Câu Hỏi Thi Vào Trạm 110kv

1) Phn bit in p dy v in p pha2)vit cng thc P tc dng,P phn khng,P biu kin trong mng xoay chiu 1 pha v 3 pha3)in k o in nng phn khng khc j vi in k o in nng tc dng,cng dng ca in k o in nng phn khng?4)Vic la chn dy chy cho cu ch phi tha mn nhng iu kin no?5)TU v TI -cn lu nhng j khi lp t TU v TI,Cho bit cc thng s k thut chnh ca TU v TI?Cu 1:Trnh by iu kin cho php MBA vn hnh vi in p cao hn in p nh mc? Vi ch vn hnh hin ti ca li in hin nay th in p nh th no? Ph ti cho php ca MBA nh th no?Tr li:a) Cho php my bin p vn hnh vi in p cao hn in p nh mc ca nc ang vn hnh:- Lu di 5% khi ph ti nh mc v 10% khi ph ti25% ph ti nh mc.- Ngn hn (6h/ngy) 10% vi ph ti khng qu nh mc.b) Ch vn hnh hin nay th in p thanh ci 22kV gi mc 23.1kV do cc MBA phn phi c in p nh mc pha s cp 12.7/22kV (NC nc 3) ang vn hnh qu p lu di 5%.c) Cc MBA trn ch c php vn hnh n ph ti nh mc.

Cu 2:iu kin vn hnh song song cc my bin p? Gii thch tng iu kin?Tr li:a) My bin p c php lm vic song song vi cc iu kin sau :- T u dy ging nhau.- T s bin p bng nhau hoc chnh lch khng qu0.5%.- in p ngn mch chnh lch khng qu10%.- Hon ton ng v pha.b) Gii thch: i vi cc MBA khi u vn hnh song song th cc pha tng ng pha s cp phi u ni tng ng vi li pha s cp, lc :khng lch pha nhau, gi tr khng lch qu qun gia cc MBA gy s c).qu 0.5% (khng xut hin dng- T u dy ging nhau, t s bin p bng nhau hoc chnh lch khng 0.5%, hon ton ng v pha: in p pha th cp gia cc MBA- in p ngn mch chnh lch khng qu10%: mc mang ti gia cc MBA khng lch nhau qu 10%.

Cu 3:Nguyn l lm vic ca recloser? Phm vi s dng? Cch thao tc?Tr li:a) Nguyn l lm vic ca recloser nh sau: khi dng in chy qua recloser vt qu gi tr ci t, recloser s tc ng, c lp cc ph ti sau n trong khong thi gian tr hon v sau s t ng li. Tuy nhin, sau mt s ln ng li m gi tr dng vn vt gi tr ci t th n s t c lp hon ton cc ph ti.b) Phm vi s dng- S dng ng, ct cc ph ti v dng ti ng ct phi khng vt qu dng nh mc ca n.- Gii tr cc s c thong qua v c lp s c vnh cu bng cch phi chc nng t ng li v lockout.- Thit k i km vi DS hoc LTD to khong cch nhn thy phc v cho cng tc sa cha, bo tr li in.c) Cch thao tc:- Bng tay: thao tc ng ct thc hin ti t iu khin, khi cn thit c th ct bng so thao tc t cn ct bng tay trn my ct (tuy nhin khng th ng li c bng so).- T ng: khi dng ti qua recloser vt qu gi tr ci t, recloser t ng ct v t ng ng li theo lp trnh ci t sn.

Cu 4:Li ch ca LBS so vi DS? Khi thao tc ct LBS th ng dy c th xem nh l c lp c cha?Tr li:a) Li ch ca LBS so vi DS :- C th thao tc ng ct dng ti trong khi DS ch ng ct dng khng ti.- C th ng ct khi hai u ng dy khng ng p, DS ch ng ct mch ng p.b) Khi thao tc ct LBS th ng dy vn c xem nh l cha c c lp v tip im LBS l tip im khng nhn thy.

Cu 5:Cho bit trnh t c lp v ti lp my ct c DS cch ly 2 pha?Tr li:a) Trnh t c lp :- Ct MC.- Ct DS cch ly 2 pha.b) Trnh t ti lp :- ng DS cch ly 2 pha.- ng MC.

