Bảo Vệ Chống Sét Đánh Trực Tiếp Vào TBA

8/6/2019 Bo V Chng St nh Trc Tip Vo TBA

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n tt nghip K thut incao p

Nguyn ng Hng Chinh H8A Trang 1

Chng 1:BO V CHNG ST NH TRC TIP VOTRM BIN P

1.1. Khi nim chung:

Trong h thng in trm bin p l mt phn t ht sc quan trng. N thc

hin nhim v truyn ti v phn phi in nng. Do khi cc thit b ca trm

b st nh trc tip th s dn n nhng hu qu rt nghim trng khng nhng

ch lm hng n cc thit b trong trm m cn c th dn n vic ngng cung

cp in ton b trong mt thi gian di lm nh hng n vic sn sut in

nng v cc ngnh kinh t quc dn khc ..Do vic tnh ton bo v chng st

nh trc tip vo trm bin p t ngoi tri l rt quan trng. Qua ta c th

a ra nhng phng n bo v trm mt cch an ton v kinh t. Nhm m

bo ton b thit b trong trm c bo v an ton chng st nh trc tip .

Ngoi vic bo v chng st nh trc tip vo cc thit b trong trm ta

cng phi ch n vic bo v cho cc on ng dy gn trm v on y

dn ni t x cui cng ca trm ra ct u tin ca ng dy .

bo v chng st nh trc tip vo trm bin p, ngi ta dng h thng

ct thu li, dy thu li.Cc ct thu li c th t c lp hoc trong nhng iukin cho php c th t trn cc kt cu ca trm v nh my,cc ct n chiu

sng ..V bo v cho on ng dy ni t x cui cng ca trm n ct

u tin ca ng dy ta dng dy chng st . Tc dng ca 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. Chiu cao ca ct v cch b tr cc ct

thu li tu thuc vo tng mt bng trm, cch b tr cc thit b trong trm,

cao bo v theo yu cu .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


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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 .

in tr ni t ca h thng thu st ca cc trm bin p c cp in p

khc nhau l khng ging nhau. Vi trm bin p c cp in p t 110KV tr

nn th h thng ni t chng st v h thng ni t an ton c th ghp

chung.

Ngoi ra khi thit k h thng bo v chng st nh trc tip vo trm bn

cnh nhng vn m bo v yu cu v k thut ta cn phi quan tm n cc

ch tiu kinh t v m thut ca h thng ni t.

Thng thng gim vn u t v cng tn dng cc cao cc trm

bin p v cc nh my in ta c gng t cc ct thu li trn cc ct n chiu

sng, trn cc kt cu ca trm, trn mi nh.. .Ct thu li c lp t hn nn

ch dng khi khng th tn dng c cc chiu cao khc.

1.2- Cc yu cu k thut khi tnh ton bo v chng st nh trc tip vo trm

bin p.

*Phi m bo tt c cc thit b cn bo v phi c nm trn trong phm

vi bo v an ton ca h thng bo v.

* i vi trm phn phi ngoi tri c in p t 110KV tr nn do c mc

cch in kh cao nn c th t cc ct thu li trn cc kt cu ca trm.Tuy

nhin cc tr ca kt cu trn c t ct thu li th phi ni vo h thng ni

t ca trm theo ng ngn nht sao cho dng in st i S khuch tn vo t

theo 3 z4 cc ni t. Ngoi ra mi tr ca kt cu phi c ni t b xung

ci thin tr s in tr ni t .

* Khu yu nht ca trm phn phi ngoi tri in p t 110KV tr nn l

cun y my bin p .V vy khi dng chng st van bo v my bi n p thyu cu khong cch gia hai im ni vo h thng ni t 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. Tuy nhin nu ta s dng h thng ni t chung th ta c th b

qua yu cu ny.


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* Khong cch trong khng kh gia kt cu ca trm c t ct thu li v b

phn mang in khng c b hn di ca chui s .

* Khi 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 c t. 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. 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.

* Khi b tr ct thu li trn x ca trm phn phi ngoi tri 110 kV tr ln

th phi thc hin yu cu :

ch ni cc kt cu trn c t ct thu li vo h thng ni t cn phi

c ni t b xung (dng ni t tp trung) nhm m bo in tr ni t

khng c qu 4; (ng vi dng in tn s cng nghip).

* Khi dng ct thu li c lp th phi th phi ch n khong cch gia

ct thu li n cc b phn ca trm trnh kh nng phng in t ct thu li

n thit b c bo v .

* Tit din cc dy dn dng in st phi ln m bo tnh n nh

nhit khi c dng in st chy qua.

* Khi s dng ct n chiu sng lm gi cho ct thu li th cc dy dnin phi c cho vo ng ch v c chn trong t.

1.3. Cc phng thc s dng tnh ton chiu cao ct v phm vi bo v .

1.3.1. Cng thc tnh chiu cao ca ct thu li .

h = hx + ha

Trong :

. h-L chiu cao ct thu li .

. hx-L cao cn c bo v .

. ha-L cao tc dng ca ct thu li .

ha:Xc nh theo nhm ct vi iu kin l h a u D/8.

. D-L ng knh ng trn ngoi tip a gic to bi cc ct .


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1.3.2. Phm vi bo v ca mt ct thu li c lp

Phm vi bo v ca mt ct thu li c lp l min gii hn bi mt ngo i

ca hnh chp trn xoay c ng sinh xc nh bi phng trnh :

rx

=

xxhh

hh1

6,1

Trong :

rx-L phm vi bo v mc cao hx ca ct thu li .

d dng thun tin trong vic tnh ton thit k thng dng phm vi bo v

dng n gin ho. c tnh ton theo cng thc :

- Nu hx h3

2e th )

h.8,0

h1.(h.5,1r xx !

- Nu hx h.3

2" th )

h

h1.(h.75,0r xx !

* Cc cng thc trn ch ng vi nhng ct thu li cao di 30 m. Hiu qu

ca ct thu li hn 30 m s gim do cao nh hng ca st l hng s .

h

rx

hx

2/3h

,75

1,5h,75

1,5h

,2h

Hnh 1.1Phm vi bo v ca mt ct thu


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Khi ct c chiu cao trn 30m th ta vn dng cng thc trn nhng phi nhn

thm vi h s hiu chnh p =h

5,5

V trn hnh v ta s dng cc honh 0,75.h.p

1,5.h.p

1.3.3. Phm vi bo v ca hai hay nhiu ct thu li

Phm vi bo v ca hai hay nhiu ct thu li th ln hn nhiu so vi tng s

phm vi bo v ca hai hay nhiu ct n .Nhng hai ct thu li c th phi

hp bo v c khong gia chng th khong cch a gia hai ct phi tho

mn iu kin a e 7.h

a) Phm vi bo v ca hai ct thu li c cng cao .

Khi hai ct thu li c cng cao h t cch nhau mt khonh a(a e 7.h) th

cao ln nht ca khu vc bo v gia hai ct thu li l h 0c xc nh :

h0 = h -7

a

Bn knh phm vi bo v ti khong gia hai ct c tnh n hu sau :

- Nu hx 0h3

2e th r0x = )

h.8,0

h1.(h.5,1

0

x0

- Nu hx 0h3

2" th r0x = )

h

h1.(h.75,0

0

x0

Trong :

. h0-L cao ln nht ca khu vc bo v gia hai ct thu li

. r0x-L bn knh phm vi bo v ti khong gia hai ct thu li

Khi cao ca ct thu li ln hn 30 m th ta cng phi thm h s hiu

chnh p nh mc 1.3.2 v tnh h 0 theo

h0 = h -p.7

a; p =

h

5,5


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0

Hnh 1.2. Tr- ng h p hai ct thu li c chiu cao bng nhau .

rx

0,2hh

1,5h

0,75h

rxorx

R

hx

a

b) Phm vi bo v ca hai ct thu li c cao khc nhau .

c xc nh nh sau :

Gi s c hai ct thu li : Ct 1 c cao h 1

Ct 2 c cao h2Khong cch gia hai ct la a v h 1 > h2

Trc tin ta v phm vi bo v ca ct cao h 1sau t nh ca ct thp h2

ging ng thng ngang sang ct h 1. Ct ng sinh ca phm vi bo v ca

ct 1 ti im 3

im ny c coi l nh ca ct thu li gi nh (ct 3) Ct thu li gi nh

ny c cng cao vi ct 2 v hnh thnh i ct c chiu cao bng nhau, cch

nhau mt khong a ,

1,5h0

0,75h0h0

0,2h0


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D dng nhn thy khong cch x t h1 (ct 1) n ct gi tng (ct 3)chnh l bn knh bo v ca ct cao h 1 i vi chiu cao cn bo v bng h 2

Do tnh khong cch x theo :

Nu :

!e

8,0

hh5,1xh3/2h 2112

2112 hh75,0xh3/2h !"

T ta tnh c a = a - x v h0 = h2 - a/7

T ta tnh c bn knh bo v r0x

Nu

!e

0

x0x00x h8,0

h1h.5,1rh3/2h

!"

0

x0x00x

h

h1h.75,0rh3/2h

h0

a'

2

h2

1

h1

3R

a

r0x

r1x

r2x

hx

1,5h

0,75h2 0,75h1 1,5h

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

x


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c) Phm vi bo v ca nhiu ct thu st

Khi cng trnh cn c bo v chim mt khu vc rng ln nu ch dng mt

vi ct th ct phi rt cao gy nhiu kh khn cho vic thi cng v lp rp.

Trong trng hp ny ta dng phi hp nhiu ct vi nhau bo v. Phn

ngoi ca phm vi bo v s c xc nh cho tng i ct mt ( vi yu cu

khong cch l a e 7h ). Cn phn bn trong a gic s c kim tra theo iu

kin an ton.

Vt c cao hx nm trong a gic s c bo v nu tho mn iu kin:

D e 8.(h - hx) = 8.h

a

Trong :

D - ng knh vng trn ngoi tip a gic hnh thnh bi cc ct thu st

ha = h - hx l cao hiu dng ca ct thu st.

Nu cao ct vt qu 30 m th iu kin an ton s c hiu chnh l

D e 8.(h - hx).p = 8.ha.p ; p =h

5,5

1.4- Cc s liu dng tnh ton thit k ct thu li bo v trm bin p

110kV.

Vi yu cu ca ti. Thit k bo v chng st nh trc tip vo trm.

- Trm c din tch l: 140 x 92m2 v bao gm:

+ Hai my bin p T 1 v T2

+ 4 l dy vo110 kV

+ S u dy ca trm l s hai thanh gp c thanh gp ng vng.- cao cc thanh x l 11 m v 7,5m.

- cao nh iu khin : 5 m


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1.5. Cc phng n b tr ct thu li bo v

1.5.1.Phng n 1.

S mt bng trm v cch b tr cc ct thu li nh hnh v , phng n

ny ta b tr 5 ct thu li (1) ; (3) ; (4) ; (5) ; (6) trn x ca trm c chiu cao

11 m v 1 ct thu li ct (2) t trn x cao 7,5 m .ng thi s dng cc c t

n chiu sng (7) ; (8) ; (9) ; (vi chiu cao ca cc ct n l 21m ) bo

v.Ta tin hnh tnh ton chiu cao ca cc ct v phm vi bo v ca h thng.

1.5.1.1. cao tc dng ca cc ct thu li .

tnh c cao tc dng ha ca cc ct thu li Trc ht ta cn xc

nh ng knh D ca ng trn ngoi tip tam gic (hoc t gic) i qua 3

(hoc 4) nh ct .

cho ton b phn din tch gii hn bi tam gic (hoc t gic ) c

bo v th : D e 8.ha hay hau8

D

T s b tr cc ct ca phng n 1 ta thy cc nhm ct b tr hnh

thnh nn nhng hnh ch nht bng nhau nh :

HCN (4),(5),(8),(7) = HCN (5), (6),(8),(9)

V cc cp tam gic bng nhau nh :

TG(1),(2),(4) = TG(2),(3),(6)

TG(2),(4),(5) = TG(2),(5),(6)

T ta ch cn xc nh ng knh ng trn ngoi tip ca mt trong cc

HCN ,TG bng nhau ny :

a) Xt nhm ct (1),(2),(4):

Nhm 3 ct ny hnh thnh mt tam gic thng c di cc cnh l :

a12 = 55,1m ; a14 = 27m ; a42 = 66,03mBn knh ng trn ngoi tip tam gic i qua 3 chn ct bt k c xc

nh bi cng thc H rng:

)cp)(bp)(ap.(p.

c.b.ar

!

