电工电子课件3

1、定子定子U1V2V1W1NSU2-+W2转子转子+eAeBeCU2U1V1V2W1W2+ 三个绕组空间位置互差三个绕组空间位置互差120o 转子装有磁极并以转子装有磁极并以 速度匀速旋转。三个速度匀速旋转。三个绕绕组组中便依次产生三个单中便依次产生三个单相正弦电动势。相正弦电动势。tEesinm1)sin(m1202tEe )sin(m1203tEe EEE01)(23211202jEEE)(23211203jEEEE1.tEe sinm1)sin(m1202tEe )sin(m1203tEe EEE011202EE1203EEe3e12 e2e0t 120 240 360.E21201201

2、20E3.对称三相电动势的瞬时值之和为对称三相电动势的瞬时值之和为 00321eee即：0321EEE或三个正弦交流电动势满足以下特征三个正弦交流电动势满足以下特征e1 e2 e3中点中点e1+e3+e2+NU1V1W1U2V2W2L1L2L3端线端线 （L1L2L3 ）中线：）中线：N端线端线 （L1L2L3 ）e1+e3+e2+NU1V1W1U2V2W2L1L2L3321UUU、312312UUU、Up、Ul3U+2U+1U12U+23U+31UNe1+e3+e2+3U2U1U12U23U31UU1V1W1U2V2W2L1L2L3+1U3U2U12U2U2112UUU3223UUU131U

3、UU3由相量图可得由相量图可得303112UU 30303223UU303331UU31U23U 30 30Ne1+e3+e2+3U2U1U12U23U31UU1V1W1U2V2W2L1L2L3+303112UU303223UU303331UUPlUU31U3U2U12U 3031U23U 30 30Ne1+e3+e2+3U2U1U12U23U31UU1V1W1U2V2W2L1L2L3+1U3U2U12U 3031U23U 30 30 p 相电压相电压线电压线电压 时时结结结论：电源结论：电源形联形联UUl +L2L1L312U31UBCU23UZ3Z2Z11U+2U+3UNL1L2L312U

4、+23U+31UZ3Z2Z11U+2U+3UNL1L2L312U+23U+31U1I2I3INI1zI2zI3zIPlUU3对称的。对称的。1U3U2U12U 3023U31UZ3Z2Z11U+2U+3UNL1L2L312U+23U+31UU1V1W1U2V2W2321IIIIN333222111ZUIZUI ZUI1I2I3INI+Z2Z3Z11U2U3U+1zI2zI3zI1I2I3INI+Z2Z3Z11U2U3U+321ZZZ对称负载：对称负载：333222111ZUI ZUI ZUI线电流：线电流：1U3U2U3I2I1I0321IIII N中中线线电电流流所以负载对称时，线所以负载对

5、称时，线电流也对称。电流也对称。 负载对称时，只负载对称时，只需计算一相电流，需计算一相电流，其它两相电流可根其它两相电流可根据对称性直接写出。据对称性直接写出。负载对称时负载对称时,中线无中线无电流电流,可省掉中线。可省掉中线。0321NIIII1I2I3INI+Z2Z3Z11U2U3U+1zI2zI3zI0321NIIII例例1 1：若若R1=R2= R3 = 5 ，求线电流及中性线电，求线电流及中性线电若若R1=5 , R2=10 , R3=20 ,求线电流及求线电流及 一星形连接的三相电路，电源电压对称。设电一星形连接的三相电路，电源电压对称。设电源线电压源线电压 。 负载为负载为Vt

6、u)sin(30314238012电灯组，电灯组，流流 IN ;中性线电流中性线电流 IN 。N+NI1U2U3UN R1R2R3NL1L2L31I2I3I中线电流中线电流0321NIIII解：解： 已知：已知：VU3038012VU02201 RUI A04450220111(1) 线电流线电流A120442 IA120443I 线电流对称线电流对称例例1 1：R1=R2= R3 = 5 ，Vtu)sin(30314238012求线电流及中性线电流求线电流及中性线电流 IN ; +NI1U2U3UN R1R2R3NL1L2L31I2I3I+NI1U2U3UN R1R2R3NL1L2L31I2

