hgdsfgsdBV1BV4BV2BV5BV3BV6

1.Cho mng in nh hnh v: (Cc BV l BV D c hng ) - Ti sao phi phi hp v nhy gia cc bo v :1,3,5 v 2,4,6 ? (Ik1 > Ik3 > Ik5; Ik6 > Ik4 > Ik2) - Phn tch hin tng khi ng khng ng thi ca BVDCH trong mng in trn(trong l thuyt)

BV12BV11BV10BV9BV1BV6BV5BV4BV2BV7BV3BV8

2. Cho mng in nh hnh v:

Tnh thi gian tc ng cho cc bo v t1 t8 v xc nh v tr t b phn nh hng cng sut (RW) ti u bit thi gian t=0.3s, thi gian tBV9 = 0,9s tBV10 = 0,5s; tBV11 = 0,3s ; tBV12 = 0,7s. (Dng nguyn tc chn thi gian tc ng cho bo v dng in cc i c c tnh thi gian c lp)

p:Phn tch chia BV thnh 2 nhm: nhm chn v nhm lNhm l gm BV 1, 3, 5, 7Chn thi gian ct ca BV7 l t7 = 0 sec Thi gian ct ca BV5 c xc nh nh sau:t5 = Max{t7,t12}+t = 0,7+0,3=1 sec Tng t:t3 = Max{t5,t11}+t = 1+0,3=1,3 sect1 = Max{t3,t10}+t = 1,3+0,3=1,6 secNhm chn gm BV 2, 4, 6, 8Chn thi gian ct ca BV2 l t2 = 0 sec Thi gian ct ca BV4 c xc nh nh sau:t4 = Max{t2,t10}+t = 0,5+0,3=0,8 sec Tng t:t6 = Max{t4,t11}+t = 0,8+0,3=1,1 sect8 = Max{t6,t12}+t = 1,1+0,3=1,4 secV tr t Role nh hng cng sut RW Xt ng dy 1-2: t1 = 1,6 sec > t2 = 0 sec nn ta t BV2 Xt ng dy 3-4: t3 = 1,3 sec > t4 = 0,8 sec nn ta t BV4 Xt ng dy 5-6: t5 = 1sec < t6 = 1,1 sec nn ta t BV5 Xt ng dy 7-8: t7 = 0 sec > t8 = 1,4 sec nn ta t BV7BV1BV9BV4BV2BV3BV8BV5BV7

BV63. Tnh thi gian tc ng cho cc bo v t1 t8 v xc nh v tr t b phn nh hng cng sut (RW) ti u bit thi gian t=0,3s, thi gian tBV9 = 0,3s .

p:Phn tch chia BV thnh 2 nhm: nhm chn v nhm lNhm l gm BV 1, 3, 5, 7Chn thi gian ct ca BV7 l t7 = 0 sec Thi gian ct ca BV5 c xc nh nh sau:Khi c s c ti ng dy 5-6: cng sut ngn mch dn theo ng ngn nht v im ngn mch nn ta c th chn t5 = 0 sec Thi gian ct ca BV4 c xc nh nh sau:t3 = Max{t5,t9}+t = 0,3+0,3=0,6 sec Tng t:t1 = t3+t = 0,6+0,3=0,9 secNhm chn gm BV 2, 4, 6, 8Chn thi gian ct ca BV2 l t2 = 0 sec Thi gian ct ca BV4 c xc nh nh BV5:t4 = 0sec Tng t:t6 = Max{t4,t9}+t = 0,3+0,3=0,6 sect8 = t6+t = 0,6+0,3=0,9 secV tr t Role nh hng cng sut RW Xt ng dy 1-2: t1 = 0,9 sec > t2 = 0 sec nn ta t BV2 Xt ng dy 3-4: t3 = 0,6 sec > t4 = 0 sec nn ta t BV4 Xt ng dy 5-6: t5 = 0 sec < t6 = 0,6 sec nn ta t BV5 Xt ng dy 7-8: t7 = 0 sec > t8 = 0,9sec nn ta t BV7

