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第5章 短路电流的计算
在电力供电系统中,对电力系统危害最大的就是短路。所谓短路是指一切不正常的相与相之间或相与地发生通路的情况。
产生短路的原因很多,主要有以下几个方面:
(1)元件损坏,例如绝缘材料的自然老化,设计、安装及维护不良所带来的设备缺陷发展成短路等;
(2)气象条件恶劣。例如雷击造成的闪络放电或避雷动作,架空线路由于大风或导线覆冰引起电杆倒塌等;
(3)人为事故,例如运行人员带负荷拉刀闸,线路和设备检修后未拆除接地线就加上电压等;
(4)其他,例如挖沟损伤电缆,鸟兽跨接在裸露的载流部分等。
在三相系统中短路的形式可以分为三相短路、两相短路、两相短路接地、单相短路接地。三相短路也叫对称短路,系统各相与运行正常时仍处与对称状态,其他类型的短路都不是对称短路。
电力系统的运行经验表明,在各种类型的短路中,单相短路占大多数,两相短路较少,三相短路的机会很少发生,但情况较严重,应给与足够的重视。况且,从短路计算方法来看,一切不对称短路的计算,在采用对称分量法后,都归结为对称短路的计算。因此,对三相短路的研究具有重要的意义。在短路电流计算过程中,便都以最严重的短路形式为依据。因此,本文的短路电流计算都以三相短路为例。
5.1 短路电流的计算目的
5.1.1 短路电流的危害
在供电系统中发生短路故障时,在短路回路中短路电流要比额定电流大几倍至几十倍,通常可达数千安,短路电流通过电气设备和导线必然要产生很大的电动力,并且使设备温度急剧上升有可能损坏设备和电缆;在短路点附近电压显著下降,造成这些地方供电中断或影响电动机正常工作;发生接地短路时所出现的不对称短路电流,将对通信线路产生干扰;当短路点离发电厂很近时,将造成发电机失去同步,而使整个电力系统的运行解列。
5.1.2 计算短路电流的目的
计算短路电流的目的是为了正确选择和校验电器设备,避免在短路电流作用下损坏电气设备,如果短路电流太大,必须采用限流措施,以及进行继电保护装置的整定计算。
为了达到上述目的,须计算出下列各短路参数:
I″— 次暂态短路电流,用来做为继电保护的整定计算和校验断路器额定断流容量。应采用(电力系统在最大运行方式下)继电保护安装处发生短路时的次暂态短路电流来计算保护装置的整定值。
— 三相短路冲击电流,用来检验电器和母线的动稳定。
I— 三相短路电流有效值,用来检验电器和母线的热稳定。
S″— 次暂态三相短路容量,用来检验断路器的遮断容量和判断母线短路容量是否超过规定值,作为选择限流电抗器的依据。
5.2 短路电流的计算
为了简化短路电流的计算方法,在保证计算精度的情况下,忽略次要因素的影响,做出一下规定:
(1) 所有的电源电动势相位角均相等,电流的频率相同,短路前,电力系统的电势和电流是对称的。
(2) 认为变压器是理想变压器,变压器的铁心始终处于不饱和状态,即电抗值不随电流的变化而变化。
(3) 输电线路的分布电容略去不计。
(4) 每一个电压级采用平均电压,这个规定在计算短路电流时,所造成的误差很小。唯一例外的是电抗器,应该采用加于电抗器端点的实际额定电压,
因为电抗器的阻抗通常比其他元件阻抗大的多,否则,误差偏大。
(5) 计算高压系统短路电流时,一般只计及发电机、变压器、电抗器、线路等元件的电抗,因为这些元件X/3>R时,可以略去电阻的影响。只有在短路点总电阻大于总电阻的1/3时才加以考虑,此时采用Z∑=X∑。
(6) 短路点离同步调相机和同步电动机较近时,应该考虑对短路电流值的影响。有关感应电动机对电力系统三相短路冲击电流的影响:在母线附近的大容量电动机正在运行时,在母线上发生三相短路,短路点的电压立即降低。电动机将变为发电机运行状态。
(7) 在简化系统阻抗时,距短路点远的电源与近的电源不能合并。
(8) 以供电电源为基准的电抗标幺值>3,可以认为电源容量为无限大容量的系统,短路电流的周期分量在短路全过程中保持不变。
5.2.1 短路电流计算结果
短路电流计算结果
短路点
I"(KA)
I(KA)
I(KA)
I(KA)
I(KA)
110KV(d)
3.32
3.32
3.32
3.32
8.45
35KV(d)
2.8
2.8
2.8
2.8
11.09
10KV(d)
12.1
12.1
12.1
12.1
30.8
5.2.2 短路计算过程
1.选择基准容量 =1000MVA 、Ud1=115KV 、Ud2=37KV Ud3=10.5KV
则基准电流
I d1= =5.02kA
Id2 = =15.60KA
Id3 = =55KA
系统短路等效图如图5-1:
系统简化等效图
2.计算系统各元件的标幺值:为了简化计算线路阻抗均采用0.4Ω/KM(注:短路电流的计算过程中忽略了甲变对短路计算的影响。)则电力系统各元件的相对应的标幺值为:
5.2.3 短路点的短路电流
则当短路时系统简化图为
系统d1点短路时的简化图
(1)110KV侧短路电流计算
由于题目要求系统是无穷大系统。
所以当s ,2s,4s,和无穷大时电流相等。
I f.t===0.33
无限大容量电源直接计算; I f.t=
由于两个系统是相同的所以;
=2×=1.66×2=3.32
=*8.45
(2)35KV侧短路电流计算:
系统等效图可转换为:图5-3
通过星三角转换可得下图;
X= X= X===1.01
X= + +X=+0+3.02=4.99
再通过星三角转化可得;
X=1.01+4.99+=10.99
X=1.01+4.99+=10.99
由于题目要求系统是无穷大系统。
所以当s ,2s,4s,和无穷大时电流相等。
I f.t===0.09
无限大容量电源直接计算; I f.t
由于两个系统是相同的所以;
=2×=1.4×2=2.8
冲击电流为;
= *11.09
(3)10KV侧短路计算:
系统等效图转换如:图5-4
通过星三角转换可得;
X= X= X===1.01
再通过星三角转换可得;
有星三角转换公式得;
X=1.01+4.09+=9.19
X=1.01+4.09+=9.19
电源是无穷大系统
当s ,2s,4s,和无穷大时电流相等。
I f.t===0.11
无限大容量电源直接计算; I f.t=6.05
由于两个系统是相同的所以;
=2×=6.05×2=12.1
冲击电流为;
= *30.8
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