CUNG CP IN 2GV: Nguyn Quang Thun

Ni dung mn hc

Ti liu tham kho1. TS. Ng Hng QuangThit k cp in, NXBKHKT- 20062. TS. Ng Hng QuangLa chn cc phn t thit b t 0,4-500kV, NXBKHKT- 20053. TS. Trn Quang Khnh H thng cung cp in, NXKHKT HN 20054. GS. Nguyn Cng Hin H thng cung cp in, NXKHKT HN 2004

Chng 7. LA CHN CC PHN T, THIT B TRONG HTCC7.1. KHI QUT CHUNG1. t vn Trong qu trnh lm vic, cc phn t, thit b c th phi chu 3 ch lm vic:Bnh thng: Um, ImQu ti: > Um, ImS c (NM): >> Im Phi ct phn t, thit b b s c ra khi ngun cng nhanh cng tt. Tuy nhin phi c thi gian PT, TB phi chu ng c trong thi gian tn ti s c ny.Bi vy cc PT, TB a vo lm vic cn phi c la chn tho mn ng thi 3 iu kin trn.2. iu kin chung la chn cc PT, TB:a. iu kin PT, TB m bo lm vic bt v qt: i vi d lv //: Ilvmax = 2Ibt= 2Icp (tc l tnh khi 1 d b t); i vi mch MBA: Ilvmax = kqtmaxIbt=kqtmaxImBA(thg kqtmax= 1,4 ); i vi mch MP: Ilvmax = kqtmaxIbt= 1,05Im

2. iu kin chung la chn cc PT, TB:b. iu kin PT, TB m bo chu ng c ch s c:Dng in NM ln sinh ra lc in v nhit ln c th ph hng v t chy phn dn/cch in ca PT, TB. Do cn kim tra theo 2 iu kin: iu kin n nh ng: I.m ixk (2)iu kin n nh nhit:(Vi tqd = tN)Lu :i vi cc PT, TB h p (U 1000V) khng cn kim tra n nh ngi vi PT, TB c Im 1000A, khng cn kim tra n nh nhiti vi dy dn v thanh dn, iu kin n nh nhit kim tra theo tit din ti thiu:

7.2. LA CHN KH C IN CAO P1. La chn MC

TTCc i lng chn v kim traCng thc chn v kim tra1in p nh mc, UmMC (kV)UmMC Um.m2Dng in nh mc, ImMC (A)ImMC Ilv.max3Dng in ct nh mc, IC.m (kA)IC.m IN4Cng sut ct nh mc, SC.m (MVA)SC.mMC SN5Dng in nh mc, I.m (kA)I.m ixk6Dng in n nh mc, Inh.m (kA)

7.2. LA CHN KH C IN CAO P2. La chn MC ph ti

TTCc i lng chn v kim traCng thc chn v kim tra1in p nh mc, UmMC (kV)UmMC Um.m2Dng in nh mc, ImMC (A)ImMC Ilv.max3Dng in nh mc, I.m (kA)I.m ixk4Dng in n nh mc, Inh.m (kA)5Dng in nh mc ca CC, ImCC (A)ImCC Ilv.max6Dng in ct nh mc ca CC, IC.mCC (A)IC.mCC IN7Cng sut ct nh mc ca CC, SC.mCC (A)SC.mCC SN

3. La chn DCL4. La chn CC cao p

TTCc i lng chn v kim traCng thc chn v kim tra1in p nh mc, UmDCL (kV)UmDCL Um.m2Dng in nh mc, ImDCL (A)ImDCL Ilv.max3Dng in nh mc, I.m (kA)I.m ixk4Dng in n nh mc, Inh.m (kA)

TTCc i lng chn v kim traCng thc chn v kim tra1in p nh mc, UmCC (kV)UmCC Um.m2Dng in nh mc, ImCC (A)ImCC Ilv.max3Dng in ct nh mc ca CC, IC.mCC (A)IC.mCC IN4Cng sut ct nh mc ca CC, SC.mCC (A)SC.mCC SN

7.3. LA CHN MY BIN PLa chn MBA in lci vi TBA c 1 my: khcSmB Stt i vi TBA c 2 my: khckqtmaxSmB SttV d: H ni nhit trung bnh 240C;Mtcva nhit trung bnh 50C;Th:- nhit mi trng s dng v nhit ch to (0C)H s hiu chnh gia mt ch to v s dng (ch s dng khc nu MBA ngoi nhp) Trong :SmB - cng sut m ca MBA, (nh ch to cho);Stt - cng sut tnh ton (cng sut ln nht ca ph ti).kqtmax - h s qu ti ln nht ca MBA, kqtmax = 1,4 (qu ti khng qu 5 ngy 5 m, mi ngy khng qu 6 gi).

7.3. LA CHN MY BIN P2. La chn MBA o lnga. My bin dng in (BI)

TTCc i lng chn v kim traCng thc chn v kim tra1in p s cp nh mc, Um.BI (kV)Um.BI Um.m2Dng in s cp nh mc, I1m.BI (A)

3Ph ti cun dy th cp, S2m.BI, (VA)S2m.BI S2tt4H s n nh ng, k 5H s n nh nhit, knh

2. La chn MBA o lngb. My bin in p (BU) - sai s tiu chun.

TTCc i lng chn v kim traCng thc chn v kim tra1in p s cp nh mc, Um.BU (kV)Um.BU Um.m2Ph ti 1 pha th cp, S2m.BI, (VA)S2m.pha S2tt.pha3Sai s cho php, N%

7.4. LA CHN KH C IN H Pmotor controlLi in h pCch lyng ctBo v ngn mchBo v qu tiiu khin cng sutCch lyng ctBo v ngn mchin-cin t

motor starterSwitch(cu dao)Cu daotiCng-tctR lenhitptmtkiu t inptmtkiut nhitThit btch hpCch lyng ctNgn mchQu tiiu khin

7.4. LA CHN KH C IN H PCc kh c mng in h p nh ptmt, cngtct, cu dao, cu ch,....c la chn theo iu kin in p v dng in, kiu loi v hon cnh lm vic khng cn kim tra iu kin n nh ng v n nh nhit do dng ngn mch. Ring chn ptmt v cu ch cn lu : i vi ATM: Phi kim tra kh nng ct dng in ngn mch v chnh nh ct dng in qu ti;i vi CC: Phi phn bit dng cho mng in sinh hot, chiu sng hay dng trong mng cng nghip m chn cho ng. Sau y, s nu cch chn cc thit b ny.

7.4. LA CHN KH C IN H P1. Chn ATM:Qu ti: chnh nh

7.4. LA CHN KH C IN H P2. Chn cu ch:a. i vi CC dng cho mng in chiu sng:

2. Chn cu ch:b. i vi CC dng cho mng in cng nghip:Trong : kt - h s ti ca ng c, nu khng bit ly kt = 1;Im.C- dng in nh mc ca ng c, Im.C = Um- in p dy nh mc, Um = 380V;Cosm - h s cng sut nh mc ca .c, thng Cosm = 0,8; - hiu sut ca c, thng = 0.8-0,95 (c th ly = 1);Kmm - h s m my .c (nh ch to cho), thng Kmm = 5; 6; 7; - h s ph thuc vo iu kin khi ng ca .c, ly nh sau:Vi .c m my nh (hoc khng ti) nh my bm, my ct gt kim loi = 2,5;Vi ng c m my nng (c ti) nh cn cu, cn trc, my nng =1,6; Nu BV cho 1 ng c:

b. i vi CC dng cho mng in cng nghip: Nu BV cho nhiu ng c:

7.5. CHN TIT DIN DY DN V CP 1. Chn tit din dy dn theo mt dng in kinh tPP ny ch dng chn tit din dd ca cc mng in cao p. iu kin chn tit din:Kim tra:Tn tht in p:Pht nng: Isc IcpNgoi ra, i vi cp bt buc phi kim tra thm iu kin n nh nhit:

- h s nhit , cu = 6 v Al = 11; tq = (0,5-1)s

2. Chn tit din dd theo tn tht in p cho php PP ny dng chn tit din dd mng in h p v mng in a phng (U 35kV) chiu di ln.Xut pht t cng thc tnh tn tht in p:Nhn thy, r0 v x0 u ph thuc vo tit din F, do ta c th chn tit din bng cch chn in khng x0 (v x0 = 0,350,45/km - khng thay i nhiu).Cc bc chn tit in dd theo phng php ny lm nh sau:B1. Chn s b x0 i vi dy h p: Chn x0 = 0,35/kmi vi dy TA p: Chn x0 = 0,38/km (vi 1022kV); x0 = 0,4/km (vi 35kV) B2. T x0 chn, xc nh c:

2. Chn tit din dd theo tn tht in p cho phpB3. T Ucp xc nh c UR: UR = Ucp - UXB4. Xc nh tit din dy dn cn chn:

T F tnh c, tra bng ph lc chn dy dn c tit din gn nhtB5. Kim tra dy dn chn theo iu kin tn tht in p cho php:

Nu iu kin c tha mn th dy dn chn t yu cu;Trng hp khng tho mn, chn dy dn c tit din ln hn 1 cp, ri kim tra li theo iu kin trn. - in dn sut, v d:

3. Chn tit din dy dn theo dng pht nng cho phpPP ny dng chn tit din dy dn li h p cng nghip v sinh hot th.Trnh t cc bc chn tit din theo phng php ny nh sau:- Bc 1: Xc nh dng in tnh ton m ng dy phi ti Itt;- Bc 2: La chn loi dy, tit din dy theo biu thc:k1k2Icp IttTrong : k1- h s hiu chnh nhit , ng vi mi trng t dy, cp;k2- h s hiu chnh nhit , k n s lng dy hoc cp i chung mt rnh; k1 v k2 tra trong ph lc.Icp- dng in lu di cho php ng vi tit din dy hoc cp nh chn (tra bng)

3. Chn tit din dy dn theo dng pht nng cho php- Bc 3: Kim tra li:* Theo iu kin kt hp vi thit b bo v:+ Nu bo v bng cu ch: (i vi mch nh lc = 3; cn mch nh sng sinh hot = 0,3)+ Nu bo v bng ATM: Ikddt - dng in khi ng in t ca ATM (dng chnh nh ct ngn mch)Ikdnh - dng in khi ng nhit ca ATM (dng chnh nh ct qu ti ca r le nhit)

3. Chn tit din dy dn theo dng pht nng cho php* Theo iu kin n nh nhit dng ngn mch:Trong : -h s ph thuc vt liu lm dd, cu = 6; Al = 11tq = tN = (0,5-1)s * Theo iu kin tn tht in p:Umax Ucp = 5%Um

7.6. CHN TIT DIN THANH DN (THANH CI)Trong : k1 h s hiu chnh, ph thuc vo vic t thanh dn.+ t ng: k1 = 1;+ t nm ngang:k1 = 0,95. k2 H s hiu chnh theo nhit mi trng (tra trong s tay KT khi nhit mi trng khc vi nhit TC). - h s ph thuc vt liu lm thanh dn: Cu = 6; Al = 11. cp - ng sut cho php ca vt liu lm thanh dn (cp.Cu= 1400 kG/cm2; cp Al= 700 kG/cm2; cp Fe= 1600 kG/cm2).

TTCc i lng chn v kim traCng thc chn v kim tra1Dng in pht nng lu di cho php, Icp (A)k1k2Icp Ilv.max2Kh nng n nh ng, cp (kG/cm2)cp tt3Kh nng n nh nhit, F (mm2)

7.6. CHN TIT DIN THANH DN (THANH CI)tt - ng sut tnh ton (kG/cm2): sut hin khi c lc in ng ca dng NM: , M - m men un tnh ton, + Khi thanh dn c t 3 nhp tr ln: + Khi thanh dn c 2 nhp: ixk- dng in ngn mch xung kch 3 pha, kA l - khong cch gia cc s trong mt pha (chiu di mt nhp thanh ci), cm; a - khong cch gia cc pha, cm; W - m men chng un ca TC, cm3 tnh c da vo hnh dng thanh gp:

W - m men chng un ca thanh dn, cm3 tnh c da vo hnh dng thanh gp:

Thanh ch nhtThanh ch nht rngThanh trnThanh trn rngt ngt ngang

7.7. CHN S THANH DNFcp - Lc cho php tc ng ln u s, (kG):(Fcp = 0,6Fph, Fph-lc ph hng s, nh ch to cho)k h s hiu chnh:H - chiu cao s;H - chiu cao t chn s n tm tit din thanh dn

Chng 8. BO V RLE V TH HTCC8.1. KHI NIM, MC CH NGHA1. Khi nim Rle l phn t chnh trong h thng thit b bo v. Thut ng rle c phin m t ting nc ngoi: RELAIS-Php, RELAY-Anh, PEE-Nga ... vi ngha ban u l phn t lm nhim v t ng ng ct mch in. Ngy nay khi nim rle thng dng ch mt t hp thit b thc hin mt hoc mt nhm chc nng bo v v t ng ho h thng in gi l BVRL2. Mc ch, nghaV mt k thut: BVRL l thit b bo v c nhim v pht hin v cch ly cc phn t b s c hoc hot ng bt thng ra khi HT.V mt kinh t: BVRL l thit b t ng ho c dng trong HT vi mc ch phng nga, ngn chn cc thit hi kinh t c th xy ra cho ch u t khi xy ra s c.

8.2. CC YU CU C BN I VI BVRL thc hin c cc chc nng v nhim v quan trng k trn, thit b bo v rle phi tho mn c cc yu cu c bn: tin cy, chn lc, tc ng nhanh v kinh t.1. Tin cy (Reliability): L tnh nng m bo cho thit b bo v rle lm vic ng, chc chn khi xy ra s c trong phm vi c xc nh. 2. Chn lc (selectivity): l kh nng ca bo v rle c th pht hin v loi tr ng phn t b s c ra khi h thng. 3. Tc ng nhanh: Bo v rle cn phi cch ly phn t b s c cng nhanh cng tt. Tuy nhin cn kt hp vi yu cu chn lc.4. nhy (sensitivity): Phn nh kh nng phn ng ca bo v vi mi mc s c. nhy c biu th bng t s i lng tc ng ti thiu vi i lng t. V d: i vi BV qu dng:

Quy nh c th vi cc loi bo v:- Bo v chnh: knh = 1,52- Bo v d phng: knh= 1,21,5.

8.2. CC YU CU C BN I VI BVRL5. Kinh t: - i vi mng cao p v siu cao p (U 110 kV): Chi ph mua sm v lp t thit b bo v thng ch chim mt vi phn trm gi tr cng trnh, mt khc yu cu phi c bo v rt chc chn, v vy gi c thit b bo v khng phi l yu t quyt nh trong la chn chng loi hoc nh phn phi thit b m 4 yu cu k thut k trn ng vai tr quyt nh.- i vi mng trung p v h p (U < 110 kV): V s lng cc phn t c bo v rt ln, mt khc cc yu cu i vi thit b bo v khng cao bng mng cao p v siu cao p cho nn khi la chn thit b bo v cn ch m bo c cc yu cu v k thut vi chi ph thp nht.

3.3. CU TRC C BN CA HTBVRL

8.3. CU TRC C BN CA HTBVRL1. BI/TICu to:L MBA lm vic ch NM

T s bin i nI (hoc kI):Chc nng:

K hiu:

Dng bin i dng ln xung dng nh 5A hoc 1A cp cho TBBV v o lng

8.3. CU TRC C BN CA HTBVRL2. BU/TUCu to:L MBA lm vic ch h mch

T s bin i nU (hoc kU):Chc nng:

K hiu:

Dng bin i p ln xung p nh cp cho TBBVRL v o lng

8.3. CU TRC C BN CA HTBVRL3. Ngun thao tc NKhi nim: Tt c cc mch ca s iu khin my ct, bo v rle, o lng, tn hiu c gi l s nh th. Ngun in cung cp cho vic thao tc cc phn t trong s ny gi l ngun thao tc. N ngun in thao tc ring c lp vi phn t c bo v.Phn loi ngun in thao tc: Ngun thao tc c th l ngun mt chiu hoc xoay chiu.C th Dng c quy (DC). Nu cn ngun AC (Nghch lu).C th dng nng lng tch sn trn t in (thng c np in DC).C th ly t BI v BU sau chnh lu thnh ngun DC. tng tin cy ca ngun thao tc, ngi ta thng dng kt hp d cc ngun k trn.

8.3. CU TRC C BN CA HTBVRL4. Rle Rle l phn t chnh trong h thng thit b bo v. Phn t Rle nhn mt (X) hoc 1 s u vo (X1, X2,Xn) tng t, bin i v so snh tn hiu ny vi ngng tc ng cho tn hiu ra Y di dng cc xung ri rc vi 2 trng thi i lp: 1 (c xung) v 0 (khng c xung):Rle chia thnh 3 nhn chnh:RLin tTnhKT s

4. Rlea. Rle in t (electromagnetic relay): Nguyn l lm vic da trn nguyn l in t (c cc tip im ng m c kh).- u im: D ch to, r tin- Nhc im: Tiu th cng sut ln, qun tnh cao i khi tc ng khng chun xc, kh m rng ghp ni vi my tnh.b. Rle tnh (static relay): L rle bn dn khng c phn ng. Rle tnh gm cc khi chnh sau:- u: Khng c tip im qun tnh nh v lm vic m du. So vi RL in t, tiu tn t nng lng, kch thc nh gn hn- Nhc: B nh hng nhiu bi mi trng xung quanh, i hi cht lng ngun thao tc cao, i hi s bo qun v chm sc chu o.

