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    Chapter 7 Chapter 7 Presentation Transcript

    • Chapter 7 Direct Current Machines 7.1 fundamentals 7.2 No Load Magnetic Field 7.3 Armature Winding and Armature Reaction 7.4 Emf and Electromagnetic Torque in Armature Winding 7.5 DC GeneratorThursday, July 22, 2010
    • 7.1 fundamentals 7.1.1 Generating an ac voltage 7.1.2 DC generator 7.1.3 DC motor 7.1.4 Construction 7.1.5 Name plate dataThursday, July 22, 2010
    • 7.1.1 Generating an ac voltageThursday, July 22, 2010
    • 7.1.2 DC generator Windin Main poles g The simplest possible 主磁极 绕组 rotating dc machine is shown in figure 7-1 commutator It consists of a single loop of s wire rotating about a fixed axis. The rotating part of this machine is called roor ,and the stationary part is called stator. brushes Figure 7-1Thursday, July 22, 2010
    • 7.1.2 DC generatorAs the coil rotates, a voltage isinduced between its terminals. WindingThis voltage is connected to Main poles 绕组the brushes A and B, and 主磁极therefore, across the load.The induced voltage is commutatorstherefore maximum when thecoil is momentarily in thevertical position; and zerowhen the coil is momentarilyin the horizontal position. brushes Figure 7-1Thursday, July 22, 2010
    • 7.1.2 DC generator Brushe This figure is that its s polarity changes every time Commutat or the coil makes half a turn. Armature The voltage can therefore core be represented as a function of the angle of Main poles rotation. The waveform depends upon the shape of the N,S poles. Figure 7-2Thursday, July 22, 2010
    • 7.1.2 DC generator We assume the poles were designed to generate the wave. Because each angle of rotation corresponds to a specific interval of time, we can represent the induced voltages as a function of time as in Figure 7-3. Figure 7-3Thursday, July 22, 2010
    • 7.1.3 DC motor In the Figure 7-1, if a power supply is provided across the brushes A and B, and a current is produced in the coil, the conductors carrying the current will subject to a force to drive the coil to rotate, which is a dc motor operation. Figure 7-4 The simplest possible dc motor is shown in Figure 7-4.Thursday, July 22, 2010
    • 7.1.4 constructionThursday, July 22, 2010
    • 7.1.4 constructionFigure 7-6 is a real rotor of a dc machine. Figure 7-6Thursday, July 22, 2010
    • 7.1.4 constructionFigure 7-7 shows a real 4-pole stator. Figure 7-7Thursday, July 22, 2010
    • 7.1.4 construction The main section of stator main pole core interpole brushThursday, July 22, 2010
    • 7.1.4 construction The main section of rotor armature corn armature winding commutatorThursday, July 22, 2010
    • 7.1.5 Name plate data  A name plate gives the rating information about a machine, andif a machine is running at the rated condition, theperformancewill be better.Rated power (w or kW): defined as output power. For agenerator, it is the output electrical power; and for a motor, it isthe output mechanical power on shaft.Rated voltage (V): defined as the armature voltage under therated condition.Rated current (A): defined as the armature current under therated condition. A dc machine can not be running for a long timeat over-rated current, which will burn the machine.Thursday, July 22, 2010
    • 7.1.5 Name plate data Rated speed (rpm): defined as the rotor speed under the ratedcondition.Rated efficiency (Hz): defined as the radio of output power toinput power under the rated condition.Excitation mode: shunt machine, series machine or compoundmachine.Rated excitation data: rated excitation voltage, excitation current,ect.Insulation class: Class A, E, B etc.Rated temperature rise: impose the temperature limit.Thursday, July 22, 2010
