bai tap dai cao ap 2

7/28/2019 bai tap dai cao ap 2

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BI TP DI MN KT CAO P

BI TP DI MN K THUT IN CAO P

H v tn sinh vin: Nguyn Tin HngLp: 1-H2 Khoa: H Thng in

Gio vin hng dn: Th.S Trn Hong Hip

TNH TON THIT K BO V CHNG ST CHOTRM BIN P 220/110KV

I. S LIU BAN U: * Trm bin p: Pha 220 kV c din tch: (120.90) m2

Pha 110 kV c din tch: (100.90) m2

* in tr sut ca t: =100 m* ng dy:

- Trm 220 kV c 5 l- Trm 110 kV c 10 l- Dy dn: ACO 240- Dy chng st: C-70

* Chiu di khong vt ca ng dy 110 kV: l = 300 m* Chiu di khong vt ca ng dy 220 kV: l = 300 m* Khi tnh ni t: Rc = 6

II. NI DUNG BI TP DI

Phn I:* Chng I: Tnh ton bo v st nh trc vo trm bin p 220/110 kVA. L thuyt chung

+ Cc yu cu i vi h thng thu st+ Cch xc nh, tnh ton PVBV ca h thng thu st ( 1 ct, 2 ct )

B. Tnh ton* Chng II: Tnh ton h thng ni t trm bin p 220/110 kV

Phn II: Cc bn v lin quan

SVTH:Bi Trung Dng 1


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I.1. C s l thuyt chung:

I.1.1. Nhng yu cu i vi h thng thu st:

Cc thit b phn phi in t ngoi tri nh ng dy, trm bin p rt d b

qu in p c th l do qu in p kh quyn (st nh trc tip, cm ng hay lan truyntrn ng dy), hoc qu in p ni b. Trong , s qu in p kh quyn do st nh

l rt nguy him v gy nhng thit hi nghim trng cho cc cng trnh trong h thng

in. V vy, bo v chng st nh trc tip l mt trong nhng yu cu hng u khi

thit k v vn hnh mt mng in.

H thng thu st l mt b phn cng trnh quan trng nhm bo v cc b phn

ca h thng in nh ng dy, trm bin p khi h hng khi b st nh. i vi

ng dy, bo v chng st nh trc tip ngi ta s dng h thng dy chng st.i vi trm bin p v nh my in ngi ta s dng cc ct thu li. Cc ct thu li c

th t c lp hoc trong iu kin cho php c th t trn cc kt cu ca trm v ca

nh my.

Yu cu chnh i vi h thng thu st l phi c in tr ni t nh m

bo tn nhanh dng in st xung t, trnh hin tng phng ngc dng in st t

thit b ny sang thit b khc, hoc t ct thu st hay dy chng st sang cc cng trnh

mang in t ln cn. Khi thit k bo v chng st th cn m bo cc yu cu v kthut, kinh t v m thut.

- i vi cc trm phn phi ngoi tri t 110 kV tr ln do c mc cch in cao

nn c th t ct thu li trn kt cu ca trm phn phi. Cc tr ct ca cc kt cu trn

c t ct thu li phi c ngn nht v sao cho dng in st I S khuch tn vo t

theo 34 thanh ci ca h thng ni t . Ngoi ra mi tr ca kt cu y phi c ni

t b sung ci thin ch s in tr ni t.

- Ni yu nht ca trm phn phi ngoi tri vi in p t 110 kV tr ln l cun

dy my bin p, v vy khi dng chng st van bo v my bin p th yu cu

khong cch gia hai im ni vo h thng ni t ca ct thu li v v my bin p

theo ng in phi ln hn 15m.

SVTH:Bi Trung Dng 2


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

thc hin cc im sau:

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

ni t b sung (dng ni t tp trung) nhm m bo in tr khuch tn khng cqu 4 (ng vi dng in tn s cng nghip).

+ Khi b tr ct thu li trn x ca trm 35kV phi tng cng cch in ca

n ln n mc cch in ca cp 110 kV.

+ Trn u ra ca cun dy 6 10kV ca my bin p phi t cc ct chng

st van (CSV), cc thit b chng st ny c th t ngay trn v my.

+ bo v cun dy 35 kV cn t cc ct chng st van. Khong cch gia

ch ni vo h thng ni t ca v my bin p v ca chng st van (theo ng in)

phi nh hn 5m. Khong cch y c th tng ln nu im ni t ca chng st van

vo gia hai im ni t ca v my bin p v ca kt cu trn c t ct thu li.

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

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

- C th ni ct thu li c lp vo h thng ni t ca trm phn phi cp inp 110kV nu nh cc yu cu trn c thc hin.

- Khng nn t ct thu li trn kt cu ca trm phn phi20 35 kV, cng nh

khng nn ni cc ct thu li vo h thng ni t ca trm 20 35 kV.

