Bi tp di Cao p

Li ni u St l hin tng phng in trong kh quyn gia cc m my v t hay gia cc m my mang cc in tch tri du.St khng nhng mang n hu qu nghim trng i vi con ngi khi b st tc ng m vi cc thit b phn phi cng vy. H thng in l mt b phn ca h thng nng lng bao gm NM - ng dy - TBA v cc h tiu th in. Trong c phn t c s lng ln v kh quan trng l cc TBA, ng dy. Trong qu trnh vn hnh cc phn t ny chu nh hng rt nhiu s tc ng ca thin nhin nh ma, gi, bo v c bit nguy him khi b nh hng ca st. Khi c s c st nh vo TBA, hoc ng dy n s gy h hng cho cc thit b trong trm dn ti vic ngng cung cp in v gy thit hi ln ti nn kinh t quc dn. nng cao mc cung cp in, gim chi ph thit hi v nng cao an ton khi vn hnh chng ta phi tnh ton v b tr bo v chng st cho HT V vy vic thit k bo v chng st l mt vic rt quan trng i vi nhng sinh vin nghnh h thng in.Di s hng dn ca thy Trn Hong Hip th em hon thnh bi tp di Cao p. Tuy nhin em khng th trnh khi nhng sai st, do vy em mong c s ng gp cho em v bi tp di ny.Mt ln na em xin chn thnh cm n thy hng dn cho em hon thnh bi tp di ny.

CHNG IBO V CHNG ST NH TRC TIP TRM BIN P

A. L thuyt chung1.1 t vn : Trm bin p l mt b phn quan trng trong h thng truyn ti v phn phi in.i vi trm bin p 220/110kV th cc thit b in ca trm c t ngoi tri nn khi c st nh trc tip vo trm s xy ra nhng hu qu nng n khng nhng ch lm hng n cc thit b trong trm m cn gy nn nhng hu qu cho nhng ngnh cng nghip khc do b ngng cung cp in . Do vy trm bin p thng c yu cu bo v cao..Hin nay bo v chng st nh trc tip cho trm bin p ngi ta dng h thng ct thu li, dy thu li. Tc dng cu h thng ny l tp trung in tch nh hng cho cc phng in st tp trung vo , to ra khu vc an ton bn di h thng ny.H thng thu st phi gm cc dy tip a dn dng st t kim thu st vo h ni t. nng cao tc dng ca h thng ny th tr s in tr ni t ca b phn thu st phi nh tn dng in mt cch nhanh nht, m bo sao cho khi c dng in st i qua th in p trn b phn thu st s khng ln gy phng in ngc n cc thit b khc gn .Ngoi ra khi thit k h thng bo v chng st nh trc tip vo trm ta cn phi quan tm n cc ch tiu kinh t sao cho hp l v m bo v yu cu v k thut, m thut. 1.2 Yu cu k thut i vi h thng thu st: + Cng trnh cn bo v an ton phi c nm gn trong phm vi bo v ca h thng thu st.H thng ny c th t ngay trn bn thn cng trnh hoc t cch ly ty thuc vo hon cnh v iu kin c th.t h thng thu st trn bn thn cng trnh s tn dng c cao ca phm vi bo v v s gim c cao ca ct thu li. Nhng mc cch in ca trm phi m bo an ton trong iu kin phng in ngc t h thng thu st sang thit b. V t kim thu st trn cc thanh x ca trm th khi c phng in st, dng in st s gy nn mt in p ging trn in tr ni t v trn mt phn in cm ca ct, phn in p ny kh ln v c th gy phng in ngc t h thng thu st n cc phn t mang in trong trm khi m mc cch in khng ln. Do iu kin t ct thu li trn h thng cc thanh x ca trm l mc cch in cao v tr s in tr tn ca b phn ni t nh.i vi trm phn phi c in p t 110kV tr ln c mc cch in kh cao (c th khong cch gia cc thit b ln v di chui s ln) do c th t cc ct thu li trn cc kt cu ca trm v cc kt cu trn c t ct thu li th phi ni t vo h thng ni t ca trm theo ng ngn nht sao cho dng in st khuych tn vo t theo 3 n 4 cc ni t, mt khc mi tr phi c ni t b xung ci thin tr s in tr ni t.Khu yu nht trong trm phn phi ngoi tri in p t 110kV tr ln l cun dy my bin p v vy khi dng ct thu li bo v my bin p th yu cu khong cch gia im ni vo h thng ca ct thu li v im ni vo h thng ni t ca v my bin p l phi ln hn 15m theo ng in.+Tit din cc dy dn dng in st phi ln m bo tnh n nh nhit khi c dng in st chy qua. chng n mn phn dn in cn phi sn hoc trng km v khng nn dng loi dy xon.Cc mi ni dc theo mch in ca h thng thu st phi m bo c tip xc tt ,nu khng ti cc ni ny c th qu nng hoc c phng in tia la.Khi s dng ct n chiu sng lm gi cho ct thu li th cc dy dn in phi c cho vo ng ch v chn trong t.1.3 Phm vi bo v ca ct thu st v b tr ct thu stCt thu st l thit b khng phi trnh st m ngc li dng thu ht phng in st v pha n bng cch s dng cc mi nhn nhn to sau dn dng in st xung t. S dng cc ct thu st vi mc ch l st nh chnh xc vo mt im nh sn trn mt t ch khng phi l vo im bt k no trn cng trnh. Ct thu st to ra mt khong khng gian gn ct thu st (trong c vt cn bo v), t c kh nng b st nh gi l phm vi bo v.1.3.1 Phm vi bo v ca mt ct thu st c lpPhm vi bo v ca mt ct thu st c lp l min c gii hn bi mt ngoi ca hnh chp trn xoay c ng knh xc nh bi phng trnh:

Trong : h: cao ct thu st. hx: cao cn bo v. h-hx = ha: cao hiu dng ct thu st, c xc nh theo tng nhm ct ( ha D/8m ). ( Vi D l ng knh ngoi tip vng trn a gic to bi cc chn ct) rx: bn knh ca phm vi bo v.

d dng v thun tin trong tnh ton thit k thng dng phm vi bo v dng dng n gin ho ng sinh ca hnh chp c dng ng gy khc nh hnh sau:

Hnh 1.1. Phm vi bo v ca mt ct thu st. Bn knh c tnh ton theo cng thc sau:

Nu th

Nu th

Cc cng thc trn ch ng khi ct thu st cao di 30m. Hiu qu ca ct thu st cao trn 30m gim i do cao nh hng ca st gi hng s. C th dng cc cng thc trn tnh ton phm vi bo v nhng phi nhn thm h s hiu chnh v trn hnh v dng honh 0,75h.p v 1,5h.p. Phm vi bo v ca hai hoc nhiu ct thu st th ln hn tng phm vi bo v cc ct n cng li. Nhng cc ct thu li c th phi hp c th khong cch a gia hai ct phi tho mn a 7h (trong h l cao ca ct thu st).1.3.2 Phm vi bo v ca hai hay nhiu ct thu sta. Phm vi bo v ca hai ct thu st c cng cao. Phn bn ngoi khong cch gia hai ct c phm vi bo v ging nh ca mt ct. Phn bn trong c gii hn bi vng cung i qua 3 im l hai nh ct v im c cao h0 - phm vi bo v cao ln nht gia hai ct c xc nh theo cng thc:

Hnh 1.2. Phm vi bo v ca hai ct thu st c cao ging nhau. Khong cch nh nht t bin ca phm vi bo v ti ng ni hai chn ct l r0x v c xc nh nh sau:

Nu th

Nu th

Khi cao ca ct thu st vt qu 30m th c cc hiu chnh h s , trn honh ly cc gi tr 0,75h.p v 1,5h.p; khi h0 ly theo cng thc:

b. Phm vi bo v ca hai ct thu st c cao khc nhau. Trng hp hai ct thu st c cao h1 v h2 khc nhau th vic xc nh phm vi bo v c xc nh nh sau: V phm vi bo v ca ct cao (ct 1) c cao h1 v ct thp (ct 2) c cao h2 ring r. Qua nh ct thp (ct 2) v ng thng ngang gp ng sinh ca phm vi bo v ct cao im 3 im ny c xem l nh ca mt ct thu st gi nh. Ct 2 v ct 3 hnh thnh i ct c cao bng nhau v bng h2 vi khong cch a. Bng cch gi s v tr x c t ct thu li 3 c cao h2. im ny c xem nh nh ca mt ct thu st gi nh. Ta xc nh c cc khong cch gia hai ct c cng cao h2 l a v x nh sau:

Hnh 1.3. Phm vi bo v ca hai ct thu st c cao khc nhau.