    • 7.2 No Load Magnetic Field 7.2.1 Magnetization Characteristic 7.2.2 Flux Density DistributionThursday, July 22, 2010
    • 7.2.1 Magnetization Characteristic Magnetization Characteristic The flux is produced by the field current, and the Ф = f(If) is the magnetization curve as shown in figure Figure 7-8Thursday, July 22, 2010
    • 7.2.2 Flux density distribution flux / pole   2 p Bav     area / pole l dl / 2 p dl Figure 7-9Thursday, July 22, 2010
    • 7.3 Armature Winding  and Armature Reaction 7.3.1 Some Terms in Armature Winding 7.3.2 Simple Lap Winding 7.3.3 Armature MMF and DC Machine Magnetic Field 7.3.4 Armature Reaction when Brushes at Geometrical Neutral LineThursday, July 22, 2010
    • 7.3.1 Some terms in armature winding  1.Pole axis: is defined as the middle line of a magnetic pole; 2.Geometric neutral line: is the middle line between two addjacent poles; 3.Number of slots Z: is the total number of slots around the full rotor circumference. 4.Pole pitch: is the distance between two adjacent geometric neutral line.Thursday, July 22, 2010
    • 7.3.1 Some terms in armature winding  5.Coil:is a basic element of armature winding. It can have one turn and multi-turns, and its two terminals, starting terminal and end terminal, will connect to two different commutator segments separately. 6.Coil pitch: represents the distance between the two effective sides, numbered by slots, because the sides of a coil must be placed in slots, so it must be an integer.Thursday, July 22, 2010
    • 7.3.2 Simple lap winding The best way to understand the simple lap winding is to design it through an example.Example:design a simple lapwinding with Z=16slots and 2p=4 poles. Figure 7-10Thursday, July 22, 2010
    • 7.3.2 Simple lap windingSolution : Z 161. Calculate:     4 2P 42. Draw the winding:1).Draw 16slots, andnumber them.2).Draw 16 coils one byone, series connected intoa loop.3).Draw commutatorsegments, connected to Figure 7-10coil terminals.Thursday, July 22, 2010
    • 7.3.2 Simple lap winding4).Draw poles, evenlydistributed, and assign thepoles by N and S.5).Assume motion direction.6).Determine the directionof induced emf.7).Place brushes:8).Connecting the brushesterminals to complete thearmature winding as infigure 7-9. Figure 7-10Thursday, July 22, 2010
    • 7.3.2 Simple lap winding Conclusion :1.The coils under each magnetic pole consist of one branch,so for a 4-pole dc machine with simple lap winding, thereare four branches. 2.The four emfs in the four branches are identical. 3.These four emfs are paralleled connected. 4.The emf appears at the brushes, which are connected to the external terminals 5.The terminal current is the sum of the currents in the four branches. Where i a ----branch current; Ia ia  a ----number of branch pair; 2aThursday, July 22, 2010
    • 7.3.2 Simple lap winding The diagram of parallel branches Figure 7-11Thursday, July 22, 2010
    • 7.3.3 Armature mmf and dc machine field AS stated above, the brushes are at the positions where themain pole flux density is zero and it has been known that thecurrent directions. At the two sides of each brush are differentwhich give rise to an armature mmf as in figure 7-12a. This time, the dc machine magnetic field consists of mainfield mmf F f and armature mmf as Fa shown in Figure 7-12b. Figure 7-12a Figure 7-12b.Thursday, July 22, 2010
