Bai Tap Dai Cao AP Cua Thanh(1)

8/2/2019 Bai Tap Dai Cao AP Cua Thanh(1)

1/25

Trng i hc in Lc Khoa H thngin

Tnh ton thit k bo v chng st cho

trm bin p 220/110 kVBo v chng st nh trc tip v ni t cho trm 220/110 kV

A. L thuyt chung1. Yu cu i vi h thng thu sta. Cng trnh cn bo v phi c nm gn trong phm vi bo v ca h thng thu st.

H thng ny c th t ngay trn bn thn cng trnh hoc t cch ly ty thuc

vo hon cnh v iu kin c th.

t h thng thu st ngay trn cng trnh c u im l tn dng c phm vi bo

v. Do gim cao ca h thng thu st, v d nh t kim thu st trn x trm bin p

hoc treo dy chng st trn ct in. Nhng khi c phng in st, dng in st s gy

nn in p ging trn in tr ni t v trn phn in cm thn ct. Tr s in p nykh ln v c th gy phng in ngc t h thng thu st ti cc b phn nu cch in

gia chng khng chu ni. Do iu kin t h thng thu st trn cc cng trnh c

mang in l phi m bo c mc cch in cao v tr s in tr tn ca b phn ni t

b. i vi trm bin p v ng dy in p trn 110 kV tr ln, cc yu cu trn c

thc hin tng i d dng. cc cp in p thp hn, vic t h thng thu st trn

cng trnh s gp nhiu kh khn v khng hp l v kinh t k thut. V vy, trong trng

hp ny, h thng thu st c t cch ly vi cng trnh.

Khi t cch ly, gia chng cng phi c khong cch nht nh. Nu khong cch

ny qu b th vn c kh nng phng in trong khng kh cng nh trong t t h thng

thu st ti cng trnh. V nh vy, cng khng km phn nguy him so vi khi c st nh

thng vo cng trnh.

in p ti im trn thn ct cch b phn ni t on di l c tnh theo cng

thc:

btr

SXKSk

dt

dILRIU

.

..

+=

Trong :

IS Bin dng st, kAL in cm ca phn dy ni di, H

Rxk in tr ni t xung kch ca b phn ni t,

btr

S

dt

dI

.

- dc trung bnh ca phn u sng dng in st (ly theo dng sng xin

gc), kA/s

Lp 1 H1 1


2/25


Trong tnh ton thng ly IS = 150 kA vbtr

S

dt

dI

.

= 30 kA/s. i vi ct thu st

kim loi c kt cu kiu mng li hoc khi dy ni t t ring, tr s in cm theo n

v di c tr s Lo = 1,7 H/m. Nh vy c th tnh c:

lRU XKk .50.150 +=V in p trn b phn ni t bng:

XKXKSdRRIU .150. ==

Chng c th t c cc tr s rt ln, v d khi Rxk = 10 v l = 10m th Uk =

2000kV v Ud = 1500kV. Do , khng th xy ra phng in t h thng thu st ti

cng trinh, cc khong cch khng kh (sk) v khong cch trong t (sd) phi ln c

mc cch in khng thp hn so vi cc tr s in p ni trn.

Cng cch in xung kch ca khng kh thng ly bng 500kV/m v tr s

trng phng in trong t ly bng 300kV/m, t suy ra cc khong cch an ton (tnhbng mt):

XKd

d

XKk

RU

s

lRU

s

.5,0300

1,0.3,0500

1

=

+=

C th xut pht t cc iu kin an ton trn xc nh in tr ni t ca h

thng thu st hoc khi b phn ni t c sn, s cn c vo kim tra kh nng

phng in ngc.

b. Phn dn in ca h thng thu st (ca b phn thu nhn st v ca dy ni t) phi

c tit din tha mn iu kin n nh nhit khi c dng in st i qua.

