in the moving member and compliance of the mechanical system Wm Energy transferred to the coupling field from the mechanical system + WfL Energy dissipated as heat (I2R) 4

55 Electromechanical Energy Conversion The electromechanical systems obey the law of conservation of energy. Energy Balance in an Electromechanical System WF = Wf WE + WfL = We + WeL WeS Wm WfL Wf WmL WM WmS 5

56 Electromechanical Energy Conversion If the losses are neglected, we will obtain the following formula, WF = We + Energy transferred to the coupling field by the electric system Wm Energy transferred to the coupling field from the mechanical system 6

57 Electromechanical Energy Conversion Consider the electromechanical system given below, f k r i + V - L + ef m f fe N - D x x0 7

58 Electromechanical Energy Conversion The equation for the electric system is- The equation for the mechanical system is- di V ri L e f dt 2 dx dx f m 2 D K ( x x0 ) fe dt dt 8

59 Electromechanical Energy Conversion The total energy supplied by the electric source is - The equation for the mechanical system is- di WE V i dt ri L e f i dt dt dx dt WM f dx f dt 9

60 Electromechanical Energy Conversion Substituting f from the equation of motion- WE dx 2 dx fe f dx m 2 D K ( x x0 ) dx dt dt Total energy Potential Energy Heat loss Kinetic energy stored in the spring transferre d to the due the friction coupling field stored in the mass (Wall) from the mechanical system 10

61 Electromechanical Energy Conversion WM f e dx * Recall W f We WM W f e f idt f e dx dw f e f idt f e dx 11

62 Electromechanical Energy Conversion If dx=0 is assumed, then d W f WE e f idt i dt dt W f id dx 0 12

63 Electromechanical Energy Conversion Recalling the normalized magnetization curve, W f id (i, x) d Wc di i 13

64 Electromechanical Energy Conversion (i, x) d Wf (i, x) i (i, x) i di (i, x) x dx i di dx 0 14

65 Electromechanical Energy Conversion i i (, x ) di i (, x) d i (, x) x dx i (, x) Wc di d dx 0 15

66 Electromechanical Energy Conversion From the previous relationship, it can be shown that for one coil, i* Wf i d L( x) i 0 i* W f i L( x)di 0 For a general case, W f i j d j 1 j dx 0 16

67 Electromechanical Energy Conversion For two coupled coils, For the general case with n-coupled coils, W f L11i 1 L12i1i2 L22i n n W f L pqi piq 2 p 1q 1 17

68 Electromagnetic Force Recalling, k r i + V - f L + ef m f fe N - D x x0 18

69 Electromagnetic Force W f We WM W f e f idt f e dx f e dx dwe dw f d ef dt 19

70 Electromagnetic Force d dwe e f idt i dt i d dt f e dx i d dw f (i, x) (i, x) d di dx i x W f (i, x) W f (i, x) dw f di dx i x Substituting for d and dwf in fedx=id dwf, it can be shown f e i, x i dw f x 20

71 Electromagnetic Force Recall, W f id d Wc i W f Wc W f i x x x W f f e (i, x) i x x Wc di i 21

72 Electromagnetic Force i W f Wc W f i Wc W f f e (i, x) i x x Wc f e (i, x) i i x x x Wc f e (i, x) x 22

73 UNIT-IV DC MACHINES

74 Introduction The direct current (dc) machine can be used as a motor or as a generator. DC Machine is most often used for a motor. The major advantages of dc machines are the easy speed and torque regulation. However, their application is limited to mills, mines and trains. As examples, trolleys and underground subway cars may use dc motors. In the past, automobiles were equipped with dc dynamos to charge their batteries.

75 Construction

76 Construction (cont.,) The stator of the dc motor has poles, which are excited by dc current to produce magnetic fields. In the neutral zone, in the middle between the poles, commutating poles are placed to reduce sparking of the commutator. The commutating poles are supplied by dc current. Compensating windings are mounted on the main poles. These short-circuited windings damp rotor oscillations..

77 Construction (cont.,) The poles are mounted on an iron core that provides a closed magnetic circuit. The motor housing supports the iron core, the brushes and the bearings. The rotor has a ring-shaped laminated iron core with slots. Coils with several turns are placed in the slots. The distance between the two legs of the coil is about 180 electric degrees.