4. Rlec. Rle KT s (digital relay):

* Cu trc v nguyn l ca RL sGm cc khi chnh sau:Khi o lng (tn hiu vo): c nhim v o lng cc tr s ca i lng tng t l dng v p (nhn c t pha th cp ca ca my bin dng in v my bin in p) lm bin u vo ca rle. Khi lc tn hiu, ly mu v chuyn i A/D: Sau khi tn hiu qua cc b lc tng t, b ly mu (cht hoc bm i lng tng t theo mt chu k no ), cc tn hiu ny s c chuyn thnh cc tn hiu s v so snh vi i lng chun.Khi x l (dng b vi x l): Sau khi so snh vi i lng chun, b VXL s cho tn hiu ng hoc m cc tip im RL v iu khin my ct (c th lu tr, kt ni vi my tnh,)Khi u ra (tn hiu ra): gm cc rle v mch iu khin ng ct MC

* Cu trc v nguyn l ca RL s

* u im ca RL KT sChc nng hot ng ca rle s c m rng rt nhiu so vi cc th h rle trc y, d dng m rng kh nng o lng, bin di tn hiu, so snh v t hp lgc trong cu trc ca rle. C th kt hp nhiu nguyn l pht hin s c v bo v trong mt h thng rle. Ngoi chc nng bo v v cnh bo, rle s hin i cn c th thc hin nhiu nhim v quan trng khc nh: ghp ni cc thng s vn hnh v s c; xc nh v tr s c; thc hin lin ng vi thit b bo v v t ng ca cc phn t ln cn; ng tr li my ct;... D dng ghp ni vi nhau v vi cc thit b bo v, t ng, thng tin v o lng khc trong h thng; d ghp ni vi h thng my tnh.Thng s ca bo v c th chnh nh n gin vi chnh xc cao v d dng thc hin vic chnh nh thng s t xa hoc chnh nh t ng theo nguyn l thch nghi. Cng sut tiu th nh, kch thc gn nh.Gi thnh tng i tnh theo tng quan gia chi ph v chc nng ca h thng bo v k thut s r hn cc h thng rle in c thng thng.

* K hiu cc phn t v chc nng bo v theo ASNI

* K hiu cc phn t v chc nng bo v theo ASNI

8.5. S NI BI VI RLBI- S sao :- S sao khuyt:- S hiu 2 dng pha:+ N(3):+ N(2) (A v C):+ N(2) (A v B):

8.6. BO V QU DNG (overcurrent protection) 1. Nguyn l tc ngBVQD l bo v tc ng khi gi tr dng in chy qua bo v IBV vt qu ngng no IK: IBV IKNh vy m bo tnh chn lc, dng IK ca bo v c th thc hin theo 2 cch (Xt v d mch hnh tia nh hnh v): BV t cng xa ngun c thi gian tc ng cng ln BVQD c thi gian BV t cng xa ngun c IK cng nh BVQD ct nhanh

8.6. BO V QU DNG tip)2. BVQD c thi gian I> (51)L loi BVQD m bo tnh chn lc bng cch chnh nh thi gian tc nga. Dng khi ng IK (pick-up current):IKR c xc nh nh sau: IK c chn theo cc iu kin sau:Trong : - Kat: h s an ton, tnh n kh nng tc ng thiu chnh xc ca BV. Thng ly: Kat 1,1 i vi rle tnh v rle s Kat 1,2 i vi rle in c Km= 2-4: h s m my ca cc ph ti C c dng in chy qua ch t BV : h s tr v (vi ITV = Kat.Km.Ilvmax), KTV 1 vi RL tnh v RL s KTV = 0,85 0,9 i vi RL in c. - Ilvmax: dng in lm vic ln nht c th chy qua bo v

2. BVQD c thi gian I> (51) th c trng chn dng khi ng ca BV qu dng c thi gian:

b. nhy (sensitivity):c nh gi bi knh:Quy nh: - Bo v chnh: knh = 1,52 - Bo v d phng: knh= 1,21,5.

2. BVQD c thi gian I> (51)c. c tnh thi gian:c tnh c lp c tnh ph thuc

c. c tnh thi gian ca BV 51:Phi hp c tuyn thi gian ca bo v qu dng trong mng in hnh tia (a), cho c tuyn c lp (b) v c tuyn ph thuc (c)

L (km)Thng t = 0,25 0,6 sec

c. c tnh thi gian ca BV 51:t = tMC (n-1) + st.t(n-1) + tqt + tdt tMC (n-1): thi gian tc ng ca MC BV trc st: tng gi tr sai s v thi gianca BV trc v bn thn BV ang xt;(RL in t st = 0,1s; RL s 0,03 0,05s) t(n-1): thi gian tc ng ca bo v trc tqt: sai s do qun tnh, thng tqt = 0,03 0,1s tdt: thi gian d tr, tdt = 0,06 0,2sV vy, trong chnh nh RL thng ly: t = 0,25 0,6 sec

d. u nhc im v phm vi p dng ca BV51u im: Ch to, lp t v thc hin BV n gin, gi thnh rNhc: Thc hin m bo tnh chn lc theo nguyn tc chn thi gian tng dn tng cp t (cp chn lc v thi gian), cng pha gn ngun ti gian tc ng cng ln do kh m bo c tnh tc ng nhanh.Phm vi p dng: Dng lm BV chnh trong cc mng in c mt ngun cp n 35kV (mng cung cp). i vi mng in p cao hn ch c dng lm BV d phng.

3. BVQD ct nhanh I>> (50)L loi BVQD m bo tnh chn lc bng cch chn dng khi ng ln hn dng ngn mch ln nht qua ch t bo v khi h hng ngoi phn t c bo v. a. Dng khi ng IK (pick-up current):IK c xc nh nh sau: IK = Kat.INng max

3. BVQD ct nhanh I>> (50)b. nhy (sensitivity):c. u nhc im v phm vi ng dng ca BV50u im: Ch to, lp t v thc hin BV n gin, gi thnh r lm vic tc thi (hoc tr rt nh c 0,1s)Nhc: Khng bo v c ton b i tng, khi NM cui phn t, BVCN khng tc ng. Hn na vng BVCN LCNc th thay i nhiu khi NM h thng thay iPhm vi p dng: Dng BV cc mng in c mt ngun cp n 35kV (mng cung cp). Khng m bo c tnh chn lc trong li in phc tp, c nhiu ngun cp.

Bi tp v dV d 1: Tnh ton BVQD c thi gian cho ng dy 22kV?Bit Ilvmax = 357A; Km = 1,6; Kat = 1,2; dng NM cui ng dy IN = 1,32kA Bi gii- Cn c vo dng in Ilvmax = 357A, ta chn BI c I1 = 400A cn I2 = 5A- Gi thit BI c u vi RL theo hnh sao khuyt, nn ks = 1.Dng RL s, nn chn Kv = 1Do :Chn RLQD 51 c dng 9A. Vy cn chnh nh dng K ca BV:- KT nhy:

V d 2: Tnh ton BVQD c thi gian cho mng in 10kV trong 2 TH: a. Dng RL s vi c tnh thi gian c lp;b. Dng RL s vi c tnh thi gian ph thuc.Bit: - H s: Km = 1,6; Kat = 1,2; thi gian tc ng ca BV1 t1 = 0,4s - Dng lm vic v dng ngn mch trn cc on ng dy:- Gi thit: tqt = tdt = tMC = 0,1s; st = 0,08s

Bi giiXc nh dng in chy trn cc on dy:Dng qua BV1: Ilv1 = I1 = 83A;Dng qua BV2: Ilv2 = I1+I1 = 83+76= 159ADng qua BV3: Ilv3 = I3 = Ilv2+I2= 159+167 = 326A

Bi gii (tip)Cn c dng lm vic chy trn cc on dy, chn cc BI u sao khuyt:BI1: n1I = 100/1; BI2: n2I = 200/1; BI3: n3I = 400/1; ks = 11. Tnh ton cho BV1:

8.7 BO V SO LCH 87 (Differential protection)1. Khi qut chung bo v cc phn t quan trng trong h thng in, cn m bo yu cu ct nhanh. Bo v qu dng ct nhanh c th m bo c yu cu ny, nhng li ch c th bo v c nhng vng nht nh trong phm vi c phn cng bo v do dng in NM c nhng gi tr khc nhau (Loi NM v v tr NM). BVSL c th khc phc c cc iu k trn: m bo tc ng trong vng c phn cng bo v v khng tc ng khi c NM ngoi vng.Theo nguyn l lm vic, BVSL c chia thnh 2 loi: BV so lch dc v so lch ngang.Bo v so lch dc ch yu dng bo v cc my in nh: MBA; MP v ng c in. Ngoi ra cng c dng bo v cc ng dy c chiu di ngn v thanh ci.Bo v so lch ngang dng bo v cc ng dy v cc cun dy trong my in song song

2. Nguyn l tc nga. Bo v so lch dcBo v so lch dc l loi bo v lm vic da trn nguyn tc so snh trc tip dng in (k c gc pha ca dng in) hai u ca phn t c bo v. Nu s so snh ny sai khc tr s nh trc th bo v s tc ng ct phn t c phn cng bo v ra khi mng in. ~AB Vng bo vN ~NIT1IT287IBI1BI2MC1MC2IS1IS2

2. Nguyn l tc ngb. Bo v so lch ngangBVSL ngang da vo vic so snh dng in ca 2 hay nhiu nhnh song song. Nu s sai khc vt qu mt gi tr nh trc BV s tc ng ct phn t b s c ra khi mng.