    • 7.3.4  Armature reaction when brushes at geometrical neutral line The armature mmf will distort the main magnetic fieldproduced by the main poles as shown in Figure 7-13 the nextpage. It must be noticed that when the peak value of combinedflux density is large enough to make the magnetic fieldsaturate , the armature reaction will reduce the total fluxunder a pole ,which is unexpected.Thursday, July 22, 2010
    • 7.3.4  Armature reaction when brushes at geometrical neutral line 电枢磁场磁通 密度分布曲线 Bx 主磁场的 B0 x 磁通密度 分布曲线 Bax 两条曲线逐点叠加后 得到负载时气隙磁场 的磁通密度分布曲线Thursday, July 22, 2010
    • 7.4 Emf and  Electromagnetic Torque in  Armature Winding 7.4.1 EMF in Armature Winding 7.4.2 Electromagnetic TorqueThursday, July 22, 2010
    • 7.4.1 Emf in Armature Winding 1.emf in a conductor An emf e c induced in a conductor can be express as: ec  B av lv (7-5) Where Bav --------average flux density weber/meter l --------effective length meter V --------tangential speed meter/secondThursday, July 22, 2010
    • 7.4.1 Emf in Armature Winding 2.The average flux densityBased on the flux curve in the figure 7-8,the average flux densityis: B av   /  l   . 2 p /  dl (7-6)Where p -------number of pole pair l -------effective length of conductor d -------diameter where the winding conditions are fixed  -------flux per poleThursday, July 22, 2010
    • 7.4.1 Emf in Armature Winding 3.The tangential speed v If n represents the rotor speed in revolution per minute, then the tangential speed v is given by: v  dn / 60 (7-7)Thursday, July 22, 2010
    • 7.4.1 Emf in Armature Winding4.Armature emf The armature emf is: N N n pN Ea  Bavlv  l  2 p  n  Ce n (7-8) 2a 2a l 60 60a Where Z-------number of conductors in the armature winding Ce  2pz/ 60 -----emf costant (7-9)Thursday, July 22, 2010
    • 7.4.1 Emf in Armature Winding• From the equations(7-8),it can be seen that the induced voltage in any given machine depend on three factors: 1.The flux  in the machine; 2.The speed n of the machine rotor 3.A constant depending on the construction of the machineThursday, July 22, 2010
    • 7.4.2  Electromagnetic TorqueEach conductor of an armature in the dc machine will subjectto a average force f av ,which is: f av  B av li a (7-10) Where i a ----Branch current in armature winding.From the equation (7-10),the average torque produced by eachconductor is Tav : T av  B av li a d / 2 (7-11)If the total number of conductors is N, the total torque will be : T  NB av li a d / 2 (7-12)Thursday, July 22, 2010
    • 7.4.2  Electromagnetic Torque 2 p  IaIn view of d  , Bav  and i a  ,  l 2athe total magnetic torque will be: D D  I a 2 p pN T  f av N  Bavlia N  l N I a  CT I a 2 2 l 2a 2 2a (7-13) pN Where C T  ,torque constant. 2 aThursday, July 22, 2010
    • 7.4.2  Electromagnetic Torque Compare CT and Ce ,we have: pN CT  2  a  60  9 . 55 (7-14) Ce pN 2 60 a It can be seen that the ratio of torque constant to emf constant is 9.55Thursday, July 22, 2010
    • 7.5   DC Generator 7.5.1 Excitation methods 7.5.2 Balance Equations 7.5.3 Characteristcs of separate excited DC generator 7.5.4 Shunt Generator 7.5.5 The Series DC Generator 7.5.6The compound DC GeneratorThursday, July 22, 2010
    • 7.5.1  Excitation MethodsAccording to the ways of the winding connection ,dc machinesare classified as separately excited machines , shunt machines,series excited machines and compound machines.1.Separately excited generator  U  A separate supply ensures that the I  Ia excitation provided by this winding G is not dependant. Therefore the field excitation current If can be easily controlled by any normal methods.  Uf Thursday, July 22, 2010