Do thi gian tn ti ca dng in rt ngn nn trong tnh ton pht nng c th b

qua tn nhit vo mi trng xung quanh v nu in tr ca phn dn in theo n v di

l r th nng lng pht nng ca dng in st s bng:

=0

2.. dtirW

s

Nh trn phn tch, trong trng hp ny, dng sng tnh ton hp l l dngsng hm s m:

t

ST

t

SSSeIeIi.

7,0

..

==Vi s l di sng 0,7.t

Lp 1 H1 2


3/25


Nh vy:4,1

.... 2

0

.7,0 S

S

t

S IreIrWS

==

Nhit pht nng ca phn dn in c xc nh theo cng thc:

2

2

...4,1

..

.. SCg

I

CSs

Wt SSo

==

Trong :

- g l mt ca vt liu dn in.

- C l nhit dung trung bnh ca vt liu dn in.

- S l tit din ca phn dn in.

- l in tr sut ca vt liu.

Ly bin dng in l IS = 150kA v di sng l S = 100S (do c xt n

pht nng ph ca cc phn phng in k tip), phn dn in thng dng thp nn c g= 7,8 g/cm3, C = 0,1 cal.g.Co v tr s trong phm vi nhit 0 400Co bng 3.10-

5.cmvi cc s liu trn s tnh c nhit pht nng:

2

4,13

Sto =

Dy thp c tit din 25mm2 s b pht nng ti nhit to = 215Co, nh vy l

hon ton cho php ngay c khi t n dc theo g. Nhng vi mc ch nng cao bn

c kh v tng thi gian s dng, thng chn tit din 50,mm 2 tr ln (thp trn 8).

chng n mn, phn dn in cn c sn hoc trng km v khng nn dng

loi dy xon.

Cc mi ni dc theo mch in ca h thng thu st phi m bo c tip xc tt,

nu khng ti cc ni ny c th qu nng hoc c phng in tia la (thng dng

phng php hn in hoc ni bulng m khng c dng cc cch buc xon thng

thng).

2. Cch xc nh phm vi bo v ca mt, hai hay nhiu ct thu sta. Phm vi bo v ca mt ct thu st

Phm vi bo v ca mt ct thu st l min gii hn bi mt ngoi ca hnh chp trn xoay

c ng sinh xc nh bi phng trnh:

( )x

x

xhh

h

hh

r +

= .5,1

Trong :

Lp 1 H1 3


4/25


- h l chiu cao ct thu st.

- hx l chiu cao cn bo v.

- rx l bn knh ca phm vi bo v cho cao hx.

- h hx = ha l cao hiu dng.

Trong tnh ton thc t, ng sinh c a v dng gy khc a, b, c.

Khi h < 30m:

Lp 1 H1 4


5/25


- Khi hhx 3

2 :

=h

hhr xx .8,0

1..5,1

- Khi hhx 3

2> :

=

h

hhr

x

x1..75,0

Trng hp ct thu st cao hn 30m c th dng cng thc trn nhng phi nhn vi h s

hiu chnh p =h

5,5v trn hnh v dng cc honh 0,75h.p v 1,5h.p.

Khi h > 30m:

- Khi hhx 3

2 :

=h

hphr xx .8,0

1...5,1

- Khi hhx 3

2> :

=h

hphr xx 1...75,0

b. Phm vi bo v ca hai ct thu st

Hai ct cao bng nhau:

Gi s c hai ct thu st c cng cao h1 = h2 = h t cch nhau mt khong l a.

Khi a = 7h th mi vt nm trn mt t khong gia hai ct khng b st nh vo.

Lp 1 H1 5


6/25


Khi a < 7h th gia hai ct thu st bo v c cho cao ln nht ho = h -7

a.

Phm vi bo v gm hai phn:

+ Phn ngoi phm vi bo v xc nh nh i vi mt ct.

+ Phn trong phm vi bo v xc nh nh i vi phm vi bo v ca ct c cao ho cho cao hx.