- Khi dng ct thu li c lp 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 vt c bo v.

- Khi dng ct n chiu sng lm gi cho cc ct thu li phi cho dy dn

in n n vo ng ch v chn vo t.

- i vi cc nh my in dng s b th ch c t ct thu li trn x my

bin p khi my pht in v my bin p c ni vi nhau bng cu bc kn v hai u

SVTH:Bi Trung Dng 3


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c ni t. Nu cu c phn on th khng c php t ct thu li trn x ca my

bin p. Vi my b ng b cng p dng iu ny.

- C th ni dy chng st bo v on n trm vo h thng ni t ca trm nu

nh khong cch t ch ni t ca trm n im ni t ca my bin p ln hn 15m.

- m bo v mt c tnh ( bn c hc) v chng n mn cn phi theo ng

quy nh v loi vt liu, tit din dy dn dng trn mt dt v di t phi theo bng

sau:

Bng 1.1 Bng qui nh qui cch loi dy dn dng dn dng in st

Loi vt liu Dy dn dng in st

dng trn mt t

Dy dn dng in st

dng di mt t

Thp trn m km 8 mm 10 mm

Thp dt m km 20 x 2,5 mm2 30 x 3,5 mm2

Cp thp Khng c dng Khng c dng

Thanh ng trn 8 mm 8 mm

Thanh ng dt 20x2,5 mm2 20x2,5 mm2

Dy ng xon Khng c dng Khng c dng

Thanh nhm trn Khng c dng Khng c dng

1.2 Cch xc nh phm vi bo v ca h thng thu st:1.2.1 Phm vi bo v ca mt ct thu st:

Phm vi bo v ca mt ct thu st c cao l h tnh cho cao h x l mt hnhchp trn xoay c ng sinh c xc nh nh sau:

SVTH:Bi Trung Dng 4


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rx

)hh(

h

h1

6,1x

x

+

=

hx

r

rx

h

Hnh 1.1. Phm vi bo v cho mt ct thu st.

Trong :- h: chiu cao ct thu st.- hx: chiu cao cn c bo v.- h hx: chiu cao hiu dng.Trong tnh ton, ng sinh c a v dng ng gy khc abc c xc nh

nh sau:

h

0,8hh x

23h

0,75h

1,5h

rx

a

b

c

Hnh 1.2. Phm vi bo v ca mt ct thu st (ng sinh gp khc).

Trong :ab: ng thng ni t nh ct n im trn mt t cch xa chn ct mt

khong l 0,75h.bc: l ng thng ni 1 im c cao trn thn ct l 0,8h n 1 im trn mt

t cch chn ct l 1,5h.

SVTH:Bi Trung Dng 5


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Khi: hx h3

2

Th: rx = 1,5h(1-h8,0

hx ) = 1,5h 1,875hx

Khi: hx

h

3

2

Th: rx = 0,75h(1-h

hx ) = 0,75h 0,75hx

Cc cng thc ch s dng cho HTTS c cao h < 30m. Khi h 30m ta cnhiu chnh cc cng thc theo h s p.

ph

5,5=

1.2.2 Phm vi bo v ca hai ct thu st:a. Hai ct thu st c cao bng nhau:

Xt 2 ct thu st c cao bng nhau h1 = h2 = h, cch nhau 1 khong a.

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

+ Khi a = 7h th mi vt nm trn mt t khong gia 2 ct khng b st nhvo.

+ Khi a < 7h th khong gia 2 ct s bo v c cho cao ln nht h 0 c xc

nh nh sau: h0 = h -7

a

Phm vi bo v:

- Phn ngoi: ging nh ca tng ct.- Phn gia: cung trn i qua 3 im 1,2,3 (im 3 l im t ct gi tng c

cao h0.+) Tnh ton phm vi bo v:- Bn knh bo v ca tng ct: rx1 = rx2 = rx- Bn knh bo v gia hai ct: r0x.

SVTH:Bi Trung Dng 6


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- cao ln nht bo v c gia hai ct: h0 = h -7

a

Nu: hx0

h3

2

Th: r0x = 1,5h0.(1 - 0x

h8,0

h

)

Nu: hx 0h3

2

Th: r0x = 0,75h0.(1 -0

x

h

h)

Cc cng thc trn c p dng khi h thng chng st c cao nh hn 30m.Nu h thng chng st c cao ln hn hoc bng 30m th cc cng thc cng cnc hiu chnh theo h s p nu mc.b. Hai ct thu st c cao khc nhau:

Xt 2 ct thu st c cao l h1 v h2, cch nhau 1 khong a c b tr nh hnh

v:

Hnh 1.4 Phm vi bo v ca 2 ct c cao khc nhau.