78 Construction (cont.,) The coils are connected in series through the commutator segments. The ends of each coil are connected to a commutator segment. The commutator consists of insulated copper segments mounted on an insulated tube. Two brushes are pressed to the commutator to permit current flow. The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing.

79 DC Machines The rotor has a ring-shaped laminated iron core with slots. The commutator consists of insulated copper segments mounted on an insulated tube. Two brushes are pressed to the commutator to permit current flow. The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing.

80 DC Machines The commutator switches the current from one rotor coil to the adjacent coil, The switching requires the interruption of the coil current. The sudden interruption of an inductive current generates high voltages. The high voltage produces flashover and arcing between the commutator segment and the brush.

81 DC Machine Construction Rotation Ir_dc/2 Brush Ir_dc/2 Ir_dc Shaft Pole winding N S 4 5 Insulation Rotor Winding Ir_dc Copper segment

82 DC Machine Construction Figure A. Details of the commutator of a dc motor.

83 DC Machine Construction Figure B. DC motor stator with poles visible.

84 DC Machine Construction Figure D. Cutaway view of a dc motor.

85 DC Generator Operation

86 DC Generator Operation v B a S N 1 30 Vdc 2 b v Ir_dc (a) Rotor current flow from segment 1 to 2 (slot a to b) B S 2 a 30 v v N Vdc 1 The N-S poles produce a dc magnetic field and the rotor coil turns in this field. A turbine or other machine drives the rotor. The conductors in the slots cut the magnetic flux lines, which induce voltage in the rotor coils. The coil has two sides: one is placed in slot a, the other in slot b. b Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a)

87 DC Generator Operation v B a S N 1 30 Vdc 2 b v Ir_dc (a) Rotor current flow from segment 1 to 2 (slot a to b) B S 2 a 30 v v N Vdc 1 The conductors in slot a are cutting the field lines entering into the rotor from the north pole, The conductors in slot b are cutting the field lines exiting from the rotor to the south pole. The cutting of the field lines generates voltage in the conductors. The voltages generated in the two sides of the coil are added. b Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a)

88 DC Generator Operation v B a S N 1 30 Vdc 2 b v Ir_dc (a) Rotor current flow from segment 1 to 2 (slot a to b) B S 2 a 30 v v N Vdc 1 The induced voltage is connected to the generator terminals through the commutator and brushes. The induced voltage in b is positive, and in a is negative. The positive terminal is connected to commutator segment 2 and to the conductors in slot b. The negative terminal is connected to segment 1 and to the conductors in slot a. b Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a)

89 DC Generator Operation v B a S N 1 30 Vdc 2 b v Ir_dc (a) Rotor current flow from segment 1 to 2 (slot a to b) B S 2 a 30 v v N Vdc 1 When the coil passes the neutral zone: Conductors in slot a are then moving toward the south pole and cut flux lines exiting from the rotor Conductors in slot b cut the flux lines entering the in slot b. This changes the polarity of the induced voltage in the coil. The voltage induced in a is now positive, and in b is negative. b Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a)

90 DC Generator Operation v B a S N 1 30 Vdc 2 b v Ir_dc (a) Rotor current flow from segment 1 to 2 (slot a to b) B S 2 a 30 v v N Vdc 1 The simultaneously the commutator reverses its terminals, which assures that the output voltage (Vdc) polarity is unchanged. the positive terminal is connected to ommutator segment 1 and to the conductors in slot a. The negative terminal is connected to segment 2 and to the conductors in slot b. b Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a)

91 UNIT-V DC Motor

92 DC Motor Operation In a dc motor, the stator poles are supplied by dc excitation current, which produces a dc magnetic field. The rotor is supplied by dc current through the brushes, commutator and coils. The interaction of the magnetic field and rotor current generates a force that drives the motor Rotation Ir_dc/2 Brush Ir_dc/2 Ir_dc Shaft Pole winding N S 4 5 Insulation Rotor Winding Ir_dc Copper segment

93 DC Motor Operation v B a S N 1 30 Vdc 2 b v Ir_dc (a) Rotor current flow from segment 1 to 2 (slot a to b) B S 2 a 30 v v N Vdc 1 The magnetic field lines enter into the rotor from the north pole (N) and exit toward the south pole (S). The poles generate a magnetic field that is perpendicular to the current carrying conductors. The interaction between the field and the current produces a Lorentz force, The force is perpendicular to both the magnetic field and conductor b Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a)