4


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2

cbap

! : l na chu vi ca tam gic.

- a,b,c: l di cc cnh ca tam gic.

2

cbap

! = m05,74

2

03,271,55

2

aaa 421412!

!

T :)cp)(bp)(ap.(p.

c.b.ar

!4

=

m67,3303,6605,74.2705,74.1,5505,74.05,74.4

03,66.27.1,55!

Suy ra ; D = 2.R = 2.33,67= 67,34m

cao tc dng nhm ct (1),(2),(4) bo v c hon ton phn din

tch gii hn bi 3 nh ct phi tho mn iu kin :

hau D/8 =8

34,67=8,42 m

b)Xt nhm ct (2),(4),(5)

Nhm 3 ct ny hnh thnh mt tam gic vung c cc cnh l

Ta c:

a25 = 38 m ; a45 =54 m

ng knh ng trn ngoi tip tam gic vung ny chnh l cnh huyn ca

tam gic v bng a24 c xc nh

D = m03,665438aa 22245252 !!

cao tc dng nhm ct (2),(4),(5) bo v c hon ton din tch gii

hn bi chng phi tho mn iu kin hau D/8 = m25,88

03,66!

c) Xt nhm ct (4),(5),(7),(8)

Nhm ct ny hnh thnh mt hnh ch nht c di cc cnh

a45 = 54 m ; a47 = 44 m

ng knh ng trn ngoi tip hnh ch nht ny chnh l di ng

cho ca hnh ch nht :


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m65,694454aa 222472

45 !!!

cao tc dng ti thiu cc ct (4),(5),(7),(8) bo v c hon ton

din tch gii hn bi chng l: m707,88

65,69

8

Dha !!u

d)Chn cao tc dng chung cho ton trm

Qua tnh ton cao tc dng ca cc ct thu li, c th ly chung mt gi

tr cao tc dng ln nht ca ct thu li cho ton trm l.

Do vy ta ly : h a =8,8m.

1.5.1.2. Tnh cao ca cc ct thu li cao ct thu li dng bo v chng st nh trc tip vo trm bin p

c xc nh bi: h = ha + hx

Trong : + h: cao ct thu li.

+ hx: cao ca vt c bo v.

+ ha: cao tc dng ca ct thu li.

i vi pha 110kV ca ti cc thanh x cn bo v c cao ln nht l

11m (hx = 11m) do cao ti thiu ca ct thu li l:h = hx + ha =11 + 8,8 = 19,8 m.

Nhng do cc ct (7),(8),(9) c t trn cc ct n chiu sng c cao

l 21m, nn ta tn dng ht cao ny lm ct thu st.

T ta chn chiu cao chung ca cc ct thu st cho ton trm l h = 22 m

1.5.1.3. Tnh phm vi bo v ca cc ct thu li:

a) Bn knh bo v ca tng ct thu li cao 11m:

h = 22 m : hx =11 m :

Ta c : hx =11 < 2/3 h =2/3.22 = 14,76 m.

Nn: m37,1222.8,0

11122.5,1

h.8,0

h1h.5,1r xx !

!

!

b) Bn knh bo v ca tng ct cao 7,5 m:


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h = 22 m : hx =7,5 m :

Nn m94,1822.8,0

5,7122.5,1

h.8,0

h1h.5,1r xx !

!

!

1.5.1.4Phm vi bo v ca cc cp ct thu li

a) Xt cp ct (1),(2).

C cao bng nhau : h 1 = h2 = 22 m

Khong cch gia hai ct l: a = 55,11 m.

- cao ln nht ca khu vc bo v gia hai ct thu li l:

m13,147

11,5522

7

ahho !!!

- Bn knh ca khu vc bo v gia hai ct thu li l:

cao 11m: hx = 11m >3

2ho = 9,42m.

Nn : .m35,213,14

111.13,14.75,0

h

h1h.75,0r

o

xoxo !

!

!

cao 7,5m: hx = 7,5m

2

1k

K

TT !

y Do ta coi h thng ni t l s ghp song song ca hai tia nn tng tr

xung kch u vo ca h thng ni t ti t = X/s l:

X

!X g

!

X

71k

T2

s/d

1

0s/d

ks/d

e1k

1T21.

lg2

1);0(

T cng thc trn ta thy tng tr xung kch tng ca ni t gm 2 thnh

phn:

- Thnh phn n nh c tr s bng tr s in tr ti xoay chiu:

lg02

1

- Thnh phn bin thin theo thi gian (thnh phn in cm):

g

!

X

X 1k

T2

s/d

1

0

ks/d

e1k

1T2.

lg2

1

Ta thy tng tr xung kch ca h thng ni t tin ti tr s n nh cng

nhanh khi chiu di in cc cng ngn (chiu di in cc cng ln th in

p u cc cng b). iu ny chng t cc phn cui ca in cc phthuy tc dng km.

xc nh ZZ ( 0, X/s) ta xt cc chui s sau:

!

!T

!!1k

2222645,1

6k

1..

2

1

1

1

k

1


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2

Tk

2

2T

2

T

1k

T2 k

e..

2

e

1

ee.

k

1s/ds/d

1s/d

ks/d

XXX

g

!

X

!

Trong chui s ny ch xt n s hng cha e -4

Khi 0k

e4

T 2

T

k

s/dk

s/d

}"X

X

Ta c th b qua cc gi tr t e -5 tr i v chng rt nh so vi cc s hng

trc.

Tc l ta tnh vi k sao cho *

k

s/d Zk;4T

eX

4

k

TT2

1

s/d

k

s/d eX!X s/d

1

s/d

12 T2kT4k Xe

Xe

*s/d Zk;m5 Q!X => 5,5

5.14,3

228.10.68,3.945,1LGl2k

2

23

s/d2

2

!!XT

e

Vy k = 1z5 ( kZ*)

T : 7,37

14,3

228.10.68,3.945,1lg.LT

2

23

2

20

1 !!

T

!

Vi k = 1z5 ta c bng kt qu sau :

Bng 2-3: Kt qu tnh ton 2

T

kek

s/dX

k 1 2 3 4 5

Tk(Qs) 37,7 9,425 4,198 2,356 1,508

k

s/d

T

X 0,133 0,531 1,194 2,122 3,316


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kTt

e1

0,875 0,588 0,303 0,119 0,036

2

T

k

e ks/dX

0,875 0,147 0,0337 0,0075 0,0015

T bng s liu trn ta tnh c : 0647,1k

e5

1k2

Tk

s/d

!!

X

Vy :

X

!X

!

X

7

1k

T2

s/d

1

0s/d

ks/d

e1k

1T21.

lg2

1);0(Z =

!

0647,1645,15

7,37.21.

228.10.68,23

13

= 5,81 ;

in p khi c dng in st i vo ni t ti thi im t = Xd/s .

Thi im dng t cc i : U = I.ZZ (0, Xd/s) = 150.5,81 = 871,6 kV

Ta thy U = 871 kV > U50%MBA =460 kV.

Ta thy vi gi tr in p U ny khng tho mn vi iu kin kim tra l :

U < U50%my bin p =460 kV.

Do ta phi tin hnh ni t b xung m bo t c gi tr in p

ny.

2.3.6. Ni t b xung

m bo yu cu v in tr ni t v khng xy ra phng in

ngc ti thit b bo v th ngoi vic ni chng vo h thng ni t chng ta

cn phi t thm mt in tr ni t ngay chn ct thu li, ni t cc thit

b. l ni t b xung.Trong ni t b xung ta s dng ni t tp chung gm thanh v cc.

Do vic xc nh Zbx bng l thuyt l rt phc tp, nn ta s chn hnh

thc ni t b xung nh sau s thun tin cho qu trnh tnh ton Z bx

y Vi thanh ni t b xung l loi thp dt c kch thc:

-Chiu di l = 10 m


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-B rng b = 0,04 m

y Dc theo chiu di thanh c chn 3 cc trn c kch thc:

-Chiu di cc lcc = 2,5 m

-ng knh cc d = 0,04 m

-Khong cch gia hai cc a = 5 m

y chn su ca thanh v cc t = 0,8 m

S ni t ca h thng khi c ni t b xung :

S ni t ca tia b xung :

t=t

lt

lc c

Hnh2-10:S ni t ca tia b xung.

y

y

y

y

y

yy yyyyyy y yy

y

yyyyyy y yy

y

y

y

y

yy

Hnh 2 9: ni t ca h thngkhi c ni t b xung


51/131



Vi ni t chng st nn h s ma Kma c xc nh :

- i vi thanh chn su t = 0,8 m ; Kma = 1,25

- i vi cc di 2,5 m chn su t = 0,8 m ; Kma = 1,15

2.3.6.1. in tr ni t ca thanh b xung :

y Cng thc s dng tnh ton :d.t

l.Kln.

l..2R

2Ttt

T T

V!

Trong :

- l : chiu di ca thanh; l = 10 m

- t : chn su ca thanh ; t = 0,8 m

- VttT : in tr sut tnh ton ca t i vi thanh

VttT =Vo.Kma = 0,9.10

4.1,25 = 1,125.104 cm = 1,125.102 m

- d : ng knh ca thanh lm tia vi thanh dt c b rng b = 0,04 m

=> m02,02

04,0

2

bd !!!

- K : h s hnh dng, vi ni t l hnh tia ngang nn ta ly K = 1

Thay cc gi tr vo cng thc tnh ton ta c :

d.tl.Kln.

l..2R

2Ttt

TT

V! = ;! 65,1502,0.8,0

10.1ln.10.24,3.2

10.125,122

2.6.6.2. in tr ni t ca cc b xung

y Cng thc :

T

V!

l't4

l't4ln

2

1

d

l2ln.

l..2R

Ctt

C

Trong :

-VttT : in tr sut ca t i vi cc chn su 0,8 m

VttT =Vo. Kma(cc)= 0,9.10

4.1,15 = 1,035.104;cm = 1,035.102;m

- l : chiu di ca cc ; l = 2,5 m


52/131



- d : ng knh ca cc d = 0,04 m

- m05,28,02

5,2t

2

l't !!!

Thay vo cng thc tnh RC ta tnh c:

;!

! 88,33

5,205,2.45,205,2.4ln

21

04,05,22ln.

5,2.14,3.210.035,1R

2

C

2.3.6.3. in tr ni t b xung

in tr ni t b xung ca h thng ni t thanh v cc c xc nh

theo cng thc:tCct

ctbx

.Rn..R

R.RR

LL!

Trong :

- Lt, Lc : h s s dng ca thanh v cc

vi n = 3 ; lcoc = 2,5 m ; a = 5 m => 25,2

5

l

a!!

Tra bng 3 v 5 Phn ph lc sch HDTKTN KTCA

ta c Lc = 0,86; Lt = 0,9

Thay cc gi tr vo cng thc ta c :

;!

!LL

! 48,79,0.88,333.86,0.65,15

88,33.65,15

.Rn..R

R.RR

tCct

ctbx

T tnh c tng tr ca h thng khi c ni t b xung.

2.3.6.4. Tng tr vo ca h thng ni t khi c ni t b xung

Ta s dng ton t Laplax tm c cng thc tnh tng tr xung kch ca

h thng ni t:

!X

X

T

1s/d

2

2k

T.

X

K2

bx

)st(NT

)st(NT

)st(NTbx

)st(NTbxs/dXK e.

xcos

1

R

R

R.2

RR

R.R);0(Z

= A + B


53/131



Trong 552,0596,048,7

596,0.48,7

RR

R.RA

)st(NTbx

)st(NTbx !

!

!

X

T

! 1

s/d

2

2k

T.

X

K2

bx

)st(NT

)st(NT e.

xcos

1

R

RR.2B

Xt chui s : g

!

X

T

1k

T.

1

s/d2

2k

e

Vi chui s ny ta ch xt n gi tr e-4 (v t e-5 tr i c gi tr rt nh)

Tng ng vi vic tnh xksao cho :

s/d

1k

1

k2

2k T.2x4

T.

x

XTee

X

Txk> 0

Do c ni t b xung nn ta xt nh hng ca ni t b xung ln in

tr xung kch

Vi vic coi mch vng ca h thng ni t l s ghp // ca hai tia c

cng di l = 228 m nn ta c s thay th:

Trong s trn :

L : in cm tng ng ca mt n v di L = L/2

G : in dn tng ng ca mt n v di G = 2.G

Do ta tnh c:

Hnh 2 11:S thay th ca 1 tia

L


54/131



25,175

7,3714,3.2

T.2x

7,3714,3

228.10.68,3.945,1l.G.Ll.G.2.