3. Tnh ton bo v so lcha. Dng khi ng IK: m bo cho bo v so lch lm vic ng khi ngn mch trong vng bo v xc nh, dng khi ng ca rle cn phi chnh nh trch khi tr s tnh ton ca dng khng cn bng tnh ton ln nht tng ng vi dng ngn mch ngoi cc i Ikcbttmax: IK = IK = Kat.IkcbttmaxTrong :Ikcbttmax= fimax.Kn.Kkck.INng max Vi: fimax - sai s ln nht cho php ca BI, fimax= 10% Kn - h s ng nht ca cc BI, thng Kn= 0 1Kn = 0 khi cc BI hon ton ging nhau v dng in qua cun s cp ca chng bng nhau; Kn = 1 khi cc BI khc nhau nhiu nht. Kkck- h s k n thnh phn khng chu k ca dng in ngn mch:Kkck = 1 i vi cc BI c bo ho t nhanh; Kkck = 2 i vi cc BI khc. INng max- thnh phn chu k ca dng in ngn mch ngoi ln nht. Ch : Ring i vi MBA: Ngoi nhng yu t k trn, Ikcb cn ph thuc vo sai s do iu chnh in p sU (thng sU = 10%) v sai s do s chnh lch

3. Tnh ton bo v so lch INmin- dng NM nh nht khi c ngn mch trc tip trong vng bo v b. nhy yu cu:gia dng th cp hai pha MBA s2i. gim bt s chnh lch v pha ca hai dng in ny, s ni cc BI phi chn i ngc vi cc t u dy ca MBA. V d: MBA c t u dy Y-, th s ni cc BI phi chn l -Y. Do vy dng KCB c xc nh nh sau:Ikcbttmax = (Kkck.Kn.fimax + sU + s2i)INng maxTrong : s2i l sai s tng i do s chnh lch cc dng in th cp ca cc BI. Xc nh nh sau:

8.8. BO V KHONG CCH (21)1. Nguyn l tc ng v phm vi p dngBo v khong cch l loi bo v tc ng da vo vic o tng tr trong phm vi bo v, nu thy tng tr o c nh hn hoc bng vi tng tr nh trc n s tc ng ct phn t h hng ra khi mng in. V th bo v khong cch l loi bo v dng rle tng tr. ZS ZK BV s tc ngBo v khong cch thng c dng bo v li in phc tp nhiu ngun cp vi hnh dng bt k. c bit dng tt cho cc ng dy ti in.

2. Tnh ton bo v khong ccha. i vi BVKC lm vic khng thi gian:ZK = K.ZD = (0,80,85).ZD K - h s k n nh hng ca Rhq ti ch NM; sai s ca BI, BU v cc sai s nh hng khc, (ly K= 0,80,85). ZD - tng tr ng dy c bo vV d:

2. Tnh ton bo v khong cchb. i vi BVKC lm vic c thi gian:BVKC dng bo v ng dy ti in thng c nhiu vng tc ng v do m bo tin cy v tnh chn lc BVKC cng c nhiu cp thi gian bo v khc nhau.c tnh thi gian ca bo v khong cch thng c dng c lp (dng bc thang) v vic chn thi gian lm vic cho cc bo v ngc vi c tnh thi gian ca BV 51. chnh lch v thi gian lm vic gia cc vng (cp) bo v lin k nhau t = 0,3 0,5s. Vy: Vi BV1 c ZK1 v thi gian lm vic t1 Vi BV2 c ZK2 v thi gian lm vic t2 = t1 + t Vic chn cc i lng ny nh sau

2. Tnh ton bo v khong cchV d: Chn ZK v thi gian lm vic ca 3 BV21 bo v ng dy c s nh hnh v:

3. nhy bo v khong cchVi:ZD l tng tr ng dy cn bo v;ZIK l tng tr khi ng ca BVKC. Quy nh: knhyc 1,2

4. Bi tp v d Tnh ton BVKC cho ng dy 110kV: ~ABCBit: Tng tr n v: z0 = 0,37/km;Dng in lm vic chy trn ng dy: Ilv = 450A;Thi gian: tc ng ca BVA l 0,03s v t = 0,5sCc h s: kat = 1,2; kmm = 1,65Bi gii:1. Chn cc BI v BU: Cn c vo dng in lm vic v in p ca mng, ta chn c cc BI v BU c t s bin i nh sau:nI = 600/5 = 120; nU = 110.103/100 = 11002. Xc nh tng tr ca cc on dy:ZAB = z0L1 = 0,38.45 = 16,65; ZBC = z0L2 = 0,38.86 = 31,82; L1 = 45kmL2 = 86km

4. Bi tp v d (tip)3. Xc nh tng tr khi ng ca cc BV:BVA: - Vng 1: ZIA = K.ZAB = 0,8.16,65 = 13,32 - Vng 2:ZIIA= 0,8(ZAB+ZIB) = 0,8(ZAB+0,8ZBC) = 0,8(16,65+0,8.31,82) = 33,68

BVB: ZIB = K.ZBC = 0,8.31,82 = 25,464. Thi gian tc ng ca cc BV:tIA = 0,03s; tIIA = tIA + t = 0,53s; tIB = 0,03s5. Kim tra nhy ca cc BV:

1. S BV ng dy v thanh cia. BV ng dy:Thermal overloadProtection against overheating due to overload currents in conductors under steady state conditions is provided by the thermal overload protection function (ANSI 49RMS), which estimates temperature buildup according to the current measurement.Phase-to-phase short circuits - Phase overcurrent protection (ANSI 51) may be used to clear the fault, the time delay being set to provide discrimination. A distant 2-phase fault creates a low level of overcurrent and an unbalance; a negative sequence / unbalance protection function (ANSI 46) is used to complete the basic protection function (fig. 1).

a. BV ng dy:- To reduce fault clearance time, a percentage-based differential protection function (ANSI 87L) may be used. It is activated when the differential current is equal to more than a certain percentage of the through current. There is a relay at either end of the link and information is exchanged by the relays via a pilot (fig. 2).Phase-to-earth short circuitsTime-delayed overcurrent protection (ANSI 51N) may be used to clear faults with a high degree of accuracy (fig. 1).For long feeders though, with high capacitive current, the directional earth fault protection function (ANSI 67N) allows the current threshold to be set lower than the capacitive current in the cable as long as system earthing is via a resistive neutral.

a. BV ng dy:

b. BV thanh ci: Phase-to-phase and phase-to-earth faults- Overcurrent protectionThe use of time-based discrimination with the overcurrent (ANSI 51) and earth fault (ANSI 51N) protection functions may quickly result in excessive fault clearing time due to the number of levels of discrimination.In the example (fig.1), protection unit B trips in 0.4 s when there is a busbar fault at point 1; when a busbar fault occurs at point 2, protection unit A trips in 0.7s, since the discrimination interval is set to 0.3 s. The use of logic discrimination (fig. 2) with overcurrent protection provides a simple solution for busbar protection. A fault at point 3 is detected by protection unit B, which sends a blocking signal to protection unit A. Protection unit B trips after 0.4 s.However, a fault at point 4 is only detected by protection unit A, which trips after 0.1 s; with backup protection provided if necessary in 0.7 s.

- Differential protectionDifferential protection (ANSI 87B) is based on the vector sum of the current entering and leaving the busbars for each phase. When the busbars are fault-free, the sum is equal to zero, but when there is a fault on the busbars, the sum is not zero and the busbar supply circuit breakers are tripped. This type of protection is sensitive, fast and selective.+ With percentage-based, low impedance differential protection, the difference is calculated directly in the relay. The threshold setting is proportional to the through current and CTs with different ratios may be used. However, the system becomes complicated when the number of inputs increases.+ With high impedance differential protection (fig. 3), the difference is calculated in the cables, and a stabilization resistor is installed in the differential circuit. The CTs are sized to account for saturation according to a rule given by the protection relay manufacturer. The threshold setting is approximately 0.5 CT In and it is necessary to use CTs with the same ratings.

Load shedding functionThe load shedding function is used when a shortage of available power in comparisonto the load demand causes an abnormal drop in voltage and frequency:certain consumer loads are disconnected according to a preset scenario,called a load shedding plan, in order to recover the required power balance.Different load shedding criteria may be chosen:+ undervoltage (ANSI 27),+ underfrequency (ANSI 81L),+ rate of change of frequency (ANSI 81R).

