Bài giảng Electromechanical energy conversion - Chapter VI: DC Machines - Nguyễn Công Phương

NguyễnCôngPhương ELECTROMECHANICAL ENERGY  CONVERSION DC Machines Contents I. Magnetic Circuits and Magnetic Materials II. Electromechanical Energy Conversion Principles III. Introduction to Rotating Machines IV. Synchronous Machines V. Polyphase Induction Machines VI. DC Machines VII.Variable – Reluctance Machines and Stepping Motors VIII.Single and Two – Phase Motors IX. Speed and Torque Control sites.google.com/site/ncpdhbkhn 2 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 3 Introduction (1) • Can be designed to display a wide variety of volt – ampere or speed – torque characteristics. • Frequently used in applications requiring a wide range of motor speeds ‐ or precise control of motor magnet‐dc‐motor‐or‐pmdc‐motor/ output. sites.google.com/site/ncpdhbkhn 4 Introduction (2) Quadrature axis 2 Direct  poles   axis TFmech d a1 sin r 22     o r  90     2    poles TFmech  d a1 22 Field coil Brushes Armature coils 8 C F  a i aa1  2 2m poles poles Field TCiKimech a  d a a d a 2m Armature sites.google.com/site/ncpdhbkhn 5 Introduction (3) Quadrature axis 1.5 Direct   axis   1   Voltage   0.5     Rectified coil voltage Rectified coil voltage Brush voltage 0 Brushes 100 200 300 400 500 600 Field coil Time Armature coils poles TCiKi mech2m a d a a d a eTamechm poles eCK  aadmadm2 m sites.google.com/site/ncpdhbkhn 6 Introduction (4) Quadrature axis Direct  ddPNi ff   axis d           Brushes 0 Ni Field coil  f f Armature coils poles TCiKieKT ;    mech2 m a d a a d a a a d m mech m eKaadm00 m n eeeaaa00  m00n sites.google.com/site/ncpdhbkhn 7 Introduction (5) Field Series field Armature To dc source Separately –excited Series Series field Shunt Compound sites.google.com/site/ncpdhbkhn 8 Introduction (6) • Generators: Field – The required field current is a very small fraction of the rated armature current; on the order of 1 to 3 Armature percent in the average generator. To dc source – A small amount of power in the field circuit may control a relatively Separately –excited large amount of power in the armature circuit, i.e., the generator is a power amplifier. – Often used in feedback control systems when control of the armature voltage over a wide range is required. – The terminal voltage decreases slightly with an increase in the load current. • Motors: – The field flux is nearly constant. sites.google.com/site/ncpdhbkhn 9 Introduction (7) • The field current is the same as the load Series field current, so that the air – gap flux & hence the Series generator voltage vary widely with load  are not often used. sites.google.com/site/ncpdhbkhn 10 Introduction (8) • Increase in load is accompanied by increases in the armature current & mmf & the stator field flux. • Because flux increases with load, Series field speed must drop in order to maintain the balance between impressed voltage & counter emf. Series motor •  A varying – speed motor with a markedly drooping speed torque characteristic. • For applications requiring heavy torque overloads, this is particularly advantageous because the corresponding power overloads are held to more reasonable values by the associated speed drops. sites.google.com/site/ncpdhbkhn 11 Introduction (9) • The voltage drops off somewhat with load, but it is acceptable. • The voltage can be Shunt generator controlled over reasonable limits by means of rheostats in the shunt field. sites.google.com/site/ncpdhbkhn 12 Introduction (10) • The field flux is nearly constant. • Increased torque must be accompanied by a very nearly proportional increase in armature current & hence by a small decrease in counter emf Ea to allow this increased current through the small Shunt motor armature resistance. • Since counter emf is determined by flux & speed, the speed must drop slightly. • A constant – speed motor having about 6 percent drop in speed from no load to full