94 DC Motor Operation v B a S N 1 30 Vdc 2 b v Ir_dc (a) Rotor current flow from segment 1 to 2 (slot a to b) B S 2 a 30 v v N Vdc 1 The generated force turns the rotor until the coil reaches the neutral point between the poles. At this point, the magnetic field becomes practically zero together with the force. However, inertia drives the motor beyond the neutral zone where the direction of the magnetic field reverses. To avoid the reversal of the force direction, the commutator changes the current direction, which maintains the counterclockwise rotation.. b Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a)

95 8.2.1 DC Motor Operation S N 30 Vdc b v Ir_dc (a) Rotor current flow from segment 1 to 2 (slot a to b) B a S 2 B a 2 v 1 Before reaching the neutral zone, the current enters in segment 1 and exits from segment 2, Therefore, current enters the coil end at slot a and exits from slot b during this stage. After passing the neutral zone, the current enters segment 2 and exits from segment 1, This reverses the current direction through the rotor coil, when the coil passes the neutral zone. The result of this current reversal is the maintenance of the rotation. 30 v v N Vdc 1 b Ir_dc (b) Rotor current flow from segment 2 to 1 (slot b to a)

96 EMF equation Let Ø= flux/pole in Wb Z = total number of armature conductors P = number of poles A = number of parallel paths = 2 for wave winding = P for lap winding N = speed of armature in r.p.m. Eg= e.m.f. of the generator = e.m.f./parallel path

97 EMF equation

98 Armature Resistance (Ra) The resistance offered by the armature circuit is known as armature resistance (Ra) and includes: (i) resistance of armature winding (ii) resistance of brushes The armature resistance depends upon the construction of machine. Except for small machines, its value is generally less than 1Ω.

99 Back EMF When the armature of a d.c. motor rotates under the influence of the driving torque, the armature conductors move through the magnetic field and hence e.m.f. is induced in them as in a generator The induced e.m.f. acts in opposite direction to the applied voltage V(Lenz s law) and in known as back or counter e.m.f. Eb. The back e.m.f. Eb (= P Φ ZN/60 A) is always less than the applied voltage V, although this difference is small when the motor is running under normal conditions. Consider a shunt wound motor shown in Fig. (4.2). When d.c. voltage V is applied across the motor terminals, the field magnets are excited and armature conductors are supplied with current. Therefore, driving torque acts on the armature which begins to rotate. As the armature rotates, back e.m.f. Eb is induced which opposes the applied voltage V. The applied voltage V has to force current through the armature against the back e.m.f. Eb. The electric work done in overcoming and causing the current to flow against Eb is converted into mechanical energy developed in the armature. It follows, therefore, that energy conversion in a d.c. motor is only possible due to the production of back e.m.f. Eb.

100 Back EMF Net voltage across armature circuit = V Eb If Ra is the armature circuit resistance, then, Ia = (V Eb)/ Ra Since V and Ra are usually fixed, the value of Eb will determine the current drawn by the motor. If the speed of the motor is high, then back e.m.f. Eb (= P Φ ZN/60 A) is large and hence the motor will draw less armature current and viceversa.

101 Significance of Back E.M.F. The presence of back e.m.f. makes the d.c. motor a self-regulating machine i.e., it makes the motor to draw as much armature current as is just sufficient to develop the torque required by the load. Armature current,ia = (V Eb)/ Ra (i) When the motor is running on no load, small torque is required to overcome the friction and windage losses. Therefore, the armature current Ia is small and the back e.m.f. is nearly equal to the applied voltage. (ii) If the motor is suddenly loaded, the first effect is to cause the armature to slow down. Therefore, the speed at which the armature conductors move through the field is reduced and hence the back e.m.f. Eb falls. The decreased back e.m.f. allows a larger current to flow through the armature and larger current means increased driving torque. Thus, the driving torque increases as the motor slows down. The motor will stop slowing down when the armature current is just sufficient to produce the increased torque required by the load. (iii) If the load on the motor is decreased, the driving torque is momentarily in excess of the requirement so that armature is accelerated. As the armature speed increases, the back e.m.f. Eb also increases and causes the armature current Ia to decrease. The motor will stop accelerating when the armature current is just sufficient to produce the reduced torque required by the load. It follows, therefore, that back e.m.f. in a d.c. motor regulates the flow of armature current i.e., it automatically changes the armature current to meet the load requirement

