《电力系统分析 》( I )
陈功贵 156********
chengonggui@https://www.doczj.com/doc/b58875321.html,
Chongqing University of Posts
and Telecommunications
国家级重点教材《电力系统分析》
何仰赞温增银著
华中科技大学出版社
第八章电力系统不对称故障的分析和计算8.1 简单不对称短路的分析
8.2 电压和电流对称分量经变压器后的相位变换8.3 非全相断线的分析
8.4 应用节点阻抗矩阵计算不对称故障
8.5 复杂故障的计算方法
第八章电力系统不对称故障的分析和计算
本章主要内容
?各种简单不对称故障的序分量边界条件
?复合序网的概念和正序等效定则
?电压电流对称分量经过变压器后的相位变换
?利用阻抗矩阵计算不对称故障的原理和方法
三个方程,六个未知量,需要补充三个方程式才能解。 根据不对称故障的具体边界条件获取这三个方程。
(1)(1)(1)fa eq ff fa V E jX I =?(2)(2)(2)
fa ff fa V jX I =?(0)(0)(0)
fa ff fa V jX I =?(2)fa
(2)
fa V (0)
fa
(0)
fa V (1)fa
(1)
fa V 8.1 简单不对称短路的分析(横向故障) (0)eq f E V =故障前故障端口开路电压或正序等效电势
复习第7章:序网方程
1. 单相(a 相)接地短路—
(选特殊相为参考相)序分量边界条件 0, 0(1), 0fa fb fc
V I I === 相量表示的边界条件:(1)
(2)
(0)
(1)(2)(0)(1)(2)(0)000
2fa
fa fa fa fb fb fb fb fc fc fc fc V
V V V I I I I I I I I =++==++
==++= ()对称分量表示的边界条件
V 0 a b
c
(1)
(2)
(0)
2(1)(2)(0)20
030
fa
fa fa fa fb fa fa fa a V V V V
I I I I I I I I αα
αα=
++
==
++=++ ()以相为参考相
8.1 简单不对称短路的分析(横向故障)
1
120
abc ?=I S I 12
2
1
11=11S αααα?
2
10αα++=
1. 单相(a 相)接地短路—联立方程求解
V 0 a b
c
(0)(1)
(1)(2)(0) (83)
()
f
fa ff ff ff V I j X X X ?++
1. 单相(a 相)接地短路—串联复合序网
V 0 a b
c
(1)(2)(0)(1)(2)(0)0(82)fa fa fa fa fa fa V V V I I I ++=
? ==
()(0)(1)(1)(1)
(2)(0)(1)(2)(2)(2)
(0)(0)(0) fa f ff fa ff ff fa fa ff fa fa ff fa V V jX I j X X I V jX I V jX I
=? =+
=? =?
——将各序网络在
故障端口连接起来
所构成的网络
1. 单相(a 相)接地短路—故障点各相电流电压及正序等效定则
(0)(1)(1)(2)(0)3
()
0, 0f
f fa
ff ff ff fb fc V I I j X X X I I ==++== (1)
(2)(0)
(1)
(3)
(3)
3
01
ff ff X X X m
X m
?
?