load. • An outstanding advantage: ease of speed control. sites.google.com/site/ncpdhbkhn 13 Introduction (11) • Are normally connected so that the mmf of the series winding aids that of the shunt winding • Then the flux per pole can Series field increase with load, resulting in a voltage output which is nearly constant or which even rises Compound generator somewhat as load increases. • The shunt winding usually contains many turns of relatively small wire. • The series winding (wound on the outside) consists of a few turns of comparatively heavy conductor. • The voltage can be controlled over reasonable limits by means of rheostats in the shunt field. sites.google.com/site/ncpdhbkhn 14 Introduction (12) • Does not have the disadvantage of very high light – load speed associated with a series Series field motor, but it retains to a considerable degree the Compound motor advantages of series excitation. sites.google.com/site/ncpdhbkhn 15 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 16 Commutator Action (1) sites.google.com/site/ncpdhbkhn 17 Commutator Action (2) sites.google.com/site/ncpdhbkhn 18 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 19 Effect of Armature MMF (1) • Armature mmf has definite effects on both the space distribution of the air – gap flux & the magnitude of the net flux per pole. • The effect on flux distribution is important because the limits of successful commutation are directly influenced. • The effect on flux magnitude is important because both the generated voltage & the torque per unit of armature current are influenced thereby. sites.google.com/site/ncpdhbkhn 20 Effect of Armature MMF (2) ‐of‐armature‐reaction‐on.html sites.google.com/site/ncpdhbkhn 21 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 22 Analytical Fundamentals: Electric – Circuit Aspects (1) poles TKIEK;;  K  C mech a d a a a a m a2 m a EIaa Tmech m Generator Generator I I a Motor L Motor Field  VERIaaaa  I f rheostat V VEta() RRI asa  Va Series Shunt t field field IIILaf   sites.google.com/site/ncpdhbkhn 23 Analytical Fundamentals: Ex Electric – Circuit Aspects (2) Generator Generator A 25-kW 110-V separately – excited DC machines is Ia I L operated at a constant speed of 2400 rpm with a Motor Motor constant field current such that the open – circuit  Field  armature voltage is 110V. R = 0.02Ω. Compute the I f a rheostat armature current, terminal power, electromagnetic V power, and torque when the terminal voltage is: Va Series Shunt t a) 114V and b) 108V? field field   VERItaaa VERItaaa  VEta114V 110V VEta 114 110 Ia  200A Ra 0.02 VIta114 200  22.8kW EIaa110 200  22.0kW 3 EIaa EI aa 22.0 10 Tmech   87.5Nm m 2 (n / 60) 2  (2400 / 60) sites.google.com/site/ncpdhbkhn 24 Analytical Fundamentals: Ex Electric – Circuit Aspects (3) Generator Generator A 25-kW 110-V separately – excited DC machines is Ia I L operated at a constant speed of 2400 rpm with a Motor Motor constant field current such that the open – circuit  Field  armature voltage is 110V. R = 0.02Ω. Compute the I f a rheostat armature current, terminal power, electromagnetic V power, and torque when the terminal voltage is: Va Series Shunt t a) 114V and b) 108V? field field   VERItaaa VERItaaa  VEta108V 110V EVat 110 108 Ia  100A Ra 0.02 VIta108 100 10.8kW EIaa110 100 11.0kW 3 EIaa EI aa 11.0 10 Tmech   43.8Nm m 2 (n / 60) 2  (2400 / 60) sites.google.com/site/ncpdhbkhn 25 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects a) Armature Reaction Neglected b) Effects of Armature Reaction Included 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 26 Armature Reaction Neglected (1) Main- field mmf Nf Ifss N I N s Gross- mmf If Is N f m n EaaaEE00 m00n sites.google.com/site/ncpdhbkhn 27 Armature Reaction Neglected (2) Ex Generator Generator A 150-kW 250-V, 400-A generator has an armature Ia I L resistance (including brushes) of 0.025Ω, a series – Motor Motor field resistance of 0.005 Ω. There are 1000 shunt –  Field  field turns per pole & three series – field turns per I f rheostat pole. The shunt – field current is 4.7A & the speed is V 1150 rpm. Compute the terminal voltage at rated Va Series Shunt t terminal current. Neglect the effects of armature field field reaction.   