102 Armature Reaction In a d.c. generator, the purpose of field winding is to produce magnetic field (called main flux) whereas the purpose of armature winding is to carry armature current. Although the armature winding is not provided for the purpose of producing a magnetic field, nevertheless the current in the armature winding will also produce magnetic flux (called armature flux). The armature flux distorts and weakens the main flux posing problems for the proper operation of the d.c. generator. The action of armature flux on the main flux is called armature reaction. It was hinted that current in the coil is reversed as the coil passes a brush. This phenomenon is termed as commutation. The criterion for good commutation is that it should be sparkless. In order to have sparkless commutation, the brushes should lie along magnetic neutral axis.

103 Armature Reaction So far we have assumed that the only flux acting in a d.c. machine is that due to the main poles called main flux. However, current flowing through armature conductors also creates a magnetic flux (called armature flux) that distorts and weakens the flux coming from the poles. This distortion and field weakening takes place in both generators and motors. The action of armature flux on the main flux is known as armature reaction. The phenomenon of armature reaction in a d.c. generator is shown in Fig. (2.1). Only one pole is shown for clarity. When the generator is on no-load, a smal1 current flowing in the armature does not appreciably affect the main flux Φ1 coming from the pole [See Fig 2.1 (i)]. When the generator is loaded, the current flowing through armature conductors sets up flux Φ1. Fig. (2.1) (ii) shows flux due to armature current alone. By superimposing Φ1 and Φ2, we obtain the resulting flux Φ3 as shown in Fig. (2.1) (iii). Referring to Fig (2.1) (iii), it is clear that flux density at; the trailing pole tip (point B) is increased while at the leading pole tip (point A) it is decreased. This unequal field distribution produces the following two effects: (i) The main flux is distorted. (ii) Due to higher flux density at pole tip B, saturation sets in. Consequently, the increase in flux at pole tip B is less than the decrease in flux under pole tip A. Flux Φ3 at full load is, therefore, less than flux Φ1 at no load. As we shall see, the weakening of flux due to armature reaction depends upon the position of brushes.

104 Armature Reaction

105 Geometrical and Magnetic Neutral Axes (i) The geometrical neutral axis (G.N.A.) is the axis that bisects the angle between the centre line of adjacent poles [See Fig. 2.2 (i)]. Clearly, it is the axis of symmetry between two adjacent poles. (ii) The magnetic neutral axis (M. N. A.) is the axis drawn perpendicular to the mean direction of the flux passing through the centre of the armature. Clearly, no e.m.f. is produced in the armature conductors along this axis because then they cut no flux. With no current in the armature conductors, the M.N.A. coincides with G, N. A. as shown in Fig. [2.2(ii)]. In order to achieve sparkless commutation, the brushes must lie along M.N.A.

106 Geometrical and Magnetic Neutral Axes With no current in armature conductors, the M.N.A. coincides with G.N.A. However, when current flows in armature conductors, the combined action of main flux and armature flux shifts the M.N.A. from G.N.A. In case of a generator, the M.N.A. is shifted in the direction of rotation of the machine. In order to achieve sparkless commutation, the brushes have to be moved along the new M.N.A. Under such a condition, the armature reaction produces the following two effects: 1. It demagnetizes or weakens the main flux. 2. It cross-magnetizes or distorts the main flux. Let us discuss these effects of armature reaction by considering a 2-pole generator (though the following remarks also hold good for a multipolar generator). (i) Fig. (2.3) (i) shows the flux due to main poles (main flux) when the armature conductors carry no current. The flux across the air gap is uniform. The m.m.f. producing the main flux is represented in magnitude and direction by the vector OFm in Fig. (2.3) (i). Note that OFm is perpendicular to G.N.A. (ii) Fig. (2.3) (ii) shows the flux due to current flowing in armature conductors alone (main poles unexcited). The armature conductors to the left of G.N.A. carry current in (x) and those to the right carry current out ( ). The direction of magnetic lines of force can be found by cork screw rule. It is clear that armature flux is directed downward parallel to the brush axis. The m.m.f. producing the armature flux is represented in magnitude and direction by the vector OFA in Fig. (2.3) (ii).