+ Xc nh phm vi bo v:

- Phn ngoi: ging nh ca tng ct.- Phn trong: t nh ct h1 dng ng thng nm ngang ct phm vi bo v cact h2 ti 3, vi 3 l v tr t ct gi tng c cao l h1.

- Phn gia: ging nh ca hai ct c cng cao h1.( xaOOOOaOO 2

'

321

''

31=== , x l bn knh bo v ca ct cao h2 cho ct

gi tng '1h .+ Tnh ton phm vi bo v:

SVTH:Bi Trung Dng 7


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- Tnh bn knh bo v tng ct rx1, rx2.- Tnh bn knh bo v gia hai ct rox .- Khong cch gia ct thp v ct gi tng 3

a = a x ( trong x l bn knh bo v ca ct cao h2 cho ct gi tng c cao h1).

- cao ln nht c bo v gia 1, 3: '301h = h1 -7

'a

I.1.2.3.Phm vi bo v cho nhiu ct thu st:

+) Phm vi bo v cho 3 ct thu st:

Phm vi bo v cho 3 ct thu st c cao h1 = h2 = h3 = h cng bo v cho cao

hx c minh ha nh hnh v di y

r0x2-3

r0x1-3 r0x1-2

rx3 rx2

rx1

Hnh 1.5. Bn knh bo v ca 3 ct c cao bng nhau

l phn din tch nm trong ng bao m cc ng trn to ra.

Trong :

- rx1 = rx2 = rx3 = rx: bn knh bo v ca tng ct.- rox1-2 = r0x1-3 = r0x3-2: bn knh bo v chung gia cc ct 1-2, 2-3, 3-1.

+) Phm vi bo v cho 4 ct thu st:

Phm vi bo v cho 3 ct thu st c cao h 1 = h2 = h3 = h4 = h cng bo v cho cao hx c minh ha nh hnh v di y.

SVTH:Bi Trung Dng 8


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1 2

34

r0x1-2rx1

Hnh 1.6. Phm vi bo v ca bn ct thu st c cao bng nhauPhm vi bo v l phn din tch cng trnh nm trong ng bao ngoi cng ca

hnh v.

Vi: rx1 = rx2 = rx3 = rx4 = rx

rox1-2 = r0x3-4: bn knh bo v chung gia cc ct 1-2, 3-4

rox1-4 = r0x2-3: bn knh bo v chung gia cc ct 1-4, 2-3

iu kin cng trnh nm trong min gii hn bi cc ct thu st c bo v

an ton l:D8.(h-hx))

- trong : D l ng trn ngoi tip phn mt bng c dng hnh tam gic, ch

nht; h: chiu cao ct thu st; hx: chiu cao cn bo v.

II. Tnh ton:

SVTH:Bi Trung Dng 9


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Theo s kt cu ca trm, ta mi ch bit din tch mt bng trong trm m

cha bit v tr ca cc thit b trong trm. V vy ch cn b tr cc ct thu st sao cho

bo v c phn din tch mt bng c cao hx.

i vi pha 220 kV: hx = 17mi vi pha 110 kV: hx = 11m

Ta thc hin tnh ton cho cc pha. Ta s xt 2 phng n t v tr ct thu st cho

trm bin p.

*) Phng n A

Trong phng n ny, ta s dng 15 ct thu st c b tr trn mt bng trm

bin p nh hnh v.

120.000 100.000

1 4 7 10 13

14

1512963

25 118

50.

000

40.

000

60.000 50.000

A-1. Tnh ton phm vi bo v cho pha 220 kV:

iu kin cn cng trnh c bo v an ton l:

D8.(h-hx)

Suy ra: hhx+8

D

*)Xt nhm ct (5,8,9,6) to thnh hnh ch nht bo v cho cao 16,5m.

+ Cnh (5-8) = 60m = a

+ Cnh (5-6) = 50 m =b

ng knh D1 ca ng trn ngoi tip qua cc nh ct l:

SVTH:Bi Trung Dng 10


11/37


D= 2 2 2 2a b 60 50 78,102+ = + = m

cao tc dng ti thiu cho cc ct bo v hon ton din tch gii hn bi

chng l:

ha = 78,1028

= 9,763 m

cao ca ct thu li l:

h = hx + ha = 17 + 11 = 28 m

A-2. Tnh ton phm vi bo v cho pha 110 kV:

Xt nhm ct (11,14,12,15) to thnh hnh ch nht bo v cho cao 11m.

