4×25ＭW水轮机组继电保护设计

.0??? 3F( 4F)I 2d )3(F3=787 .02 573 .12 22???B Id 系统 I、 II 供给的短路电流 I 2d )3(L=I 2d )3(B1—I 2d )3(B2=577 .23904 .0 1? L 1:I 2d )3(1L=?????)28.0()12.1( )28.0( XL XL XL I 2d )3(L=)178 .02.0()148 .02.1( 004 .1)178 .08.0(????? 10 =0.422 L 2:I 2d )3(2L=582 .0)178 .08.0()148 .02.1( 004 .1)148 .02.1()28.0()12.1( 577 .2)12.1(???????????? XL XL XL (4 )求 d 3 点的短路电流由于发电机、变压器是对称的,故 d 3 点的短路电流同 d 2点(5 )求 d 4 点的短路电流由图 1 变换 X 1 =X 2 =X B1+362 .02 3904 .0167 .02 1??? FX X 2 =0.8+X 1L =0.8+0.178=0.978 X 3=301 .0978 .0181 .0 148 .0)978 .0181 .0(148 .0978 .0 //181 .0 //2 1 3 1???????? LXX X 由系统 II 和四台机组提供 d 4 点的短路电流 I 4d )3(=4 1X =322 .3301 .0 1?流经各元件的短路电流 X ?1 =1.2//X 4=136 .0153 .02.1 153 .02.1???线路 I:I 4d )3(1L=I 4d )3( =3.322 线路 I:I 4d )3(2L=??32/1 2/1XX X I 4d )3(1L=519 .0322 .3978 .02/362 .0 2/362 .0???