2

L

l'.G'.LT

s/d

1k

2

23

2

2

2

2

2

2

1

!!X

Te

!!T

!T

!T

!

Xd/s = 5Qs

Trong xk l nghim ca phng trnh :

kkkkbx

)s

t(NTk x.08,0x.

48,7

596,0tgxx.

R

Rtgx !!!

Tng ng vi vic gii h phng trnh k1 tgx!

y2 = - 0,08.xk

Ta dng phng php th gii phng trnh ny:

Ta c cc nghim :

X1 = 2,75 rad => cosX1 = - 0,924

X2 = 5,694 rad => cosX2 = 0,831

X3 = 8,737 rad => cosX3 = - 0,773X4 = 11,56 rad => cosX4 = 0,535

X4 = 14,68 rad => cosX5 = - 0,555

Khi k = 6 th xk> 17,25 nn ta ch ly 5 nghim trn.

Sau ta lp bng tnh gi tr ca chui s vi cc gi tr xk,; k = (1z5)

X3X1

4/2-1

1X4 X5X2

114/294/74/54/34/

y=-0,08.XK

Hnh 2 12:Gii h phng trnh bng phng php th


55/131



Ta c T1 = 37,7 Qs => 133,07,37

5

T1

s/d!!

X

V Rbx = 7,48 ; 08,048,7

596,0

R

R

bx

)s

t(NT!!

Bng 2-4:Bng kt qu tnh ton Bk

K 1 2 3 4 5

xk 2,75 5,694 8,737 11,56 14,73

cos xk - 0,924 0,831 - 0,773 0,535 - 0,515

1/ cos2 xk 1,171 1,448 1,674 3,494 3,77

k2

bx

)st(NT

xcos

1

R

R 1,251 1,528 1,754 3,574 3,85

1

s/d

T

X 0,133 0,133 0,133 0,133 0,133

1

s/d2

k

T.

x X

T 0,102 0,437 1,028 1,8 2,924

X

T

1

s/d2kT

.

e

0,903 0,646 0,358 0,165 0,054

Bk 0,86 0,504 0,243 0,055 0,017

X

T

!

1

s/d2

2k

T.

k2

bx

)s

t(NT

)s

t(NTk e.

xcos

1

R

R

R.2B

T bng trn ta c : ;!! 679,1BB5

1k

Do Zxk(0,Xd/s) = A + B = 0,552 + 1,697 = 2,23 ;

Vy in p khi c dng in st i vo h thng ni t ti thi im

t = X /s (thi im dng in st t gi tr cc i ) l :

U = I.ZXK(0; X /s) = 150.2,23 = 334,5 KV


56/131


57/131



Chng III:

BO V CHNG ST NG DY TI IN 110 KV

3.1. Yu cu chung i vi bo v chng st ng dy.

y ng dy l mt phn t di nht trong h thng in.Chng b trdn tri .Nn thng b st nh v thng chu tc dng ca qu inp kh quyn.

y Qu in p khng ch gy nn phng in trn cch in ng dym cn truyn sng vo trm bin p c th gy ct my ct ng dyv s c ph hoi cch in cc thit b trong trm nh hng n scung cp in ca li. Do khi gii quyt vn bo v chng stng dy phi c quan im tng hp , phi c kt hp cht ch vivic bo v chng st cho trm ,c bit l on ng dy gn trm

phi c bo v tuyt i v khi st nh khu vc ny s a votrm cc qu in p vi tham s ln, rt nguy him cho cch in cacc thit b trong trm.

y Qu in p kh quyn c th do st nh trc tip vo ng dy hocdo st nh xung mt t gn gy nn qu in p cm ng trnng dy . Ta thy trng hp st nh thng vo ng dy l nguyhim nht v ng dy phi chu ton b nng lng ca dng inst. V vy trng hp ny c chn tnh ton chng st chong dy.

y Do tr s qu in p kh quyn l rt ln nn ta khng th chn mccch in ng dy p ng hon ton yu cu ca qu in p khquyn m ch chn theo mc hp l v kinh t v k thut. Do yu cu i vi bo v chng st ng dy khng phi loi tr honton kh nng s c do st m ch l gim bt s ln s c do st tigii hn hp l. ( xut pht t yu cu v s cung cp in ca phti, s ln ct dng in ngn mch cho php ca my ct in, ngdy c hoc khng c trang thit b t ng ng li).Tc l phitm c phng thc bo v ng dy sao cho tn hao do st gy ral thp nht.

y Trong vic tnh ton ca bo v chng st ng dy do st nh ta stnh ton cho mt nm v vi chiu di ng dy l 100 km .


58/131



3.2. L thuyt tnh ton:

3.2.1.Phm vi bo v ca 1 dy chng st:

y Chiu rng ca phm vi bo v b x cao hx c tnh theo cngthc sau:

- Nu hx > 2/3 h th :

!

hxhhxb 1..6,0

- Nu hxe 2/3 h th :

y

!h

xhhxb 8.01..2,1

y Phm vi bo v bo v ca mt dy thu st nh hnh v:

y Dy thu st c dng bo v chng st cho ng dy in

cao p. V treo cao trung bnh ca dy dn thng ln hn 2/3

treo cao ca dy thu st (t l hx/h bng khong 0,8) nn c th

khng cn cp n phm vi bo v m biu th bng gc bo v

E (l gc gia ng thng ng vi ng thng ni lin dy thu

st v dy dn).

j Gc bo v ca 1 dy thu st:

h

h

0,2.h

0,6.h

2.bx1,2.

Hnh 3.1. Phm vi bo v ca 1 y thu li


59/131



C B

A

E 2

E 1

Hnh 3.2. Gc bo vc a mt y thu s t.

y C th tnh ton c tr s gii hn ca gc E l 31o

(tgE = 0.6),trong thc t thng ly E = (20 z 25)o.

3.2.2. Ch tiu chng st ca ng dy:

y Vi treo cao trung bnh ca dy trn cng (c th l dy dn

hoc dy chng st) l h tb, ng dy s thu ht v pha mnh cc

phng in st trn di t c chiu rng l 6.h tb v chiu di chnh

bng chiu di ng dy L. T s ln c phng in st xung

din tch 1 Km2 ng vi 1 ngy st l 0,1 z 0,15 ln c th tnh

c:

jTng s ln st nh trc tip vo ng dy:

* Tng s ln st nh trc tip vo ng dy c tnh theo cng thc sau:

N = ( 0,1 z 0,15).6.htb.10-3.L. nngs (ln/LKm.nm).

Trong :

htb: L treo cao trung bnh ca dy trn cng tnh bng m.

L : Chiu di ng dy c tnh bng Km.

nngs: S ngy st trong mt nm (tnh cho khu vc c ng dy i qua).


60/131



y Tu theo v tr st nh, qu in p xut hin trn cch in ca

ng dy c tr s khc nhau. Ngi ta phn bit s ln st nh

trc tip vo ng dy c treo dy chng st gm:

- S ln st nh vo nh ct (K c s ln st nh vo on dy chng

st gn nh ct):

NC} N/2

- S ln st nh vng qua dy chng st vo dy dn:

NE = N.RE

*Vi REl xc sut st nh vng qua chng st vo dy dn. Kinh

nghim vn hnh cho thy RE khng ch ph thuc vo gc bo v E m cn

tng theo chiu cao ct in, xc sut ny c biu th bng cng thc kinh

nghim:

490

.lg ! c

hEER

Trong :

E(o) : L gc bo v ca dy chng st.

hc(m) : L chiu cao ca ct in.

- S ln st nh vo im gia khong vt (bao gm s ln st nh

vo dy chng st khong cch xa ct).

NKV = N - NC - NE} N/2

Trong :

NC: L s ln st nh vo nh ct.

NE: L s ln st nh vng qua dy chng st vo dy dn.

NKV: L s ln st nh vo khong vt.

j S ln xy ra phng in trn cch in ng dy:


61/131



yV tham s ca dng in st (bin Is, dc a = dis/dt) c nhiu tr

s khc nhau. Do khng phi tt c nhng ln c st nh thng vo

ng dy u dn n phng in trn cch in ng dy. iu kin

c phng in trn cch in ng dy l qu in p kh quyn phi

c tr s ln hn mc cch in ca ng dy, kh nng ny c biu

th bi xc sut phng in Rp v nh vy s ln xy ra phng in trn

cch in ng dy s l:

Np = N.Rp= (0,6 )9,0z .htb.10-3.L.nngs.Vpd

j S ln xy ra ct in ng dy:

y Do thi gian tc dng ca qu in p kh quyn rt ngn (kho ng 100Qs), trong khi thi gian lm vic ca h thng bo v Rle thng khng

nh hn 1 na chu k tn s cng nghip (bng 0,01s) nn cc Rle bo

v cha kp tc ng. V vy Np cha phi l s ln nhy my ct do st

nh hng nm. Phng in trn cch in ng dy ch gy nn ct

in ng dy khi tia la phng in xung kch trn cch in chuyn

thnh h quang duy tr bi in p lm vic ca li in. Xc sut

chuyn t tia la phng in xung kch thnh h quang k hiu l , ph

thuc vo nhiu yu t, trong yu t quan trng nht l Grain in

p lm vic dc theo ng phng in . S ph thuc ny c biu

din nh sau:

= f (Elv);

vi Elv = Ulv / lp (KV/m).

Trong :L : L xc sut hnh thnh h quang.

Ulv (KV): l in p lm vic ca ng dy.

lp(m): L chiu di ng phng in.

yDo s ln ct in ng dy c tnh bng cng thc sau:


62/131



Nc = (0,6 z 0,9).htb.10 3.L.nngs.Rp.L (ln/LKm.nm)

j Sut ct ng dy:

* so snh kh nng chu st ca cc ng dy c cc tham s khc

nhau, i qua nhng vng c cng hot ng ca st khc nhau ta dng tr s

t Sut ct ng dy , chnh l s ln ct din khi ng dy c chiu di

L = 100Km. ta c:

nc = (0,06 0,09).6.htb.nngs.Rp.L (ln/ 100Km.nm)

T ta xc nh c ch tiu chng st cho ng dy :

N l kh nng vn hnh ton ca ng dy i vi qu in p kh quyn,l

khong thi gian gia hai ln ct ca ng dy .

cdnM

1!

tlnc

mn

1

* T cng thc trn ta c 2 hng khc nhau gim s ln ct in do st

nh thng vo ng dy: gim L hoc gim Rp.

y Gim xc sut hnh thnh h quang L: c thc hin bng cch gin

cng in trng dc theo ng phng in bng vic tng chiu di

ng phng in .

y Gim xc sut phng in Rp: c thc hin bng cch treo dy chng

st v tng cng cch in ca ng dy. Treo dy chng st l bin php rt

hiu qu trong vic gim s ln ct in ng dy:

3.3. Tnh ton ch tiu chng st cho ng dy 110 kv

3.3.1. Cc tham s s dng tnh ton:

A:Kt cu ct:


63/131



+Vi ng dy truyn ti cp in p 110KV thuc loi quan trng, dng

truyn ti in i xa vi cng sut ln. Loi ct s dng l loi ct st, kt

cu ca ct hnh Thp :

+ Dy chng st (1 dy, treo ti nh ct) dng dy thp loi C - 70 c ng

knh d = 11mm bn knh r = 5,5mm.(tra o ti lieeeuh no)

+ Dy dn cc pha dng loi dy nhm li thp AC - 120 c ng knh

d = 19,2mm bn knh r = 9,6 mm.(tra ti liu no)

+ Cch in chui s dng loi T - 4,59(tra u) c 7 bt mi bt cao

170mm do chiu di chui s l = 7.170 = 1190mm.

+ treo cao ca dy chng st l: h cs = 18,4m.

+ treo cao ca dy dn pha A l: h A = 14,6m.

+ treo cao ca dy dn pha B,C l: h B = hC = 11,3m.

+ vng dy dn fdd =4m.

+ vng dy chng st fcs = 2,5m.

14,6

2,6

2

3,8

18,4

11,3

1,1

Hnh 3.3. Kt cu ct ca ng y

DC

A

B


64/131



+ X pha A di lx A = 2m.

+ X pha B,C di lx B = lx C = 2,6 m.

+ Khong vt ca ng dy l =200m.

+ U50% ca cch in 110 KV l 660KV/ 7 bt s.Khong cch gia cc pha c cho trong hnh 3 -3.

B:Xc nh gc bo v E:

+ Gc bo v E ca dy chng st i vi cc pha c tnh da vo kt cu

ct:

- Vi pha A: o7,27A526,08,3

2

3,114,18

2Atg !E!!