=+===令附加电抗
比例系数三相对称短路时
因为X ff (1) ≈X ff (2),当X ff (0) 1. 单相(a 相)接地短路—相量图 fb I fb I (0) fb I (0)fa I (0)fc I V (1) fb V fc V V (0)fa 1. 单相(a 相)接地短路—特例分析(I&II) (0)(0(1)(1)(2)(0)(1) )(()3)0)1(3Case I >() f f fa ff ff ff f f f f f f f f V V I j X X X I X jX I X ==?=+>+ :(1)(2)(0)(),ff ff ff X X X ≈与故障点离电源的电气距离有关(与系统中性点接地情况有关) fb V 1. 单相(a 相)接地短路—特例分析 (0)Case III ff X →∞:中性点,不接地系统()(0)(0)(1)(2)(0)(1)(2)(0)(2)(2)(1)(0)(0)(0)(1)(/()) 0fa ff ff f ff ff ff f fa ff fa fa ff fa f V j X X V j X X X V V jX I V jX I V =+++==?==?=?(1)(2)(0)22(1)(2)(0)(1) 2(1)(2)(0)(1) =0 (1)(1)fa fa fa fa fb fa fa fa fa fc fa fa fa fa V V V V V V V V V V V V V V αααααα=++=++=?=++=? = 小接地系统(中性点不直接接地或经消弧线圈或经高阻抗接地), 当单相接地短路时,故障相短路电流为0,非故障相电压上升到线电压。所以可以不立即跳闸,最高允许运行2小时,但非故障 1. 单相(a 相)接地短路—(2)(0)(2)(1)Case IV ()ff ff ff ff X X X X =≈:()(1)(2)(0)(1)(2)(2)(1)(0) ((0) (0) 0(01) ))(2313 13fa ff ff fa fa ff fa fa f f f f f fa V j X X I V V jX I V j V X V I ?+=?==?=?= =(0) V V V ==(1)(2)(0) 22(1)(2)(0)22(1)(2)(0(0)(0)(0)(0)(0)(0))211333 211333=0fa fa fa fa fb fa fa fa fc fa fa f f f f f f f a V V V V V V V V V V V V V V V V V V αααααααα???=++=++=++?2 (0)(0)(02)(0) 21332133 =0f f fa fc f f fb V V V V V V V αααα=++= 2. 两相(b 相和c 相)短路— (选特殊相为参考相)序分量边界条件 , 0(1), 0fb fc fa fb fc V V I I I ==+= 相量表示的边界条件:(1) (2) (0) (1) (2)(0) (1)(2)(0)(1)(2)(0)(1)(2)(0)002fb fb fb fc fc fc fa fa fa fb fb fb fc fc fc V V V V V V I I I I I I I I I ++=++++= +++++= ()对称分量表示的边界条件 fa V fb fc V V =a b c 22(1) (2) (1)(2)(0)22(1)(2)(0)()()0 0()()203fa fa fa fa fa fa fa fa V V I I I I I a I αααααααα?+?=++= ++++= ()以相为参考相 两相短路没 2. 两相(b 相和c 相)短路—联立方程求解 (0)(1)(1)(1) (2)(2)(2)(0)(0)(0) (81) fa f ff fa fa ff fa fa ff fa V V jX I V jX I V jX I =? = ?? =? (0)(1) (1)(2) (88) () f fa ff ff V I j X X ?+ (1)(2)(1)(2)(0)0(87)0fa fa fa fa fa V V I I I = +=? = fa V fb fc V V =a b c 2. 两相(b 相和c 相)短路—并联复合序网 (0)(1)(1)(2) (88) () f fa ff ff V I j X X =?+ fa V fb fc V V =a b c (1)(2)(1)(2)(0)0(87)0 fa fa fa fa fa V V I I I = +=? = (2)(1) (1)(2) (2)(1)(0)(0)(0)(89)0fa fa fa fa ff fa fa ff fa I I V V jX I V jX I =? ==? =?= 两相短路没 有零序分量 2. 两相(b 相和c 相)短路— 故障点各相电流电压及正序等效定则 (1)(2)(0)(1)(2)(1) 2(1)(2)(0)(1)222fa fa fa fa fa ff fa fb fa fa fa fa fa V V V V V j X I V V V V V V αα=++===++=?=? (2) (1)(0)(1)(2)(2)(1)(0), 0, , 0fa fa fa fa fa ff fa fa I I I V V jX I V =?==== fa V fb fc V V =a b c (0)(0)(1) 2(1)(2)(1)(88)= () () f f fa ff ff ff V V I j X X j X X ? ?++ 2. 两相(b 相和c 相)短路—相量图 (1) fa I (2)fa I (1) fc V (1) fb fb V (2) fc fb V fc V (1) fa I 以为参考相量(2)(1)(0)(1)(2)(2)(1)(0), 0, , 0fa fa fa fa fa ff fa fa I I I V V jX I V =?==== 3. 两相(b 相和c 相)短路接地. (选特殊相为参考相)序分量边界条件 (1)0, 0fb fc fa V V I === 相量表示的边界条件:(1) (2) (0) (1)(2)(0)(1)(2)(0)0 002fb fb fb fc fc fc fa fa fa V V V V V V I I I ++=++=++= () 对称分量表示的边界条件 a b c 2(1) (2) (0) 2 (1)(2)(0)0 003fa fa fa fa fa fa V V V V V V I I a I αααα++=++=++= ()以相为参考相 3. 两相(b 相和c 相)短路接地—联立方程求解 a b c (2)(0)(1)(2)(0)(1) (2)(0) ff ff fa fa fa fa ff ff X X V V V j I X X ===+