Breaker failureThe breaker failure function (ANSI 50BF) provides backup when a faulty breaker fails to trip after it has been sent a trip order: the adjacent incoming circuit breakers are tripped. The example (fig. 1) shows that when a fault occurs at point 1 and the breaker that has been sent the trip order fails, the breaker failure protection function is faster than action by upstream protection time-based discrimination: 0.6 s instead of 0.7 s.

Examples of applications

Chng 9. BV QU IN P KH QUYN9.1. ST V QU TRNH PHNG IN ST1. St: L s phng in tia la trong kh quyn gia cc m my tch in tri du hoc gia cc m my vi t. Khi bo v chng st cho ngi, cc cng trnh v thit b trn mt t chng ta cn quan tm n s phng in gia cc m my v t.2. S hnh thnh st: S hnh thnh st gn lin vi s hnh thnh cc m my ging. Cc m my ging to thnh do cc lung khngkhis nng m t mt t bc ln i vo vng nhit m, hi nc ngng t thnh cc tinh th bng. Cc m my mang in l do kt qu ca cc lung khng kh mnh lit tch ri nhau to ra cc in tch tri du v tp trung chng trong cc phn khc nhau ca m my. Cc kt qu quan trc cho thy, 80% phn di ca my c cc tnh m, cn phn trn ca m my thng tch cc in tch dng.

9.1. ST V QU TRNH PHNG IN ST3. Qu trnh phng in ca stPhn di cc m my ging c tch in m, do cm ng trn mt t nhng in tch dng tng ng v to nn mt t in khng kh khng l. Theo tch lu cc in tch m ca m my, cng in trng ca t my-t s tng dn ln v nu ti ch no cng in trng t ti tr s ti hn 25 30 KV/cm th khng kh s b ion ho to thnh dng plasma v bt u tr nn dn in, m u cho qu trnh phng in ca st.Phng in st c th chia lm 3 giai on chnh:Phng in tin oPhng in ngc (phng in ch yu)Kt thc qu trnh phng in Cc giai on phng in c th hnh dung qua dng in st bin thin theo thi gian nh hnh v (trang bn).

3. Qu trnh phng in ca st G phng in ngc v = 6.104 105 km/s Kt thc P

9.2. THAM S CA PHNG IN STiS, KAt, ss sISDng in st theo thi gianIS/2 Dng in st c ghi li bi cc my hin sng cc nhanh c dng ng hnh v. Hai tham s quan trng nht ca phng in st l bin dng in st IS v dc u sng a. 1. Bin dng in stKt qu o lng cho thy bin st IS bin thin trong phm vi rng t vi kA n hng trm kA v c phn b theo quy lut thc nghim: Hay:IS: Bin dng in st, kAvI: Xc sut xut hin st c bin IS Vng ng bng: Vng trung du v min ni:Hay:

9.2. THAM S CA PHNG IN ST 2. dc u sngTrong trng hp tng qut, dc u sng a c nh ngha l o hm ca dng in st theo thi gian:Hay: Khi tnh ton, u sng dng in st thng c thay bng ng thng xin gc c dc trung bnh: Xc sut xut hin dc u sng (xc nh theo thc nghim): Cho vng ng bng: Cho min ni:Hay:

9.2. THAM S CA PHNG IN STTrong tnh ton c khi cn phi ng thi xt n c hai yu t: Bin dng in st v dc u sng, ta dng xc sut phi hp: i vi vng ng bng:

i vi vng min ni:

9.3. CNG HOT NG CA STCng hot ng ca st ti cc vng lnh th (hoc kh hu) c th c biu th thng qua 2 i lng nngs v mS. S ngy st trong nm nngs Theo ti KC.03.07 nc ta c: nngs = 100; nngsmax = 114 Mt st mS (l s ln c st nh trn 1km2 din tch ng vi 1 ngy c st).Thng mS = 0,1 0,15.Vy s ln st nh trn din tch 1km2 mt t trong 1 nm s l:Nj =mS nngs = (0,1 0,15) nngs

Vng lnh thNngs (ngy/nm)Vng xch o100 150Vng kh hu nhit i60 100Vng kh hu n i30 50Vng kh hu hn i< 5

9.4. TC HI CA ST V ND C BN BO V CS1. Tc hi ca stKhi st nh trc tipKhi st nh gin tip2. ND c bn bo v chng stBo v chng st nh trc tipBo v chng st lan truyn v cm ng

9.5. B.V CHNG ST NH TRC TIP 1. Khi qut chung hn ch thit hi v ngi v ca do st nh trc tip c nhiu bin php ngy cng hon thin nhng u da vo nguyn l c in do Franklin pht minh ra vo nm 1752, l: dng vt thu st (kim thu st, dy thu st,...) t cao hn vt cn bo v ri ni vi h thng ni t c in tr nh bng cc dy (hoc thanh) dn kim loi c tit din hp l tn dng in st. Mc ch dng cc vt t cao hn cng trnh, thit b l khi xut hin hin my ging, cc vt thu ny s tp trung in tch t mt t, to nn mt cng in trng ln gia vt thu st v my s nh hng phng in v pha mnh to nn mt khng gian an ton cho cng trnh, thit b cn bo v.

Nh vy, BVCS nh trc tip th HTCS s c 3 b phn:

1. Khi qut chung (tip)Ct thu st thng dng bo v cc cng trnh, thit b, nh xng chng st nh thng, ngoi ra ngi ta cn c th dng phi hp vi dy chng st. i vi cc ng dy ti in trn khng dng DCS. bo v chng st cc ng dy ti in, nn treo dy chng st trn ton b tuyn ng dy l tt nht trong vic bo m vn hnh an ton v lin tc cung cp in nhng lm nh vy rt tn km.Trong thc t, tu theo tm quan trng ca ng dy m ngi ta c th b tr dy chng st trn ton tuyn hay khng: Thng cc ng dy c in p 110 KV tr ln c bo v trn ton tuyn ng thi c phi hp vi khe h phng in, chng st ng hoc tng s lng bt s nhng ni hay b st nh, ct vt cao v ch giao cho vi ng dy khc hay nhng on ni vi trm. Cc ng dy in p n 35KV t c b tr bo v trn ton tuyn m bo v trn cc on hay b st nh v on 1 2 km trc khi ni vi trm bin p.

2. Phm vi bo v ca ct thu st (Franklin)a. Nguyn l chung:MFXtCt TSin cc (u tia tin o)M hnh xc nh phm vi bo v ca ct TS:Tm KL Khong khng gian gn ct thu st m vt c bo v t trong , rt t kh nng b st nh gi l vng hay phm vi bo v ca ct thu st.

hr2h/3Phm vi bo v ca 1 ct thu st theo thc nghimTrn c s nghin cu cc m hnh, ngi ta thy rng phm vi bo v ca mt ct thu st c gii hn bi hnh nn trn xoay c ng sinh gy khc cao 2h/3 (hv). Bn knh bo v ca ct thu st rx bo v vt cao hx c xc nh bi cng thc sau: P = 1 khi h 30m

b. Phm vi bo v ca 1 ct thu st

c. Phm vi bo v ca 2 ct thu st PVBV ca hai ct thu st cao bng nhau: Phm vi bo v ca hai ct thu st c kch thc ln hn nhiu so vi tng sPVBV ca hai ct thu st n nu hai ct t cch nhau mt khong a < 7h.

Hai ct thu st c cao khc nhau:

d. Phm vi bo v ca nhiu ct thu st D 8haD 8ha Phm vi bo v ca 4 ct thu stPhm vi bo v ca 3 ct thu st

3. Phm vi bo v ca dy thu sthx h 2h/30,6h1,2h0,2hhaPhm vi bo v ca mt ct thu st 2bx1,2h0,6h1. Phm vi bo v ca 1 dy thu st

2. Phm vi bo v ca 2 dy thu st

Lu :DTSah = a/41. V treo cao trung bnh ca dy dn thng ln hn 2/3 treo cao ca dy thu st (hx > 2h/3) nn c th khng cn cp ti phm vi bo v m biu th bng gc bo v .C th tnh ton c tr s ca gc l 310 (tg = 0,6) Thc t ng dy s c bo v khi: 200 < gh < 300.

2. Dy chng st cng cao, phm vi bo v cng ln. Tuy nhin xc sut nh vng v qua dy chng st li cao. Theo kinh nghim vn hnh, xc sut ny c xc nh nh sau: l gc bo v; h l chiu cao dy thu st

9.6. MT S U THU ST TH H MIHin nay, ngi ta ci tin kim thu st nhn kiu Franklin thnh cc loi chnh sau:u thu phng in smu thu Laseru thu dng cht phng x (b cm s dng nm 95)3.4.1. u thu phng in smXut hin vo khong nhng nm 60-70 ca th k 20. Nguyn l chung ca loi ny nh sau: Khi tia tin o ca m my ging xut pht hng v pha mt t th u thu phng in sm c in tch cm ng v c mt in trng tch lu mt nng lng trong b phn ion ho.