!E

- Vi pha B v C:

o20CB366,01,7

6,2

8,33,3

6,2CtgBtg !E!E!!

!E!E

+ Vy gc bo v ca dy chng st i vi cc pha nh trn tho mn yu

cu k thut (Ee 31o).

C:Xc nh treo cao trung bnh ca dy chng st v dy dn:+ treo cao trung bnh ca dy c xc nh theo cng thc sau:

htb = h 2.f/3

y Trong :

h(m): L cao ca dy ti nh ct hay ti kho no ca chui s.

f(m): l vng ca dy.

+ Vi dy chng st ta c hcs = 18,4m; fcs = 2,5m, nn:

.m73,165,2.3

24,18f

3

2hh cscs

tbcs !!! y

+ Vi dy dn pha A ta c h A = 14,6m; fdd =4m nn:


65/131



.m93,114.3

26,14f

3

2hh ddA

tbA !!! y

+ Vi dy dn cc pha B,C ta c h B = hC = 11,3m; fdd =4m nn:

.m63,84.323,11f

32hhh ddB

tbC

tbB !!!! y

D:Xc nh tng tr sng ca dy dn v dy chng st:

Tng tr sng ca dy dn c xc nh theo cng thc

r

hlnZ

tby

y!2

60

Trong : htb : L treo cao trung bnh ca dy.

r : L bn knh ca dy

+ Tng tr sng ca dy dn pha A:

.1,469106,9

93,112ln60

r

h2ln60Z

3dd

tbAA

dd;!!!

y

y

y

y

y

+ Tng tr sng ca dy dn pha B v pha C:

.66,449106,9

63,8.2ln60

r

h2ln60ZZ

3dd

tbC,BC

ddBdd

;!!!!

y

y

y

y

+ Tng tr sng ca dy chng st khi khng xt ti nh hng ca h quang

.8,522

105,5

73,16.2ln60

r

h2ln60csZ 3

cs

tbcs ;!!!

y

y

y

y

+ Tng tr sng ca dy chng st khi c xt n nh hng ca vng quang:

- Khi c xt n nh hng ca vng th thnh phn in dung tng ln

lm cho tng tr sng gin.Do tng tr sng s gim i P ln : P-H s hiu

chnh vng quang.


66/131



- Vi in p 110KV, s dng 1 dy chng st ta c P = 1,3.(Tra trong

bng 3 3 sch hng dn thit k k thut cao pnguyen minh chc,trang

43,). .2,4023,1

8,522ZZ cscsvq ;!!

P!

E:H s ngu hp gia chng st vi dy dn cc pha:

H s ngu hp gia dy dn pha A v dy chng st

+ Khi cha xt n nh hng ca vng quang th h s ngu hp hnh hc

Ko c xc nh theo cng thc sau:

2r2h2ln

12d12Dln

oK y!

y Trong :

h2: L cao trung bnh ca dy chng st: .m4,18hhtbcs2 !!

r2: l bn knh ca dy chng st: r2 = 5,5mm.

d12: L khong cch gia dy chng st v dy dn pha A.

D12: L khong cch gia dy chng st v nh ca dy dn pha qua

php chiu gng phng qua mt t.

+ Php chiu gng phng qua mt t nh hnh v di y:

2'

h2

D12

1

d12

2

h1

h1

(h=h2 h1

1


67/131



+ Trn hnh v ta c:

h1: L treo cao trung bnh ca dy dn pha A.

h1=htb

A =11,93 m

*h2 :L treo cao trung bnh ca dy chng st

h2= htbcs=16,73 m

(h = h2 h1 =16,73-11,93 =4,8 m

2212xa12 )hh(lD ! =

2266,282 =28,73 m

22xa12 hld (! =22 8,42 =5,2 m

+ H s ngu hp hnh hc: K0

196,0

105,573,16.2ln

2,5

73,28ln

2r2h2ln

12d12Dln

oK

3

!!!

y

y

+ Khi xt n nh hng ca vng quang ta xt thm h s hiu chnh vng

quang P ,vi P =1,3

255,0196,03,1KK ovq

csA !!P! yy

y Tnh h s ngu hp gia dy dn pha B, pha C v dy chng st:

Khi cha xt n nh hng ca vng quang

+ Ta c: lx B,C = 2,6.

;m1,863,873,16hhh;m63,8hh12

tbB1

!!!(!!

* r2 =rcs=5,5.10-3m


68/131



.m49,25)73,1663,8(6,2)hh(lD 222212xa12 !!!

.m507,81,86,2hld 2222xa12 !!(!

12607138

09731

1055

731625078

4925

2

2212

12

3

,,

,

,

,.ln

,

,

ln

r

hln

d

Dln

oK !!!!

y

y

+ Khi xt n nh hng ca vng quang:

16380126031 ,,,Kvq

cscKK o

vqcsB !!P!

! yy

*Nhn xt:

Qua qu trnh tnh ton trn ta thy rng h s ngu hp gia dy chng

st v dy dn pha A ln hn h s ngu hp gia dy chng st v dy pha

B,pha C do ta c kt lun:

+ tnh ton sut ct do st nh vng qua d y chng st vo dy dn ta

ch xt cho pha c gc bo v ln hn - L pha A.

+ tnh ton sut ct do st nh vo khong vt dy chng st ta s

tnh ton cho pha c qu in p kh quyn t ln cch in ln hn tc l phac h s ngu hp nh hn.Ta tnh cho pha B,C.

+ tnh sut ct do st nh vo nh ct v ln cn nh ct ta phi xc

nh qu in p kh quyn t ln cch in cc pha v s tnh ton vi trng

hp nguy him nht tc l pha c U cd(a,t) ln hn

3.3.2. Xc nh tng s ln st nh vo ng dy hng nm

j Tnh s ln st nh vo ng dy hng nm+ Cng thc tnh ton

N = ( 0,1 z 0,15).6.hcstb.10-3.L.nngs ( ln/LKm.nm ).

y Trong :

(0,1 z 0,15): L s ln st nh vo din tch 1 Km2 trong mt ngy st.


69/131



hcstb(m): L treo cao trung bnh ca dy chng st hcs

tb = 16,73m.

L(m): L chiu di ng dy, y ta tnh s ln ct in ng dy

trn 100Km L= 100Km.

nngs: L s ngy st trong mt nm: nngs = 80 (ngy/nm).+ Thay vo cng thc trn ta c:

N = ( 0,1 z 0,15).6.16,73.10 -3.100.80 = (80 z 120) (Ln/100 Km.nm).

+ Trong tnh ton thit k nhm nng cao an ton ta chn :

N = 120 (Ln/100 Km nm).

y T c s l thuyt v cc tham s tnh c trn, ta tin hnh tnh

sut ct cho ng dy 110KV trong cc trng hp:

St nh vng qua dy chng st vo dy dn.

St nh vo khong vt ng dy.

St nh vo nh ct.

3.3.3. Tnh sut ct ca ng dy 110KVdo st nh vng qua dy chng

st vo dy dn:

+ ng dy 110KV c bo v bng dy chng st tuy vy vn c trnghp st nh vng qua dy chng st vo dy dn.

+ S ln st nh vng qua dy chng st vo dy dn c tnh bng cng

thc: NE = N .RE (Ln/100Km.nm).

+ Sut ct ca ng dy do st nh vng qua dy chng st vo dy dn

xc nh bi: nE = N.RE.Rp.L (Ln/100Km.nm).

y Trong

N (Ln/100Km.nm): L tng s ln st nh vo dy dn trong mt

nm.

RE : L xc sut st nh vng qua dy chng st vo dy dn.


70/131



Rp: L xc sut xy ra phng in trn cch in.

L: L xc sut hnh thnh h quang.

+ RE c tnh theo cng thc vn hnh kinh nghim nh sau:

490

h.lg c

E!ER

y Trong :

E(o) : L gc bo v ca dy chng st.

hc(m) : L chiu cao ca ct in.

+ Khi dy dn b st nh c th xem ti ni st nh mch ca khe st cghp ni tip vi tng tr sng ca dy dn c tr s bng Z dd/2 (dy dn 2 pha

ghp song song). V tng tr sng ca dy dn kh ln, khong (400 z 500);,

nn dng in st gim i rt nhiu so vi khi st nh vo ni c ni t tt. Ta

c dng in ni st nh ch cn l:

2

I

2

ZZ

ZII S

ddo

oS }

! y

+ in p lc trn ng dy l: ddS

dd Z4

IU y!

Zdd/2 Zdd/ 2

Is/ 4 Is/4

Is/ 2

IsZ0


71/131



+ Khi Uddu U0,5 ca chui s th c phng in trn cch in gy s c

ngn mch 1 pha N(1). T cng thc trn ta c th vit:

5,0ddS U

4

ZIu

y

Dng in st :dd

5,0S

Z

4I

y

!

Xc sut phng in trn cch in l:

dd

5,0SZ1,26

U4

e1,26I

epdy

y

!

!R

+ Xc sut hnh thnh h quang L ph thuc vo Grain ca in p

dc theo ng phng in: L = f (Elv).vi Elv = Ulv / lp (KV/m).

y Trong :

L : L xc sut hnh thnh h quang.

Ulv (KV): l in p lm vic ca ng dy bng in p pha ca

ng dy:

.KV5,633

110U lv !!

lp(m): L chiu di ng phng in, ly bng chiu di chui s.

lp = 1,19m.

+ Do ta c:

Elv = Ulv / lp = 63,5/1,19 = 53,37 KV/m.

+ Bng 21 - 1 sch k thut in cao p cho ta quan h L = f (Elv) nh sau:

Elv = Ulv / lp (KV/m) 50 30 20 10

L(n v tng i) 0,6 0,45 0,25 0,1

Bng 3.1. Quan h L = f(Elv)


72/131



+ V th v s dng phng php ni suy tnh L:

L0,1 0,2 0,3 0,4 0,5 0,6 0,7

60

50

40

30

20

10

0

Elv(kV/m)

0,625

53,37

Hnh 3.5 th biu din quan h L = f (Elv)

+ Bng phng php v th v ni suy nh trn ta c: L = 0,625.

y Nhn xt:

T cc cng thc tnh RE v Rp:

490

h.

lg

c

E

!ER

dd

5,0

Z1,26U4

epdy

y!R

+ Ta nhn thy RE v Rp u ph thuc t l chiu cao ct h c hay treo cao

dy dn v gc bo v E. Chiu cao ct hc hay gc bo v E tng th RE tng,

treo cao dy dn tng th Zdd tng pRp tng, do ta chn dy dn pha A tnh ton v dy dn pha A c treo cao v gc bo v E ln hn treo cao v

gc bo v ca dy dn pha B,C. C ngha l ta coi khi x st nh vng qua

dy chng st vo dy dn th ch nh vo dy dn pha A.

y Vi dy dn pha A ta c:


73/131



.o7,27;1,469Z;6,14hA

Add

Add m !E;!!

U0,5 = 660KV (Tra trong bng 9 5 sch k thut in cao p).

` + Thay cc s liu trn vo cng thc tnh RE v Rp ta c:

002,068,2490

4,18.7,274

90

h.lg cA !EE RR !!

E!

806,01,4691,266604

eZ1,26

U4

epd

Add

5,0

!

!

! yy

y

y

R

j S ln st nh vng qua dy chng st vo dy dn:

NE = N .RE = 120 . 0,002 = 0,24 (Ln/100Km.nm).j Sut ct ca ng dy do st nh vng qua dy chng st vo dy dn:

nE = N.RE.Rp.L = 120 . 0,002 .0,806 .0,625 = 0,1209 (ln/100Km .nm).

3.3.4. Tnh sut ct ca ng dy 10KVdo s t nh vo khong vt:

+ S ln st nh vo khong vt ca ng dy c tnh theo cng thc:

2NNKV !

N: L tng s ln st nh vo 100 Km ng dy, trong mt nm.

N = 120 ln/100Km.nm

NKV =120/2 = 60 ln/100Km.nm.

+ Sut ct ca ng dy 110KV do st nh vo khong vt c tnh

bng cng thc sau:

L!L! .v.2N.v.Nn pdpdKVKV

y Trong xc sut hnh thnh h quang L = 0,625 ( tnh c phn

trc).

+ Nh vy tnh c n KV ta phi xc nh c vpd.


74/131



1. Phng php xc nh vpd:

+Vi gi thit coi dng in st c dng xin gc vi bin I s = a.t.