+ Cnh (11-14) = 50m = c

+ Cnh (11-12) = 50 m = d

ng knh D ca ng trn ngoi tip i qua cc nh ct l:

D = 2 2 2 2c d 50 50+ = + = 70,71 m

cao tc dng ti thiu cho cc ct bo v hon ton din tch gii hn bi

chng l:

ha =70,71

8= 8,84 m

cao ca ct thu li l:

h = hx + ha = 11 + 8 = 19 m

*)Tnh bn knh bo v ca ct thu li c b tr nh hnh v:

- Tnh bn knh bo v ca mt ct thu li

+Xt ct 5:

Ta c: hx=17(m) 2

3h0=

2

3.19,357=12,905(m)

Vy bn knh ca khu vc bo v gia hai ct thu li l:

rx2-3=0,75.h0(1-x

0

h

h)=0,75.19,357.(1-

16,5

19,357)=2,143 (m)

+Xt 2 ct (1-2): khong cch gia 2 ct l 40 m

Ta c: h0=h-a

7=26,5-

40

7= 20,786 (m)

hx=16,5(m) >2

3h0=

2

3.20,786=13,857(m)



13/37



rx1-2=0,75.h0(1-x

0

h

h)=0,75.20,786.(1-

16,5

20,786)=3,215 (m)

Pha 110kV:chiu cao cn bo v hx=11m+Xt 2 ct (11-14): khong cch gia 2 ct l 50 m

Ta c: h0=h-a

7=20-

50

7= 12,857 (m)

hx=11(m) >2

3h0=

2

3.12,857=8,571(m)


rx11-14=0,75.h0(1-x

0

hh

)=0,75.12,857.(1- 1112,857

)=1,393 (m)

+Xt 2 ct (10-11): khong cch gia 2 ct l 40 m

Ta c: h0=h-a

7=20-

40

7= 14,286 (m)

hx=11(m) >2

3h0=

2

3.14,286=9,524(m)


rx10-11=0,75.h0(1-x

0

h

h)=0,75.14,286.(1-

11

14,286)=2,464 (m)

Hai ct ct chiu cao khc nhau:

+Xt hai ct (7-10) vi khong cch gia 2 ct a=50m

Chiu cao ct cao h7=26,5m

Chiu cao ct thp h10 =20m.

Ta tnh bn knh bo v ca tng ct cho cao 11 m. Ct 10 theo kt qu phn trn ta

tnh c l rx10 = 9,375 m.

Ta tnh cho ct 7.

- Bn knh bo v ca ct 7 l rx7.

Ta c:



14/37


hx = 11m < 72 2

h 26,5 17,673 3

= = m. Nn:

Bn knh bo v ca ct 7 l:

rx7 = 1,5.h(1- xh0,8.h )=1,5.26,5(1- 110,8.26,5 )= 19,125 m.

Bn knh bo v ca ct cao h7=26,5m cho cao cn bo v l chiu cao ca ct thp

h10 =20m.

Ta c: hx=h7=20m >2

3.26,5=17,67m

x=0,75.h(1- xh

h

)=0,75.26,5.(1-20

26,5

)=4,875 (m)

Ta c: a= 50-4,875=45,125m

'0710h =h10-,a

7=20-

45,125

7= 13,553(m)

hx=11(m) >2

3'0710

h =2

3.13,553=9,035 (m)

Vy bn knh ca khu vc bo v gia hai ct thu li c cao hx=11m l:

rx7-10=0,75. '0710h (1-x

0

hh

)=0,75.13,553.(1- 1113,553

)=1,915 (m)

Tnh ton tng t cho cc ct cn li ta c kt qu ghi trong bng sau:NHM CT C CAO BNG NHAU

PHA 220KV hx = 16,5m



15/37


Cp ct cao ct(m)

a(m)

h0(m)

2

3.h0

(m)

r0x(m)

3-6 26,5 60 17,928 11,952 1,071

2-3 26,5 50 19,357 12,905 2,1434-5 26,5 40 20,786 13,857 3,215PHA 110KV hx = 11m

Cp ct cao ct(m)

a(m)

h0(m)

2

3.h0

(m)

r0x(m)

13-14 20 40 14,286 9,524 2,46411-12 20 50 12,857 8,571 1,393

NHM CT C CAO KHC NHAU hx = 11mCp ct cao ct

(m)

a

(m)

x

(m)

a

(m)

h0

(m)

r0x

(m)8-11 20 50 4,875 45,125 13,553 1,915

Phm vi bo v ca cc ct thu st trong phng n 1

1 4 7 10 13

14

1512

963

2 5118

R8.812 R9.3751.071

3.215

2.143

1.0711.915 1.393

1.393

2.464

R19.125

R9.375

R8.812

Kt lun.Phng n bo v tha mn yu cu t ra.Tng s ct l 15 ct, trong c 9 ct cao 26,5m v 6 ct cao 20mTng chiu di : L1 = 9.(26,5 16,5) + 6.(20 11) =144 (m).

*) Phng n B



16/37


Trong phng n ny, ta s dng 21 ct thu st c b tr trn mt bng trm

bin p nh hnh v.

120.000

40.000

100.000

1 4 7 10 13

14

1512963

2 5 118

16

17

18

19

21

90.