Nu th

Nu th

Phn cn li ging phm vi bo v ca hai ct c cao bng nhau h2c. Phm vi bo v ca mt nhm ct thu st ( s ct > 2 ). Khi cng trnh cn bo v chim mt din tch rng ln th ngi ta thng dng mt h thng nhiu ct thu st bo v. xc nh phm vi bo v, ta chia h thng ct thu st thnh tng nhm ba hoc bn ct thu st gn nhau. Bn ngoi din tch ca a gic i qua chn cc ct thu st, phm vi bo v c xc nh nh gia hai ct thu st vi nhau.

Hnh1.4: Phm vi bo v ca nhm ct to thnh tam gic v ch nht.

bo v c mt din tch gii hn bi mt a gic th cao ca ct thu li phi tho mn:

Trong : D l ng knh vng trn ngoi tip a gic to bi cc chn ct. Nhm ct tam gic c ba cnh l a, b,c c:

Vi p l na chu vi: Nhm ct to thnh hnh ch nht:

Vi a, b l di hai cnh hnh ch nht. Chiu cao tc dng ca ct thu st ha phi tho mn iu kin:

B. Tnh ton1.1 B tr ct thu stTheo s kt cu ca trm th ta bit c mt bng m cha bit c th v tr t cc thit b trong trm. Vi thng tin ny ta ch cn b tr ct chng st sao cho cc ct c th bo v c phn din tch mt bng ca trm vi cao hX l c.Vi hx l chiu cao ca trm cn c bo v.i vi trm 220 kV x cao: 16,5m v 11,5m nn chn cao cn bo v hx = 16,5mi vi trm 110 kV x cao: 11m v 7,5m nn chn cao cn bo v hx = 11mDin tch mt bng: l1 =78+136 = 214 m. l2 = 104 min tr sut ca t 80 .mVi kt cu ca trm ta c phng n b tr cc ct c khong cch gia cc ct nh hnh v sau.

Trm 220 kV ng knh ca vng trn qua 4 nh ca cc nhm ct 1, 2, 6, 7 l:

Chiu cao cn bo v hx = 16,5 m.Nn chiu cao ti thiu ca ct chng st l:

Trm 110 kV ng knh ca vng trn qua 4 nh ca cc nhm ct 3, 4, 9, 8 l:

Chiu cao cn bo v hx = 11 m.Nn chiu cao ti thiu ca ct chng st l:

Vy phng n 1 c:Ct 1, 2, 3, 6, 7, 8, 11, 12, 13 cao 25 m. Cn cc ct cn li cao 22 m

Ct123456789101112131415

h(m)25252522 22 25252522 22 25252522 22

Tng 357 m

1.2 Tnh ton v v phm vi bo v ca cc ct thu stPhm vi bo v ca cc ct pha 220kv cao 25m.+ Bn knh bo v ca tng ct cao 16,5m:

Do hx=16,5 m

Bn knh bo v ca tng ct c lp pha 220kv l:

==6,6( m)+ Bn knh bo v ca cc cp ct Tnh ton cho cp ct (1,6)Bn knh bo v ca tng ct c lp l 6,6m.Chiu cao ln nht ca khu vc bo v gia hai ct (1,6) cao hx=16,5m

= (m)

Do hx=16,5 mBn knh bo v ca khu vc bo v gia hai ct thu li (1,6) cao hx=16,5m l:

==0,8( m)

cao ln nht ca khu vc bo v gia hai ct thu li cao hX = 16,5 m gm nhm hai ct (1, 2) :

=Bn knh ca khu vc bo v gia 2 ct thu st (1,2) hX = 16,5m

Do hX = 16,5 > 2/3h0 = 13,52 nn:

==2,2( m)

Tnh ton tng t vi cc cp ct khc ta c bng sau:NHM CT CNH NHAU

220kV

Cp ct

cao ct(m)a(m)h0(m)2/3*h0(m)r0x(m)

hx=16.5m

1-6255217,611,70,8

3-8255217,611,70,8

6-11255217,611,70,8

8-13255217,611,70,8

1-2253912,912,92,2

2-3253912,912,92,2

11-12253912,912,92,2

12-13253912,912,92,2

Phm vi bo v ca cc ct pha 110kv cao 22m.+ Bn knh bo v ca tng ct cao 11m:

Do hx=11 mBn knh bo v ca tng ct c lp pha 110kv l:

==12,4( m)+ Bn knh bo v ca cc cp ct Tnh ton cho cp ct 4-5Bn knh bo v ca tng ct l 12,4m.Chiu cao ln nht ca khu vc bo v gia hai ct thu li (4-5) cao hx=11m.

= (m)

Do hx=11m

Bn knh bo v ca khu vc bo v gia hai ct thu li (4-5) cao hx=11m l:

=1( m).Tnh ton cho cp ct 5-10Bn knh bo v ca tng ct l 12,4m.Chiu cao ln nht ca khu vc bo v gia hai ct thu li (5-10) cao hx=11m.

=( m)

Do hx=11 mBn knh bo v ca khu vc bo v gia hai ct thu li (5-10) cao hx=11m l:

==2,7( m).

Tnh ton tng t vi cc cp ct khc ta c bng sau:

NHM CT CNH NHAU

110kV

Cp ct

cao ct(m)a(m)h0(m)2/3*h0(m)r0x(m)

hx=11m

4-5226812,38,21

5-1023,57014,69,72,7

10-1523,57014,69,72,7

14-15226812,38,21

Bn knh bo v cho cp ct c cao khc nhau:

Tnh ton cho cp ct 3-4:Ct 3 c cao 26m,bn knh bo v ca ct 3 l: r03=6,6 mCt 4 c cao 20m, bn knh bo v ca ct 4 l: r04=12,4 mBn knh bo v ca ct 3 cho ct 4.

Do hx=20m

=>==2,3( m).Khong cch t ct 4 ti ct gi tng l:a=a-rx=68 2,3=65,7mChiu cao ln nht ca khu vc bo v gia ct 4 v ct gi tng

h0=h4 -22-12,6 mBn knh bo v gia hai ct l:

Do hx=11 m

=>==1,2( m).

Tnh ton tng t i vi cp ct 8-9,13-14 ta c bng sau:

NHM CT C CAO KHC NHAU

Cp ct

cao ct(m)a(m)x(m)a'(m)ho(m)r0x(m)

hx=11 m

8-926-20682,365,712,61,2

13-1426-20682,365,712,61,2

Qua tnh ton cho thy phng n trn tha mn yu cu k thut ra,tha mn iu kin bo v cho trm bin p Bi tp di mn Cao p