4. Phng thc bo v cho MBA 2 cun dy B1 v ng dy D1:(v s nguyn l bo v)

SmB = 30MVAHT110kV35kVL1B1

Phng thc BV MBA 2 cun dy:Bo v chnh:- bo v so lch 87T- bo v bng role du- bo v role gas- bo v dng chm tBo v d phng:- Bo v khong cch 21- bo v role nhit 49- bo v chng s c, h hng MC (50REF)Phng thc bo v ng dyBo v chnh- Bo v khong cchBo v d phng- bo v qu dng ct nhanh 50- bo v chng s c, h hng MC (50REF)5. Tnh thi gian tc ng v v c tnh thi gian (c lp) cho cc bo v t ti my ct MC1, MC2, MC3, MC4, MC5, MC6, MC7, MC8 bit thi gian t=0.2s, thi gian tA = tB1 = 0,5; tC1 = tB2 =0,8s; tD1 = tE = 0,7; tD2 =1,1. Xc nh ti u v tr trn ng dy cn t b phn nh hng cng sut.HT1ABL1CL2L3DHT2tD1tB2tC1EtAtB1tEL4tD2MCAMCB1MCC1MCD1MC1MC2MC3MC4MC5MC6MC7MC8MCEMCB2MCD2

p:Phn tch chia BV thnh 2 nhm: nhm chn v nhm lNhm l gm BV 1, 3, 5, 7 Thi gian ct ca BV7 c xc nh nh sau:t7 = tE+t = 0,7+0,2=0,9 sec Thi gian ct ca BV5 c xc nh nh sau:t5 = Max{tD1, tD2,t7}+t = 1,1+0,2=1,3 sec Tng t:t3 = Max{t5,tC1}+t = 1,3+0,2=1,5sect1 = Max{t3,tB1, tB2}+t = 1,5+0,2=1,7 secNhm chn gm BV 2, 4, 6, 8 Thi gian ct ca BV2 c xc nh nh sau:t2 = tA+t = 0,5+0,2=0,7 sec Tng t:t4 = Max{t2, tB1, tB2}+t = 0,8+0,2=1 sect6 = Max{t4,tC1}+t = 1+0,2=1,2 sect8 = Max{t6, tD1, tD2}+t = 1,2+0,2=1,4 secV tr t Role nh hng cng sut RW Xt ng dy 1-2: t1 = 1,7 sec > t2 = 0,7 sec nn ta t BV2 Xt ng dy 3-4: t3 = 1,5 sec > t4 = 1 sec nn ta t BV4 Xt ng dy 5-6: t5 = 1,3sec < t6 = 1,2 sec nn ta t BV6 Xt ng dy 7-8: t7 = 0,9 sec > t8 = 1,4 sec nn ta t BV7MCP6. Tnh thi gian tc ng cho cc bo v t1 t7 v xc nh v tr t b phn nh hng cng sut (RW) ti u bit thi gian t=0,2s, thi gian tBV8 = 0,5s .

BV1BV8BV4BV2BV3BV5BV7BV6

p:MCP thng ngPhn tch chia BV thnh 2 nhm: nhm chn v nhm lNhm l gm BV 1, 3, 5, 7Chn thi gian ct ca BV7 l t7 = 0 sec Thi gian ct ca BV5 c xc nh nh sau:Khi c s c ti ng dy 5-6: cng sut ngn mch dn theo ng ngn nht v im ngn mch nn ta c th chn t5 = 0 sec Thi gian ct ca BV4 c xc nh nh sau:t3 = Max{t5,t8}+t = 0,5+0,2=0,7sec Tng t:t1 = t3+t = 0,7+0,2=0,9 secNhm chn gm BV 2, 4, 6Chn thi gian ct ca BV2 l t2 = 0 sec Thi gian ct ca BV4 c xc nh nh BV5:t4 = 0sec Tng t:t6 = Max{t4,t8}+t = 0,5+0,2=0,7 secV tr t Role nh hng cng sut RW Xt ng dy 1-2: t1 = 0,9 sec > t2 = 0 sec nn ta t BV2 Xt ng dy 3-4: t3 = 0,7 sec > t4 = 0 sec nn ta t BV4 Xt ng dy 5-6: t5 = 0 sec < t6 = 0,7 sec nn ta t BV5