1. u thu phng in smKhi tia tin o xung gn, nng lng trong b phn ion ho tng nhanh v t ngt. B phn ion ho gii phng nng lng to ra nhiu ion v pht trin thnh tia m ng i ln ch ng n tia tin o ca m my ging. Nh PVBV c m rng.Ct TSPVBV ct TS PVBV ca u phng in sm

1. u thu phng in smThuc loi ny c cc u thu:Prevercton-21. H thng thu st trung tm2. H thng cc in cc pha trn3. Hp bo v bng ng v thit b to ion4. H thng cc in cc pha diV mt s loi u thu khc gii thiu sau y:

CC LOI U THU ST DO INDELEC CH TOprevec1 Prevec2 Prevec3.40

CC LOI U THU ST DO INDELEC CH TO prevects3.40 prevects2.25prevec4.50

1. u thu phng in smha - chiu cao hiu dng ca u thu st D = 20, 45, 60m tu theo cp cp BV (I, II hay III)L = V (m/s).T (s) = 106. TVi T l li v thi gian ca tng loi u TS. V d: Loi S4.50 c T = 50; Loi TS2.25 c T = 25,

2. u thu LaserLoi u thu ny do k s Leonard v Ball thit k, th nghim vo nhng nm 70 th k 20.Khi c ging st, u thu Laser c kh nng to ra mt dng ion hng tia tin o t m may ging t cao hng nghn mt i xung.Loi u ny tuy c hiu qu thu st cao, nhng gi thnh cao v b phn laser d b hng khi st nh vo.Ngoi cc ci tin u thu bo v chng st nh trc tip nh nu, hin nay ngi ta cn s dng bin php khc l dng dn tiu hu st (dng thit b chng st bng cch trung ho ion): Khi xut hin my ging n gn, mi nhn ca thit b trung ho ion phng ln m my trung ho in tch ca m my ny, lm gim cng in trng ca m my ngn khng cho m my ging hnh thnh tia tin o phng xung cng trnh.Gi thnh: Truyn thng X; Thu st ci tin 2X; Dn tiu hu 3X.

9.7. CC YU CU I VI BVCS NH TRC TIPCng trnh cn BVCS phi nm trong PVBV ca HT thu st.H thng ni t phi c in tr tn nh.Dy dn (thanh dn) ni h thng thu st vi h thng ni t phi c tit din ln tho mn iu kin n nh nhit khi c dng st chy qua v n cng cn c bo v chng an mn.Khong cch t h thng thu st n cc cng trnh cng nh khong cch trong t ca HT ni t chng st v ni t an ton phi ln khng gy phng in ngc.C th t u thu st trn cc kt cu ca cc cng trnh cn bo v v HTN AT dng chung vi HTNCS khi tha mn 2 iu kin:- TBV c cch in xung cao- Vng c in tr sut v in tr ni t nh (

CP BO V THEO TC: 20TCN-46-84Theo tiu chun 20 TCN-46-84 cc cng trnh, nh c phn thnh 3 cp. Khi thit k chng st, ty theo cp ca cng trnh m p dng cc phng thc bo v tng ng.- Cng trnh cp I: l nhng cng trnh, nh trong c ta ra cc loi hi, kh bi chy; ng thi khi kt hp vi khng kh c th to thnh hn hp n trong ch vn hnh bnh thng. i vi cc cng trnh loi ny, phi p dng phng thc bo v ton b, tc ton b cng trnh phi nm trong phm vi bo v ca b phn thu st; ng thi phi t vt thu st c lp hay cch ly vi cng trnh.- Cng trnh cp II: l nhng cng trnh, nh c tnh cht nh cp I nhng ch xy ra n khi c s c hay lm sai quy trnh. Cc kho vt liu d chy, n thuc cng trnh cp II.Cng trnh thuc cp II, phi p dng phng thc bo v ton b v c th dng loi thu st c lp, cch ly hoc t trc tip trn cng trnh. - Cng trnh cp III: l cc cng trnh, nh cn li. Bo v chng st nh trc tip cho cc cng trnh thuc loi ny, yu cu:+ nhng ni tp trung ng ngi: cn phi c bo v ton b;+ nhng ni tp trung t ngi: cho php bo v trng im.Bo v trng im l t b phn thu st nhng ch nh cao, nh ra ca cng trnh nh nc nh, dim mi, chn mi, dim hi, ng khi, ng thng gi, i nc, cc ct cao trn mi.

9.8. BO V CHNG ST LAN TRUYN1. M u bo v chng st lan truyn, ngi ta dng cc thit b chng st (lightning arrester) nh: khe h phng in, chng st ng v chng st van. Cc thit b chng st c ni song song vi thit b cn bo v n sng qu in p kh quyn truyn t cc ng dy vo thit b cn bo v. Khi c sng qu in p, cc thit b chng st s phng in lm gim bin qu in p t ln cch in khng gy hng cch in v do s an ton cho thit b cn bo v.2. Cc yu cu i vi thit b chng st thc hin c nhim v trn, cc thit b chng st phi m bo mt s yu cu sau:

2. Cc yu cu i vi thit b chng st (tip)TB CS phi c c tnh V-S nm di c tnh V-S ca cch in thit b cn bo v.Sau khi cc thit b chng st phng in (lm vic), cn phi m bo in p d nh khng gy nh hng n cch in ca cc phn t trong mng in.Thit b chng st c kh nng dp tt nhanh h quang ca dng in xoay chiu khi ht qu in p h quang b dp tt trc khi bo v rle tc ng m bo tnh cung cp in lin tc.Thit b bo v chng st khng c lm vic vi a s cc loi qu in p ni b (v khi lm vic thit b chng st thng s hng).

3. KHE H PHNG INa. Cu to v nguyn l lm vicKhe h phng in (cn gi l chng st sng) c cu to gm hai in cc kim loi (thng l thp) t cch nhau mt khong s. Mt in cc ni vi mch cn bo v cn cc kia ni vi t. Khe h s gia hai in cc c chn sao cho vi in p bnh thng khng gy phng in nhng khi c qu in p th s gy ra phng in tn dng in st xung t.

in p nh mc, kVS (mm)Bo v chnhBo v ph620401030502212018035140200

3. KHE H PHNG INb. u nhc im v phm vi ng dng u im: n gin, d ch to, gi thnh h Nhc im: - Khng c b phn dp h quang, nn khi lm vic vi dng st ln h quang duy tr lu tr thnh ngn mch, thit b bo v rle s tc ng ct mch khng cn thit. c tnh V-S rt dc nn khng bo v c cc my in c cch in thp nh: my bin p, my pht in, Phm vi ng dng: - Thng c t nhng ni xung yu ca ng dy nh ch giao nhau gia cc ng dy, on ng dy trc khi ni vi trm bin p. - Ch c dng lm bo v ph trong cc s chng st cc phn t h thng in v c s dng lm mt b phn trong cc thit b chng st khc.

4. CHNG ST NG (CS)a. Cu to v nguyn l lm vicCu to: Gm 2 khe h phng in S1 v S2. Trong S2 c t trong ng lm bng vt liu sinh kh nh Phibr Baklt hoc Phiniplt Nguyn l lm vic: Khi xut hin sng qu in p th c 2 khe h phng in S1 v S2 u phng in dn dng in st xung t. Di tc ng ca h quang, cht sinh kh b pht nng v sn sinh ra rt nhiu kh lm cho p sut trong ng tng cao (ti hng chc at) thi tt h quang.

Um, kVS1 (mm)Bo v phi hpBo v c lp615101020152280403512060

4. CHNG ST NG (CS)b. u nhc im v phm vi ng dng u im: Ch to d dng, gi thnh tng i thp Nhc im: Kh nng dp tt h quang hn ch khi dng st ln, h quang khng c dp tt gy ra ngn mch tm thi, thit b bo v rle c th tc ng ct mch khng cn thit. Phm vi ng dng: - Dng bo v cc ng dy ti in khng treo dy chng st. c dng lm bo v ph trong cc s bo v chng st trm bin p.

5. CHNG ST VAN (CSV)SiO2 dy ~ 10-5cm( = 104-106 m)SiC( = 10-2m)a. Cu to: Gm 2 phn t chnh:Chui khe h phng in v chui intr phi tuyn (in tr lm vic) ct trong v s cch in. Chui khe h phng in c lmbng ng. in tr phi tuyn c ch to bngVt liu Vilt.