    • 7.5.1  Excitation Methods2.Shunt generator 3.Series excite generatorWhen the excitation winding When the excitation winding ofof a dc machine is paralleled a dc machine is seriesconnected with the winding, connected with the winding,thethe generator is called shunt generator is called seriesgenerator. excited generator Ia = I = I I a = I+ I f f  U   U  I If I Ea Ea Ia IaThursday, July 22, 2010
    • 7.5.1  Excitation Methods4.Compound generator There are two parts of Excitation windings, one is series connected to armature winding ,and the other is shunted connected.  U   U  I I If 1 If 1 Ea Ea Ia IaThursday, July 22, 2010
    • 7.5.2  Balance Equations1.Voltage balance equations The reference directions are shown as the Figure Ea  U  I a Ra  2 ΔU b  U  I a Ra (7-15)  U  T1 T0 2.Torque balance equations Ia n Tem T1  Tem  T0 (7-16)  Ea  Where T1 ------mechanical torque  If Tem ------electromagnetic torque Uf Φ T0 ------no-load torque Thursday, July 22, 2010
    • 7.5.2  Balance Equations 3.Power balance equation pN 2 pNPem  T    CT I a  I a n nI a  E a I a 2a 60 60a For the equations(7-16), each item times regular frequency  then: T1  T0   Te Thursday, July 22, 2010
    • 7.5.2  Balance Equations Therefore p1  P0  Pem (7-17) P0  Pm  Pf  Pfe  Pco  P2 (7-18) Where P ------input power 1 Pem ------electromagnetic power P0 ------no-load losses Pm ------mechanical power Pfe ------iron losses Pf ------magnetic field losses Pco ------armature copper losses P2 ------output powerThursday, July 22, 2010
    • 7.5.2  Balance Equations Therefore the energy flow diagram is shown in Figure 7-19. Figure 7-19Thursday, July 22, 2010
    • 7.5.3 Characteristic of separate excited DC generator1.No-load characteristicn=nN, U0=E0=ƒ(If) when Ia=0.Thursday, July 22, 2010
    • 7.5.3 Characteristic of separate excited DC generator 2.Output characteristic or External Characteristic ofseparate excited Generator U Separate • When excite U0 n=nN, If=IfN, U=ƒ(I) 0 IThursday, July 22, 2010
    • 7.5.4 Shunt Generator • 1.The process of self excitationThursday, July 22, 2010
    • 7.5.4 Shunt Generator U 他励 U0 并励 0 I Critical resistance of a shunt dc generator Terminal characteristic •2.Three Essential conditions for the self-excitation (1) Residual flux density (2) Correct connection (3) Critical resistance larger than field resistance •3.Terminal characteristic of a shunt generatorThursday, July 22, 2010
    • 7.5.5 THE SERIES DC GENERATOR  U  I If Ea Ia U=Ea-(Ra+Rs)IaThursday, July 22, 2010
    • 7.5.6 The Compound DC Generator 1.Few series turns 2.More series turns 3.Even more series turns are addedThursday, July 22, 2010
    • 7.6 DC Motor 7.6.1 Basic Balance Equations 7.6.2 Reversible Principle of DC Machine 7.6.3 Characteristic of Shunt Motor 7.6.4 Series DC Motor 7.6.5 The Compounded DC Motor 7.6.6 Modified Torque‐speed CharacteristicsThursday, July 22, 2010
    • 7.6 DC Motor Four major types of dc motors The separately excited dc motor The shunt dc motor The series dc motor The compound dc motor If Ia If Uf M U U M U M U M 他励 并励 串励 复励Thursday, July 22, 2010
    • 7.6.1 Basic Balance Equations 1.Voltage balance equation U  Ea  R a I a I Ia If E U MThursday, July 22, 2010
    • 7.6.1 Basic Balance Equations2.Torque balance equation T em  T 2  T 0 T em  C T Φ N I aThursday, July 22, 2010
    • 7.6.1 Basic Balance Equations 3.Power balance equation EaIa=ΩT pm+pad pco pfe P1  p co  PM  p co  p m  p fe  p ad  P2Thursday, July 22, 2010
