《工业与民用供配电设计手册》 （第四版）疑问表 8 1. P16 倒数第 1 行，改为“t——导体达到稳定温升的时长，h” 2. P40 表 1.12-1 点焊机： 222 1 53.720.123691.71 e P 2 37699.71 e P 2 2 () 29.6 e eq e P n P 1.10 m K 3. P256 表 4.4-1 中线路 L1 为 24240 解： 元件阻抗 短路点阻抗 序号 元件 短路点 R(mΩ) X(mΩ) Z(mΩ) 三相短路 电流(kA) k 冲击电流 (kA) 1 系统 0.101 1.011 2 T1 2.436 9.601 3 1+2 F1 2.537 10.612 10.911 21.113 1.498 44.738 4 L1 0.385 0.395 5 3+4 F2 2.922 11.007 11.388 20.228 1.462 41.821 6 L2 5.420 1.740 7 L4 18.500 14.850 8 5+6+7 F3 26.842 27.597 38.498 5.984 1.073 9.080 （1）正序阻抗计算 1）电网阻抗：根据式 4.2-3 2 2 1.1 10 1.016() 10 3 3 10 () 0.4 nQ Q kQ cU Zm I t 0.9951.011() XZm0.10.101() RXm 0.1011.011() Q Zjm 2）变压器阻抗：根据式 4.2-5~7、4.2-11 22 12 1 1 4%400 10.159() 100%100%630 T T T uU Zm S 22 12 1 22 1 6.2 400 2.449() 630 rT T T P U Rm S 22 111 9.847() TTT XZRm 1 11 22 2 630 9.8470.03877 400 T TT S xX U max 1 1 0.95 0.975 1 0.6 T T c K x 111 2.4369.601() T kTT ZK Zjm 3）线路阻抗： 1 0.5 (0.0770.079) 100.3850.395() L Zjjm 2 (0.2710.087) 205.4201.740() L Zjjm 4 1 0.3704(/) 54 50 CU L n Rkm q 1.144.55 n q rmm 4 3 110.4 0.0628ln0.0628ln0.297(/) 444.55 10 L d Xkm r 4 (0.3700.297) 5018.5014.85() L Zjjm （2）三相短路电流计算 1）F1 点短路 11 0.1011.0112.4369.6012.53710.612() FQT k ZZZjjjm 1 10.911() F Zm 1 1 1.05 380 21.113() 33 10.911 n kF F cU IkA Z 3 /3 2.537/10.612 1.020.981.020.981.498 R X kee 11 22 1.498 21.11344.728() pFkF ikIkA 2）F2 点短路 211 2.92211.007() FQT kL ZZZZjm 2 11.388() F Zm 2 2 1.05 380 20.228() 33 11.388 n kF F cU IkA Z 3 /3 2.922/11.007 1.020.981.020.981.462 R X kee 22 22 1.462 20.22841.821() pFkF ikIkA 3）F3 点短路 31134 26.84227.597() FQT kLLL ZZZZZZjm 3 38.498() F Zm 3 3 1.05 380 5.984() 33 38.498 n kF F cU IkA Z 3 /3 26.842/27.597 1.020.981.020.981.073 R X kee 33 22 1.073 5.9849.080() pFkF ikIkA 4. P265 SA I计算方法有误。
解： 1）电网阻抗：根据式 4.2-3 2 2 2 2 1.1 110 0.03667( ) 110 3000 () 10 nQ Q kQ cU Z I t 0.9950.03648( ) XZ0.10.00365( ) RX 0.003650.0348( ) Q Zj 2）变压器阻抗：根据式 4.2-5~7、4.2-11 22 2 12% 10.5 0.42( ) 100%100% 31.5 KT T T uU Z S 22 2 1%10.5 0.035( ) 100%100% 31.5 RT T T uU R S 22 111 0.4185( ) TTT XZR 22 2 31.5 0.41850.1196 10.5 T TT S xX U max 0.95 0.975 1 0.6 T T c K x 0.3410.408( ) TkTT ZK Zj 01 1.60.54560.6528( ) TkT ZZj 3）线路阻抗： 1 0.196(/) 34 150 AL L n Rkm q 6.91 n q rmm 1122.38 0.0628ln0.0628ln0.0895(/) 446.910 L d Xkm r 1 0.741(/) 54 25 CU S S Rkm q 0 1.851 931.592( )m 4 0 2 50 4100.3948 0 0.0628 2 00 ln0.049350.6646 82 S Ajj r 2 2 3 0.049350.66463 0.8540.967 0.741 0.049350.6646 S jA Bj RAj 00 0 2 3 1 33ln 824 0.1960.14800.0628 (0.253 11.0282)0.8540.967 1.1981.1264(/) LSEL L ZRjB r d jj jkm 1 00 0.741 0.54920.46830.717640.06 0.741 0.049350.6646 ln 82 S S R rj j Rj r 4）短路电流计算 三相短路电流 maxmax 3()3() nn k QTkLkL cUcU I ZZZ LZZ L 接地故障电流 maxmax 1 00 33 22() nn k SE QTkTkLLSESELSE cUcU I ZZZZZLZZL 0.037780.4448( ) kQTk ZZZj 0 220.62121.5424( ) SEQTkTk ZZZZj 0 22(0.1960.0895) 1.1981.12641.591.3054(/) LSELLSE ZZZjjjkm ————————————————————————————————————————————— 0Lkm maxmax 1.1 10 1.20414.176()14.22785.14 () 3()33(0.037780.4448) nn k kLk cUcU IjkAkA ZZ LZj maxmax 1 333 1.1 10 4.28110.629()11.45868.06 () 0.62121.5424 nn k SE SELSESE cUcU IjkAkA ZZLZj ————————————————————————————————————————————— 1Lkm max 1.1 10 4.3659.976()10.88966.374 () 3()3(0.037780.44480.1960.0895) n k kL cU IjkAkA ZZ Ljj max 1 33 1.1 10 3.2414.174()5.28452.17 () 0.62121.5424 1.591.3054 n k SE SELSE cU IjkAkA ZZLjj 1 (0.54920.4683) (3.2414.174)0.17473.8101()3.81492.63 () E Ak SE IrIjjjkAkA 1 (1)(1 0.54920.4683) (3.2414.174)3.41570.3639()3.4356.08 () SAk SE Ir IjjjkAkA ————————————————————————————————————————————— 5Lkm max 1.1 10 3.9691.205()4.14716.89 () 3()3[0.037780.44485 (0.1960.0895)] n k kL cU IjkAkA ZZ Ljj max 1 33 1.1 10 1.1781.109()1.61843.27 () 0.62121.54245 (1.591.3054) n k SE SELSE cU IjkAkA ZZLjj 1 (0.54920.4683) (1.1781.109)0.12761.1611()1.16883.73 () E Ak SE IrIjjjkAkA 1 (1)(1 0.54920.4683) (1.1781.109)1.0500.052()1.0522.82 () SAk SE Ir IjjjkAkA ————————————————————————————————————————————— 10Lkm max 1.1 10 2.19271.470。