Cu 6:T b trn li in phn phi c cng dng g? Ti sao khi lp t b vi dung lng ph hp th gim c TTN? T b trn li phn phi thuc loi b no? Nu phn theo cch iu khin gm my loi?Tr li:a) T b l thit b pht ra cng sut phn khng b vo cng sut phn khng do ti tiu th trn li, nh iu chnh c in p, nng cao c h s cng sut v gim tn tht in nng trn li.b) T b lp t vi cng sut ph hp th ton b cng sut ca t b pht ra s thay th cho lng cng sut ca ngun do on ng dy t ngun n v tr m ph ti nhn cng sut phn khng ca t b s gim mt lng cng sut chuyn ti bng vi cng sut ca t v tng ng s gim tn tht in nng chuyn ti lng cng sut ny.c) T b trn li phn phi thuc loi b ngang, nu phn theo cch iu khin gm c hai loi: T b tnh v t b ng ng.Cu 7:Tc dng ca vic lp LA trn li in phn phi? Gii thch ngha ca k hiu MCOV ghi trn nhn ca LA? Li 22kV s chn LA c c tnh nh th no?Tr li:a) Chng qu in p do xung st lan truyn trn ng dy; chng qu in p do ng ct thit b hoc trng thi qu khc ca li.b) MCOV (kV): in p lm vic lin tc ln nht ca LA (MCOV: Maximum Continuous Operating Voltage), n v l kVc) Li 22kV chn LA c c tnh nh sau:- Um: 18kV.- MCOV: 15,3kV.- Class: 1.- Ix nng lng: 10kA.

Cu 8:Hy lit k nhm thit b c mc ch ci thin cht lng p? Nhm c mc ch bo v s c.Tr li:a) Nhm thit b ci thin cht lng in p.- T b.- AVR (my iu p trung th).b) Nhm thit b bo v s c.- Aptomat.- FCO, LBFCO.- Recloser.- My ct (kt hp vi h thng r le bo v).

Cu 9:Ch qu ti ngn hn cho php ca my bin p nh p no khi c s c?Tr li- Nu c quy nh c th ca nh sn xut th thc theo quy nh, cc trng hp cn li khi c s c my bin p c php qu ti ngn hn theo bng sau:

Qu ti theo dng in, % 30 45 60 75 100Thi gian qu ti, pht 120 80 45 20 10

Cu 10:B iu p di ti cc tc dng g, c lp t u?Tr li:a) B iu p di ti (OLTC) c tc dng iu chnh in p pha th cp ca MBT trong phm vi cho php khi my bin th ang mang ti.b) OLTC c lp t pha cun s cp (cun cao) ca MBT.

Cu 11:Hy cho bit nh hng ca vic my bin p lm vic khng i xng khi ti phn b khng u?Tr li:- My bin p lm vic khng i xng khi ti phn b khng u cho ba pha. Khi dng in khng bng nhau cc pha gy nh hng xu n tnh trng lm vic bnh thng ca my bin p nh in p pha v dy s khng i xng, tn hao ph trong dy qun v li thp tng ln, chnh lch nhit ca my vt qu quy nh.

Cu 12:iu kin cn thit thc hin ng kt vng h 2 pht tuyn trung th? iu kin cn thit thc hin kt vng kn 2 pht tuyn trung th?Tr li:a) iu kin cn thit thc hin ng kt vng h- Phi s dng thit b ng ct c ti c dng nh mc ph hp.- 2 pht tuyn phi cng cp in p vn hnh.- im u ni kt vng phi m bo 2 pht tuyn cng th t pha.b) iu kin cn thit thc hin kt vng ng th- Phi s dng thit b ng ct c ti c dng nh mc ph hp.- 2 pht tuyn phi cng cp in p vn hnh.- im u ni kt vng phi m bo 2 pht tuyn cng ng v pha.- Gc lch pha gia 2 im ha: 30o.- Chnh lch tn s gia 2 im ha:f 0.25Hz.- Chnh lch in p gia 2 im ha:U 10%.

Cu 13:Kim tra nh k m ng dy trung th ch yu ni dung g?Tr li:- Pht nhit cc mi ni.- Phng in, r in.- Cc nguy c v hin tng bt thng xy ra.

Cu 14:My bin th 3 pha c t u dy Dyn11 ni t s trung tnh h th c tc dng?Tr li:- m bo in p pha t gi tr ng thit k.- m bo my vn hnh an ton.