+ Khi ng dy ti in b st nh vo khong vt s sinh ra cc in p

l:a/ in p tc dng ln cch in khng kh gia dy chng st v dy dn:

+ Ta ch xt vi dy dn pha A c khong cch n dy chng st l nh

nht. Khi gi tr ny ln s gy phng in trn cch in khng kh ca

ng dy.

+ in p tc dng ln cch in khng kh ca ng dy c tnh bng

cng thc: Ucd

3

l.a).k1( !

y Trong :

k: h s ngu hp gia dy dn v dy chng st.

a: dc dng in st.

l: khong vt ng dy.

+ Vi cng cch in xung kch ca khe h kh (trong phm vi thi gian

u sng) khong 750KV/1m v gi S l khong cch gia dy dn v dy

chng st th iu kin phng in vit theo biu thc:l).k1(

S.2250a

u

+ T c th tnh c xc sut phng in ,s ln ct in ,sut ct ca

ng dy do phng in trn cch in khng kh ca ng dy .

+ Tuy nhin trong thit k v thi cng ng dy thng chn khong cch

ln trnh chm dy nn kh nng phng in gia 2 khong vt rt t

xy ra v d c xy ra th xc sut hnh thnh h quang rt b cho nn kh nng

ct in ng dy l khng ng k. Do ta s b qua khng tnh ton nh

hng ca loi in p ny.

b/ in p t ln cch in chui s U cd.

+ Qu in p st xut hin trn cch in ng dy p h thuc vo tham s

ca dng in xt c phn lm 2 thnh phn, c tnh theo cng thc sau:


75/131



Uc(t) = Uc(I,a) + Ulv

y Trong :

Uc(I,a) l thnh phn qu in p do dng st gy ra ph thuc bin

I v dc ca dng in st a.

Ulv l in p lm vic ca ng dy.

+ Trong tnh ton I v a c coi nh nhng tham s vi xc sut cc dng

in st c bin I u IS v dc a u aS l:

9,10

a

1,26

I

a,I

SS

eV

!

+ Nu in p trn cch in ti thi im t i no ln hn in p chu

ng cho php ca cch in thi im tng ng, ly theo c tnh Vn -giy (V - S) th phng in xy ra, vy trng thi ti hn:

Uc(ti) = Up(ti)

hay Uc(Ii, ai) + Ulv = Up(ti) vi Ii = ai.ti (*)

y Trong :

Up(ti): in p phng in ly theo c tnh V-S ti thi im t i.

Ii, ai l bin v dc ca dng in st tng ng vi trng thi

nguy him thi im t i.

+ Gii h phng trnh (*) vi nhng thi im ti khc nhau ta c nhng

cp tham s ca nhng dng in st nguy him trn h to ( I, a) chng to

thnh mt ng cong gi l ng cong nguy him nh hnh v :

Minnguyhim

I

a


76/131



Hnh 3.6. ng cong nguy him

+ ng cong nguy him chia mt phng thnh hai min :

* Min pha bn phi ng vi cc dng in st gy ra phng in trn

chui s y l min nguy him.

* Min pha bn tri ng vi cc dng in st an ton.

+ ng cong nguy him l c s xc nh xc sut phng in trn cch

in ng dy. Tht vy xc sut phng in Rp ca ng dy chnh bng xc

sut cc dng in st c tham s nm trong min nguy him ( min gch cho).

+Phng php xc nh Vvd nh sau.

Do vic coi dng in st c dng xin gc Is=a.t nn thnh phn qu in p

trn cch in do dng in st gy ra U c(i,a) t l vi dc a :V vy c th

t Uc(i,a) =Z.a

Trong : Z L mt hng s i vi i v a

Do : Upd(ti) =Z.ai + Ulv

Hay ta c dc nguy him ca dng in st thi im t i :

Z

U-(ti)Ua lvpdi !

Trong qu trnh tnh ton ta xc nh c trc U c(I,a) nn Z c xc nhtheo:

+ Bin dng in st nguy him: I i = ai.ti

+ Xc sut phng in c tnh bng cng thc:

!1

0

.aIpd dvvV

Vi910126 ,i

a

ia,,i

I

I evev !!

- Bng phng php gn ng v tuyn tnh ho ng cong nguy him :

Chia ng cong thnh n khong (n = 10 z15) ta c :


77/131



!

(!

n

iaIpd ii

vvv1

.

vi

910

1

910 ,ia

,ia

iaeev

!(;

126,iI

iIev

!

+ Sau khi xc nh c vp , ta s tnh c sut ct do st nh vo khong

vt ng dy 110KV .

2. Trnh t tnh ton:

+ n gin trong tnh ton coi nh st nh vo khong gia ca dy

chng st trong khong vt, khi dng in st c chia u cho 2 pha ca

dy chng st nh hnh 3.7.

a.t/2 a.t/2

RCRC

RC

200m

Hnh 3.7. St nh vo dy chng st gia khong vt.

+ Gi thit dng in st c dng xin gc :

XuX

X!

dsds

ds

Stkhi.a

tkhit.aI

+ Ta s tnh ton IS ng vi cc gi tr trong bng sau:

a(KA/Qs) 10 20 30 40 50 60 70 80 90 100

t(Qs) 1 2 3 4 5 6 7 8 9 10


78/131



+ in p tc dng ln chui cch in khi st nh vo khong vt ca

ng dy l:

lv

icsc

sccd U)K1(dt2

)t(d.L2

)t(i.R)t,a(U

s

!

- Dng in st c dng iS(t) = a. t do ta tnh c:

lvvqcs

cClvvqcs

cc

cd U)K1).(Lt.R(2

aU)K1(

2

a.L

2

t.a.R)t,a(U !

!

+ T cng thc tnh Uc ta thy nu h s ngu hp K nh th U c ln, do

theo ti liu hng dn thit k tt nghip k thut in cao p th khi tnh Uc(t)

phi tnh vi pha c h s ngu hp (c xt n nh hng ca vng quang) Kvqnh nht. Trong phn trc ta tnh c:

16380

2550

,KK

,Kvq

csCvq

csB

vqcsA

!!

!

Tnh Uc(t) vi Kvq = 0,1638

Vi V = 0,9.104;.cm. Tra bng 19.6 sch k thut in cao p ta c

RC=10;.

csC l in cm ca thn ct in tnh t mt t n im treo dy

chng st,csC c tnh theo cng thc sau:

csC = l0.hC

- l0 l in cm n v ca thn ct l 0=0,6 QH/m

- hC l chiu cao ca ct h c= 18,4m.

cs

L = 0,6.18,4 = 11,04QH

+ Xc nh in p U lv.

- Cng thc tnh ton: maxphlv U.2

UT

!

y Trong Uphmax l bin in p pha ca ng dy.3

U.2U maxph !


79/131



KV18,573

2.110.

14,3

2

3

2.U.

2U lv !!

T!

+ Thay cc gi tr trn vo cng thc tnh U c(t) ta c:

)KV(,),)(,t(a

)t,a(Ucs

18571638010411102

!

hay Uc(a,t) = 0,4181a.(10.t+11,04) + 57,18

Tnh Uc(t) vi dc a thay i v cc thi im khc nhau ta c bng kt

qu trang sau:

+ T cc gi tr ca Uc(t) ta v c cc ng Uc=f(t)vi a thay i trn hnh

3 - 8

+ ng c tnh V-S ca chui cch in cng c biu din trn hnh 3 - 8

- c tnh V-S ca chui cch in tra trong bng 25 sch hng dn thit

k tt nghip k thut in cao p nh sau:

Bng 3.3. c tnh V-Sca chui cch in

t(Qs) 1 2 3 4 5 6 7 8 9 10

Up(K) 1020 930 860 815 790 780 780 770 770 740

+ Khi in p t trn chui s ln hn in p phng in ca chui s th sc phng in . T th trn hnh 3 - 8 ta xc nh c cc cp gi tr (a i, ti) l

giao im ca Uc(a, t) vi c tnh V-S ca chui s. Sau ta xc nh c

cp thng s nguy him (I i, ai) vi quan h : Ii = ai.ti (KA)

Bng 3.4:Cc cp gi tr ai.ti

ai(KA/Qs) 10 20 30 40 50 60 70 80 90 100

ti(Qs) 16 7.44 4.8 3.5 2.74 2.25 1.89 1.6 1.38 1.17

Ii=ai.ti(KA) 160 148.8 144 140 137 135 132.3 128 124.2 117

+ Vi cc cp s (Ii, ai ) tnh c trong bng trn ta xy dng c ng

cong nguy him c dng nh hnh 3 9 :


80/131



j Xc nh xc sut phng in vp:

+ Xc sut phng in Rp chnh l xc sut cp thng s (I, a) ca dng

in st nm trong min nguy him :

dvp = p {a = ai}. p{I u I i}

Ta c p{I u I i} = RI = 126,I i

e

p {a = ai}= p _ aaiai

daada ee = dRa

p{ a u a i} = Ra = 910,iI

e

RI l xc sut cho dng in st I ln hn mt gi tr I i no .

Ra l xc sut cho dc dng in st a ln hn dc a i no .

T ta c :

dRp = RI. dRa hay Rp = 1

0aIdvv

+ Bng cng thc sai phn ta xc nh c :

!

(!

n

i

aIpd iivvv

1

.

vi9,10

1a

9,10

a

a

ii

ieev

!( ; 1,26I

I

i

iev

!

+ Gi tr Rp c tnh trong bng 3 - 5:

T bng 3 - 5 ta c: Rp = 0,00113.

3. Sut ct do st nh vo khong vt ng dy 110KV:

+ Sut ct do st nh vo khong vt c tnh theo cng thc :

nKV = NKV.Rp.L (ln/100Km.nm)

Vi:

NKV = N/2 (ln/100Km.nm)

N = 120

L= 0,625


81/131



Rp = 0,00113

+ Thay vo cng thc tnh sut ct ta c:

0424062500011302

120

2,,.,..v.

Nn pdkv !!L! (ln/100Km.nm).

3.3.5. Tnh sut ct ca ng dy 110KV do st nh vo nh ct (bao

gm c trng hp st nh vo on dy chng st gn ct):

+ n gin trong qu trnh tnh ton coi st nh ngay trn nh ct. Lc

ny phn ln dng in st s i vo t qua b phn ni t ct in, phn cn

li s theo dy chng st i vo cc b phn ni t ca cc ct ln cn. S

m phng ng dy nh sau:

3.3.5.1. L thuyt tnh ton:

j in p trn cch in ng dy khi c st nh khu vc nh ct:

+ in p trn cch in ng dy khi c st nh vo nh ct c xc

nh theo cng thc:

! yyydt

di)t(M)t,a(

dt

diL)t,a(UR)t,a(i)t,a(U sddCdd

cd-cCCcd

lvcsvq U)t,a(UK y

ic

IcsIcs

I}0

ic

RcRcRc

Hnh 3.10 S m phng ng ! y.


82/131



+ in p ny c phn thnh cc thnh phn in p nh sau:

1: in p ging trn b phn ni t ca ct bng: iC (a,t) . RC

y Trong :

RC(;): L in tr ct. iC(a,t): L dng in i trong ct.

2:in p cm ng phn t: ),(- taUtc

+ Xut hin trn dy dn do h cm gia dy dn v knh st gy ra.

dt

di).t(M)t,a(

dt

di.L)t,a(U sddCddc

t

-c !

y Trong :

dt

diC (a,t): L tc bin thin ca dng in i trong ct.

adt

dis! (KA/Qs): L dc ca dng in st (dng sng xin gc).

ddCL : L in cm ca thn ct k t t ti treo cao ca dy pha :

ddodd

C hlL y! (lo l in cm n v ca thn ct )

Mdd(t): L h cm gia mch khe st v mch dy dn, n ph thuc

vo s pht trin chiu di ca knh st v c xc nh bng cng thc sau:

(

(

! 1

2120

h

H

h

h

H

HtvhtM

dddd

dd ln..).(

.ln.,)(

F

y Vi:

hdd : L treo cao ca dy dn.

H = hdd + hc

h = hc - hdd

hc : L cao ca ct.

F : Tc tng i phng in ngc ca dng in st:


83/131



F = 0,3

v : Vn tc pht trin ca phng in ngc ca khe st:

v =F.c

c l tc nh sng: c = 300 m/ Qs.+ S biu din:

3:in p cm ng phn in: )t,a(Ud-c

+ in p cm ng phn in c xc nh bi :

(

(

!

Hhh

htvHtvhtvha

h

hKtaU

C

C" "

" "

csvq"c

...)(

).).(.()..(ln.