    • 7.6.2  Reversible Principle of dc machine Question: 我们已经做过实验,直流电动机可以直 接拿来做发电机,反之亦然。 那前面学过的三相交流电机可以吗?Thursday, July 22, 2010
    • 7.6.3  Characteristics of shunt motor 1.Speed Characteristic When U=UN, If=IfN, the relationship of between speed and armature current is UN Ra n  I a = n0-βIa Ce  Ce Thursday, July 22, 2010
    • 7.6.3  Characteristics of shunt motor 2.Torque Characteristic When U=Un, I=Ifn, we can know How to increase  T  K T ΦI a Ia?(U=Un) TL  n E T Ia Thursday, July 22, 2010
    • 7.6.3  Characteristics of shunt motor 3.Efficiency P1  P Pco+Pf+Pfe+Pm+Pad  1 P1 U(Ia+If) Neglect Pad and If,then  P1   P P0  Ra I a2 η  1 P1 U N Ia P0=Pf+Pm+Pfe=constant When P0=RaIa2,  is max Ia 0Thursday, July 22, 2010
    • 7.6.3  Characteristics of shunt motor4.Terminal Characteristic or Torque-Speed Curve When U=UN If=IfN U  E  Ia Ra U Ra E  KCe n   T E Φn K EΦ 2 Ce Ce E Φ K T KCT T  KCΦI a T T U When T=0, n0= (theoretical no-load speed ) C eΦ U R  When T=T0 , n 0   2 T 0 (practical no- C eΦ C eC T Φ load speed)Thursday, July 22, 2010
    • 7.6.3  Characteristics of shunt motor If we consider armature reaction: n n n0 n0 With AR n 0 n 0 nN nN No AR T0 TN Tem T0 TN Tem Without armature reaction With armature reactionThursday, July 22, 2010
    • 7.6.4  Series DC Moter If 1.Torque Characteristic The electromagnetic torque is: T= CT Ia U Ea Ia 1)The magnetic circuit is not saturated, =KIf T= CT Ia=CTKIa2 2)The magnetic circuit is saturated, =constant T= CTK’IaThursday, July 22, 2010
    • 7.6.4  Series DC Moter 2.Torque-speed curve 1)The magnetic circuit is not saturated, =KIf E  C e Φn  C e K I a n T  C T ΦIa  C T K I a 2 U  E  I a RaThursday, July 22, 2010
    • 7.6.6  Modified Torque –Speed Characteristics1.Inherent torque-speed curve n n0 n 0 U Ra nN n  2 T K e Φ K T K ET CE CeC Φ T0 TN TemThursday, July 22, 2010
    • 7.6.6  Modified Torque –Speed Characteristics2.Modified T-n curve with series resistance in the armature circuit n n0 Ra UN Ra  RSc n  2 Tem C e Φ N C e CT Φ N Ra  RS TemThursday, July 22, 2010
    • 7.6.6  Modified Torque –Speed Characteristics 3.Modified curve with terminal voltage varying n 01 n U2  U N n0 U2 UN U R n  T Ce  CeCT  2 TemThursday, July 22, 2010
    • 7.6.6  Modified Torque –Speed Characteristics4.Modified curve when the field current If is reduced n Rf2  Rf1n01 If=U/Rf Rf If Φ Rf2n0 E Ia T n T >TL Rf1 TemThursday, July 22, 2010
    • 7.6.6  Modified Torque –Speed Characteristics5.Inherent torque –speed curve from name plate data We know Pn, Nn, Un, IN There are two points very important (n0, T0), (nN, TN) Summary: Summary 1 UN I N  P103 1)Calculate Ra Ra  2 2 IN U N  I N Ra 2)Calculate Ceφ and CTφ: Ce   , CT   9.55Ce  nN UN 3)Calculate n0 n0  CeN 4)Calculate TN TN  CT  N I NThursday, July 22, 2010
    • Transient  process Speed  7.7  control Drives of Separately  Excited DC Motor breaking startingThursday, July 22, 2010
    • 7.7.1 Starting a DC Motor • Starting a DC motor is to increase its speed from standstill to required speed If we supply full voltage ti a stationary shunt motor,the starting current in the armature will be very high and we run the risk of: 1)Burning out the armature 2)damaging the commutator and brushes 3)overloading the feeder 4)snapping off the shaft due to mechanical shock 5)damaging the driven equipment because of the sudden mechanical hammerblowThursday, July 22, 2010
