（完整版）电力系统分析第三版 - 何仰赞-课后习题答案第2章

2020-6-2 - 1 -

第二章

2—1 110 kV架空线路长70 km，导线采用LGJ—120型钢芯铝线，计算半径r＝7.6mm，相间距离为3.3m，

导线分别按等边三角形和水平排列，试计算输电线路的等值电路参数，并比较分析排列方式对参数

的影响。 解：取DS?0.8r

(1) 等边三角形排列，Deq?D?3.3m R?rOl??Sl?31.5?70??18.375? 1203.3?103 X?xOl?0.1445lg?l?0.1445?lg?70??27.66?

Ds0.8?7.6 B?bOl?Deq7.587.58?6?4?10?6?l??10?70??2.012?10? 3Deq3.3?10lglg7.6req(2) 水平排列，Deq?1.26D?4.158m R?rOl??Sl?31.5?70??18.375? 1204.158?103 X?xOl?0.1445lg?l?0.1445?lg?70??28.676?

Ds0.8?7.6 B?bOl?Deq7.587.58?6?4?10?6?l??10?70??1.938?10? Deq4.158?103lglg7.6req 电阻与导线的排列方式无关，相间几何均距越大，等值电抗越大，等值电纳略有越小。

2—4

有一台SFLl—31500／35型双绕组三相变压器，额定变比为35／11，查得

?P0?30kW,I0?1.2%,?PS?177.2kW,VS?8%,求变压器参数归算到高、低压侧的有名值。

解：由题意知V1N?35kV,V2N?11kV,SN?31.5MVA

(1) 归算到高压侧的参数

?PSV12177.2352N RT?????0.2188? 21000SN100031.52V128352N XT?VS????3.11?

SN10031.5 GT??P01301??S?2.449?10?5S 221000V1N100035 BT?I0SN1.231.5?4??S?3.086?10S 22V1N100352020-6-2 - 2 -

(2) 归算到低压侧的参数

?PSV22N177.2112 RT?????0.0216? 221000SN100031.5V22N8112 XT?VS????0.3073?

SN10031.5 GT??P01301??S?2.479?10?4S 221000V2N100011 BT?I02—5

SN1.231.5??2S?3.124?10?3S 2V2N10011有一台型号为SFL-40000／220的三相三绕组变压器，容量比为100／l00／100，额定变比 220／38.5／11，查得?P0?46.8kW,I0?0.9%,?PS(1?2)?217kW,?PS(1?3)?200.7kW,

?PS(2?3)?158.6kW,VS(1?2)?17%,VS(1?3)?10.5%,VS(2?3)?6%,试求归算到高压侧的变压器参数有

名值。

解： 由题意知V1N?220kV,V2N?38.5kV,,V3N?11kV,SN?40MVA,三个容量相等。先求等 值电阻。

所以：?PS1?11(?PS(1?2)??PS(1?3)??PS(2?3))?(217?200.7?158.6)kW?129.55kW2211?PS2?(?PS(1?2)??PS(2?3)??PS(1?3))?(217?158.6?200.7)kW?87.45kW

2211?PS3?(?PS(1?3)??PS(2?3)??PS(1?2))?(200.7?158.6?217)kW?71.15kW22?PS1V12129.552202N那么：R1?????3.919?21000SN1000402?PS2V1287.452202NR2?????2.645? 221000SN100040?PS3V1271.152202NR3?????2.152?221000SN100040 同理求出等值电抗。

VS1?VS2VS311(VS(1?2)?VS(1?3)?VS(2?3))?(17%?10.5%?6%)?10.75"11?(VS(1?2)?VS(2?3)?VS(1?3))?(17%?6%?10.5%)?6.25% 2211?(?PS(1?3)?VS(2?3)?VS(1?2))?(10.5%?6%?17%)??0.25"2020-6-2 - 3 - V1210.752202NX1?VS1????130.075?10040SNV126.252202NX2?VS2????75.625?

10040SNV12?0.252202NX3?VS3?????3.025?10040SN

最后求出等值导纳。

GT??P0146.81??S?9.669?10?7S 221000V1N1000220 BT?I02—6

SN0.940?6??S?7.438?10S 22V2N100220 有一台型号为SFSL-31500／110的三相三绕组变压器，容量比为100／l00／66.7，额定变比 110

??／38.5／11，?P0?80kW,励磁功率850kvar,?PS(1?2)?450kW,?PS(1?3)?240kW,

?P?S(2?3)?270kW,VS(1?2)?11.55%,VS(1?3)?21%,VS(2?3)?8.5%,试求归算到各电压级的参数。

解：由题意知V1N?110kV,V2N?38.5kV,,V3N?11kV,SN?31.5MVA,三个容量不相等，短路 损耗要折算。

?PS(1?2) ?PS(2?3)?SN???PS(1?2)??S?2N??100???450???kW?450kW?100???2222??SN?100????P?S(2?3)??270???kW?606.893kW ?min{S,S}??66.7?2N3N???SN???PS(1?3)??S?3N??100???240???kW?539.46kW?66.7???22?PS(1?3)

所以：?PS1?11(?PS(1?2)??PS(1?3)??PS(2?3))?(450?539.46?606.893)kW?191.2835kW2211?PS2?(?PS(1?2)??PS(2?3)??PS(1?3))?(450?606.893?539.46)kW?258.7165kW2211?PS3?(?PS(1?3)??PS(2?3)??PS(1?2))?(539.46?606.893?450)kW?348.1765kW22VS1?11(VS(1?2)?VS(1?3)?VS(2?3))?(11.55%?21%?8.5%)?12.025"11?(VS(1?2)?VS(2?3)?VS(1?3))?(11.55%?8.5%?21%)??0.475% 2211?(?PS(1?3)?VS(2?3)?VS(1?2))?(21%?8.5%?11.55%)?8.975" VS2VS3

2020-6-2 - 4 - （1） 折算到高压侧。

?PS1V12191.28351102N那么：R1?????2.333?21000SN100031.52?PS2V12258.71651102NR2?????3.155? 21000SN100031.52?PS3V12348.17651102NR3?????4.246?21000SN100031.52同理求出等值电抗。

V1212.0251102NX1?VS1????46.191?10031.5SNV12?0.4751102NX2?VS2?????1.825?

10031.5SNV128.9751102NX3?VS3????34.475?10031.5SN最后求出等值导纳。

GT??P01801?6??S?6.612?10S 221000V1N1000110?Q018501??S?7.025?10?5S 221000V1N1000110BT?（2） 折算到中压侧。

?PS1V22N191.283538.52那么：R1?????0.2857?221000SN100031.5?PS2V22N258.716538.52R2?????0.3865? 21000SN100031.52?PS3V22N348.176538.52R3?????0.52?21000SN100031.52同理求出等值电抗。

V22N12.02538.52X1?VS1????5.658?10031.5SNV22N?0.47538.52X2?VS2?????0.2235?

10031.5SNV22N8.97538.52X3?VS3????4.223?10031.5SN最后求出等值导纳。

GT??P01801?5??S?5.397?10S 221000V2N100038.5