Cu 15:Trm bin th 3 pha gm t hp 3 my bin th 1 pha c t u dy Ynyn0, khi 1 my b s c th c th cung cp in bnh thng cho cc khch hng 1 pha thuc 2 pha cn li khng? Ti sao?Tr li:C th cung cp in bnh thng cho cc khch hng 1 pha thuc 2 pha cn li do i vi khch hng 1 pha th ngun cung cp in thuc tng my c lp khng b nh hng bi cc my cn li.

Cu 16:Cng dng ca bin dng in (TI)? Trong bin dng cun dy c cp chnh xc 0.5 s dng cho mc ch g, cun dy c cp chnh xc 5P20 s dng cho mc ch g?Tr li:- Cng dng ca bin dng in (TI): bin i gi tr dng in trong mch s cp c gi tr ln v gi tr danh nh ca mch th cp phc v cho cng tc o lng cc i lng in (cp danh nh mch th cp thng l 1A hoc 5A).- Cun dy c cp chnh xc 0.5 (CL 0.5): s dng cho mch th cp thc hin chc nng o m in nng.- Cun dy c cp chnh xc 5P20 (CL 5P20): s dng cho mch th cp thc hin chc nng bo v r le.- Cu 17:Cng dng ca bin in p TU?Tr li:Bin in p (TU) dng bin i gi tr in p ca li ang vn hnh v gi tr danh nh ca mch th cp phc v cho cng tc o lng cc i lng in hoc cp tn hiu in p cho h thng r le bo v.V l thuyt th MBA, TU, TI tng i ging nhau v cu to v nguyn l lm vic (chng th m gi l MBA lc v MBA o lng), MBA th c th lm vic bnh thng c cc ch : Khng ti, non ti, y ti, qu ti (c h s qu ti v thi gian quy inh) nhng TU li lm vic bnh thng ch h mch v TI lm vic bnh thng ch ngn mch?xem tit din v s vng dy qun ca TU v TI th bc bit riA/ My bin p o lng - bin in p TU

Cu to th TU v TI l tng t nh my bin p thng thng gm mch t trn c qun cun s cp v th cp. Cun s ni ti mch c in p cn o, th th ni ti vn k, hoc cun nhn tn hiu ca cc thit b o c hin th s hin nay... m bo an ton th cun th lun ni t.V dng ti cun th cc nh nn c th ni l TU lm vic trng thi h mch ( khng ti ).in p nh mc cun th thng n nh l 100V bt k u vo l cp p no. 10 KV/100V or 35 KV/100V.V vy lu khi lp TU l phi c cu ch v ni t cho v thit b.DKH

B/ My bin p o lng - bin dng in TI

Tng t cng vn l mt mch t, cun s cp t 1 vng n vi vng, cun th th nhiu vng hn v u vo ampe k hoc cc thit b hin th s... Cun th cng c ni t.V tr khng ampe k c gi tr nh nn bin dng lun lun lm vic trong trng thi gn ngn mch, t thng qua mch t rt nh, trong trng hp b h mch tng thng tng ln d lm hng mch t, ng thi in p th cp tng cao gy h hng thit b o lng v thVi TI khng c lm h mch cng nh t cu ch bo v- TU th dng s cp b nn t thng sinh ra trong mch t t, v vy c th hot ng ch h mch.Cn TI th dng s cp ln nn khi hot ng ch h mch t thng sinh ra ln, lm hng mch t, v vy n phi c ngn mch li nu ko c gn vo thit b o dng cn bng lng t thng trong mch t.- A/ My bin p o lng - bin in p TU



Tng t cng vn l mt mch t, cun s cp t 1 vng n vi vng, cun th th nhiu vng hn v u vo ampe k hoc cc thit b hin th s... Cun th cng c ni t.V tr khng ampe k c gi tr nh nn bin dng lun lun lm vic trong trng thi gn ngn mch, t thng qua mch t rt nh, trong trng hp b h mch tng thng tng ln d lm hng mch t, ng thi in p th cp tng cao gy h hng thit b o lng v thVi TI khng c lm h mch cng nh t cu ch bo v- TU th dng s cp b nn t thng sinh ra trong mch t t, v vy c th hot ng ch h mch.Cn TI th dng s cp ln nn khi hot ng ch h mch t thng sinh ra ln, lm hng mch t, v vy n phi c ngn mch li nu ko c gn vo thit b o dng cn bng lng t thng trong mch t.theo em hiu l vy, ko bit c chnh xc ko? nu c g sai mong bc b qua. heheKhng phi nh vy u pc i.

