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1、CHAPTER 3 THREE-PHASE CIRCUITS (三相电路 ),1. Introduction,A balanced three-phase system is produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120. Three-phase systems are important for at least three reasons: Nearly all electr
2、ic power is generated and distributed in three-phase. The instantaneous power in a three-phase system can be constant. For the same amount of power, the three-phase system is more economical than the single-phase.,CHAPTER 3 THREE-PHASE CIRCUITS (三相电路 ),3.1 Balanced Three_Phase Voltages 3.2 Balanced
3、Connection 3.3 Unbalanced System 3.4 Power Analysis in Three_phase Circuits,3.1.1 对称三相电动势的产生,在两磁极中间,放一个线圈, 让线圈以的速度顺时针旋转。,根据右手定则可知,线圈中产生感应电动势,其方向由AX。,合理设计磁极形状,使磁通按正弦规律分布,线圈两端便可得到单相交流电动势。,3.1 BALANCED THREE_PHASE VOLTAGES,三相交流电动势的产生,三线圈空间位置各差120o,铁心(作为导磁路经),三相绕组,尺寸、匝数相同,空间排列互差120,: 直流励磁的电磁铁,The coils
4、are placed 120apart, the induced voltages across the coils are equal in magnitude but out of phase by 120.,1. Balanced Three-Phase Voltages,Phase voltage,(1).Phasor representation:,(2). character:,(3).The phase sequence is the time order in which the voltages pass through their respective maximum va
5、lues. (相序),The abc sequence or positive sequence. The acb sequence or negative sequence.,发电机三相绕组通常将三个末端接在一起。 三个绕组末端的连接点称为中性点 ( N )。 三个首端A、B、 C称为端点。 由中性点N引出的供电线称为中性线或零线。 由三个首端引出的供电线称为端线或火线。,1、三相交流电源的连接-星形(Y)接法,3.1.2 Y connection and connection:,3.1.2 三相交流电源的连接-星形(Y)接法,(1) 联接方式,中性线(零线、地线),中性点,端线(火线),相
6、电压:端线与中性线间的电压,线电压:端线与端线间的电压,N,B,C,A,三相四线制 供电,(X.Y.Z),一般用UP表示,一般用UL表示,(1) Line(火线),(2) Neutral line(中线),(3) Line current(线电 流),(5) Phase current(相电流),(4) Line voltage(线电压),(6) Phase voltage相电压,(2)线电压和相电压的关系,根据KVL定律,相量图:,(1) 相电压对称 线电压也对称, 结论:,(大小关系),U AB超前 UA 30o (相位关系),在日常生活与工农业生产中,多数用户的电压等级为,conclus
7、ion:,(1) Line voltage is balanced. Phase voltage is balanced,(3) Line voltage leads corresponding phase voltage 30o。,例:已知星形联结的对称三相电源的相 电压为6kV,求它的线电压是多少, 如以A相为参考相,试写出所有相 电压与线电压的向量表达式。,解:,=10.4kV,三相电源的三角形连接,将三相电源中每相绕组的首端依次与另一相 绕组的末端连接在一起,形成闭合回路。,3.2 Connection 三相负载及三相电路的计算,三相负载分为两种:,2. 不对称负载: 使用时平均分配到
8、三个相上。,三相负载的连接方式有两种:,星形接法,三角形接法,三相负载的两种接法:,具体采用何种接法,应根据电源电压和负载 额定电压的大小来决定。 原则上应使负载的实际电压等于其额定相电 压。,3.2.1 星形接法及计算,线电流:流过端线的电流,相电流:流过每相负载的电流,(1) 联结形式,N 电源中性点,N负载中性点,Line currentphase current,三相四线制电路,星形接法特点 相电流=线电流,相电压:每相负载两端的电压用UP 表示。 线电压:各端线之间的电压用UL表示. 相电流:每相负载流过的电流用IP 表示. 线电流:电源端线流过的电流用IL表示.,三相四线制(有中性