107 Geometrical and Magnetic Neutral Axes (iii) Fig. (2.3) (iii) shows the flux due to the main poles and that due to current in armature conductors acting together. The resultant m.m.f. OF is the vector sum of OFm and OFA as shown in Fig. (2.3) (iii). Since M.N.A. is always perpendicular to the resultant m.m.f., the M.N.A. is shifted through an angle θ. Note that M.N.A. is shifted in the direction of rotation of the generator. (iv) In order to achieve sparkless commutation, the brushes must lie along the M.N.A. Consequently, the brushes are shifted through an angle θ so as to lie along the new M.N.A. as shown in Fig. (2.3) (iv). Due to brush shift, the m.m.f. FA of the armature is also rotated through the same angle θ. It is because some of the conductors which were earlier under N-pole now come under S-pole and viceversa. The result is that armature m.m.f. FA will no longer be vertically downward but will be rotated in the direction of rotation through an angle θ as shown in Fig. (2.3) (iv). Now FA can be resolved into rectangular components Fc and Fd.

108 Geometrical and Magnetic Neutral Axes

109 Geometrical and Magnetic Neutral Axes

110 Geometrical and Magnetic Neutral Axes (a) The component Fd is in direct opposition to the m.m.f. OFm due to main poles. It has a demagnetizing effect on the flux due to main poles. For this reason, it is called the demagnetizing or weakening component of armature reaction. (b) The component Fc is at right angles to the m.m.f. OFm due to main poles. It distorts the main field. For this reason, it is called the crossmagnetizing or distorting component of armature reaction. It may be noted that with the increase of armature current, both demagnetizing and distorting effects will increase. Conclusions (i) With brushes located along G.N.A. (i.e., θ = 0 ), there is no demagnetizing component of armature reaction (Fd = 0). There is only distorting or crossmagnetizing effect of armature reaction. (ii) With the brushes shifted from G.N.A., armature reaction will have both demagnetizing and distorting effects. Their relative magnitudes depend on the amount of shift. This shift is directly proportional to the armature current. (iii) The demagnetizing component of armature reaction weakens the main flux. On the other hand, the distorting component of armature reaction distorts the main flux. (iv) The demagnetizing effect leads to reduced generated voltage while crossmagnetizing effect leads to sparking at the brushes

111 Demagnetizing and cross magnetizing conductors With the brushes in the G.N.A. position, there is only cross-magnetizing effect of armature reaction. However, when the brushes are shifted from the G.N.A. position, the armature reaction will have both demagnetizing and crossmagnetizing effects. Consider a 2-pole generator with brushes shifted (lead) θm mechanical degrees from G.N.A. We shall identify the armature conductors that produce demagnetizing effect and those that produce cross-magnetizing effect. (i) The armature conductors θ m on either side of G.N.A. produce flux in direct opposition to main flux as shown in Fig. (2.4) (i). Thus the conductors lying within angles AOC = BOD = 2θm at the top and bottom of the armature produce demagnetizing effect. These are called demagnetizing armature conductors and constitute the demagnetizing ampere-turns of armature reaction (Remember two conductors constitute a turn). ii) The axis of magnetization of the remaining armature conductors lying between angles AOD and COB is at right angles to the main flux as shown in Fig. (2.4) (ii). These conductors produce the cross-magnetizing (or distorting) effect i.e., they produce uneven flux distribution on each pole. Therefore, they are called cross-magnetizing conductors and constitute the cross-magnetizing ampere-turns of armature reaction.

112 Demagnetizing and cross magnetizing conductors

113 Calculation of Demagnetizing Ampere-Turns Per Pole (ATd/Pole) It is sometimes desirable to neutralize the demagnetizing ampere-turns of armature reaction. This is achieved by adding extra ampere-turns to the main field winding. We shall now calculate the demagnetizing ampere-turns per pole (ATd/pole). Let Z = total number of armature conductors I = current in each armature conductor = Ia/2 for simplex wave winding = Ia/P for simplex lap winding θm = forward lead in mechanical degrees Referring to Fig. (2.4) (i) above, we have, Total demagnetizing armature conductors= Conductors in angles AOC and BOD = 4θm/360 Z

114 Calculation of Demagnetizing Ampere-Turns Per Pole (ATd/Pole) Since two conductors constitute one turn, Therefore Total demagnetizing ampere-turns =

115 Demagnetising ampere-turns Per Pole (ATd/Pole)

116 Demagnetising ampere-turns Per Pole (ATd/Pole) Note. When a conductor passes a pair of poles, one cycle of voltage is generated. We say one cycle contains 360 electrical degrees. Suppose there are P poles in a generator. In one revolution, there are 360 mechanical degrees and 360 P/2 electrical degrees.