Cch xc nh phm vi bo v ca hai ct cao bng nhau:

+ Bn knh bo v ca tng ct: rx1 = rx2 = rx

+ Bn knh bo v gia hai ct: rox

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

a

Nu hx oh.3

2 th rox =

o

xo h

hh

.8,01..5,1

Nu hx oh.3

2> th rox =

o

xo h

hh 1..75,0

Lp 1 H1 6


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Hai ct c cao khc nhau

Gi s c hai ct c cao khc nhau vi h1 > h2 v cch nhau mt khong l a.

Cch xc nh phm vi bo v:

- Xc nh phm vi bo v ca ct cao h1.

- T nh ct h2 ging ng thng nm ngang ct phm vi bo v ca h1 ti 3. Ti

3 ta t mt ct gi tng vi cao h2.

- Xc nh phm vi bo v ca hai ct c cao h2.Tnh ton phm vi bo v:

- Xc nh bn knh bo v ca tng ct: rx1 v rx2.

- Xc nh bn knh bo v gia hai ct:

aOO =21,

3132 aOOaOO ==

Lp 1 H1 7


8/25


xOO =31 l bn knh bo v ca ct h1 cho ct h2.

cao ln nht c bo v gia hai ct 2 v 3: ho = h2 -7

,a

Xc nh bn knh bo v rox ca cao ho cho cao hx tng t nh trng hp hai ct

cao bng nhau.c. Phm vi bo v ca nhiu ct thu st

Phm vi bo v ca ba ct thu st

Lp 1 H1 8


9/25


Phm vi bo v ca bn ct thu st

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

ton:

( )xhhD .8

Trong :

- D l ng knh ng trn ngoi tip tam gic, t gic.

- h l chiu cao ct.

- hx l chiu cao cn bo v.

Cch xc nh ng knh ng trn ngoi tip tam gic:

( ) ( ) ( )

2

....2

..

cbap

cpbpapp

cbaD

++=

=

Vi a, b,c l ba cnh ca tam gic.Sau xc nh phm vi bo v ca tng cp ct bin tng t nh xc nh phm vi bo

v ca hai ct.

Lp 1 H1 9


10/25


B. Tnh ton bo v chng st nh trc tip v ni t cho trm bin

p 220/110 kV

Chng I: Bo v chng st nh trc tip vo trm 220/110 kVTheo s kt cu ca trm th th mi ch bit din tch mt bng m cha bit c th v

tr t cc thit b trong trm. Vi thng tin ny ta ch cn b tr ct chng st sao cho cc

ct c th bo v c phn din tch mt bng ca trm vi cao hx.

Din tch mt bng pha 220 kV l 87x95 m v pha 110 kV l 95x92 m.

cao ln nht cn bo v pha 110 kV l 11 m v pha 220 kV l 16,5 m.

1. Phng n 1a. Tnh ton chiu cao ct

Cc ct c b tr nh hnh v:

Pha 220 kV:Xt nhm ct (1, 2, 6, 7): nhm ny to thnh hnh ch nht.

Cnh 1,2 = 43,5 m ; Cnh 1,6 = 47,5 m.

ng knh ng trn ngoi tip: D = =+ 22 5,475,43 64,41 m

iu kin phn trong cc ct thu st c bo v an ton l:

D ax hhh .8).(8 =

Lp 1 H1 10


11/25


=> 05,88

41,64

8==

Dha m

=> 55,2405,85,16 =+h m

Tnh ton tng t i vi cc nhm ct (2, 3, 7, 8), (6, 7, 11, 12) v (7, 8, 12, 13).

Pha 110 kV:Xt nhm ct (3, 4, 8, 9): nhm ny to thnh hnh ch nht.

Cnh 3,4 = 46 m ; Cnh 3,8 = 47,5 m.

ng knh ng trn ngoi tip: D = =+ 22 5,4746 66,12 m


D ax hhh .8).(8 =

=> 27,88

12,66

8==

Dha m

=> 27,1927,811 =+h mTnh ton tng t vi cc nhm ct (4, 5, 9, 10), (8, 9, 13, 14), v (9, 10, 14, 15).