5. CHNG ST VAN (CSV)b. Nguyn l lm vic: Khi xut hin sng qu in p kh quyn th chui khe h s phng in, dng in st c dn qua cc in tr phi tuyn dn dng in st xung t. in tr phi tuyn c c im khi t in p ln th in tr c tr s rt nh cho dng in qua mt cch d dng, nhng khi in p t nh th in tr c tr s rt ln ngn cn dng in khng cho qua. Hay ni cch khc, chng st van cho dng in ln (khi in p cao) qua nhng ngn cn dng in nh (khi in p thp). T c im ny m n c tn gi l chng st van. c. u nhc im v phm vi ng dng: u im: - Duy in p d tng i n nh khi dng in ln- Dp tt h quang mt cch d dng nh c in tr phi tuyn Nhc im: Ch to phc tp, gi thnh cao Phm vi ng dng: c dng bo v qu in p kh quyn thit b v trm quan trng (c bit l cc trm bin p in lc v cc my pht in).

6. CHNG ST VAN (CSV) TH H MIChng st van c khe h phng in nh nu, hin nay c thay th bi CSV khng c khe h (ch gm chui cc in tr phi tuyn). Vt liu ch to chui in tr phi tuyn trong CSV khng c khe h phng in c thay th bng in tr oxt kim loi MxOy (thng 99,9% ZnO; 0,05% Bi2O3 v 0,05% MnO2) c c tnh V-A hon ton phi tuyn v c kh nng hp th nng lng cao.Khi in p tng, van chng st chuyn ngay t in tr c tr s rt ln 1,5 M sang in tr c tr s rt nh 15 .u im ca cc loi CSV khng khe h l nh, t v, chng m tt, hn ch in p v tin cy cao hn so vi CSV c khe h.u s cpThit b x p sutChui in tr MxOyS cch inu ni tu thot p sut1223456

7. Mt s s nguyn l c bn BVCS TBADng DCS bo v CS nh trc tip trn on 1-2kmCS1: t ti ct u tin t DCS hn ch bin ca sng qu in p truyn vo TBA. Theo quy phm, in tr ni t ca CS ny:+ R 10 khi 103 m;+ R 15 khi > 103 m;CS2: t cui ng dy (ct cui trc khi ni vi MBA) bo v MC ng dy khi n h mch (in p tng cao do phn x ni h mch). Phi chnh nh sao cho, CS2 khng c lm vic khi MC ang ng mchCSV: t ti thanh ci TBA bo v MBA.a. i vi TBA t 35-110kV:

7. S nguyn l bo v chng st TBAc. TBA ni vi cpb. TBA t 110kV tr ln

Chng 10. NNG CAO H S CNG SUT10.1. KHI QUT CHUNG1. Khi nim v h s cng sut:

a. H s cos tc thi:

Xc nh c nh dng c o ti thi im no Cos bin thin theo thi gian nn khng c ngha trong tnh tonb. H s cng sut trung bnh costb: L h s cos trong mt khong thi gian no (1 ca, 1 ngy m, 1 thng,...):

costb dng nh gi mc s dng in tit kim v hp l ca XN

10.1. KHI QUT CHUNGc. H s cng sut cos t nhin: L h s cng sut trung bnh tnh trong mt nm (8760h) khi khng c thit b b. H s cng sut t nhin c dng lm cn c xc nh ph ti tnh ton, nng cao h s cng sut v b cng sut phn khng. i vi CKB c cos thp (cos = 0,50,7), do CKB tiu th cng sut phn khng nhiu nht, chim (6570)%, sau l my bin p.2. B cng sut phn khng trong cc XN cng nghip: L s dng cc thit b b (t b, my b ng b) t song song vi thit b tiu th cng sut phn khng cung cp 1 phn hoc ton b lng cng sut phn khng m thit b ny tiu th.

10.2. NGHA CA VIC NNG CAO COS Lm gim c tn tht in p

Lm gim tn tht cng sut Lm gim tn tht in nng Tng kh nng truyn ti

10.2. NGHA CA VIC NNG CAO COSTM LI: Vic nng cao cos c 2 li ch c bn:1. Li ch to ln v kinh t cho ngnh in v doanh nghip

2. Li ch v k thut: Nng cao cht lng in p

10.3. CC GII PHP NNG CAO COSC hai nhm gii php nng cao cos1. Nhm cc gii php nng cao cos t nhin: L cc gii php khng dng cc thit b b. C cc gii php c bn sau:Thay th cc CKB lm vic non ti bng C c CS nh hnThng xuyn bo dng v nng cao cht lng sa cha CSp xp, s dng hp l cc qu trnh cng ngh ca cc my mc, thit b in.S dng CB thay cho CKBThay th cc MBA lm vic non ti bng MBA c dung lng nh hn.S dng chn lu in t hoc chn lu st t hiu sut cao cho cc chn lu st t thng thng. 2. Nhm cc gii php nng cao cos nhn to: L gii php s dng cc thit b b (t b hoc my b ng b). Cc thit b b pht ra Q cung cp 1 phn hoc ton b nhu cu Q trong XN. Lm nh vy gi l B cng sut phn khng.

So snh kinh t - k thut ca my b v t b

10.4. TNH TON B CNG SUT PHN KHNGB1. Xc nh dung lng b:Qb = P( tg1 - tg2)B2. Xc nh v tr t t b: V l thuyt c th t t pha cao p hay h p hay bt c u ca mng XN.t t b phn tn ti cc ng c l c li nht v mt tn tht in p v in nng. Tuy nhin t t kiu ny chi ph cao v kh khn trong qun l, vn hnh.V vy, t t b pha in p cao hay h p, tp trung hay phn tn n mc no cn phi so snh KT-KT. Qua kinh nghim thc t, nn t t b nh sau:Vi my bm v xng c kh: t t b cnh t phn phiVi XN nh: t tp trung ti thanh ci h p TBA. Ngoi ra vi cc px c C cng sut ln, t c lp nn t ring 1 b t b. Vi XN ln: t t b phn tn ti cc phn xng. B3. Phn b ti u dung lng b:Khi b phn tn, p dng cng thc phn b ti u cng sut nh sau:

10.4. TNH TON B CNG SUT PHN KHNGB3. Phn b ti u dung lng b (tip):Nu mng in XN hnh tia:

Nu mng in XN phn nhnh: Cn bin i cc nhnh song song thnh nhnh tng ng ri p dng cng thc trn. B4. iu khin dung lng b: Bng tay hoc t ngQb1Mng in hnh phn nhnhMng in hnh tia

Mt x nghip c cng sut tng nh sau:S = 100 + j152 KVA. V dTnh ton iu khin dung lng b nng h s cng sut ln 0,65; 0,75 v 0,85.

T cng sut ph ti cho xc nh c h s cng sut ca x nghipTnh c tg = 1,51- Khi yu cu cos1 = 0,65 tnh c tg 1 = 1,17- Khi yu cu cos2 = 0,75 tnh c tg 2 = 0,88- Khi yu cu cos3 = 0,85 tnh c tg 3 = 0,62Li gii

nng h s cng sut ln cos1 th cng sut phn khng cn b ca nhm 1 l:Qb1 = P( tg - tg1) = 100(1,51- 1,17) = 34 KVAr nng h s cng sut ln cos2 th cng sut phn khng cn b ca nhm 2 l:Qb2 = P( tg1 - tg2) = 100( 1,17 0,88) = 29 KVAr nng h s cng sut ln cos3 th cng sut phn khng cn b ca nhm 3 l:Qb3 = P( tg2 - tg3) = 100( 0,88- 0,62) = 26 KVAr

S iu khin dung lng b nng h s cng sut XN ln 0,65; 0,75 v 0,85.

Vn iu chnh t ng dung lng b iu chnh t ng dung lng b c thc hin nh b S6-Q hoc S12-Q hoc b PDCF.

Chng 11. K THUT CHIU SNG11.1. S LC V LCH S CHIU SNGC th coi vic pht hin v ch ng ngn la l bc khi u cho K thut chiu sng.Ngun sng nhn to u tin l ngn nn c s dng t 5000 nm v trc.1669 Newton pht hin nh sng trng l t hp ca nh sng by mu khi cho tia sng mt tri chiu qua lng knh. Nm 1756 M. Lomonosov ln u tin pht hin cc loi t bo th gic v xut hc thuyt ba mu ca nh sng. Th k 19, Maxwell xut L thuyt trng in t thng nht v tin on s tn ti ca sng in t. Nm 1888 Henry Hertz thu c sng in t u tin. Cng c phn tch ph do R.Bunsen v G. Kirchhoff pht trin, nh mn b mt ca nh sng c pht hin.Cui th k 19, Albert Einstein tc gi ca c hc lng t v l thuyt tng i l ngi u tin xut bn cht sng-ht ca nh sng v gii thch nh sng gm v s ht nh mang nng lng l cc phton.Nm 1879, n si t u tin do Thomas Edison ch to t si cc bon, to nn nhit 39000K, hiu qu nh sng 2lm/W, tui th 600 gi. Nm 1908 Siemens s dng si t Vonfram-Nicken.