..,.),(- 21

101

FF

yTrong :

Kvq: L h s ngu hp gia dy dn v dy chng st c xt n nh

hng ca vng quang.

hC : l chiu cao ca ct.

hdd : L treo cao dy dn.

hcs : L treo cao ca dy chng st .

H = hcs + hdd

(h = hcs - hdd

v : Vn tc pht trin ca phng in ngc ca khe st.

Mtdd Zdcs

Lcdd

ZddZddic

R


84/131



v =F.c vi c l tc nh sng (c = 300 m/ Qs).

c: L tc nh sng c = 300 m/ Qs.

F : Tc tng i ca phng in ngc dng in st

F = 0,3 a: L dc dng in st.

4:in p lm vic ca ng dy:

+ Ulv c ly l in trung bnh pha:

.KV,UtdtsinUU dmmaxphalv 18571103

22

3

2210

!T

!T

![[T

! yyyyyy

T

5:in p do dng in i trong dy chng st gy ra: Kvq

.Ucs(a,t)+ Thnh phn ny lm gim in p trn cch in v t l vi in p

trn dy chng st qua h s ngu hp Kvq gia dy dn v dy chng st.

+ Ucs(a,t) c tnh theo cng thc:

dt

di)t(M

dt

diLRi)t,a(U scsCcsCCCcs yyy !

y Trong :

iC(a,t): L dng in i trong ct.

dt


adt

dis ! (KA/Qs): L dc ca dng in st (dng sng xin gc).

csCL : L in cm ca thn ct k t mt t ti treo cao ca dy

chng st: cs#csC hlL y! (lo l in cm n v ca thn ct )

Mcs(t): L h cm gia knh st v mch vng tDy chng st - t,

Mcs(t), c tnh bng cng thc:

F

!

yy

yy

yy 112

220

cs

cscs

cs

h)(

htvlnh,)t(M


85/131



j Xc nh iC(a,t) v ),( tadt

diC :

Trng hp1: Trc khi c sng phn x t ct ln cn tr v ng

vi thi gian lc

l2t kv

y

e

y Trong :

lkv : L chiu di khong vt: l kv = 200 m.

c : L vn tc truyn sng c = 300 m/ Qs.

Trng hp ny ng vi: .s33,1300

2002t Q!$

y

+ S tng ng ca mch dn dng in st nh hnh v:

+ Trong s thay th trn ,dng in st c coi nh mt ngun dng ,dychng st c biu th bi tng tr sng ca dy chng st c xt n nh

hng ca vng quang. Thnh phn t ca in p cm ng c coi nh 1

ngun p.

+ Mcs(t): L h cm gia knh st v mch vng tDy chng st - t,

Mcs(t) c xc nh bng cng thc sau:

ic

ics i

dt

ditM scs ).(

2ics

Rc

cscL

is

ic

Hnh 3.11. S thay th ngdy trc khi c

sng phn x t ct ln cn tr v.


86/131



!yy

yy

yy 112

220

cs

cscs

cs

h

htvhtM

)(ln,)(

F

+ csCL : L in cm ca thn ct k t mt t ti treo cao ca dy

chng st: cs%

csC hlL y! (lo l in cm n v ca thn ct )

+ T s tng ng vi gi thit dng ca dng in st i s =a.t vi a

l dc u sng .Ta c phng trnh:

- Phng trnh mch vng :

)(..)(.),(.).,( 102

2 !vqcs

cstcs

CcscCC

ZitMata

dt

diLRtai

- Phng trnh th nt :

)(.. 22 taiii scsc !!

+ Gii h phng trnh (1) v (2) ta c :

E

! yyy

y 1

vqcscsvq

csC

vqcs

C

Z)t(M2tZ

R2Z

a)t,a(i

y Trong :

csc

vq

cs1

L.2R.2Z C!E ;

Cvqcs

vq

csC

R.2ZZ.a)t,a(

dtdi

!

v&

cs' : L tng tr sng ca dy chng st c xt n nh hng

ca vng quang.

Trng hp 2 : Sau khi c sng phn x t ct ln cn tr v ng vi

thi gian lc

lt kv

y

"2

y Trong :

lkv : L chiu di khong vt: l kv = 200 m.

c : L vn tc truyn sng c = 300 m/Qs.


87/131



Trng hp ny ng vi: .s33,1300

2002t Q!"

y

+ Ta c s thay th nh sau :

2ics

2

csL

dt

ditM scs ).(

2ics

Rc

cscL

is

ic

2

csR

n gin ta c th thay th dy chng st bng in cm tp trung ni tip

vi in tr ni t ca hai ct bn cnh trn s thay th:

Lcs L in cm ca mt khong vt dy chng st khng k n nh hng

ca vng quang .

c xc nh :

c

l.ZL kv

ocs

cs !

y Trong :

ocs: L tng tr sng ca dy chng st khi khng xt n nh

hng ca vng quang.

*lkv :L chiu di khong vt.

*c : L tc nh sng c =300 m/Qs.

+ Vit phng trnh mch vng v th nt gii ra ta c :

t.C

cscs

c2e1.

R.2

)t(M.2L.a)t,a(i E

!

Hnh 3.12. S thay th ngdy sau khi csng phn x


88/131



t.2

C

cscsC 2e..

R.2

)t(M.2L.a)t,a(

dt

di EE

!

y Trong : csCcs

C2

L.2L

R.2

!E

j in p do dng in i trong dy chng st gy ra: Kvq.Ucs(a,t)

+ Thnh phn ny lm gim in p trn cch in v t l vi in p

trn dy chng st qua h s ngu hp Kvq gia dy dn v dy chng st.

+ Ucs(a,t) c tnh theo cng thc:

dt

ditM

dt

diLRitaU scsCcsCCCcs yyy ! )(),(

y Trong :

iC(a,t): L dng in i trong ct.

dt


adt

dis! (KA/Qs): L dc ca dng in st (dng sng xin gc).

csCL : L in cm ca thn ct k t mt t ti treo cao ca dy

chng st: csocsC hlL y! (lo l in cm n v ca thn ct )

Mcs(t): L h cm gia knh st v mch vng tDy chng st - t

F

!

yy

yy

yy 1h)1(2

h2tvlnh2,0)t(M

cs

cscs

cs

3.3.5.2. Trnh t tnh ton :

+ Sut ct do st nh vo nh ct hoc ln cn nh ct c tnh bng

cng thc sau :

nc = Nc .Vp .L (ln /100Km.nm).

y Trong :


89/131



Nc : L s ln st nh vo nh ct hoc ln cn nh ct trong mt

nm ng vi on ng dy dy di 100 Km.

602

120

2

NNC !!! (ln/100 Km.nm).

L : L xc sut hnh thnh h quang : L = 0,625.

Rp : L xc sut phng in do qu in p ng dy khi c st

nh vo nh ct.

xc nh sut ct ca ng dy 110 KV do st nh vo nh ct

hoc ln cn nh ct ta phi xc nh Rp

T cc gi tr in p ging trn chui cch in v t c tuyn V - Sca chui cch in vi cc gi tr thi gian xy ra phng in t i .Bin dng

in st s l : I i =ai.ti

Sau khi xc nh c cc cp gi tr nguy him ( Ii,ai)ta s xc nh

c xc sut phng in Vpd

!

(!n

iiaiIpd

v.vv1

1/Xc nh pha tnh xc xut phng in :

+ i vi nhng pha khc nhau, khi c st nh vo nh ct hoc ln cn

nh ct, in p ging trn cch in ng dy ca cc pha l khc nhau. Vi

cng dc v cng thi gian tc ng, chui cch in ca pha no chu in

p ging ln hn th pha c xc sut phng in ln hn.

+ Chn 1 gi tr c th ca dng in st tnh ton tr s in p ging

trn cch in mi pha :

- Chn dng in st dng xin gc c : a = 10 KA/ Qs v thi gian tc

ng t = 3 Qs.

a. Xc nh in p tc dng ln cch in pha A :

j Tnh in p cm ng phn in d-cU :


90/131



+ Vi pha A ta c :

hcs = 18,4m; Kvq = v

(

csA) = 0,255; hC = 18,4m.

hdd = hA = 14,6m.

H = hcs + hdd = 14,6 + 18,4 = 33m.(h = hcs hdd =18,4 14,6 =3,8m.

F = 0,3; c = 300m/Qs.

v =F . c = 0,3 . 300 = 90 m/Qs.

+ Thay vo cng thc tnh d-cU ta c :

!

33.8,3.4,18.)3,01(

)33t.90).(8,3t.90().4,18t.90(ln.

3,0

6,14.a.1,0.

6,14

4,18.255,01)t,a(U

2

d-c

Hay

!22,348

)33t.90).(8,3t.90().4,18t.90(ln.a.302,3)t,a(Ud-c

+ Vi a = 10 KA//Qs; t = 3/Qs )t,a(Ud-c = 180,8 KV.

j Tnh iC(a,t) v )t,a(dt

diC :

+ V thi gian tc ng c chn l t = 3/Qs > !!300200.2

cl.2 kv 1,33Qs

nn ta p dng cc cng thc tnh i C(a,t) v )t,a(dt

diC trong trng hp sau khi c

sng phn x t ct ln cn tr v :

Dng in i trong thn ct iC(a,t) :

t.

C

cscs

c e.R.

)t(M.L.a

)t,a(i

2

12

2 E

!

y Trong :

c

l.ZL kv

ocs

cs !


91/131



ocs0 : L tng tr sng ca dy chng st khi khng xt n nh

hng ca vng quang. ocs0 = 522,8 ;.

lkv = 200m; c = 300m/Qs.

H53,348300

200.8,522L cs Q!!

csCcs

C2

L.2L

R.2

!E

csC

L : L in cm ca thn ct k t mt t ti treo cao ca dy

chng st: csocs

C hlL y! = 0,6 . 18,4 = 11,04QH.

(lo l in cm n v ca thn ct l o = 0,6QH/m.) RC = 10;.

054,004,11253,348

1022 !

!Ey

y

Mcs(t): L h cm gia knh st v mch vng tDy chng st - t:

!

F

!

y

y

yy

yy

yy 14,18).3,01(2

4,18.2t90ln.4,18.2,01

h)1(2

h2tvlnh2,0)t(M

cs

cscs

cs

68,384,47

8,36t90ln68,3)t(M cs

!

y

y

+Vi t = 3Qm Mcs(3) = 10,52

+ Thay cc gi tr va tnh c vo cng thc tnh i C(a,t) ta c :

t.054,0e1.)t(M.253,34810.2

a)t,a(i csC

!

+ Vi a = 10 KV/Qm; t = 3Qs iC(10,3) = 24,49 KA. bin thin ca dng din i trong thn ct :

t.2

C

cscsC 2e..

R.2

)t(M.2L.a)t,a(

dt

di EE

!

+ Thay cc gi tr va tnh c vo cng thc trn ta c :


92/131



t.054,0e.054,0.)t(M.253,34820

a)t,a(

dt

di csC !

+ Vi a = 10 KA/Qs; t = 3Qs dt

diC (10,3)= 7,25A/Qs

j Xc nh in p cm ng phn t: )t,a(U t -c

dt

di).t(M)t,a(

dt

di.L)t,a(U sddCddc

t

-c !

y Trong :

dt


adtdi

s ! (KA/Qs): L dc ca dng in st (dng sng xin gc).

ddC

L : L in cm ca thn ct k t t ti treo cao ca dy pha :

ddo

dd

ChlL y! = 0,6 . 14,6 = 8,76QH.

(lo = 0,6 l in cm n v ca thn ct; h dd = hA = 14,6m)


vo s pht trin chiu di ca kn h st v c xc nh bng cng thc sau:

(

(

! 121

20h

H

h

h

H

HtvhtM

dd

dd

dd ln..).(

.ln.,)(

F

y Vi:

hdd = hA =14,6 m : L treo cao ca dy dn pha A.

H = hdd + hc =14,6 + 18,4 = 33 m.

h = hc - hdd = 18,4 14,6 = 3,8 m. hc = 18,4 m : L cao ca ct.


F = 0,3



93/131



v = F . c = 0,3 . 300 = 90 m/ Qs.

c l tc nh sng: c = 300 m/ Qs.

+ Thay vo cng thc tnh M dd(t) ta c :

0982942 3390922 ,,t.ln,)t(Mdd !

+ Vi t = 3Qs Mdd(3) = 7,807QH.

y Vy )t(M.a)t,a(dt

di76,8)t,a(U ddCt -c !