T kt qu tnh ton trn, ta c th chn chiu cao thi cng ca cc ct l:

h1 = h2 = h3 = h6 = h7 = h8 = h11 = h12 = h13 = 25 m

h4 = h5 = h9 = h10 = h14 = h15 = 19,5 m.

b. Tnh bn knh bo v ca cc cp ct bin

Pha 110 kV:

cao cn bo v pha 110 kV l 11 m.

Xt cp ct (3, 4): h3 = 25 m ; h4 = 19,5 m ; a34 = 46 m.Trc ht ta xt bn knh bo v ca ct h3 cho ct h4 : '33OO

Do h4 > 3.3

2h nn =

=

==

25

5,191.25.75,01..75,0

3

4

3

'33

h

hhxOO 4,13 m

Khong cch t ct h4 n ct gi tng h3 l : a = a34 x = 46 4,13 = 41,87 m

cao ln nht c bo v gia ct h4 v ct h3 l :

ho4-3 = ==7

87,415,19

7

,

4

ah 13,52 m

Do hx > '34.32 oh nn rox4-3 = 0,75. =

=

52,13

111.52,13.75,01.'34

'34o

xo h

hh 1,89 m

Bn knh bo v ca ct 3 v ct 4 cho cao h x l: rx3 v rx4.

Do hx < 4.3

2h nn rx4 = rx3 =1,5. =

=

5,19.8,0

111.5,19.5,1

.8,01.

44 h

hh x 8,63 m

Lp 1 H1 11


12/25


Do hx < 3.3

2h nn rx3 =1,5. =

=

25.8,0

111.25.5,1

.8,01.

33 h

hh x 16,88 m

Tnh ton tng t cho cp ct (13, 14).

Xt cp ct (4, 5): h4 = h5 = 19,5 m; a45 = 46 m.

cao ln nht c bo v gia hai ct l: ho4-5 = ==7465,19

745

4ah 12,93 m

Do hx > 54.3

2oh nn rox4-5 = 0,75. =

=

93,12

111.93,12.75,01.

5454

o

xo h

hh 1,45 m


Do hx < 4.3

2h nn rx4 = rx5 =1,5. =

=

5,19.8,0

111.5,19.5,1

.8,01.

44 h

hh x 8,63 m


Xt cp ct (5, 10): h5 = h10 = 19,5 m; a5-10 = 47,5 m.

cao ln nht c bo v gia hai ct l: ho5-10 = == 7

5,475,19

7

1055

ah 12,71 m

Do hx > 105.3

2oh nn rox5-10 = 0,75. =

=

71,12

111.71,12.75,01.

105105

o

xo h

hh 1,28 m

Bn knh bo v ca ct 5 v ct 10 cho cao hx l: rx5 v rx10.

Do hx < 5.3

2h nn rx5 = rx10 =1,5. =

=

5,19.8,0

111.5,19.5,1

.8,01.

55 h

hh x 8,63 m


Pha 220 kV:

cao cn bo v pha 220 kV l 16,5 m.

Xt cp ct (1, 2): h1 = h2 = 25 m ; a12 = 43,5 m.

cao ln nht c bo v gia hai ct l: ho1-2 = ==7

5,4325

712

1

ah 18,79 m

Do hx > 21.3

2oh nn rox1-2 = 0,75. =

=

79,18

5,161.79,18.75,01.

2121

o

xo h

hh 1,72 m


Do hx < 1.3

2h nn rx1 = rx2 =1,5. =

=

25.8,0

5,161.25.5,1

.8,01.

11 h

hh x 6,56 m

Tnh ton tng t cho cp ct (2, 3), (11, 12) v (12, 13).

Xt cp ct (1, 6): h1 = h6 = 25 m ; a16 = 47,5 m.