    • 7.7.1 Starting a DC Motor• All DC motors mus, therefor, be provided with a means to limit the starting current to reasons values, usually between 1.5-2 full-load current. One solution is to connect a rheostat is series with the armature. The resistance is gradually reduced as the motor accelerates and is eventually eliminated entirely, when the machine has attained full speed. Today electronic methods are often used to limited starting current and to provide speed control.Thursday, July 22, 2010
    • 7.7.1 Starting a DC Motor• 1)Decrease armature voltage Its not allowed to supply the rated voltage directly with out any protection method,but when the supplied voltage is reduced to the extend to allow only the limited current to flow,the motor can be started from zero speed.With the increasing of the speed,increase the supplied voltage gradually to meet the requirement of the starting.This has the same effect with the speed control. Thursday, July 22, 2010
    • 7.7.1 Starting a DC Motor• 2)Increase resistance in armature circuit In order to protect the motor and other equipment in a DC drive system,a starting resistor can be inserted in the armature circuit to limit the current flow until Ea can built up to do the limiting.The resistor must not be in the circuit permanently,because it would result in excessive losses and would cause the motors torque-speed characteristic to drop off excessively with an increase in load. Thursday, July 22, 2010
    • 7.7.1 Starting a DC Motor•Starting circuit of a DC motor  U  UN Ra  R S n   2 T em n C eΦ N C eC T Φ N S S1 S 2 S3 n0 nN h M Ra R st 1 R st 2 R st 3 n3Tem f 3 g RaI Ra  Rst 1  R1 d e 2 n1 c R a  R st 1  R st 2  R 2 b 1 a Ra  Rst 1  Rst 2  Rst 3  R3 TL T2 T1 Tem IL I2 I1 IThursday, July 22, 2010
    • 7.7.2 Speed Control 1.Demands Rapid control and response of the direction and speed of a driven system is particularly required in many important equipment,such as : steelworks,mill,planner,plastic filming machine,lathe.Thursday, July 22, 2010
    • 7.7.2 Speed Control 2.Speed adjustment methods • decrease armature voltage Speed control is by armature voltage up to base speed with the flux maintained at its rated value,thus giving the maximum torque per ampere. • weakening flux motor field weakening is employed beyond base speed up to maximum permitted by mechanical or commutation consideration.Thursday, July 22, 2010
    • 7.7.2 Speed Control 3.Specifications of speed control1)speed adjustment range D nmax nmin2)speed regulation n0  n N SR  nN  100%Thursday, July 22, 2010
    • 7.7.2 Speed Control 4.Permissible torque and power If a motor is operating at its rated terminal voltage,power,and field current,then it will be running at rated speed.Field resistance control can control the speed of the motor for speeds above base speed but not for speed by field circuit control would require excessive field current,possibly burning up the field windings.Thursday, July 22, 2010
    • 7.7.3 Braking of a DC Motor Dynamic braking Plugging Generation Braking 能耗 制动 反接 回馈 制动 制动Thursday, July 22, 2010
    • 7.7.3 Braking of a DC Motor Dynamic   N braking U 0 能耗 U Ra 制动 n   2 T KeΦ CE K CeCTΦ T K EThursday, July 22, 2010
    • 7.7.3 Braking of a DC Motor Ra  RB n 2 Tem  0  βTem C eCT Φ N n 制动瞬间 n0 电动机状 B A 工作点 R a 态工作点 制动过程工作段 位能性 0 TL Tem 负载, 电动机拖动反抗性负 稳定工 载,电机停转。 作点 CThursday, July 22, 2010
    • 7.7.3 Braking of a DC Motor Voltage reverse Plugging   N U  U N 反接 U Ra n   T 制动 KeΦ CE K CK E Φ T eCT 2Thursday, July 22, 2010
    • 7.7.3 Braking of a DC Motor n n0 AB Ra C点:转速为零, TL 电磁转矩T不为零 C 0 TL Tem 且大于反抗性负载转矩。 D 应立即断电, 否则反向启动,  n0 在D稳定运行。Thursday, July 22, 2010