000

20

40.000

50.

000

30.000 40.000

B-1 Tnh ton phm vi bo v cho pha 220 kV:

iu kin cn cng trnh c bo v an ton l:

D8.(h-hx)

Suy ra: h

hx+ 8D

Xt nhm ct (2,3,5,6) to thnh hnh ch nht bo v cho cao 16,5 m

+Cnh (2-3)=50m=e

+Cnh (3-6)=40m=f


D = 2 2 2 2e f 50 40+ = + = 64,031 m

cao tc dng ti thiu cho cc ct bo v hon ton din tch gii hn bi chng l:

ha =64,031

8= 8,004 m

cao ca ct thu li l: h = hx + ha = 16,5 + 8 = 24,5 m

B-2 Tnh ton phm vi bo v cho pha 110 kV:

Xt nhm ct (17,18,20,21) to thnh hnh ch nht bo v cho cao 11 m



17/37


+Cnh (17-18)=50m=k

+Cnh (18-21)=40m=l


D =2 2 2 2

k l 50 40+ = + = 64,031 m cao tc dng ti thiu cho cc ct bo v hon ton din tch gii hn bi chng l:

ha =64,031

8= 8,004 m

cao ca ct thu li l: h = hx + ha = 11 + 8 = 19 m

*)Tnh bn knh bo v ca ct thu li c b tr nh hnh v:

- Tnh bn knh bo v ca mt ct thu li

+Xt ct 6:

Ta c: hx=16,5(m) >2

3h=

2

3.24,5=16,33(m)

Vy bn knh bo v ca 1 ct thu li vi cao cn bo v hx=16,5m

rx6=0,75.h(1- xh

h)=0,75.24,5.(1-

16,5

24,5)=6 (m)

+Xt ct 20:

Ta c: hx=11(m)2

3h0=

2

3.17,357=11,571(m)


rx2-3=0,75.h0(1- x0

hh )=0,75.17,357.(1- 16,517,357 )=0,643 (m)

+Xt 2 ct (3-6): khong cch gia 2 ct l a=40 m

Ta c: h0=h-a

7=24,5-

40

7= 18,786 (m)

hx=16,5(m) >2

3h0=

2

3.18,786=12,524(m)


rx3-6=0,75.h0(1-x

0

h

h)=0,75.18,786.(1-

16,5

18,786)=1,714 (m)

Pha 110kV:chiu cao cn bo v hx=11m


Ta c: h0=h-a

7=19-

50

7= 11,857 (m)

hx=11(m) > 23

h0= 23

.11,857=7,905(m)


rx20-21=0,75.h0(1-x

0

h

h)=0,75.11,857.(1-

11

11,857)=0,643 (m)


Ta c: h0=h-a

7=19-

40

7= 13,286 (m)

hx=11(m) >2

3h0=

2

3.13,286=8,857(m)




19/37


rx18-21=0,75.h0(1-x

0

h

h)=0,75.13,286.(1-

11

13,286)=1,714 (m)


Ta c: h0=h- a7

=19- 307

= 14,714 (m)

hx=11(m) >2

3h0=

2

3.14,714=9,809(m)


rx15-18=0,75.h0(1-x

0

h

h)=0,75.14,714.(1-

11

14,714)=2,785 (m)

Hai ct ct chiu cao khc nhau:+Xt hai ct (10-13) vi khong cch gia 2 ct a=30m

Chiu cao ct cao h10=24,5m

Chiu cao ct thp h13 =19m.

Ta tnh bn knh bo v ca tng ct cho cao 11 m. Ct 13 theo kt qu phn trn ta

tnh c l rx13 =7,875 m.

Ta tnh cho ct 10.

- Bn knh bo v ca ct 10 l rx10.Ta c:

hx = 11m < 102 2

h 24,5 16,333 3

= = m. Nn:

Bn knh bo v ca ct 10 l:

rx10 = 1,5.h(1-xh

0,8.h)=1,5.24,5(1-

11

0,8.24,5)= 16,125 m.

Bn knh bo v ca ct cao h10=24,5m cho cao cn bo v l chiu cao ca ct

thp h13 =19m.