Lp 1 H1 12


13/25



5,4725

716

1

ah 18,21 m

Do hx > 61.3

2oh nn rox1-6 = 0,75. =

=

21,18

5,161.21,18.75,01.

6161

o

xo h

hh 1,28 m


Do hx < 1.3

2h nn rx1 = rx6 =1,5. =

=

25.8,0

5,161.25.5,1

.8,01.

11 h

hh x 6,56 m


Theo kt qu tnh ton trn, ta c bng tng kt:

Pha 110 kV:

rx3 rx4 rx5 rx10 rx13 rx14 rx1516,88 8,63 8,63 8,63 16,88 8,63 8,63

rox4-3 rox4-5 rox14-15 rox14-13 rox5-10 rox10-151,89 1,45 1,45 1,89 1,28 1,28

Pha 220 kV:

rx1 rx2 rx3 rx6 rx11 rx12 rx136,56 6,56 6,56 6,56 6,56 6,56 6,56


c. V phm vi bo v

Lp 1 H1 13


14/25


2. Phng n 2

Lp 1 H1 14


15/25


a. Tnh ton chiu cao ct

Cc ct c b tr nh hnh v:

Pha 220 kV:

Xt nhm ct (1, 2, 6, 7): nhm ny to thnh hnh ch nht.Cnh 1,2 = 43,5 m ; Cnh 1,6 = 47,5 m.

ng knh ng trn ngoi tip: D = =+ 22 5,475,43 64,41 m


D ax hhh .8).(8 =

=> 05,88

41,64

8==

Dha m

=> 55,2405,85,16 =+h m

Tnh ton tng t i vi cc nhm ct (2, 3, 7, 8), (6, 7, 11, 12) v (7, 8, 12, 13).Pha 110 kV:

Xt nhm ct (3, 4, 8, 9): nhm ny to thnh hnh ch nht.

Cnh 3,4 = 62 m ; Cnh 3,8 = 47,5 m.



Lp 1 H1 15


16/25


D ax hhh .8).(8 =

=> 76,98

1,78

8==

Dha m

=> 76,2076,911 =+h m

Tnh ton tng t i vi nhm ct (8, 9, 13, 14).Xt nhm ct (4, 5, 9, 10): nhm ny to thnh hnh ch nht.

Cnh 4,5 = 30 m ; Cnh 4,9 = 47,5 m.



D ax hhh .8).(8 =

=> 02,78

18,56

8==

Dha m

=> 02,1802,711 =+h m

Tnh ton tng t i vi nhm ct (9, 10, 14, 15).

Theo kt qu tnh ton trn, ta c th chn chiu cao ct thi cng l:

h1 = h2 = h3 = h6 = h7 = h8 = h11 = h12 = h13 = 25 m

h4 = h9 = h14 = 21 m

h5 = h10 = h15 = 18,5 m

b. Tnh bn knh bo v ca cc cp ct bin

Pha 110 kV:

Xt cp ct (3, 4): h3 = 25 m ; h4 = 21 m ; a34 = 62 m.Trc ht ta xt bn knh bo v ca ct h3 cho ct h4 : '33OO

Do h4 > 2.3

2h nn =

=

==

25

211.25.75,01..75,0

3

43

'33

h

hhxOO 3 m

Khong cch t ct h4 n ct gi tng h3 l : a = a34 x = 62 3 = 59 m


ho4-3 = ==7

5921

7

,

4

ah 12,57 m

Do hx > '34.32

oh nn rox4-3 = 0,75. =

=

57,12

111.57,12.75,01.

'34

'34

o

x

o

h

hh 1,78 m


Do hx < 4.3

2h nn rx4 = rx3 =1,5. =

=

21.8,0

111.21.5,1

.8,01.

44 h

hh x 10,88 m

Lp 1 H1 16


17/25


Do hx < 3.3

2h nn rx3 =1,5. =

=

25.8,0

111.25.5,1

.8,01.