7、I3I(2) 三相负载不对称三相负载不对称 R1=5 、R2=10 、R3=20 分别计算分别计算各线电流各线电流ARUI1201120120220333ARUI1202210120220222ARUI04450220111中线电流中线电流3211IIIIA19291201112022044Z12Z23Z31L1L2L3+12U+23U+31U321III、12I23I31I23I12I31I1I2I3IZ12Z23Z31+31U12U23UL1L2L3(2) 相电流相电流121212ZUI232323ZUI313131ZUI(1) 相电压相电压即即: UP = UL即：即：23I12I31I

8、1I2I3IZ12Z23Z31+31U12U23UL1L2L3负载相电压负载相电压=电源线电压电源线电压负载对称时负载对称时, III2312U23U12U3123I12I31I1I2I3IZ12Z23Z31+31U12U23UL1L2L3负载对称时负载对称时(3) 线电流线电流由相量图可求得：由相量图可求得：31121III12232III23313III301I31II2I3III2312U23U12U31PP330cos2IIIl 所以负载对称时，线所以负载对称时，线电流也对称。电流也对称。当负载对称时：当负载对称时：P = 3Up Ipcos p23I12I31I1I2I3IZ12Z2

9、3Z31+31U12U23UL1L2L3Z3Z2Z11U+2U+3UNL1L2L312U+23U+31UllIIUU31 ,pp 接：接： PPPPcos3cos3 llIUIUP 同理同理ppppsin3 sin3 llIUIUQ llIIUU pp ,31Y接接：相电压与相相电压与相电流的相位差电流的相位差llPPIU IUQPS 3322所以所以23I12I31I1I2I3IZ12Z23Z31+31U12U23UL1L2L3Z3Z2Z11U+2U+3UNL1L2L312U+23U+31U负载对称时负载对称时, , P = 3Up Ipcos p判断对错判断对错ZUIll 21121ZZU

10、I21121ZZUIZUIlPP3IIl PUUl P3UUl ZUIPP 11213ZUI 1121330ZUI1121330ZUI1I2I3INI+Z2Z3Z11U2U3U+ 1zI2zI3zINL1L2L3判断对错判断对错Wcos23202203803 PWcos53202203803 P Wcos53202202203 P 53203038012ZVU已知：已知：1I2I3INI+Z2Z3Z11U2U3U+ 1zI2zI3zINL1L2L330 UUl相相位位超超前前,3pY接接：NU1V1W1U2V2W2L1L2L33U+2U+1U12U+23U+31U1U3U2U12U 3023U

11、31UlUU p 接：接： +L2L1L312U31UBCU23U31U23U12Upp3,3II UUllY接接：Z3Z2Z11U+2U+3UNL1L2L312U+23U+31U1U3U2U12U 3023U31U3I2I1IPllIIUU3p、 接：接： 23I12I31I1I2I3IZ12Z23Z31+31U12U23UL1L2L31II2I330III2312U23U12U31Z3Z2Z11U+2U+3UNL1L2L312U+23U+31U23I12I31I1I2I3IZ12Z23Z31+31U12U23UL1L2L3Pcos3llIUP psin3llIUQ llIUS 3321PP

12、PP321QQQQ 22QPS 321ZZZ练习练习1：?PPIIUZYLLYL求求：相相同同，、如如图图两两对对称称电电路路，Z3Z2Z11U+2U+3UNL1L2L312U+23U+31U23I12I31I1I2I3IZ12Z23Z31+31U12U23UL1L2L33133ZUZUIILLLLY解：解：练习练习1：?PPUZYL求求：相相同同，、如如图图两两对对称称电电路路，Z3Z2Z11U+2U+3UNL1L2L312U+23U+31U23I12I31I1I2I3IZ12Z23Z31+31U12U23UL1L2L331LLYII31cos33cos3cos3cos3PLYLPYLYLP