11.1. S LC V LCH S CHIU SNGNm 1910, n hunh quang ra i, nm 1933 xut hin n ng hunh quang u tin. Nm 1960, cc n halogen kim loi (Metal Halide) do cng ty General electric ch to ra i.Gn y cc ngun sng da trn hin tng pht quang trong cht bn dn c ng dng trong cc it pht quang (LED).T nm 1990 vi s ra i v hon thin ca cc ngun sng mi, ca cc phng php tnh ton v cng c phn mm chiu sng mi, k thut chiu sng chuyn t giai on chiu sng tin nghi sang chiu sng hiu qu tit kim in nng gi tt l chiu sng tin ch.Chiu sng tin ch c ni dung c bn l ti u ha ton b k thut chiu sng t vic s dng ngun sng c hiu qu cao, loi b v thay th cc loi n si t bng n compact, s dng rng ri cc n hunh quang th h mi, iu chnh nh sng theo mc ch v yu cu s dng, s dng c hiu qu ca chiu sng t nhin. Kt qu ca chiu sng tin ch phi t tin nghi nhn tt nht, tit kim in, gp phn bo v mi trng.

11.2. BN CHT SNH SNGnh sng c hai thuc tnh c bn l sng v ht:Sng nh sng l sng in t pht x khi c s chuyn mc nng lng ca cc in t trong cc ngun sng. Trong vt cht nh sng c vn tc: v = c/n (km/s) c-vn tc nh sng trong chn khng, c = 300.000 km/s n-chit sut ca mi trng.Gia tn s f v bc sng lin h bi biu thc: = v/f Tnh cht ht ca nh sng th hin qua tng tc ca nh sng vi mi trng cht. nh sng gm v s cc ht nh mang nng lng gi l phton.Mt ngi l b cm bin quang v cng tinh t v linh hot cm nhn c nh sng trong di bc sng i t 380 n 760 nm (1 nm = 10-9 m).

1.2. BN CHT SNH SNG

6. Trc mt7. Thy tinh th8. im vng9. Vng mc

Tm

Lam

Lc

Vng

Da cam

Hng ngoi

Cc tm

0,358

0,435

0,480

0,575

0,580

0,650

0,760

nh sng nhn thy

11.3. CC I LNG O NH SNGQuang thng /F (lumen/lm): l thng lng nh sng do ngun pht ra trong khng gian. Cng sng I (Candela/cd): c xc nh bng lng nh sng pht ra t ngun sng theo mt phng nht nh. Ph thuc vo hnh dng v tnh i xng ca cho n, cho n c chia thnh 2 loi: loi chm tia hp v loi chm tia rng.

11.3. CC I LNG O NH SNG ri E (lux/lx): ri xc nh mt khu vc sng nh th no khi c chiu sng bng mt ngun sng. N l t s gia quang thng v din tch c chiu sng: E = /S (1lm/m2 = 1lx). V d:

11.3. CC I LNG O NH SNG chi L (cd/m2 hoc cd/cm2): l n tng nh sng m ngi quan st c c khu vc c chiu sng:

chi thng c ngha hn ri khi xc nh cht lng chiu sng.Hiu sut pht quang/Quang hiu K (lm/W): th hin hiu qu ca s chuyn i in nng thnh nh sng. N ng thi l n v o hiu sut ca bng n.Quan h gia ri v chi: Lambert (1728-1777) chng minh quan h gia ri E nhn c trn mt c h s phn x v chi L theo biu thc: E = L

11.3. CC I LNG O NH SNGH s phn x , hp th v xuyn sng nh sng:Nu c mt lng quang thng Fi ti p vo b mt vt liu th c th xy ra cc trng hp sau:Mt phn quang thng ti s phn x t b mt , k hiu F;Mt phn quang thng ti s b vt liu hp th, k hiu F;Mt phn quang thng ti s xuyn qua vt liu, k hiu F.Khi : Fi = F + F + FNu gi: F /Fi = l h s phn x nh sng;F /Fi = l h s hp th nh sng; F /Fi = l h s xuyn sngTh: + + = 1Cc tr s ca , , thay i ty thuc c tnh quang hc ca vt liu (tra trong s tay thit k chiu sng).

11.4. PHN LOI CC HNH THC CHIU SNG

11.5. THIT B CHIU SNG CC LOI N

CC LOI NGUN SNG

PHM VI NG DNGCA MT S LOI N THNG DNG

1. N SI T

1. N SI T

1. N SI Tc im ca n si t: u im:C ch s th hin mu rt cao ( 100) cho php s dng trong chiu sng cht lng cao. Ni trc tip vo li in; kch thc nh; bt sng tc thi v gi thnh thp. Nhc im:Hiu qu nng lng thp, t 10-20lm/W; Pht nng;Tui th thp, ph thuc vo in p: trung bnh 1000h nhng khi U tng 5%Um tui th ch cn 500h.T nm 1960, ngoi kh tr ngi ta cn b sung Halogen (It, Brom) khi vonfram bc hi lng ng trn si t m khng b ngng ng trn thnh bng n cho php t nhit 31000K, hiu qu nh sng t 20-27lm/W tui th trung bnh 2000h.

2. N HUNH QUANG

2. N HUNH QUANGChn lu in t-Tc-te c ni vi bng n nh hnh v.Khi t vo in p, xy ra phng in trong tc-te thanh lng kim bin dng do nhit v tip xc vi in cc kia. Dng in chy qua tc-te v t nng cc in cc ca n. Sau khi xy ra h quang gia cc in cc ca tc-te, thanh lng kim ngui i v "m mch". H mch dn n to nn qu in p cm ng (do chn lu) lm n thp sng. Khi lm vic binh thng chn lu hn ch dng in v n nh phng in.

2. N HUNH QUANGn HQ vi ni vi chn lu in t

11.6. THIT K CHIU SNG CS cc b v chiu sng s c cn cn c vo hon cnh c th quyt nh. y s trnh by cch thit k chiu sng chung. 1. Cc yu cu c bn khi thit k chiu sngNm c cc tiu chun trong chiu chiu sng ri ng u trn b mt cn chiu sngThu thp thng tin, s liu Mt bng x nghip, phn xng, v tr cc my t trn mt bng phn xng;Mt bng v mt ct nh xng thit k xc nh v tr treo n;Nhng c im ca qu trnh cng ngh (lm vic chnh xc, cn phn bit mu sc, v.v...). Cc tiu chun v ri ca cc khu vc lm vic;S liu v ngun in, ngun vt t

2. THIT K CHIU SNG NHNG NI KHNG I HI CHNH XC CAO Bc 1: Cn c vo tnh cht ca i tng cn chiu sng, chn sut ph ti chiu sng p0 (W/m2) thch hp (P.lc chiu sng)Bc 2: Cn c sut chiu sng p0, xc nh tng cng sut cn chiu sng cho khu vc c din tch S (m2):PCS = p0.S , W Bc 3: Chn loi n (n si t hoc n hunh quang), cng sut mi bng n Pd, ri xc nh tng s bng n n cn dng chiu sng cho khu vc:

Bc 4: Cn c vo din tch S ca khu vc cn chiu sng; s bng n n v tnh cht, yu cu ca cng vic b tr n hp l trong khu vc chiu sng.Bc 5: Thit k mng in chiu sng: v s mt bng u dy t bng in n tng bng n; s nguyn l mng in chiu sng v tin hnh chn cc phn t trn s (loi bng in, dy dn, cng tc, ptmt, cu ch bo v,...).

2. THIT K CHIU SNG NHNG NI I HI RI TIU CHUN VI CHNH XC CAOa. Khi dng n si t:Bc 1: Xc nh treo cao nBc 2: Da vo t s L/H hp l(P.lc CS), xc nh khong cch gia hai n k nhau L (m)Bc 3: Cn c vo b tr n trn mt bng, mt ct xc nh h s phn x tng, trn t, tr, (%). Bc 4: Xc nh ch s phng (c kch thc ab): Bc 5: T t, tr, tra bng tm h s s dng KsdBc 6: Xc nh quang thng tnh ton ca n: Trong : K - h s d tr, tra PLE - ri yu cu ca nh xng (lx);S - din tch nh xng (m2);Z - h s tnh ton, Z= 0,81,4;n - s bng n sau khi b tr n trn mt bngBc 7: Tra s tay tm cng sut bng c quang thng F Ftt;Bc 8: Thit k mng cp in chiu sng

2. THIT K CHIU SNG NHNG NI I HI RI TIU CHUN VI CHNH XC CAOb. Khi dng n hunh quang:Giai on 1: Thit k s b (lm nh trng hp chiu sng khi khng i hi chnh xc cao). Giai on 2: Kim tra ri yu cu:Trong : F - quang thng trn n v ngun sng: n - s bng n trong ngun sng; F0 - quang thng mt bng n, lm;L - chiu di ngun sng, m. - tng ri tng i trn im cn kim tra, tr s ei tm c bng cch tra trn th da vo t s v . Tra th trang sau

th xc nh ri theo v