33 h

hh x 16,88 m


Xt cp ct (4, 5): h4 = 21 m ; h5 = 18,5 m ; a45 = 30 m.

Trc ht ta xt bn knh bo v ca ct h4 cho ct h5 : '44OO

Do h5 > 4.3

2h nn =

=

==

21

5,181.21.75,01..75,0

4

54

'44

h

hhxOO 1,88 m

Khong cch t ct h5 n ct gi tng h4 l : a = a45 x = 30 1,88 = 28,12 m


ho5-4 = ==7

12,285,18

7

,

5

ah 14,48 m

Do hx > '45.3

2oh nn rox5-4 = 0,75. =

=

48,14

111.48,14.75,01.

'45

'45

o

x

o

h

h

h 2,61 m


Do hx < 5.3

2h nn rx5 = rx4 =1,5. =

=

5,18.8,0

111.5,18.5,1

.8,01.

55 h

hh x 7,13 m

Do hx < 4.3

2h nn rx4 =1,5. =

=

21.8,0

111.21.5,1

.8,01.

44 h

hh x 10,88 m


Xt cp ct (5, 10): h5 = h10 = 18,5 m; a5-10 = 47,5 m.

cao ln nht c bo v gia hai ct l: ho5-10 = == 7

5,475,18

7105

5

ah 11,71 m

Do hx > 105.3

2oh nn rox5-10 = 0,75. =

=

71,11

111.71,11.75,01.

105

105

o

x

o

h

hh 0,53 m

Bn knh bo v ca ct 5 v ct 10 cho cao hx l: rx5 v rx10.

Do hx < 1.3

2h nn rx5 = rx10 =1,5. =

=

5,18.8,0

111.5,18.5,1

.8,01.

55 h

hh x 7,13 m

Tnh ton tng t cho cp ct (10, 15).Pha 220 kV:

Xt cp ct (1, 2): h1 = h2 = 25 m ; a12 = 43,5 m.


5,4325

7

121

ah 18,79 m

Lp 1 H1 17


18/25


Do hx > 21.3

2oh nn rox1-2 = 0,75. =

=

79,18

5,161.79,18.75,01.

2121

o

xo h

hh 1,72 m


Do hx < 1.3

2h nn rx1 = rx2 =1,5. =

=

25.8,0

5,16

1.25.5,1.8,01. 11 h

h

hx

6,56 m

Tnh ton tng t cho cp ct (2, 3), (11, 12) v (12, 13).

Xt cp ct (1, 6): h1 = h6 = 25 m ; a16 = 47,5 m.


5,4725

716

1

ah 18,21 m

Do hx > 61.3

2oh nn rox1-6 = 0,75. =

=

21,18

5,161.21,18.75,01.

6161

o

xo h

hh 1,28 m


Do hx < 1.3

2h nn rx1 = rx6 =1,5. =

=

25.8,0

5,161.25.5,1

.8,01.

11 h

hh x 6,56 m


Theo kt qu tnh ton trn, ta c bng tng kt:

Pha 110 kV:

rx3 rx4 rx5 rx10 rx13 rx14 rx1516,88 10,88 7,13 7,13 16,88 10,88 7,13


Pha 220 kV:

rx1 rx2 rx3 rx6 rx11 rx12 rx136,56 6,56 6,56 6,56 6,56 6,56 6,56


c. V phm vi bo v

Lp 1 H1 18


19/25


3. La chn phng n ti uDa vo tng chiu cao ct st m ta s dng lm ct thu st cho trm, ta s chn phng

n ti u l phng n c tng chiu cao ct st nh hn.

Phng n 1:

Chiu cao, m S ct Tng, m19,5 6 11725 9 225

Tng chiu cao ct 342

Phng n 2:

Chiu cao, m S ct Tng, m18,5 3 55,5

21 3 6325 9 225Tng chiu cao ct 343,5

Ta thy phng n 1 c tng chiu cao ct thp hn. Do , ta chn phng n 1 l

phng n ti u thi cng.