Ta c: hx=h10=19m >2

3.24,5=16,33m



20/37


rx=0,75.h(1- xh

h)=0,75.24,5.(1-

19

24,5)=4,125 (m)

Ta c: a= 30-4,125=25,875m

'01013h =h10-,

a7

=19- 25,8757

= 15,303(m)

hx=11(m) >2

3'01013

h =2

3.15,303=10,202 (m)

Vy bn knh ca khu vc bo v gia hai ct thu li c cao hx=11m l:

rx10-13=0,75. '01013h (1-'

x

01013

h

h)=0,75.15,303.(1-

11

15,303)=3,227 (m)

Tnh ton tng t cho cc ct cn li ta co kt qu ghi trong bng sau:NHM CT C CAO BNG NHAU

PHA 220KV hx = 16,5mCp ct cao ct

(m)a

(m)h0

(m)2

3.h0

(m)

r0x(m)

5-6 24,5 50 17,357 11,571 0,6431-2 24,5 40 18,786 12,524 1,714

PHA 110KV hx = 11m

Cp ct cao ct(m) a(m) h0(m) 23.h0

(m)

r0x(m)

20-21 19 50 11,857 7,905 0,64318-21 19 40 13,286 8,857 1,71415-18 19 30 14,714 9,809 2,785

NHM CT C CAO KHC NHAU hx = 11mCp ct cao ct

(m)a

(m)x

(m)a

(m)h0

(m)r0x(m)

10-13 24,5-19 30 4,125 25,875 15,303 3,227



21/37


Phm vi bo v ca cc ct thu st trong phng n 2

1 47

1013

14

15

96

3

2 5 118

16

17

18

19

21

20

1.714 1.714

0.643

2.785 1.714

0.643

1.714

3.227R6.000

R7.875R6.000

R7.8751.714

R16.125 R7.875R6.000

Kt lun.Phng n bo v tha mn yu cu t ra.Tng s ct l 21 ct, trong c 12 ct cao 24,5m v 9 ct cao 19mTng chiu di : L1 = 12.(24,5 16,5) + 9.(19 11) =168 (m).

Tng kt 2 phng n.

C hai phng n trn u tha mn nhu cu k thut ra. Qua tnh ton ta

thy phng n A c chiu di ct tnh ton nh hn do vy chi ph xy dng cng

nh hn v vy ta chn phng n A lm phng n thit k.



22/37


CHNG II

TNH TON H THNG NI T CHO TRM BIN P 220/110kV

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.



23/37


*)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:tt

250R ( ).

I

+ 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 :

tt

125R ( ).

I



24/37


+ 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 tnhn 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: tt d mua.k = Trong : tt l in tr sut tnh ton ca t.

d l in tr sut o c ca t.

muak l h s ma ca t.

H s muak 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 d =100 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:



25/37


Pha 220kV: l220 = 300m.

Pha 110kV: l110 = 300m.

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

( )

( )

3

cs.220 o 220

3

cs.110 o 110

R r .l 2,38.300.10 0,714

R r .l 2,38.300.10 0,714

= = =

= = =

S l trong trm:

Trm 220kV : n = 3 l.

Trm 110kV : n = 5 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 :

( )TN NTHT TN NTTN NT

R .RR R / /R 0,5

R R= =

+

Trong : TNR : in tr ni t t nhin.

NTR : in tr ni t nhn to, NTR 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:

cTN

c

cs

R1R .

n R1 1

2 R 4

=+ +

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.



26/37


- Pha 220 kV.

( )cTN.220c

cs.220

R1 1 6R . . 0,581

n 3R 1 6 11 1

2 0,714 42 R 4

= = = + ++ +

- Pha 110 kV.

( )cTN.110c

cs.110

R1 1 6R . . 0, 349

n 5R 1 6 11 1

2 0, 714 42 R 4

= = = + ++ +

( )1

TN TN.220 TN.110

1 1R R / /R 0,208

0, 581 0, 349

= = + =

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.

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

ttMV

K.LR .ln

2. .l d.t

=

Trong L l chu vi mch vng L=2.(l1+l2)=2.(90+220)=620(m)

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:

tt d mua.k =

Tra bng vi thanh ngang chn su 0,8m ta c kmua = 1,6



27/37


tt d mua.k 100.1,6 160( m) = = =

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

( )

22b 4.10

d 2.10 m2 2

= = =

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/L2 1 1,5 2 3 4

K 5,53 5,81 6,42 8,17 10,4

th

T th ng vi t s l1/l2

=220/90=2,444 ta ni suy

c K=7,144

Nh vy in tr mch vngl :

( )2 2

ttMV 2

K.L 160 7,144.620R .ln .ln 0,7792. .l d.t 2. .620 2.10 .0,8= = =

MV NTR R 0,779 1= = < nh vy in tr ca h thng ni t nhn to t yu

cu. Nn khng phi ng thm cc vo h thng ni t

Ta c in tr ni t ca h thng:


5,535,53

5,81

6,42

8,17

10,4

7,144

l1/l2

K

1 1,5 2 2,444 3 4


28/37


HT NT TN

0,208.0,779R R / /R 0,164( )

0,208 0,779= = =

+

HTR 0,164 0,5= < Nn h thng ni t chn l tha mn yu

cu ca ni t lm vic v an ton.

2.2.2 Ni t chng st.

y cp n c hai qu trnh ng thi xy ra khi c dng in tn trong t :

- Qu trnh qu ca s phn b in p dc theo chiu di in cc.