117 Cross-Magnetizing Ampere-Turns Per Pole (ATc/Pole) Demagnetizing ampere-turns per pole is given by

118 Cross-Magnetizing Ampere-Turns Per Pole (ATc/Pole) since two conductors make one turn) Demagnetizing ampere-turns per pole is given by

119 Cross-Magnetizing Ampere-Turns Per Pole (ATc/Pole) Therefore Cross-magnetizing ampereturns/pole are

120 COMMUTATION Fig. (2.6) shows the schematic diagram of 2-pole lapwound generator. There are two parallel paths between the brushes. Therefore, each coil of the winding carries one half (Ia/2 in this case) of the total current (Ia) entering or leaving the armature. Note that the currents in the coils connected to a brush are either all towards the brush (positive brush) or all directed away from the brush (negative brush). Therefore, current in a coil will reverse as the coil passes a brush. This reversal of current as the coil passes & brush is called commutation.

121 COMMUTATION

122 COMMUTATION The reversal of current in a coil as the coil passes the brush axis is called commutation. When commutation takes place, the coil undergoing commutation is shortcircuited by the brush. The brief period during which the coil remains shortcircuited is known as commutation period Tc. If the current reversal is completed by the end of commutation period, it is called ideal commutation. If the current reversal is not completed by that time, then sparking occurs between the brush and the commutator which results in progressive damage to both.

123 Ideal commutation Let us discuss the phenomenon of ideal commutation (i.e., coil has no inductance) in one coil in the armature winding shown in Fig. (2.6) above. For this purpose, we consider the coil A. The brush width is equal to the width of one commutator segment and one mica insulation. Suppose the total armature current is 40 A. Since there are two parallel paths, each coil carries a current of 20 A. (i) In Fig. (2.7) (i), the brush is in contact with segment 1 of the commutator. The commutator segment 1 conducts a current of 40 A to the brush; 20 A from coil A and 20 A from the adjacent coil as shown. The coil A has yet to undergo commutation.

124 Ideal commutation

125 Ideal commutation (ii) As the armature rotates, the brush will make contact with segment 2 and thus short-circuits the coil A as shown in Fig. (2.7) (ii). There are now two parallel paths into the brush as long as the short-circuit of coil A exists. Fig. (2.7) (ii) shows the instant when the brush is onefourth on segment 2 and three-fourth on segment 1. For this condition, the resistance of the path through segment 2 is three times the resistance of the path through segment 1 (since contact resistance varies inversely as the area of contact of brush with the segment). The brush again conducts a current of 40 A; 30 A through segment 1 and 10 A through segment 2. Note that current in coil A (the coil undergoing commutation) is reduced from 20 A to 10 A.

126 Ideal commutation iii) Fig. (2.7) (iii) shows the instant when the brush is one-half on segment 2 and one-half on segment 1. The brush again conducts 40 A; 20 A through segment 1 and 20 A through segment 2 (since now the resistances of the two parallel paths are equal). Note that now. current in coil A is zero. (iv) Fig. (2.7) (iv) shows the instant when the brush is three-fourth on segment 2 and one-fourth on segment 1. The brush conducts a current of 40 A; 30 A through segment 2 and 10 A through segment 1. Note that current in coil A is 10 A but in the reverse direction to that before the start of commutation. The reader may see the action of the commutator in reversing the current in a coil as the coil passes the brush axis. (v) Fig. (2.7) (v) shows the instant when the brush is in contact only with segment 2. The brush again conducts 40 A; 20 A from coil A and 20 A from the adjacent coil to coil A. Note that now current in coil A is 20 A but in the reverse direction. Thus the coil A has undergone commutation. Each coil undergoes commutation in this way as it passes the brush axis. Note that during commutation, the coil under consideration remains shortcircuited by the brush.