13、LLPYLYLYYIUIUIUIUPP练习练习2：。、及及功功率率、求求各各线线电电流流，已已知知，线线电电压压SQPIIIXXR380VCL321355i1i2iC解：解：XLL1L2L3Ri1i2XLRXLRXCi3VU02201VU3038012设：设：6010LjXRZVU1202202VU1202203练习练习2：。、及及功功率率、求求各各线线电电流流，已已知知，线线电电压压SQPIIIXXR380VCL321355解：解：i1i2iCXLL1L2L3Ri1i2XLRXLRXCi3VU022016010ZAZUIN602211AIN180222AI60223AjXUICC120441

14、2AIIIN120221121练习练习2：。、及及功功率率、求求各各线线电电流流，已已知知，线线电电压压SQPIIIXXR380VCL321355解：解：i1iCi2XLL1L2L3Ri1i2XLRXLRXCi3AIN60221AIN180222AI60223AIC12044AI120221AIIICN901 .3822AI221AI1 .382AI223练习练习2：。、，求求功功率率，已已知知，线线电电压压SQPXXR380VCL355解：解：AIN60221AIN180222AI60223AIC12044AI120221AI901 .382WIUPll7240cos3Pvar6 .4191

15、3221CCLNXIXIQVAQPS1 .838322i1iCi2XLL1L2L3Ri1i2XLRXLRXCi3练习练习3：。及及、相相负负载载发发生生短短路路时时，求求：，已已知知，PIIILVtuXXRCL321112314sin238010解：解：4525/2LjXRZXLL1L2L3Ri1i2XLRRXCi3452103CjXRZAZUI1652193313AZUI1352382212AZUI1352382212AIII352 .78321WRURIP216602122323I12I1I2I3IL1L2L3XCRXLRR31I练习练习4：电电流流。间间负负载载发发生生断断路路后后的的线

16、线、）（图图；）线线电电压压、线线电电流流相相量量；（及及、求求：，已已知知，323211232) 1 (314sin238038LLPiiiVtuXXRCL45219/12LjXRZ4523831CjXRZARUI120102323AZUI45210121212解：解：23I12I31I1I2I3IL1L2L3XCRXLRR练习练习4：；及及、求求：，已已知知，PiiiVtu32112) 1 (314sin23804523831ZAI1201023AI4521012AZUI16525313131AIII1 .352131121AIII1751 .1512232AIII1006 .102331

17、3解：解：23I12I31I1I2I3IL1L2L3XCRXLRR练习练习4：；及及、求求：，已已知知，PiiiVtu32112) 1 (314sin2380AI1 .35211AI1751 .152AI1006 .103Ati)1 .35sin(2211Ati)175sin(21 .152Ati)100sin(26 .103WRIRIRUP9500231223212解：解：练习练习4：图图；）线线电电压压、线线电电流流相相量量求求：（2AI1 .35211AI1751 .152AI1006 .103VU038012VU12038023VU12038031U12U23U311I2I3I1201

18、201 .35175100解：解：23I12I1I2I3IL1L2L3XCRXLRR31I练习练习4：电电流流。间间负负载载发发生生断断路路后后的的线线、）（求求：323LL解：解：3112II、不不变变23I0AI1201023AI4521012AIII1 .352131121AII135210122AII16525313 三相对称负载作三角形连接，三相对称负载作三角形连接，UL =220V，当，当S1、 S2 均闭合时，各电流表读数均为均闭合时，各电流表读数均为17.3A，三相功率，三相功率 P = 4.5 kW，试求，试求: 1) 每相负载的电阻和感抗；每相负载的电阻和感抗； 2) S1合、合、S2断开时断开时, 各电流表读数和有功功率各电流表读数和有功功率P； 3) S 1断、断、S 2闭合时闭合时, 各电流表读数和有功功率各电流表读数和有功功率P。练习练习5：A2A1A3S2Z12Z23或：或：P =I 2R P =UIcos tg =XL / R 223/32.17220PPIUZLL3co