+ Vi a = 10 KA/Qs; t = 3Qs;dt

diC (10,3) = 7,52 KA/Qs;

Mdd

(3) =7,807 Q H ),(Ut

-c 310 = 143,94 KV.j Xc nh in p do dng in i trong dy chng st gy ra :

[Kvq.Ucs(a,t)]

+ Ucs(a,t) c tnh bng cng thc :

)t(M.a)t,a(dt

diLR)t,a(i)t,a(U csCcsCCCcs ! yy : (a=

dt

dis )

+ Thay RC = 10 ;;csCL = 11,04QH; vo ta c :

)t(M.a)t,a(dt

di04,11)t,a(i.10)t,a(U csCCcs ! y


diC (10,3) = 7,52 KA/Qs;

Mcs(3) =10,52QH; iC(10,3) = 24,49 KA/Qs ),(U cs 310 = 433,12 KV.

j Xc nh in p tc dng trn cch in pha A : AcdU

lvcsvq

csAd-c

t

-cCCAcd U)t,a(U.K)t,a(U)t,a(UR).t,a(i)t,a(U ! + Vi cc gi tr va tnh c trn ta c :

KV37,51618,5712,433.255,08,18094,14310.49,24)3,10(UAcd !!

b. Xc nh in p tc dng ln cch in pha B,C :

j Tnh in p cm ng phn in d-cU :


94/131



+ Vi pha B,C ta c :

hcs = 18,4m; Kvq = vq csBK =

vq

csCK = 0,1638 ; hC = 18,4m.

hdd = hB = hC = 11,3 m.

H = hcs + hdd = 18,4 + 11,3 = 29,7 m.

(h = hcs hdd = 18,4 11,3 = 7,1m.

F = 0,3; c = 300m/Qs.

v =F . c = 0,3 . 300 = 90 m/Qs.

+ Thay vo cng thc tnh d-cU ta c :

! y

7,29.1,7.4,18.)3,01(

)1,7t.90).(7,29t.90().4,18t.90(ln.

3,0

3,11.a.1,0.

3,11

4,181638,01)t,a(U

2d-c

Hay

!56,451

)1,7t.90).(7,29t.90().4,18t.90(ln.a.762,2)t,a(Ud-c

+ Vi a = 10 KA/Qs; t = 3/Qs ),(- taUdc = 144,04 KV.

j Tnh iC(a,t) v )t,a(dt

diC :

+ V thi gian tc ng c chn l t = 3/Qs > !!300

200.2cl.2 kv 1,33Qs

nn ta p dng cc cng thc tnh i C(a,t) v )t,a(dt

diC trong trng hp sau khi c

sng phn x t ct ln cn tr v :

Dng in i trong thn ct iC(a,t) :

t.

C

cscs

c

2

e1.R.2

)t(M.2L.a

)t,a(i

E

!

y Trong :

c

l.ZL kv

ocs

cs !


95/131



ocs1 : L tng tr sng ca dy chng st khi khng xt n nh

hng ca vng quang. ocs1 = 522,8 ;.

lkv = 200m; c = 300m/Qs.

H53,348300

2008,522L cs Q!!y

cs

Ccs

C

L.L

R.

2

22

!E

csC

L : L in cm ca thn ct k t mt t ti treo cao ca dy

chng st: csocs

ChlL y! = 0,6 . 18,4 = 11,04QH.

(lo l in cm n v ca thn ct l o = 0,6QH/m.) RC = 10;.

054,004,11253,348

1022 !

!Ey

y

Mcs(t): L h cm gia knh st v mch vng tDy chng st - t:

!

!yy

yy

y

yy

yy

yy 1213012

2129021201

12

220

),(ln,

)(ln,)(

t

h

htvhtM

cs

cscs

cs

F

68,384,47

8,36t90ln68,3)t(Mcs

!

y

y

+Vi t = 3Qm Mcs(3) = 10,52 Q H

+ Thay cc gi tr va tnh c vo cng thc tnh i C(a,t) ta c :

t.054,0e1.)t(M.253,34810.2

a)t,a(i csC

!

+ Vi a = 10 KV/Qm; t = 3Qs iC = 24,49 KA. bin thin ca dng din i trong thn ct :

tC

cscsC

eR

tMLata

dt

di ...

.

)(..),( 222

2 EE

!

+ Thay cc gi tr va tnh c vo cng thc trn ta c :


96/131



t.054,0e.054,0.)t(M.253,34820

a)t,a(

dt

di csC !

+ Vi a = 10 KA/Qs; t = 3Qs dt

diC (10,3)= 7,52 KA/Qs

j Xc nh in p cm ng phn t: ),(- taUtc

dtdi

tMtadtdi

LtaU sddCddctc ).(),(.),(- !

y Trong :

dt

diC (a,t) (KA/Qs) : L tc bin thin ca dng in i trong ct.

adtdis ! (KA/Qs) : L dc ca dng in st (dng sng xin gc).

ddC

L : L in cm ca thn ct k t t ti treo cao ca dy pha :

ddo

dd

ChlL y! = 0,6 . 11,3 = 6,78 QH.

(lo = 0,6 l in cm n v ca thn ct; h dd = hB = hC = 11,3m)


vo s pht trin chiu di ca knh st v c xc nh bng cng thc sau:

(

(

! 1

2120

h

H

h

h

H

HtvhtM

dd

dd

dd ln..).(

.ln.,)(

F

y Vi:

hdd = hA = hA = 11,3 m : L treo cao ca dy dn pha B,C.

H = hdd + hc =11,3 + 18,4 = 29,7 m.

h = hc - hdd = 18,4 - 11,3 = 7,1 m. hc = 18,4 m : L cao ca ct.


F = 0,3



97/131



v = F . c = 0,3 . 300 = 90 m/ Qs.

c l tc nh sng: c = 300 m/ Qs.

+ Thay vo cng thc tnh M dd(t) ta c :

244,161,38

7,29t.90ln26,2)t(Mdd

!

+ Vi t = 3Qs Mdd(3) = 5,775 QH.

y Vy )t(M.a)t,a(dt

di78,6)t,a(U ddCt -c !


diC (10,3) = 7,52 KA/Qs;

Mdd

(3) =5,775 QH ),(- 310tcU = 108,74 KV.

j Xc nh in p do dng in i trong dy chng st gy ra :

[Kvq.Ucs(a,t)]

+ Ucs(a,t) c tnh bng cng thc :

)t(M.a)t,a(dt

diLR)t,a(i)t,a(U csCcsCCCcs ! yy

Cc thng s trn cng thc trn khng thay i so vi khi tnh ton vi pha A

Ta c kt qu tng t:

Khi a =10 KA/Qs

t =3 Qs ),(U cs 310 = 433,12 KV

j Xc nh tc dng trn cch in pha B,C : CBcdU,

lvcsvq

csAd-c

t

-cCCC,B

cd U)t,a(U.K)t,a(U)t,a(UR).t,a(i)t,a(U !

vi Kvq = vqcsBK =vq

csCK = 0,1638.

+ Vi cc gi tr va tnh c trn ta c :

9,48318,5712,433.1638,004,14474,10810.49,24)t(U C,Bcd !! KV.

Nhn xt : 37,516)3,10(UAcd ! KV 9483310 ,),(UC,B

cd !" KV.


98/131



+ V in p tc dng ln cch in pha A ln hn in p tc dng ln

cch in pha B,C nn ta chn pha A tnh ton in p tc dng l n chui

cch in khi c st nh vo nh ct hoc ln cn nh ct.

2. Tnh in p tc dng ln chui cch in )t,a(U cd :

+ Trong phn trc, ta tnh in p tc dng ln chui cch in ca

cc pha vi mt gi tr c th ca cp th ng s (a,t) so snh 2 gi tr in p

cc pha A v B,C . Sau khi xc nh c xc nh c gi tr qu in

p kh quyn tc dng ln cch in pha A l nguy him nht. Th ta s i tnh

ton in p t ln chui cch in ca pha A : )t,a(U cd theo gi tr ca cp

thng s (ai,ti) thay i.

j in p tc dng ln chui cch in c tnh bng cng thc :

! yyy atMtadt

diLtaURtaitaU ddCdd

cdcCCcd )(),(),(),(),( -

lvcsv2 UtaUK y ),( ; (a =

dt

dis )

+ Vi : )(.),(),(),( tMatadt

diLRtaitaU csCcsCCCcs ! yy

! y ),()..(),().(),(),( - tadt

dicscLvKddcLtaUvKRtaitaU CdcCCcd 1

lvcsvqdd UatMKtM .)(.)( + Thay cc gi tr : RC = 10 ;; Ulv = 57,18 KV;

255,0KK vq csAvq

!!

; ;H76,8LddC Q! H04,11LcsC Q! vo cng thc trn ta c :

! y )t,a(dt

di).04,11.255,076,8()t,a(U)255,01.(10)t,a(i)t,a(U CCcd

d-c

lvcsdd Ua.)t(M.255,0)t(M

Hay: ! )t,a(dt

di.945,5)t,a(U)t,a(i.45,7)t,a(U CCcd

d-c

lvcsdd Ua.)t(M.255,0)t(M y Trong :


99/131



098,29,42

33t.90ln92,2)t(M dd

!

68,384,47

8,36t90ln68,3)t(M cs

!

y

y

(Kt qu tnh Mdd(t) v Mcs(t) vi cc gi tr ti thay i nh trong bng

(3 - 6 )

!22,348

)33t.90).(8,3t.90().4,18t.90(ln.a.302,3)t,a(Ud-c

(Kt qu tnh ),(- taUdc vi s thay i ca a v t c tnh trong bng tnh 3.7 )

- Cc cng thc tnh Mdd(t); Mcs(t) v ),(-

taUdc

c xc nh trong

mc 1/Chn pha tnh cc sut phng in.

Gi tr iC(a,t) v )t,a(dt

diC c xc nh trong 2 trng hp nh sau :

+ Trng hp 1: Trc khi c sng phn x t ct ln cn tr v ng

vi khong thi gian t < 2.lkv/c = !300

200.21,33Qs. Trong trng hp ny iC(a,t)

v )t,a(dt

diC c tnh bng cc cng thc sau:

E

! yyy

y 1

22

vqcscsvq

cs

Cvqcs

C

Z)t(MtZ

RZ

a)t,a(i

- Vi :


100/131



12,1904,11.2

10.22,402

L.2

R.2Zcsc

vqcs

1C !

!

!E

vqcsZ : L tng tr sng ca dy chng st c xt n nh hng

ca vng quang: vqcsZ = 402,2;.

Thay vo cng thc ta c :

! yy12,19

2,402)t(M2t.2,402

10.22,402

a)t,a(i csC

? A04,21)t(M2t.2,4022,422

a)t,a(i csC ! yy

a.952,0

10.22,402

2,402.a

R.2Z

Z.a)t,a(

dt

di

C

vq

cs

vqcsC

!

!

!

+ Trng hp 2: Sau khi c sng phn x t ct ln cn tr v, ng vi khong

thi gian t u 2.lkv/c = !300

200.21,33Qs. Trong trng hp ny iC(a,t) v ),( ta

dt

diC

c tnh bng cc cng thc sau:

t.054,0e1.)t(M.253,34810.2

a)t,a(i csC

!

t.054,0e.054,0.)t(M.253,34820a)t,a(

dt

di csC !

- Hai cng thc trn c xc nh trong mc 1/Chn pha tnh cc

sut phng in.

Vi s thay i ca cp gi tr (a i,ti) ta tnh c cc gi tr iC(a,t) trong

bng tnh 3.8 v cc gi tr )t,a(dt

diC trong bng tnh 3.9.

j T cc gi tr Mdd(t); Mcs(t); ),(- taUdc ; iC(a,t) v )t,a(

dt

diC vi s thay i ca

a v t tnh c trn, ta tnh c U c(a,t) theo cng thc:


101/131



! )t,a(dt

di.945,5)t,a(U)t,a(i.45,7)t,a(U CCcd

d-c

lvcsdd Ua.)t(M.255,0)t(M + Cc gi tr ca Uc(a,t) c tnh trong bng tnh 3 - 10.

T cc gi tr ca Uc(a,t) trong bng tnh 3 - 10 ta v c cc ng

Uc(a,t) = f(t) vi a thay i nh trn hnh 3 - 13. Trn hnh 3 - 13 cng biu

din ng c tnh V S ca chui cch in.

3/ Tnh xc sut phng in:

+ Khi in p t trn chui cch in ln hn in p phng in ca chui

cch in th s xy ra phng in trn chui cch in. T th trn hnh

3 - 13 ta xc nh c cc cp gi tr (a i,ti) l giao im ca Uc(a,t) v c tnh

V S ca chui cch in. T xc nh c cp thng s nguy him (I i,ai)

vi quan h Ii = ai . ti (KA).