    • 7.7.3 Braking of a DC Motor 初始状态: 电机工作在A点, Potential reverse 以转速 n A n 提升重物。 n0 A B 现想慢速放下重物, 故须先使重物停止。 Ra 在电枢回路中串如大电阻, 特性曲线斜率增加 C 0 TB TK TL Tem 停止与于C点, D 但电磁转矩小于位能转矩 Ra  RB 故电机反转, Potential reverse 最终稳定运行于D 倒拉反接Thursday, July 22, 2010
    • 7.7.3 Braking of a DC Motor Another point to notice is that the characteristics Generation Generation run into the negative Braking Braking speed,positive torque quadrant at high values 回馈 of torque.This generating 回馈 制动 condition can arise if 制动 there is an activecomponent in the load torque,as in ahoist.If this is sufficient to stop themachine and reverse itsrotation,electrical braking thenoccurs,the generating torque opposingthe load torque,see the figure:Thursday, July 22, 2010
    • n 正向回馈制动 UN Ra  RS n  2 Tem n0 CeΦN CeCT ΦN A 正向电动运行 能耗 制动 电压反接 过程 制动过程 0 C Tem 能耗制动运行反向电动运行B D 倒拉反接  n0 制动运行 反向回馈E 制动运行Thursday, July 22, 2010
    • 7.7.4  Transient  process dIa U  E a  R a Ia  L a dt If the flux of a GD 2 dn separately excited dc motor T  TL  375 dt is kept constant, then the U Ra n  T  n0   transient process can be Ce CeCT expressed by the following Ea  Cen equations: T  CT I aThursday, July 22, 2010
    • 7.7.4  Transient  process 1.Mathematical analysis• In the figure below, suppose the motor is running at point A at the beginning,in order• to increase the speed to point C,switch T—n curve from 1— 2,and the transient process will be from A,to B,and then to C. Transient processThursday, July 22, 2010
    • 7.7.4  Transient  process1) General expression: GD2 dn We have n  n0  T  n0   (  TL) 375 dt GD2 dn   n  ns 375 dt Where ns=n0- βT is the speed at the stable operation point. dn Re-write the equation as: TM  n  ns dt GD 2 R a  Rc GD 2 Where TM    375 C e  C T  375 is the mechanical time constant.Thursday, July 22, 2010
    • 7.7.4  Transient  process t Its solution is: n ( t )  n s  Ce T MWhen t=0 ,the speed equals the initial speed ni , t so C=ni-ns,we have n ( t )  n s  ( n i  n s ) e T MSubstitute them ,then t  n0     n0  TL)  [(n0  Ti )  (n0  TL)]e TM t And: T ( t )  T s  (T i  T s ) e T M t  I a ( t )  I as  ( I ai  I as ) e T MThursday, July 22, 2010
    • 7.7.4  Transient  process2) Discussion of the transient expression From the responses of the equations, it can be known that if only are the three key elements known, the responses can be determined.For the first-order constant-coefficient different equation, the three key elements are: initial value ni Ti Iai,stable-state valuensTs Ias,and the time constant GD R a  Rc 2 TM    375 C e CT Thursday, July 22, 2010
    • 7.7.4  Transient  process3) Time calculation of a transient process From equations above ,the speed at any time can be calculated as:  t n x  n s  (n i  n s)e T M t e T M (n x  n s)  (n i  n s) (n i  n s) tx  T M ln (n x  n s) (T i  T s ) tx  T M ln (T x  T s ) ( I ai  I as ) tx  T M ln ( I ax  I as )Thursday, July 22, 2010
    • 7.7.4  Transient  process2.Transient process at starting 1) Starting process The starting characteristic of a separately dc motor is shown below. Transient process of startingThursday, July 22, 2010
    • 7.7.4  Transient  processThe mechanical time constant: nA GD2 n0 GD2 n0  nA GD2 TM    Tst  TL 375 Tst 375 TL 375 If   ns=nA,ni=0,     Ts=TL,Ti=Tst  tthen,    n  n A (1  e T M )  t T  T L  ( T st  T L ) e T MThursday, July 22, 2010
    • 7.7.4  Transient  process3.Transient process when dynamic braking 1) When load is constant resistant ni  nA, ns  nC  t n(t)  nC  (nA  nC)e TM , (n  0)  t T(t)  TL  (TB TL)e TM , (T  0) nA  nC t0  TM ln nC TB TL t0  TM lnTransient process of dynamic braking TLThursday, July 22, 2010