Lp 1 H1 19


20/25


Chng II: Tnh ton h thng ni t cho trm 220/110 kV- Nhim v ca ni t l tn dng in st xung t m bo cho th trn vt ni t

c gi tr b.

- Trong vic bo v qu in p, ni t ca trm, ca cc ct thu st, ca ng dy v

ca thit b chng st rt quan trng. V trm trm bin p c cp in p t 110kV v220kV nn ni t an ton v ni t chng st chung nhau.

1.Ni t an ton:- Ni t an ton vi mc ch bo v con ngi bng cch ni t cc b phn kim loi c

th mang in p khi ngn mch. V d nh: ni t v my in, ct st ca ng dy ti

in.

- Ni t an ton vi trm c in p t 110kV tr ln phi tha mn 2 iu kin sau:

1

5,0//

NT

NTTNHT

R

RRR

Trong : - RTN: in tr ni t t nhin.

- RNT: in tr ni t nhn to.

Vi ni t t nhin (RTN) ta li dng h thng dy chng st - ct:

4

1

2

1

1

++

=

cs

c

c

TN

R

R

R

nR

Trong : - n: s tuyn ng dy ca trm.

- Rcs: in tr ca ng dy chng st trong mt khong vt.

Dy chng st C-70 c Ro = 2,38 /km.

Rcs = Ro.l = 2,38. l

- Rc: in tr ni t ca ct. Rc = 8

- l: chiu di khong vt.

Chiu di khong vt ca ng dy 110 kV: l = 300 m = 0,3 km.

Chiu di khong vt ca ng dy 220 kV: l = 250 m = 0,25 km.

Vi cc gi tr trn, ta c in tr ni t ca tng cp in p:

- Trm 110 kV c s l l n = 4 l:

=++

= 515,0

4

1

3,0.38,2

8

2

1

8

4

1TNR

- Trm 220 kV c s l l n = 2 l:

Lp 1 H1 20


21/25


=++

= 952,0

4

1

25,0.38,2

8

2

1

8

2

1TNR

Vy in tr ni t t nhin ca h thng c hai cp in p 110 kV v 220 kV l:

=+

=+

== 325,0952,0515,0

952,0.515,0.//

)220()110(

)220()110(

)220()110(

TNTN

TNTN

TNTN

HT

TNRR

RRRRR

Vi RNT: in tr ni t nhn to.

TNNT

TNNT

RR

RR

+.

0,5

V iu kin RNT 1 nn ta ly RNT = 1 . Ta thit k h thng ni t theo in tr ni

t nhn to.

in tr tn ca h thng ni t mch vng quanh trm:

..

.ln

..2

2

dh

Lk

LRR

ttTMV

==

Trong : - RT: in tr ca thanh ni t.

- tt : in tr sut tnh ton.

tt = o .kma = 75. 1,6 = 120 .m

- h: chn su. h = 0,8 m.

- L: chu vi mch vng ni t. L = 2. (l1 + l2) = 2. (179 + 95) = 548 m

- d: ng knh in cc trn. Nu l thp dt th d = b/2 vi b l b rng thanh

S dng thp dt 4x40 mm2 ,do d = 20 mm = 2.10-2 m- k: h s hnh dng. k = f (l1/l2) = f (179/95) = f (1,88) = 6,27. (Xc nh bng

ni suy).

Lp 1 H1 21


22/25


Vi cc s liu trn, ta c gi tr in tr ni t mch vng l:

.65,02.10.8,0

548.27,6ln.

548..2

1202

2

==

MV

R

Do RMV < RNT nn ta khng phi ng cc vo h thng ni t gim in tr ni t

ca h thng.