127 Ideal commutation Fig. (2.8) shows the current-time graph for the coil A undergoing commutation. The horizontal line AB represents a constant current of 20 A upto the beginning of commutation. From the finish of commutation, it is represented by another horizontal line CD on the opposite side of the zero line and the same distance from it as AB i.e., the current has exactly reversed (- 20 A). The way in which current changes from B to C depends upon the conditions under which the coil undergoes commutation. If the current changes at a uniform rate (i.e., BC is a straight line), then it is called ideal commutation as shown in Fig. (2.8). Under such conditions, no sparking will take place between the brush and the commutator

128 Commutation Practical difficulties This is illustrated in Fig. (2.9). The straight line RC represents thethe ideal commutation (i.e., straight line change of current) cannot be attained in practice. This is mainly due to the fact that the armature coils have appreciable inductance. When the current in the coil undergoing commutation changes, self-induced e.m.f. is produced in the coil. This is generally called reactance voltage. This reactance voltage opposes the change of current in the coil undergoing commutation. The result is that the change of current in the coil undergoing commutation occurs more slowly than it would be under ideal commutation whereas the curve BE represents the change in current when self-inductance of the coil is taken into account. Note that current CE (= 8A in Fig. 2.9) is flowing from the commutator segment 1 to the brush at the instant when they part company. This results in sparking just as when any other current carrying circuit is broken. The sparking results in overheating of commutatorbrush contact and causing damage to both. Fig. (2.10) illustrates how sparking takes place between the commutator segment and the brush. At the end of commutation or short-circuit period, the current in coil A is reversed to a value of 12 A (instead of 20 A) due to inductance of the coil. When the brush breaks contact with segment 1, the remaining 8 A current jumps from segment 1 to the brush through air causing sparking between segment 1 and the brush.

129 Commutation Practical difficulties

130 Calculation of Reactance Voltage Reactance voltage = Coefficient of self-inductance Rate of change of current When a coil undergoes commutation, two commutator segments remain shortcircuited by the brush. Therefore, the time of short circuit (or commutation period Tc) is equal to the time required by the commutator to move a distance equal to the circumferential thickness of the brush minus the thickness of one insultating strip of mica. Let Wb = brush width in cm,wm = mica thickness in cm,v = peripheral speed of commutator in cm/s Commutation period, Tc = ( Wb Wm )/v seconds The commutation period is very small, say of the order of 1/500 second. Let the current in the coil undergoing commutation change from + I to I (amperes) during the commutation. If L is the inductance of the coil, then reactance voltage is given by; Reactance Voltage,(for linear commutation)

131 Methods of improving commutation mproving commutation means to make current reversal in the short-circuited coil as sparkless as possible. The following are the two principal methods of improving commutation: (i) Resistance commutation (ii) E.M.F. commutation We shall discuss each method in turn.

132 Resistance Commutation Resistance Commutation The reversal of current in a coil (i.e., commutation) takes place while the coil is short-circuited by the brush. Therefore, there are two parallel paths for the current as long as the short circuit exists. If the contact resistance between the brush and the commutator is made large, then current would divide in the inverse ratio of contact resistances (as for any two resistances in parallel). This is the key point in improving commutation. This is achieved by using carbon brushes (instead of Cu brushes) which have high contact resistance. This method of improving commutation is called resistance commutation. Figs. (2.11) and (2.12) illustrates how high contact resistance of carbon brush improves commutation (i.e., reversal of current) in coil A. In Fig. (2.11) (i), the brush is entirely on segment 1 and, therefore, the current in coil A is 20 A. The coil A is yet to undergo commutation. As the armature rotates, the brush shortcircuits the coil A and there are two parallel paths for the current into the brush. Fig. (2.11) (ii) shows the instant when the brush is one-fourth on segment 2 and three-fourth on segment 1. The equivalent electric circuit is shown in Fig. (2.11) (iii) where R1 and R2 represent the brush contact resistances on segments 1 and 2.