    • 7.7.4  Transient  process 2) When load is constant potential plugging operation after transient processThursday, July 22, 2010
    • 7.7.4  Transient  process (1)The first section of the transient process n i1  n A , n s 1  n C nA  nC GD 2 TM 1  TL1  T B 375 t n (t )  nC  ( nA  nB )e T M , (t  t 1), ( n  0) T i1  T B , T s 1  T L 1 t  T (t )  TL1  (T B  TL1)e T M , (T  0)Thursday, July 22, 2010
    • 7.7.4  Transient  process (2)The second section of the transient process nD GD 2 ni 2  0, ns 2  nD, TM 2  TL 2 375  t  t n(t )  nD  (0  nD)e TM 2  nD(1  e TM 2 ) Ti 2  0, Ts 2  TL 2  t T (t )  TL 2(1  e TM 2 )Thursday, July 22, 2010
    • 7.7.4  Transient  process 4.Transient process of the voltage reverse connection braking 1) When load is constant resistant Voltage reverse with resistant loadThursday, July 22, 2010
    • 7.7.4  Transient  process • (1)The first section of the transient process ni1  nA, ns1  nC nA  nC GD 2 TM 1  TL  TB 375  t n(t )  nC  (nA  nC )e TM 1 , (t  t ), (n  0) Ti1  TB, Ts1  TL1 t  T (t )  TL1  (TB  TL1)e TM 1Thursday, July 22, 2010
    • 7.7.4  Transient  process (2)The second section of the transient process nA  nD GD 2 ni 2  0, ns 2  nD, TM 2  TD  TB 375  t n(t )  nD (1  e TM 2 ) T i 2  T E , Ts 2  T D  t T (t )  TD  (TE  TD )e TM 2Thursday, July 22, 2010
    • 7.7.4  Transient  process 2) When load is constant potential Volatage reverse with gravitational loadThursday, July 22, 2010
    • 7.7.4  Transient  process(1)The first section of the transient process n i1  n A , n s 1  n C n A  n C GD 2 TM 1  T L 1  T B 375  t n (t )  n C  ( n A  n C ) e T M 1 , ( n  0 ) T i1  T B , T s 1  T L 1  t T ( t )  T L 1  (T B  T L 1 ) e T M 1Thursday, July 22, 2010
    • 7.7.4  Transient  process (2) The second section of the transint process nD GD 2 ni 2  0, ns 2  nD , TM 2  TL 2  TE 375  t n (t )  nD (1  e TM 2 ) Ti 2  TE , Ts 2  TD  t T (t )  TD  (TE  TD )e TM 2Thursday, July 22, 2010
    • 7.7.4  Transient  processExample 7-2When a generator is being driven at 1200 rpm, thegenerated EMF is 125v. Determine the generated EMF(a) If the field flux is decreased by 10 percent with the speed remaining unchanged(b) If the speed is reduced to 1100 rpm, the field flux remaining unchangedSolution: Ea 125(a) Ce   n 1200 125 Ea  0.9Ce N  0.9 1200  112.5V 1200(b) E  C  N  125  1100  114.6V a e 1200Thursday, July 22, 2010
    • 7.7.4  Transient  process Example7-3 A 4.5kw 125v 1150 rpm shunt generator has an armature resistance of 0.37Ω and a field current of 2.5 A. When the machine is driven at rated speed.Calculate:rated output current IN ,armature current Ia,rated induced armature voltage Ea. Solution: I  I  I  2.5  36  38.5 a f N PN 4500 IN    36 A UN 125 U a  U N  Ra I a  125  0.37  38.5  139.3vThursday, July 22, 2010
    • 7.7.4  Transient  process Example 7-4 A separate excited motor with PN=18kw , UN=220V , IN=94A , nN=1000rpm, Ra=0.15Ω, calculate: 1) Ce , CT  2)Rated electromagnetic torque TN 3)Rated output torque T2N 4)No-load torque T0 5)Theoretically no-load speed n0 6)Practical no-load speed n0 7)Direct start current IstartThursday, July 22, 2010
    • 7.7.4  Transient  process Solution: 1) C   U N  I N Ra  220  94  0.15  0.2059V / rpm e nN 1000 CT   9.55Ce  1.966 2) TN  CT  I N  1.966  94  184.8 N  m PN 3) T2 N  9550  171.9 N  m nN 4) T0  TN  T2 N  184.8  171.9  12.9 N  m UN 5) n0  C   1068.5rpm e U Ra 230 0.15 6) n0  N  T0   12.9  1063.7 rpm Ce Ce CT  0.2059 0.2059  9.55 2 7) U 220 I srart  N   1466.7 A Ra 0.15Thursday, July 22, 2010