IIIIsaLf400  4.7  404.7A N s 3 Gross- mmf Ifs I  4.7  404.7  5.9 equivalent shunt-field amperes N f 1000 sites.google.com/site/ncpdhbkhn 28 Armature Reaction Neglected (3) Gross-5.9 mmf  Ea0  275V sites.google.com/site/ncpdhbkhn 29 Armature Reaction Neglected (4) Ex Generator Generator A 150-kW 250-V, 400-A generator has an armature Ia I L resistance (including brushes) of 0.025Ω, a series – Motor Motor field resistance of 0.005 Ω. There are 1000 shunt –  Field  field turns per pole & three series – field turns per I f rheostat pole. The shunt – field current is 4.7A & the speed is V 1150 rpm. Compute the terminal voltage at rated Va Series Shunt t terminal current. Neglect the effects of armature field field reaction.   IIIIsaLf400  4.7  404.7A N s 3 Gross- mmf Ifs I  4.7  404.7  5.9 equivalent shunt-field amperes N f 1000 Ea0  275V n 1150 EEaa0 275  263.5V n0 1200 VEta( RRI asa  )  263.5  (0.025  0.005)404.7  253.2V sites.google.com/site/ncpdhbkhn 30 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects a) Armature Reaction Neglected b) Effects of Armature Reaction Included 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 31 Effects of Armature Reaction Included (1) Net mmf gross mmf Far NIf fssar NI F sites.google.com/site/ncpdhbkhn 32 Effects of Armature Reaction Ex Included (2) Generator Generator A 150-kW 250-V, 400-A generator has an armature Ia I L resistance (including brushes) of 0.025Ω, a series – Motor Motor field resistance of 0.005 Ω. There are 1000 shunt –  Field  field turns per pole & three series – field turns per I f rheostat pole. The shunt – field current is 4.7A & the speed is V 1150 rpm. Compute the terminal voltage at rated Va Series Shunt t terminal current. Include the effects of armature field field reaction.   IIIIsaLf400  4.7  404.7A N s 3 Gross- mmf Ifs I  4.7  404.7  5.9 equivalent shunt-field amperes N f 1000 sites.google.com/site/ncpdhbkhn 33 Effects of Armature Reaction Included (3) Gross-5.9 mmf  Ea0  260V sites.google.com/site/ncpdhbkhn 34 Effects of Armature Reaction Ex Included (4) Generator Generator A 150-kW 250-V, 400-A generator has an armature Ia I L resistance (including brushes) of 0.025Ω, a series – Motor Motor field resistance of 0.005 Ω. There are 1000 shunt –  Field  field turns per pole & three series – field turns per I f rheostat pole. The shunt – field current is 4.7A & the speed is V 1150 rpm. Compute the terminal voltage at rated Va Series Shunt t terminal current. Include the effects of armature field field reaction.   IIIIsaLf400  4.7  404.7A N s 3 Gross- mmf Ifs I  4.7  404.7  5.9 equivalent shunt-field amperes N f 1000 Ea0  260V n 1150 EEaa0 260  249.2V n0 1200 VEta( RRI asa  )  249.2  (0.025  0.005)404.7  237V sites.google.com/site/ncpdhbkhn 35 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance a) Generator Analysis b) Motor Analysis 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 36 Generator Analysis (1) • For a generator, the speed is usually fixed by the prime mover. • Problems often encountered: – To determine the terminal voltage corresponding to a specified load & excitation, or – To find the excitation required for a specified load & terminal voltage. sites.google.com/site/ncpdhbkhn 37 Generator Analysis (2) • Since the main – field Field current is independent of the generator voltage, Armature they are the simplest to To dc source analyze. Separately –excited • For a given load, the equivalent main – field excitation is: N s Gross- mmf If Is N f • Ea is determined by the appropriate magnetization curve. sites.google.com/site/ncpdhbkhn 38 Generator Analysis (3) • Found to self – excite under properly chosen