2. Ni t chng st:V trm s dng h thng ni t phn b di (ni t mch vng nn ng thi phi xt

n c hai qu trnh l: qu trnh qu v qu trnh phng in trong t. Do trm c hai

cp in p 110 kV v 220 kV nn ta ch cn tnh ni t chng st cho cp 110 kV l

c. cp in p ny, ni t an ton v ni t chng st chung nhau nn ta dng ni

t an ton kim tra li cho ni t chng st.

a. Xt qu trnh phng in trong t:

Quy i cc in tr ni t an ton sang ma st:

m

m

tt

kkRR

'

' .=

Trong : - Rt: in tr tn ca thanh vo ma kh.

- km v km ln lt l h s ma ca thanh chn su 0,8 m trong loi ni t

chng st, ni t an ton, lm vic vo ma kh.

Vi: Rt = RMV = 0,65 .

Lp 1 H1 22


23/25


km = 1,2

km = 1,6

Vy Rt = 0,65. =6,12,1

0,49 .

b. Xt qu trnh qu :S ng tr ca h thng ni t:

Cc tham s L, G c xc nh theo cng thc:

Lo = 0,2.

31,02/ln

r

LH/m

Trong : - L: chiu di in cc. L = 548 m

- r: bn knh tit din thanh in cc.

r = ==4

40

4

b10 mm = 0,01 m

Suy ra:

Lo = 0,2.

31,001,02

548

ln = 1,98 H/m

in dn: Go =LR.

11/m

Trong : - R: in tr ni t n nh ca cc.

- L: chiu di in cc.

Suy ra: Go = 548.49,0

1= 0,0037 (1/m)

T s ng tr c th thnh lp c h phng trnh vi phn:

=

=

uGx

it

iL

x

u

o

o

.

.

Lp 1 H1 23


24/25


Sau khi gii cc phng trnh vi phn trn vi dng sng ca dng in u vo ca b

phn ni t c dng sng xin gc i(0,t) = a.t, ta s c tng tr xung kch u vo.

( )( )

( )

+==

=

12

1 1.1

..21

.

1

,0

,0,0

k

T

t

o

ke

kt

T

LGtI

tUtZ

xc nh c Z(0,s), ta xt cc chui s sau:

Chui s: 643,16

...1

...3

1

2

1

1

112

22221

2==+++++=

=

kkk

Chui s: ......321

.1

2222

1

2

321

+++++=

=

k

eeeee

k

k

dsdsdsds

k

ds TTTT

T

k

Trong chui s ny, ta ch xt n s hng cha e-4 (t s hng cha e-5 tr i c gi tr rt

nh so vi cc s hng trc nn ta c th b qua). Tc l ta tnh vi k sao cho:

4k

ds

T

k Z+

Ta c: T1 = =

=

2

2

2

2

1.

2

548.0037,0.98,1

2..

LGL oo 55,78 s

ds = 5 s

4

21

=

k

TTds

k

ds

hayds

Tk

12 .4

Suy ra: 68,65

78,55.2.2 1 ==

ds

Tk

Ta chn k trong khong t 1 6 (k Z+)Ta c bng kt qu sau:

k 1 2 3 4 5 6

Tk 55,78 13,945 6,198 3,486 2,231 1,549

k

ds

Te

0,914 0,7 0,446 0,238 0,106 0,04

Lp 1 H1 24


25/25


2k

e kds

T

0,914 0,175 0,05 0,015 0,004 0,001

=

4

12

k

T

k

e kds

1,159

Tng tr xung kch ca h thng ni t l:

( )( )( )

+===

=

1

2

1

'

1.1

..212,0

,0,0.

2

1

k

T

t

t

XK

kekt

TR

ti

tutZZ

( )

+= 159,1643,1.586,29.21.

249,0

661,1=

iu kin ni t chng st tha mn khi:

UXK= ZXK. IS < UT50%

Vi dng in st: IS = 150 kA

in p 50% ca trm, i vi trm 110 kV, U50% = 460 kV

Suy ra: UXK= 1,661. 150 = 249,15 kV < 460 kV.

Vy h thng ni t an ton cng tha mn cho ni t chng st ca trm. Do takhng cn ni t b sung.

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