- Qu trnh phng in trong t.

Khi chiu di in cc ngn (ni t tp trung) th khng cn xt qu trnh qu

m ch cn xt qu trnh phng in trong t. Ngc li khi ni t dng hnh thc phn

b di (tia di hoc mch vng) th ng thi phi xt c hai qu trnh c nh hng khc

nhau n hiu qu ni t. V hnh thc ni t chng ta chn cho trm bin p l ni t

di (mch vng theo chu vi trm) nn s ng thi xt c hai qu trnh. V h thng ni

t ca chng ta c dng cho c ba nhim v nn ta s dng ni t an ton v lm

vic tnh ton cho ni t chng st.

*)Tnh ton ni t phn b di khng xt n qu trnh phng in trong t.S ng tr ca ni t c th hin nh sau:

L R

G C

L R L R L R

G C G C G C

Hnh 2.2 : S ng tr ca h thng ni t.

Trong mi trng hp u c th b qua in tr tc dng R v n b so vi tr s

in tr tn, ng thi cng khng cn xt ti phn in dung C v ngay c trong trng

hp sng xung kch, dng in dung cng rt nh so vi dng in tr tn.



29/37


L

G

L

G

L

G

L

G

Hnh 2.3: S ng tr rt gn.

Trong : L0 - in cm ca in cc trn mt n v di.

G0 - in dn ca in cc trn mt n v di.

0

lL 0,2. ln 0,31 H / m

r =

Vi: l - chiu di in cc, l = L=620m (ni t mch vng theo chu vi ca trm)

r - bn knh cc phn trc nu cc l thp dt c b rng b (m).

Do r = b/4 = 0,01m

Thay s ta c :

0

l 620L 0, 2. ln 0,31 0, 2. ln 0,31 2,145 H / m

r 0, 01

= = =

Khi ta c in dn ca in cc:

0

1 1G

R.l m =

Trong : R: in tr xoay chiu tnh cho ma st:

Trong ni t chng st khi dng thanh ngang chn su 0,8 m th kma = 1,25.at

set setMVMV muaat

mua

RR .kk

=

V ta thc hin vic ni t bng cch dng mt mch vng bao quanh trm m

khng ng thm cc nn gi tr in tr nhn to ma st c tnh theo cng thc:



30/37


setset set setmuaNT MV muaat

mua

R 0,779.1,25R R .k 0,608

k 1,6= = = =

Thay s vo ta c in dn ca in cc l:

3

01 1 1G 2,653.10 ( )

R.l 0,608.620 m= = =

*)Tnh tng tr xung kch ca h thng ni t.

T s ng tr ta c:

Gi Z(x,t) l in trng xung kch ca ni t ko di, n l hm s ca khng

gian v thi gian:U(x,t)

Z(x,t)

I(x,t)

=

Trong U(x,t) ; I(x,t) l dng in v in p xc nh t h phng trnh vi phn:

U IL.

x tI

G.Ux

= =

Gii h phng trnh vi phn trn vi dng sng ca dng in u vo ca h

thng ni t c dng sng xin gc i(o,t) = a.t ta s c in p ti im bt k trn

in cc:

( )Kt/ T1 2k 1

1 k. .xU(x, t) a.R t 2.T 1 e .cos

k l

=

= +

Suy ra tng tr xung kch u vo ca ni t:

( )kt /T1

2k 1

2.TU(0, t) 1 1Z(0, t) 1 1 e

I(0, t) G.l t k

=

= = + Trong

2 2

0 0 0 01 k ds2 2 2

L .G .l L .G .lT ,T , t 5 s

k .= = = =



31/37


32/37


Do vic xc nh Zbs l phc tp nn thun tin cho qu trnh tnh ton ta chn hnh

thc ni t b sung nh sau:

- Thanh ni t b sung l thp dt trn c chiu di l 10 m, b rng b = 0,04 m.; dc

theo chiu di thanh c chn ba in cc trn c kch thc: chiu di cc l 2,5 m;

ng knh d = 0,04 m, khong cch gia hai cc l 5 m. Cc v thanh c chn

su 0,8 m.

Ta c trong ni t chng st:

- Vi thanh ngang chn su 0,8 m th h s kma = 1,25.

- Vi cc di 2,5 m chn su 0,8 m th h s kma = 1,15.

Nh vy, in tr ca thanh ni t b xung bng:T 2 2

ttT

K.l 1,25.100 1.10R .ln ln

0,042. .l t.d 2.3,14.10 0,8.2

= =

= 17,388

in tr ni t ca cc b sung:

C

ttC

2l 1 4t ' l 1,15.100 2.2,5 1 4.2,05 1R . ln ln ln ln

2. .l d 2 4t ' l 2 .2,5 0,04 2 4.2,05 1

+ + = + = + = 36,246 .

in tr ni t b sung ca h thng thanh cc c xc nh theo cng thc:

t cbs

t c C t

R .RR

R . .n R .=

+

trong :

:, CT h s s dng ca thanh v cc.