15 Sinh vin : Anh Tn- Lp 2H5B -Trng i Hc in Lc

CHNG II :TNH TON NI T CHO TRM BIN P.2.1 YU CU K THUT KHI NI T TRM BIN P.Ni t l em cc b phn bng kim loi c nguy c b tip xc vi dng in(h hng cch in) ni vi h thng ni t. Nhim v ca ni t l tn dng in xung t dm bo cho in th trn vt ni t c tr s b. H thng ni t l mt phn quan trng trong vic bo v qu in p. Ty theo nhim v v hiu qu m h thng ni t c chia l ba loi. Ni t lm vic. Ni t an ton. Ni t chng st.Ni t lm vic. Nhim v chnh l m bo s l vic bnh thng ca thit b, hoc mt s b phn ca thit b yu cu phi lm vic ch lm vic c quy nh sn.+ Ni t im trung tnh my bin p.+ H thng in c im trung tnh trc tip ni t.+ Ni t ca my bin p o lng v cc khng in dng trong b ngang trn cc ng dy cao p truyn ti in.Ni t an ton.C nhim v m bo an ton cho con ngi khi cch in b h hng. Thc hin ni t an ton bng cch ni t cc b phn kim loi khng mang in nh v my, thng du my bin p, cc gi kim loi. Khi cch in b h hng do lo ha th trn cc b phn kim loi s c mt in th nhng do ni t nn in th ny c gi tr nh khng nguy him cho ngi tip xc.Ni t chng st.C tc dng lm tn dng in st vo trong t khi st nh vo ct thu li hay ng dy. Hn ch hnh thnh v lan truyn ca sng in p do phng in st gy nn. Ni t chng st cn c nhim v hn ch hiu in th gia hai im bt k trn ct in v t. Nu khng, mi khi co st nh vo ct chng st hoc trn ng dy, sng in p c kh nng phng in ngc ti cc thit b v cng trnh cn bo v, ph hy cc thit b in v my bin p.V nguyn tc l phi tch ri cc h thng ni t ni trn nhng trong thc t ta ch dng mt h thng ni t chung cho cc nhim v. Song h thng ni t chung phi m bo yu cu ca thit b khi c dng ngn mch chm t ln do vy yu cu in tr ni t phi nh.Khi in tr ni t cng nh th c th tn dng in vi mt ln, tc dng ca ni t tt hn an ton. Nhng t c tr s in tr ni t nh th rt tn km do vy trong tnh ton ta phi thit k sao cho kt hp c c hai c c hai yu t l m bo v k thut v hp l v kinh t.Mt s yu cu k thut ca in tr ni t.Tr s in tr ni t ca ni t an ton c chn sao cho cc tr s in p bc v tip xc trong mi trng hp u khng vt qu gii hn cho php.+ i vi cc thit b in c im trung tnh trc tip ni t yu cu in tr ni t phi tha mn : R 0,5 .

+ i vi cc thit b c im trung tnh cch in th: + i vi h thng c im trung tnh cch in vi t v ch c mt h thng ni t dng chung cho c thit b cao p v h p th :

+ Khi dng ni t t nhin nu in tr ni t t nhin tha mn yu cu ca cc thit b c dng ngn mch chm t b th khng cn ni t nhn to na. Cn nu in tr ni t t nhin khng tha mn i vi cc thit b cao p c dng ngn mch chm t ln th ta phi tin hnh ni t nhn to v yu cu tr s ca in tr ni t nhn to l: R 1 . Thc t d RTN 0,5 th vn phi ni t nhn to v RTN c th xy ra bin ng nh t dy chng st ti khong vt gn trm.+ Trong khi thc hin ni t c th tn dng cc hnh thc ni t sn c nh cc ng ng v cc kt cu kim loi ca cng trnh chn trong t Vic tnh ton in tr tn ca cc ng ng chon trong t hon ton ging vi in cc hnh tia.+ V t l mt trng khng ng nht, kh phc tp do in tr sut ca t ph thuc vo nhiu yu t: thnh phn ca t nh cc loi mui, axitcha trong t, m, nhit v iu kin kh hu. Vit Nam kh hu thay i theo tng ma, m ca t cng thay i theo dn n in tr sut ca t cng bin i trong phm vi rng. Do vy trong tnh ton thit k v ni t th tr s in tr ca t da theo kt qu o lng thc a v sau phi hiu chnh theo h s ma, mc ch l tng cng an ton.

Cng thc hi chnh nh sau: (2.1)

Trong : l in tr sut tnh ton ca t.

l in tr sut o c ca t.

l h s ma ca t.

H s ca t ph thuc vo dng in cc v chn su ca in cc.2.2 CC S LIU DNG TNH TON NI T.

in tr sut o c ca t = 80 m.in tr ni t ct ng dy : Rc = 6 .Dy chng st s dng loi C-70 c in tr n v l r0 = 2,38 /km.Chiu di khong vt ng dy: Pha 220kV: l220 = 250m.Pha 110kV: l110 = 250m.

in tr tc dng ca dy chng st trong mt khong vt l:

S l trong trm: Trm 220kV : n = 3 l.Trm 110kV : n = 9 l.2.2.1 Ni t an ton.Cho php s dng ni t an ton vi ni t lm vic thnh mt h thng. in tr ni t ca h thng l :

(2.2)

Trong : : in tr ni t t nhin.