133 Resistance Commutation

134 Resistance Commutation A resistor is not shown for coil A since it is assumed that the coil resistance is negligible as compared to the brush contact resistance. The values of current in the parallel paths of the equivalent circuit are determined by the respective resistances of the paths. For the condition shown in Fig. (2.11) (ii), resistor R2 has three times the resistance of resistor R1. Therefore, the current distribution in the paths will be as shown. Note that current in coil A is reduced from 20 A to 10 A due to division of current in (he inverse ratio of contact resistances. If the Cu brush is used (which has low contact resistance), R1 R2 and the current in coil A would not have reduced to 10 A. As the carbon brush passes over the commutator, the contact area with segment 2 increases and that with segment 1 decreases i.e., R2 decreases and R1 increases. Therefore, more and more current passes to the brush through segment 2. This is illustrated in Figs. (2.12) (i) and (2.12) (ii), When the break between the brush and the segment 1 finally occurs [See Fig (iii)], the current in the coil is reversed and commutation is achieved. It may be noted that the main cause of sparking during commutation is the production of reactance voltage and carbon brushes cannot prevent it. Nevertheless, the carbon brushes do help in improving commutation. The other minor advantages of carbon brushes are: (i) The carbon lubricates and polishes the commutator. (ii) If sparking occurs, it damages the commutator less than with copper brushes and the damage to the brush itself is of little importance.

135 EMF commutation In this method, an arrangement is made to neutralize the reactance voltage by producing a reversing voltage in the coil undergoing commutation. The reversing voltage acts in opposition to the reactance voltage and neutralizes it to some extent. If the reversing voltage is equal to the reactance voltage, the effect of the latter is completely wiped out and we get sparkless commutation. The reversing voltage may be produced in the following two ways: (i) By brush shifting (ii) By using interpoles or compoles (i) By brush shifting In this method, the brushes are given sufficient forward lead (for a generator) to bring the short-circuited coil (i.e., coil undergoing commutation) under the influence of the next pole of opposite polarity. Since the short-circuited coil is now in the reversing field, the reversing voltage produced cancels the reactance voltage. This method suffers from the following drawbacks: (a) The reactance voltage depends upon armature current. Therefore, the brush shift will depend on the magnitude of armature current which keeps on changing. This necessitates frequent shifting of brushes. (b) The greater the armature current, the greater must be the forward lead for a generator. This increases the demagnetizing effect of armature reaction and further weakens the main field. (ii) By using interpoles or compoles The best method of neutralizing reactance voltage is by, using interpoles or compoles.

136 Interpoles or Compoles The best way to produce reversing voltage to neutralize the reactance voltage is by using interpoles or compoles. These are small poles fixed to the yoke and spaced mid-way between the main poles (See Fig. 2.13). They are wound with comparatively few turns and connected in series with the armature so that they carry armature current. Their polarity is the same as the next main pole ahead in the direction of rotation for a generator (See Fig. 2.13). Connections for a d.c. generator with interpoles is shown in Fig. (2.14).

137 Interpoles or Compoles

138 Functions of Interpoles The machines fitted with interpoles have their brushes set on geometrical neutral axis (no lead). The interpoles perform the following two functions: (i) As their polarity is the same as the main pole ahead (for a generator), they induce an e.m.f. in the coil (undergoing commutation) which opposes reactance voltage. This leads to sparkless commutation. The e.m.f. induced by compoles is known as commutating or reversing e.m.f. Since the interpoles carry the armature current and the reactance voltage is also proportional to armature current, the neutralization of reactance voltage is automatic. (ii) The m.m.f. of the compoles neutralizes the cross-magnetizing effect of armature reaction in small region in the space between the main poles. It is because the two m.m.f.s oppose each other in this region. Fig. (2.15) shows the circuit diagram of a shunt generator with commutating winding and compensating winding. Both these windings are connected in series with the armature and so they carry the armature current. However, the functions they perform must be understood clearly. The main function of commutating winding is to produce reversing (or commutating) e.m.f. in order to cancel the reactance voltage. In addition to this, the m.m.f. of the commutating winding neutralizes the crossmagnetizing ampere-turns in the space between the main poles. The compensating winding neutralizes the cross-magnetizing effect of armature reaction under the pole faces.

139 Functions of Interpoles

140 DC Generator Equivalent circuit The magnetic field produced by the stator poles induces a voltage in the rotor (or armature) coils when the generator is rotated. This induced voltage is represented by a voltage source. The stator coil has resistance, which is connected in series. The pole flux is produced by the DC excitation/field current, which is magnetically coupled to the rotor The field circuit has resistance and a source The voltage drop on the brushes represented by a battery