Vi n = 3; lcc = 2,5 m; a = 5 m =>2

5,2

5

l

a==

. Tra bng 3 v 5 phn ph lc schHng dn thit k tt nghip k thut in cao p ca Nguyn Minh Chc, ta c

9,0T = ; 86,0C =

Thay cc gi tr vo cng thc ta tnh c: Rbs17,388.36,246

17,388.0,86.3 36,246.0,9=

+ = 8,134 .



33/37


* Ta tnh tng tr ca h thng khi c ni t b sung:

Ta cng thc tnh ton gi tr tng tr ni t khi c ni t b xung l:

2

d/ sk21

X. Tbs NT (st ) NT (s t )

XK d /sNT (st )bs NT (s t )

2

bs K

R .R 2.R Z (0; ) .e

R 1R R

R cos x

= ++ +

= A + B

Trong :

A bs ND

bs ND

R .R 8,134.0,6080,566

R R 8,134 0,608= = =

+ + .

2d/sk

21

X.

TNT( st)

NT( st )

2

bs K

2.RB .e

R 1

R cos x

=

+

Xt chui s:

=

1k

T.

X

1

s/d2

2k

e

Ta ch xt n gi tr e-4 (v t e-5 tr i c tr s rt nh).

iu ny tng ng vi vic tnh xk sao cho:

2

d/ sk 1k2

1 d/s

x T. 4 x 2.

T

xk> 0

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

kch.

Ta coi mch vng ca h thng l s ghp song song 2 tia c cng di l 614/2

= 307 m v ta c s thay th cho mt tia:

Trong :

- 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.



34/37


Do , ta tnh c:

T1

' ' 2 2 3 2

2 2 2

L.g .l L.G.l 2,145.2, 653.10 .31055,41

3,14

= = =

=> xk 1xk

T 55,412. . 2.3,14.5

= = 20,916

d/s = 5s.

Trong :

xk l nghim ca phng trnh:

tgxk

NDMVk k k k

BS

R 0,608x tgx x 0, 075.x

R 8,134= = =

Phng trnh ny tng ng vi h sau:y = tgxk

y = - 0,075.xk



35/37


Ta dng phng php th (v bng phn mm matlab) gii phng trnh ny:

Ta c cc nghim:

- X1 = 0,023 rad suy ra cos X1 =0,999

- X2 = 1,473 rad suy ra cos X2 = 0,097- X3 = 3,015 rad suy ra cos X3 = -0,992- X4 = 4,33 rad suy ra cos X4 = -0,373- X5 = 5,962 rad suy ra cos X5 = 0,949- X6 = 7,232 rad suy ra cos X6 = 0,583- X7 = 8,955 rad suy ra cos X7 = -0,892- X8 = 9,997 rad suy ra cos X8 = -0,84



36/37


- X9= 11,947 rad suy ra cos X9 = 0,814

Khi k = 10 th xk= 21,015 > 20,916 nn ta ch ly 9 nghim nh trn. Sau ta lp

bng tnh cc gi tr ca chui s vi cc xk trn.

Ta c T1 = 55,41 S =>ds

1

50,09

T 55,41

= =

V: Rbx = 8,134 =>MV

bx

R 0,6080,075

R 8,134= =

Ta c bng tnh chui s nh sau:

Bng . Bng tnh cc chui s

K 1 2 3 4 5 6 7 8 9

Xk 0,023 1,473 3,015 4,33 5,962 7,232 8,955 9,997 11,947

CosXk 0,999 0,097 -0,992 -0,373 0,949 0,583 -0,892 -0,84 0,814

k2 xcos

11,002 106,281 1,016 7,187 1,11 2,942 1,257 1,417 1,509

k2

BS

MV

xcos

1

R

R+ 1,077 106,356 1,091 7,262 1,185 3,017 1,332 1,492 1,584

1

dsT 0,09 0,09 0,09 0,09 0,09 0,09 0,09 0,09 0,09

2

dsk

1

x.

T

0,09 0,31 1,011 1,989 3,691 5,389 8,215 10,216 14,552

Ta c:

+

= 1s/d

2

2k

T.

X

k2

bx)st(NT

)st(NTk e.

xcos1

R

R

R.2B

T bng trn ta c:

==

9

1ikBB = 1,496

Do , tng tr xung kch:



37/37


Zxk= A + B= 1,496 + 0,566 = 2,062

Vy in p khi c dng in st i vo h thng ti thi im t = ds l:

U = Zxk.IS = 2,062.150 = 309,3 kV < U50%MBA=460kV.Nh vy h thng c ni t

b xung nh trn l tha mn.

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