Tng t nh trong trng hp st nh vo khong vt, ta c:

!

(!n

i aid I

1

RRR .


102/131



y Trong : 1,26I

I

i

eV

! ; 9,10a

ai

i

eV

!

+ Gi tr ca Rp c tnh trong bng 3 - 11.

3.3.5.3. Tnh sut ct do st nh vo nh ct hoc ln cn nh ct ng dy

110KV

nc = N.Rp. n = 3R .2 pdN

nc = 742,1625.0046452.02

120!vv ln /100Km. nm

3.4. Tnh ch tiu chng st ca ng dy 110KV

+ Sut ct tng trn 100Km ng dy do st nh l :

n = nE + nkv + nc = 0.1209 + 0.0424 + 1,742

n = 1,9053 ln /nm

Ch tiu chng st ca ng dy : 5248,09053,1

1

n

1m !!! nm/1ln ct.


103/131



Chng IV:TNH TON BO V CHNG SNG TRUYN

VO TRM PHA 110KV.

4.1 Khi nim chung.

Bo v chng qu p do st nh t ng dy truyn vo trm c yu cu rt

cao v :

+ Trm bin p l 1 phn t quan trng trong h thng truyn ti cng nh phnphi in nng , hn na trm bin p l mt ni t cc thit b quan trng ,t

tin nh : my bin p, my bin in p, my ct . . .Cch in ca cc thit b

trong trm li yu hn nhiu so vi cch in ca ng dy .Do nhng sng

xut hin bi sng nh trn ng dy khng gy ra phng in nhng khi

truyn vo trm li l nhng sng nguy him vi thit b .

+ Khi c st nh vo ng dy ti in, trn ng dy (ti im st nh)

s xut hin sng qu in p v cc sng ny s lan truyn t im st nh vo

trm, gy ra cc phng in trn cch in ca cc thit b, mc d cch in ca

cc thit b c thc hin sao cho mc cch in trong ln h n mc cch

in ngoi. Khi c phng in trn cch in ca trm th tng ng nh c


104/131



phng in trn thanh ci, d c nhng phng tin bo v hin i vn gy ra

nhng s c trm trng trong h thng.

+ Trn thc t cc thit b ca trm thng c t ngoi tri do phi

chu qu trnh l ho kh phc tp v qu trnh gi ci cch in din ra nhanh

v mnh hn dn n di tc dng ca qu in p th c th xy ra phng in

chc thng cch in ca thit b.

+ Tuy nhin trong qu trnh tnh ton bo v khng th m bo mc an ton

cho cc thit b ca trm mt cch tuyt i .Nhng khi tnh ton chn cc bin

php chng sng truyn phi thc hin gim xc sut s c ti gii hn thp

nht .m bo sao cho ch tiu chu st ca trm phi t ti hng trm nm

hoc hn na .+ Khi c sng qu in p truyn vo trm th cc qu in p ny s tc

dng ln cch in ca cc phn t mang in nh : my bin p, my ct . . .

gy ra phng in chc thng cch in khi cc sng qu in p ny c bin

ln hn U50% ca cc thit b .

+Mc cch in xung kch ca trm c chn theo tr s in p d ca

chng st van v c chiu hng ngy cng gim thp do cht lng ca thit b

ngy cng nng cao. m bo iu kin lm vic bnh thng ca chng st

van ta phi hn ch dng in qua n khng vt qu gii hn 5 z 10 KA.

Dng in st qu ln s lm cho in p d tng cao, nh hng ti phi hp

cch in trong ni b trm v cn c th lm hng chng st van.

*) c im ca qu trnh tnh ton :

Ch tiu bo v chng sng truyn vo trm l mt s liu rt quan trng, n

cho php nh gi kh nng lm vic an ton ca trm i vi sng qu in p.Do tham s ca sng truyn vo trm ph thuc rt nhiu tham s, n n qu trnh

tnh ton rt phc tp. Da vo qu trnh tnh ton tm ra tham s gii hn

nguy him ca sng truyn vo trm . Nu vt qu gi tr ny s xy ra phng

in t nht mt thit b trong trm. Vi tr s ti hn ca tham s sng st, bit


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phn b xc sut ca chng th c th tm c ch tiu bo v chng sng

truyn vo trm .

Qu trnh tnh ton sng truyn vo trm nhm xc nh chiu di cn thit

ca on ti trm cn bo v. V trn c s nhng s liu tnh ton, theo ch tiu

kinh t - k thut xc nh s lng, v tr t chng st van v cc thit b cn

bo v khc mt cch hp l .

Do khi lng ca qu trnh tnh ton kh ln v phc tp nn trong tnh ton

thit k tt nghip ta ch i xc nh qu in p xut hin trn cch in ca

thit b, theo mt sng truyn vo trm cho trc. Sau so snh qu in p

ny vi c tnh phng in ca thit b trnh kh nng gy phng in. V

trm s an ton nu tt c cc ng cong in p xut hin trn cch in unm di c tnh V-S ca thit b .

+ Qu trnh lan truyn sng trn ng dy trong trng hp tng qut c

m t bi h phng trnh vi phn :

4

x

x!

x

x

x

x!x

x

t

uCug

x

iti

Lrixu

..

..

yTrong :

r : L gi tr in tr ng vi n v chiu di ca ng dy.

L : L gi tr in cm ng vi n v chiu di ca ng dy.

g : L gi tr in dn ng vi n v chiu di ca ng dy.

C : L gi tr in dung ng vi n v chiu di ca ng dy.

+ Gii h phng trnh vi phn vi c 4 tham s l rt phc tp .Do n gin ho ta gi thit rng qu trnh truyn sng trong trm khng b bin

dng (v qung ng truyn sng trong trm ngn). Tc l b qua s nh

hng ca in tr r. Mt khc trm bin p 110KV c mc cch in rt cao


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dn n dng r khng ng k nn ta b qua nh hng ca in dn g. H

phng trnh vi phn m t qu trnh truyn sng c vit li nh sau :

xx!

xx

x

x!

x

x

tC

xi

ti

Lx

.

.

+ Nghim ca h phng trnh ny l s tng hp ca hai thnh phn sng

ti v sng phn x n c biu din nh sau :

U = f1(x - v.t) + f2(x + v.t)

i =Z

1.{ f1(x - v.t) - f2(x + v.t)}

+ Trong tnh ton cc i lng c tnh trn s tng ng v thng

s dng hai qui tc chnh tnh. l quy tc sng ng tr v quy tc

Petecxen.

4.1.1. Quy tc petecxen :

+ Khi mi trng truyn sng thay i th s c hin tng phn x v khc

x ca sng ti im nt .Mt sng truyn trn ng dy c tng tr Z 1n mt

tng tr tp trung cui ng dy Z 2 th s c thnh phn sng khc x sangmi trng mi c tng tr Z 2 v thnh phn phn x tr v mi trng c c

tng tr Z 1

j Xt trng hp sng truyn t mi trng Z1 sang mi trng Z2:

+ C th ta xt trng hp sng truyn t mi trng Z 1 sang mi trng Z2.

Ut-L sng ti t mi trng 1

Uk-L sng khc x sang mi trng 2

U(t)

Z1 A Z2Uf

Uk

Hnh 4.1 . Sng truyn t mi trng Z1 sang mi


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Uf-L sng phn x tr v mi trng 1

+ Gi s mi trng Z2 di v tn tc l khng c sng phn x t cui mi

trng Z2 tr v. Phng trnh in p v dng in c vit nh sau :

ut + uf= uk (1)

it + if= ik (2)Nhn c hai v phng trnh (2) vi Z 1 ta c :

it.Z1 + if.Z1 = ik.Z1

Mt khc :

ut = it.Z1

uf= - if.Z1

Do ta c : ut - uf= ik.Z1 (3)

Cng phng trnh (1) vi phng trnh (3) ta c :

2.ut = uk+ ik.Z1

Vi uk l sng khc x sang mi trng 2 do mi trng 2 di v tn nn :

i2 = ik

u2 = uk=i2.Z2 = ik.Z2

Suy ra : 2.ut = ik.(Z1 + Z2)

+ Phng trnh ny ng vi s thay th nh sau :

4.1.2. qui tc sng ng tr :

+ Trong thc t gp nhiu trng hp c nhiu phn t cng ni vo mt

im nt . Mi phn t u c mt tng tr sng Z1 , Z2 , ... , Zn v dc trn mi

phn t u c cc dng sng bt k U 1x, U2x ,...Unx. lan truyn v pha im nt

.S c th nh sau :

2.Ut

Z1

Z2

A

Hnh 4.2 S Petecxen.


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+ Ti nt X c phn t Zx, gi s gia cc phn t khng c pht sinh ngu

hp v qui c chiu dng in i v pha im nt l chiu dng ta c th

vit:

ux = u1x + ux1 = umx + uxm = unx + uxn (m = 1,2,3....n)

Trong : umx = Zm.imx

uxm = - Zm.ixm

Suy ra : ! !!!

!!

n

1m

n

1m m

x

n

1m m

mx

m

xmn

1m m

mxx

1..2i

!

!n

m

xxmmx i)ii(

1

U2x

Umx

Uxn

Zm

Z2

ZnZ1

Zx

U1x

Ux1

Unx

Ux2

UxmX

m

n1

2

Hnh 4.3. S sng ngtr


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Chia c hai v cho !

n

1m m

1ta s c

Ux + ix. t = 2.Ut

Vi : t = 1 // 2 //. . . // m

Vy 2.Ut = !

E

n

1m

mx,

m U. ;m

dtm

.2!E l h s khc x

2.Ut = ix.( t + x)

=> Ta c s thay th nh hnh v :

4.1.3.Xc nh in p trn Zx khi n l in dung :Trong trng hp ny ta c s thay th :

+ S thay th theo qui tc petecxen nh hnh 4 - 5

mx

n

1m m

dtdt U.Z

ZU !

!

2.Udt

Zdt

Zx

X

Hnh 4.4. S thay th quy tc sngng tr.

U(t) Zt

CiC(t)

Hnh 4.5. S thay th

Z

C2.Ut

Zt

UC

Hnh 4.5. S Petecxen


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Khi tng tr ZX ch c t in vi in dung C th :

+ T s thay th ta c th vit

(*)

Trong :

UC(t)-L in p trn t in C

iC(t)-L dng in i qua t in C

Zt-L tng tr sng ng tr ca n ng dy ti nt x

Do :

q = UC.C

dq = iC.dt

Nn ta c :

iC(t) =dt

dU.C

dt

dq C!

Thay vo cng thc (*) trn ta c

2.Ut(t) = UC(t) + Zt.C.dt

dUC

T ta rt ra c dng sai phn

C

)t(C)t(dt

dt

)t(C)t(dtC

T

UU.

Z.C

UU.

t

U !

!

(

( 22

Vi Tc = C.Zt

+ Cui cng ta c: Uc.(t + (.t) = {Uc.(t) + (Uc.(t)}

Vi iu kin ban u l 0CU( = 0

C

)t(C)t(dtC Tt).UU.2(U (!(

)t(cdt)t(C)t(dt i.ZUU.2 !


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4.1.4. Xc nh in p v dng in trn chng st van :

Ta chn chng st van khng khe h bo v chng sng truyn vo trm .Bi

v loi ny c nhiu u im hn chng st van c khe h .Thnh phn ch yu

ca in tr chng st van khng khe h c lm t ZnO(Chim 99,9%)

H s phi tuyn ca ZnO ch bng 1/10so vi ca SiC (Loi c khe h )!E 0,02 z 0,03.Trong khi loi c khe h !E 0,18 z 0,24

c tnh ca chng st van (V-A) c vit di dng U = A.I E

Xt min lm vic ca chng st van Icsv u 1 KA . Th in p d ca loi

chng st van c in tr phi tuyn lm bng ZnO thp hn loi chng st van

c in tr lm bng SiC : U SiCd- > UZnOd- . Nh vy ta s dng loi chng st van

khng khe h s c an ton cao hn. Ngoi ra n cn em li hiu qu kinh t

do vic lm gim thp mc cch in xung kch trong trm .

Khi khng c qu in p chng st van lm vic vi in p bnh thng th

dng in trn in tr ZnO ch l dng r c tr s rt nh so vi dng trn in

tr SiC : IZn

5

r


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020,csvI

+ S ng tr nh trn hnh 4.7

+ S thay th

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