9、线),三相四线制电路,电源的相、线电压就是负载上的相、线电压。电源对称,则:线电压超前相电压30,且,(2) 负载Y形联结三相电路的计算,负载端的线电压电源线电压 负载的相电压电源相电压,线电流相电流,中线电流, 结论: (1)若负载对称 (balanced) 则三相电流也对称。,*只需计算一相电流,其它两相电流可根据对称性直接写出。,*零线可以取消-三相三线制 The neutral line can be removed without affecting the system.,(2)负载不对称(unbalanced)时,各相需单独计算。,例1:,中性线电流,(2) 三相负载不对称(RA
10、=5 、RB=10 、RC=20 ) 分别计算各线电流,中性线电流,例2 .有一对称三相负载星形联结的电路. 三相电源对称.每相负载中,L=25.5mH, R=6.uAB = 3802sin(314t+30)(V) 求各相电流 iA ,iB ,iC 。,ZA=ZB=ZC= R+jXL = R + jL =6+ j31425.5103 = 6+j8 =1053.1 ,解:,根据对称关系有:,iA=222sin(314t53.1) A,例3:照明系统故障分析,解:,A相断路: 中性线未断时,求各相负载电压; 中性线断开时,求各相负载电压。,A相断路,B、C相灯仍承受220V 电压, 正常工作。,1
11、) 中性线未断,2) 中性线断开,变为单相电路,如图(b) 所示, 由图可求得,Unbalanced Three-Phase Systems,在三相四线制中,各项负载电压为电源的相电压,各相需单独计算。但中线电流不再为零,不对称程度越大,中线电流越大。,在三相三线制中,负载中点与电源中点脱开,不能保证各项负载电压为电源的相电压,不能用各相单独计算的方法。只能把三相电路看成是多电源共同作用的复杂电路来计算。,Application: electric light circuits,connection: Electric light of the same flour are in parall
12、el, powered by one phase.,Each bulb can work at the rated voltage 220v。,照明电路的一般画法,Fuse or circuit breaker must be connected into hot wire.,Electric light cant be supplied by a three-wire ac system,A,C,B,.,.,.,一层楼,二层楼,三层楼,N,设线电压为380V。 A相断开后,B、C两相串连,电压UBC (380V)加在B、C负载上。如果两相负载对称,则每相负载上的电压为190V。,问题1:若一
13、楼全部断开,二、三楼仍然接通, 情况如何?,分析:,结果二、三楼电灯全部变暗,不能正常工作。,问题2:若一楼断开,二、三楼接通。但两层楼 灯的数量不等(设二楼灯的数量为三层的 1/4 结果如何?,结果:二楼灯泡上的电压超过额定电压, 灯泡被烧毁;三楼的灯不亮。,分析,Neutral Line,1.To unbalanced loads,loads of each phase cant work as rated voltage without the neutral line. In order to get the equivalent phase voltage from the sour
14、ce, neutral line must be connected to the unbalanced Y connection loads. 2. To make sure that neutral line is working, fuse and circuit breakers can not be put on the neutral line.,Drill:20 bulbs are powered by each phase, all of these bulbs are rated as 220V, 40W. Line voltage of the source is 380V
15、. Calculate: All the bulbs of the three phases are working, determine line current,phase current and neutral line current. Half of the bulbs of phase A , all of the bulbs of phase B, C are working, determine line current,phase current and neutral line current.,(1) All the bulbs are working,Ul=380V 2
16、20V/40W,(中线电流不为零!),结论,(1)不对称负载Y联结又未接中性线时,负载相电压不再对称,且负载电阻越大,负载承受的电压越高。 (2) 中线的作用:保证星形联结三相不对称负载的相电压对称。 (3)照明负载三相不对称,必须采用三相四线制供电方式,且中性线内不允许接熔断器或刀闸开关。,Phase currents:,3.2.2 三角形接法及计算,特点:线电压=相电压,Line current:,Relation between the line current and phase current:,令, 若负载对称(balanced), 相电流和线电流对称。, 若负载不对称(unbalanced),由于电源电压对称,故负载的相电压对称,但相电流和线电流不对称。, 结论:,I