: in tr ni t nhn to, 1 .a/ in tr ni t t nhin.Ni t t nhin ca trm l h thng chng st ng dy v ct in 110kV v 220kV ti trm.

Ta c cng thc sau: (2.3)Trong : n : l s l dy. Rcs:l in tr tc dng ca dy chng st trong mt khong vt. Rc : l in tr ni t ca ct in. Pha 220 kV.

Pha 110 kV.

Ta thy gi tr in tr RTN < 0,5 t yu cu v l thuyt. Tuy vy ni t t nhin c nhiu thay i v vy m bo an ton ta phi ni t nhn to.b/ in tr ni t nhn to.Ni t c cc hnh thc cc di 2-3m bng st trn hay st chn thng ng. Thanh di chon nm ngang su 0,5 0,8 m t theo hnh tia, mch vng hoc t hp ca hai hnh thc trn. i vi ni t chn nm ngang c th dng cng thc chung sau:

(2.4)Trong :L: chiu di tng ca in cc.d: ng knh in cc khi din cc dng st trn. Nu dng st dt th tr s d thay bng b/2 vi b l chiu rng ca st dt.t: chon su.K: h s hnh dng ph thuc s ni t.- H thng ni t gm nhiu cc b tr dc theo chiu di hnh tia hoc theo chu vi mch vng:

(2.5)Trong :RC : in tr tn ca mt cc.RT : in tr tn ca thanh.n : s cc.T : h s s dng ca tia di hoc ca mch vng.C : h s s dng ca cc.i vi trm bin p khi thit k h thng ni t nhn to ta s dng hnh thc ni t mch vng xung quanh trm bng cc thanh dt.in tr mch vng ca trm l:

(2.6)Trong L l chu vi mch vng.Vi mt bng ca chm ta c mch vng hnh ch nht c cc cnh l L1 = 214 m v L2 = 104m.t : chon su ca thanh ly t = 0,8m.

tt : in tr sut tnh ton ca t i vi thanh lm mch vng chon su t: Tra bng vi thanh ngang chn su 0,8m ta c Kmua = 1,6

d: ng knh thanh lm mch vng. Chn thanh c b rng 4cm.

K : h s hnh dng ph thuc hnh dng ca h thng ni t.Gi tr ca K ph thuc vo kch thc mch vng v c cho bng sau:L1/L211.5234

K5.535.816.428.1710.4

Vi t s L1/L2 = 214/104 = 2,06 bng phng php ni suy:

Nh vy in tr mch vng l :

nh vy t yu cu.Vy ta thit k h thng ni t theo iu kin in tr ni t nhn to l:

Ta c in tr ni t ca h thng:

nh vy t yu cu.2.2.2 Ni t chng st.Khi c dng in st i vo b phn ni t, nu tc bin thin ca dng in theo thi gian rt ln th trong thi gian u in cm s ngn cn khng cho dng in i ti cc phn cui ca in cc khin cho in p phn b khng u, sau mt thi gian nh hng ca in cm mt dn v in p phn b s u hn.Thi gian ca qu trnh qu ni trn ph thuc vo hng s thi gian T = L.g.l2T ta thy T t l vi tr s in cm tng L v in dn g.l = 1/R ca in cc.T biu thc trn ta cng thy rng : khi dng in tn trong t l dng in 1 chiu hoc xoay chiu tn s cng nghip th nh hng ca L khng ng k v bt k hnh thc ni t no cng u biu th bi tr s in tr tn.Khi dng in tn trong t l dng in st, tham s biu th ca ni t ty thuc vo tng quan gia hng s thi gian T v thi gian u sng ca dng in. Khi T U50% MBA = 460 kV.(khng t yu cu)Nhn xt : Do in tr ni t khng t yu cu nn ta phi tin hnh ni t b sung cho h thng chng st.*/ Tnh ton ni t b sung.Trong ni t b sung ta s dng dng ni t tp trung gm thanh v cc ti chn cc ct thu st. Do vic xc nh Zbs bng l thuyt li rt kh khn nn ta chn hnh thc ni t b sung nh sau:Chn thanh ni t b sung l loi thp dt c:Chiu di lT = 12(m).B rng bT = 0,04 (m).Dc trn chiu di ca thanh c chn 5 cc:Chiu di cc lcc = 3 (m).ng knh d = 0,04 (m).Khong cch gia hai cc l a = 3 (m). chn su t = 0,8 (m).Ni t c tnh ton cho chng st nn ta ly h s Kma nh sau:i vi thanh ngang chn su t = 0,8(m): Kma = 1, 2.i vi cc di 3m chn su t = 0,8(m) : Kma = 1,15.S ni t ca h thng khi c ni t b sung nh sau:

Hnh 2.4 : S ni t b sung*/ in tr thanh c xc nh nh sau:

Trong : L : chiu di ca thanh l = 12(m)t : chn su ca thanh lm tia t = 0,8 (m).

tt : in tr sut tnh ton ca t i vi thanh lm mch vng chn su t: Tra bng vi ni t chng st thanh ngang chn su 0,8m ta c Kmua = 1,2

d: ng knh thanh lm mch vng. Chn thanh c b rng 4cm.

K : h s hnh dng ph thuc hnh dng ca h thng ni t, v ni t l tia ngang nn ly K = 1.Vy in tr ca thanh b sung l :

*/ in tr cc c xc nh nh sau:

Trong :l : l chiu di ca cc, l = 3m.Khong cch t = h + l/2=0,8 + 3/2 = 2,3 m (cc chn su h = 0,8m)d : ng knh cc, chn d = 40mmtt : in tr sut tnh ton ca t i vi cc chn su t,

Tra bng vi cc chn su 0,8m v tnh ton chng st ta c Kmua = 1,15

Thay s vo cng thc trn ta c:

*/ in tr b sung:in tr b sung c tnh theo cng thc sau:

(2.16)Trong :RC : in tr ca cc.RT : in tr ca thanh.n : s cc.T , C : h s s dng ca thanh v cc.Vi s cc cn thit l: n = 5 cc, lcc = 3 (m), a = 6(m) Nh vy ta c t s a/l = 1.Tra bng 3 ph lc (trang 82) sch hng dn thit k tt nghip k thut in cao p ta c h s s dng ca cc l C = 0,72. Tra bng 5 ph lc (trang 84) sch hng dn thit k tt nghip k thut in cao p ta c h s s dng ca cc l T = 0,74.Thay s vo cng thc (2.16) ta c:

*/ Tng tr ca h thng khi c ni t b sung.Ta c cng thc tnh tng tr xung kch khi c ni t b sung nh sau:

(2.17)Trong :

Xt chui s :

Tng t nh trn ta ch xt n s hng e -4 vi T1 = 66,8 v s = 5sTnh n XK sao cho:

Trong XK l nghim ca phng trnh:

Hnh 2.5 : th xc nh nghim phng trnh tgXK = - 0,14.XKGii phng trnh trn bng phng th ta xc nh c nghim phng trnh nh sau:

V XK5,XK6 ,XK7, XK8, XK9 XK10 > 23 nn ta ch xt n XK4 ta c bng sau:

Vi T1 = 66,8 (s) ; RBS = 4,4 () ; RNTms = 0,6 () ; K1234

XK4,49,614,519,2

1/cos2(XK)11,031,071,12

1,141,111,071,03

0,860,50,20,06

BK0,910,540,220,07

T kt qu ca bng trn ta tnh c:

Nh vy ta tnh c tng tr xung kch khi c ni t b sung:

in p khi c dng in i vo ni t ti thi im dng in st t gi tr cc i l: Us = I.Z(0;s) = 150.2,24 = 336 kV. Vy Us = 336 kV < U50% MBA = 460 kV.( t yu cu)Nhn xt : Sau khi ni t b sung th yu cu chng st m bo in p ging trn my bin p khi c dng st i vo Us = 336 kV < U50% (110kV) = 460 kV.

I (kA)

t (s)

s

I = a.s

0,8m

lcc

t

lT

a

XK2

XK3

XK4

XK5

Y

X

tgXK

XK1

XK6

a

b

c

D

a

b

D