operating conditions. • The generated voltage will build up spontaneously to a Shunt value ultimately limited by magnetic saturation. • The shunt – field excitation depends on the terminal voltage. • The series – field excitation depends on the armature current. sites.google.com/site/ncpdhbkhn 39 Generator Analysis (4) Ex. A 150-kW 250-V, 400-A, 1200-rpm dc shunt generator has the magnetization curves (including armature – reaction effects) of below figure. The armature – circuit resistance (including brushes) is 0.025Ω. The generator is driven at a constant speed of 1200-rpm, and the excitation is adjusted (by varying the shunt – field rheostat) to give rated voltage at no load. Find the terminal voltage at an armature current of 400 A? Shunt RIaa400 0.025  10V 10V Vt 50  4 200V sites.google.com/site/ncpdhbkhn 40 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance a) Generator Analysis b) Motor Analysis 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 41 Motor Analysis (1) • For a motor, terminal voltage is often fixed at the value of the available source. • Problems frequently encountered: – To determine the speed corresponding to a specific load & excitation, or – To find the excitation required for specified load & speed conditions. • The terminal motor is often held substantially constant or controlled to a specific value. sites.google.com/site/ncpdhbkhn 42 Motor Analysis (2) Ex. A 100-hp 250-V dc shunt motor has the magnetization curves (including armature – reaction effects) of below figure. The armature – circuit resistance (including brushes) is 0.025Ω. The field rheostat is adjusted for a no – load speed of 1100 rpm. Find the speed if the armature current is 400A? n 1200 Shunt EEa,0, no load a no load 250 273V n0 1100 I f, no load  5.9A E  260V aI0,a  400 E 250 400 0.025  240V aI,a  400 n EE aI,aa 400 a 0, I 400 n0 E aI,a  400 240 nn0 1200  1108rpm EaI0, 400 260 a sites.google.com/site/ncpdhbkhn 43 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 44 Permanent – Magnet DC Machines (1) • The principal difference between permanent – magnet dc machines & those discussed previously is that they have a fixed source of field – winding flux which is supplied by a permanent magnet. • Widely found in a wide variety of low-power applications. • The field winding is replaced by a permanent magnet. • Advantages: – Simple construction – Do not require external excitation & power to create magnetic fields. – Maybe smaller & cheaper. • Disadvantages: – The risk of demagnetization due to excessive currents in the motor windings or due to overheating of the magnet. – Permanent magnets are somewhat limited in the magnitude of air – gap flux density that they can produce. sites.google.com/site/ncpdhbkhn 45 Permanent – Magnet DC Ex. 1 Machines (2) t Both the rotor and the outer shell are made of infinitely m t permeable magnetic material (μ →∞). The magnet is made of g neodymium-iron-boron. tg = 0.5mm, tm = 3.5mmm. Ignoring the effects of rotor slots, find the magnetic flux density in the air gap. tm 773.5 BHmm0 410  H m  2810   H m tg 0.5 BBmg1.09T   2tm 2tg sites.google.com/site/ncpdhbkhn 46 Permanent – Magnet DC Machines (3) Ia  EKamm  Vt  EIaa EKaadmmech ;; T  K mad  K m sites.google.com/site/ncpdhbkhn 47 Permanent – Magnet DC Ex. 2 Machines (4) A permanent – magnet dc motor has an armature resistance of 1.03Ω. Ia When operated at no load from a dc source of 50V, it is observed to operate at a speed of 2100 rpm & to draw a current of 1.25A. Find:  a) The torque constant Km? b) The no – load rotational losses of the motor? Ea Vt c) The power output of the motor when it is operating at 1700 rpm from a 48-V source  VERItaaa EVRIataa 50 1.03  1.25  48.7V 2100 2  220 rad/s m 60 Ea 48.7 V Km  0.22 m 220 r/s Rotational losses Eaa I 48.7 1.25 61W sites.google.com/site/ncpdhbkhn 48 Permanent – Magnet DC Ex. 2 Machines (5) A permanent – magnet dc motor has an armature resistance of 