T thng ca mt my bin p ch yu ph thuc vo in p. dng in li ph thuc ti. V th my bin in p h mch s mang ti t nht (=0). Ngn mch s b qu dng.

Ngc li, my bin dng mc ni tip vi mch in nht th, nn dng in s cp ca n khng ph thuc ti bn th cp, m ph thuc ch yu vo mch nht th. Nu c dng th cp, th 2 dng ny ngc nhau s trit h ln nhau, v t thng tng khi cn rt b. nu khng c dng th cp, th ton b dng s cp s t ha li t.

Nhng li t rt kh h, m ch b bo ha. T thng s c dng hnh thang. V mi cnh ln xung ca hnh thang s to ra xung in p rt cao trong mch th cp, lm cho hng cch in ca mch th cp.

C th, 1 ci TI trong trm 110 kV, t s 600/5, khi l tay lm h mch bin dng, n gy ra tia la di khong 4cm bn mch th cp, ngay khi dng ch be b.- theo em thy thc t ci TI n c s cp l chnh dy dn dng in m mnh cn , th cp n qun rt nhiu vng dy tit din nh hn nhiu so vi s cp. So snh tng i vi s vng dy bn s cp( 1 vng) th n kh ln. v n l my bin p nn ta c:n1/n2 = u1/u2( n1, u1 l s vng dy, in p s cp; n2, u2 l s vng dy, in p bn th cp) . t cng thc trn ta thy u2 = (n2 * u1)/ n1 c gi tr rt ln, in p ny nu ko c ti s lm cch in ca dy qun b nh thng v qu p >>chp cc vng dy>>>chy xo>>h TI.

Khi c ti ( ngn mch th cp, hay ni ti no vo) th th cp kn mch, sinh dng in, dng ny li sinh t thng c tc dng chng li/lm suy yu/ lm gim ln ca t thng s cp v to thnh t thng tng gy sc in ng trn cun th cp. sdd ny nh nn TI khi c ti s chy bnh thng.- V li t b bo ha, pc .

Khi c dng th cp, th ta c:

I1n1 = I2n2.Thc ra nu cng vecto th:

I1n1 + I2n2 =I0.

y, I0 l t thng tng ca li thp, to t trng v sc in ng cm ng nhm sinh ra dng th cp. Thng thng I0 rt b. Vi bin dng c sai s 0,5%, th I0 khong bng 0,5% I1.

Ngi ta s thit k my bin dng sao cho khi I0 tng gp 10 ln th bin dng mi i vo bo ha.

Nhng khi h mch bin dng, th I2 =0. Khi I1 = I0.

Ngha l I0 tng ln n 200 ln, gp 20 ln tr s bo ha.

Khi dng in I0 th hnh sin, nhng t thng th b xn u, v thnh hnh thang.- TI m khng ni t th cp th h thng tan d hon ton bn .Vi bin dng TI th dng in pha s cp ln so vi dng th cp.N l thit b in t tnh ging nh BA.khi dng th cp ca TI nh th p th cp s rt ln.nu khng ni t th cp ca TI th cun dy th cp c chu ni in p ny khng.- A/ My bin p o lng - bin in p TU


Tri i - mi c 10 pht m cc Bc vit khe gh. Trong khi Hi vit cha xong m cc bc chi lun 3 bi ri .


Tng t cng vn l mt mch t, cun s cp t 1 vng n vi vng, cun th th nhiu vng hn v u vo ampe k hoc cc thit b hin th s... Cun th cng c ni t.V tr khng ampe k c gi tr nh nn bin dng lun lun lm vic trong trng thi gn ngn mch, t thng qua mch t rt nh, trong trng hp b h mch tng thng tng ln d lm hng mch t, ng thi in p th cp tng cao gy h hng thit b o lng v thVi TI khng c lm h mch cng nh t cu ch bo v- 1. Thng th nht: Nguyn l lm vic, cu to, cng dng ca Dao cch ly.2. Thng th hai: Nguyn l lm vic, cu to, cng dng ca My ct in.

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Câu Hỏi Thi Vào Trạm 110kv