1.03Ω. Ia When operated at no load from a dc source of 50V, it is observed to operate at a speed of 2100 rpm & to draw a current of 1.25A. Find:  a) The torque constant Km? b) The no – load rotational losses of the motor? Ea Vt c) The power output of the motor when it is operating at 1700 rpm from a 48-V source  1700 2  178 rad/s m 60 EKamm 0.22 178 39.2V VEta 48 39.2 VERIItaaaa   8.54A Ra 1.03 PEImech a a 39.2 8.54 335W Pshaft P mech rotational losses 335 61 274W sites.google.com/site/ncpdhbkhn 49 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 50 Commutation and Interpoles (1) • The most important: the ability to transfer the necessary armature current through the brush contact at the commutator without sparking & without excessive local losses & heating of the brushes & commutator. • Sparking causes destructive blackening, pitting, and wear of both the commutator & the brushes, conditions which rapidly become worse & burn away the copper & carbon. • Sparking may be caused by faulty mechanical conditions, such as chattering of the brushes or a rough, unevenly worn commutator. sites.google.com/site/ncpdhbkhn 51 Commutation and Interpoles (2) • The attainment of good commutation is more an empirical art than a quantitative science. • The principal obstacle to quantitative analysis lies in the electrical behavior of the carbon-copper (brush-commutator) contact film. • The film’s resistance is nonlinear and is a function of current density, current direction, temperature, brush material, moisture, and atmosphere pressure. • The film’s behavior in some respects is like that of an ionized gas or plasma. sites.google.com/site/ncpdhbkhn 52 Commutation and Interpoles (3) sites.google.com/site/ncpdhbkhn 53 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 54 Compensating Windings (1) • For machines subjected to heavy overloads, rapidly changing loads, or operation with a weak main field, there is the possibility of trouble other than simply sparking at the brushes. • At the instant when an armature coil is located at the peak of a badly distorted flux wave, the coil voltage may be high enough to break down the air between the adjacent segments to which the coil is connected and result in flashover, or arcing, between segments. • The breakdown voltage here is not high, because the air near the commutator is in a condition favorable to breakdown, due to the presence of the plasma carrying the armature current between the brushes & the commutator. sites.google.com/site/ncpdhbkhn 55 Compensating Windings (2) • Flashing between segments may quickly spread around the entire commutator and constitutes a direct short circuit on the line. • Solution: compensate or neutralize the armature mmf under the pole faces. • Disadvantage: expensive Commutating winding Shunt field Armature Series field Field rheostat Compensating winding sites.google.com/site/ncpdhbkhn 56 DC Machines 1. Introduction 2. Commutator Action 3. Effect of Armature MMF 4. Analytical Fundamentals: Electric – Circuit Aspects 5. Analytical Fundamentals: Magnetic – Circuit Aspects 6. Analysis of Steady – State Performance 7. Permanent – Magnet DC Machines 8. Commutation and Interpoles 9. Compensating Windings 10. Serial Universal Motors sites.google.com/site/ncpdhbkhn 57 Serial Universal Motors • Series universal motor: the rotor & stator structures of a series connected motor are properly laminated to reduce ac-eddy current losses. • It has the convenient ability to run on either Series field ac or dc current & with similar   characteristics. Ia • Universal motor: a single – phase series V Ea a motor. Armature • Used where light weight is important (e.g.,   vacuum cleaner, kitchen appliances, portable tools) and usually operate at high speed (1500 to 15000 rpm). • Advantage: highest horsepower per dollar in the fractional – horsepower range. • Disadvantages: noise, relative short life. sites.google.com/site/ncpdhbkhn 58