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n : Tnh ton thit k chng st cho ng
dy v trm bin p 110/22KV
H thng in l mt h thng quan trng ca h thng nng lng
Vit Nam v khng th thiu c trc cng cuc hin i ho, cng nghip
ho t nc. Do ngun in thng t xa ni tiu thin nng nn phi
chuyn qua cc trm bin p tng hoc gim in p. i vi nc ta l nc
c kh hu nhit i gi ma, m h thng in li ko di t Bc vo Nam do
phi i qua nhiu vng kh hu khc nhau c bit l nhng ni c m
cao, mt ging st nhiu. Thit hi do st gy ra cho ngnh in v nn
kinh t quc dn l rt ln.
V vy chng ta phi u t vo nghin cu, tm ra nhng gii php
chng st nh vo cc nh my, trm bin p, ng dy gim n mc
ti thiu thit hi do st gy ra cho nn kinh t. Vi yu cu nh vy, n
thit k ca em gm bn thuyt minh ny v km theo bn v thit k v bov chng st cho trm bin p 110/22/0,4KV v ng dy 110KV.
Do thi gian c hn nn vic thit k ca em khng trnh khi nhng
sai st, em mong c s ch bo gip ca cc thy c gio b mn h
thng in. ng thi em xin chn thnh cm n thy gio Nguyn nh
Thng tn tnh hng dn em hon thnh n thit k na v em cng
xin cm n cc thy c gio b mn quan tm ch bo, hng dn chng
em trong vic thit k n tt nghip.
Mt ln na em xin chn thnh cm n!
Ngy 29 thng 4 nm 2003Sinh vin thit k
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on Vn Minh
CHNG MU
------o0o-----
TNH HNH GING ST VIT NAM V NH HNG CA NTI LI IN
ic nghin cu ging st v bin php chng st c lch s
lu di cng vi s pht trin ca ngnh in. Ngy nay
ngi ta tm ra c cc phng php, nhng h thng thit b v k thut
cao phng st nh mt cch hu hiu v an ton.
Tuy nhin mt , thi gian xy ra st nh khng th don c
trc nn vic nghin cu chng st l rt quan trng c bit l ngnh in .
I - TNH HNH GING ST VIT NAM.
- Theo ti KC - 03 - 07 vin Nng Lng, trong mt nm s ngy st
nh mim Bc khong t 70 - 100 ngy v s ln c ging t 150 - 300
ln. Vng c ging nhiu nht trn min Bc l khu vc: Mng Ci, Tin Yn
(Qung Ninh) hng nm c t 100 - 110 ngy ging st, thng 7 - 8 c th c
n 25 ngy ging trn mt thng.
- Mt s vng c a hnh chuyn tip gia cc vng ni v vng ng
bng, s ln ging cng n 200 ln/ 1 nm, vi s ngy cng n 100
ngy/nm. Ni t ging nht l Qung Bnh, hng nm ch c khong 80 ngy
ging. Xt v din bin ca ma ging trong nm, ma ging khng honton ng nht gia cc vng. Ni chung mim Bc ma ging tp trung t
thng 4 n thng 9, min Ty Bc tp trung khong t thng 5 n thng 8
trong nm.
V
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- Trn vng duyn hi Trung Bpha Bc n Qung Ngi l khu vc
tng i nhiu ging trong thng 4 n thng 8 s ngy ging xp x 10
ngy/thng. Thng nhiu nht l thng 5 c th t 12 n 15 ngy. Nhng
ngy u ma v cui ma ch c 2 n 5 ngy/ thng.
- T Bnh nh trvo l khu vc t ging nht thng ch c vo thng 5
vi s ngy xp x bng 10 ngy (Tuy ho: 10 ngy; Nha Trang 8 ngy; Phan
Thit: 13 ngy), cn cc thng khc ca ma ng ch quan st c t 5 n
7 ngy ging st.
- Min Nam cng c kh nhiu ging hng nm quan st c t 40 - 50
ngy tu tng ni. Khu vc nhiu nht l ng bng Nam B, s ngy ging
st c th ln ti 120 n 140 ngy. Ma ng Nam B t thng 4 nthng 11 s ngy ging trung bnh 10 ngy/thng cn t thng 5 n thng 10
c khong trn 20 ngy ging ( Si Gn: 22 ngy; H Tin: 28 ngy)
- Ty Nguyn ma dng thng ch c t cc thng 4,5 v thng 9. Thng
cc i (thng 5) trung bnh quan st c 15 ngy ging v Ty Nguyn
trung bnh s nga ging t 10 - 12 ngy(Ply cu: 17 ngy; Kon Tum: 14
ngy, Lt: 10 ngy) cn cc thng khc ca Ma ng trung bnh c t 5
n 7 ngy/thng.
- Qua s liu kho st ta thy rng trung bnh ging st 3 min Bc - Trung -
Nam, nhng vng ln cn li c mt tng i ging nhau. Kt qu nghin
cu ngi ta lp c bn - phn vng st Vit Nam (cc thng s cho
trong bng 1-1).
Bng 1-1:
VngNgy ging
trung bnh(ngy/nm)
Giging trungbnh(h/nm)
Mt sttrung bnh
Thngging cc
i
ngBngvenbin Mim Bc
81,1 215,6 6,47 8
Min ni trung 61,6 219,1 6,33 7
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du Bc B.
Cao nguynMinTrung
47,6 216,21 3,31 5,8
Ven binMinTrung
40,0 95,2 3,55 5,8
ng BngMin Nam
60,1 89,32 5,37 5,9
- T cc s liu v ngy giging, s lng o lng nghin cu thc
hin qua cc giai on c th tnh ton a ra cc s liu d kin v mt
phng in xung cc khu vc (sliu dbo trong bng 1-2 ).
Bng 1-2:
S ngy
dng
ng bng
ven bin
Min ni
trung du Bc
B
Cao nguyn
min trung
Ven bin
min trung
ng bng
Min Nam
20 40 2,43 4,68 2,1 4,2 1,2 2,4 1,22 2,44 1,26 2,52
40 60 4,68 7,92 4,2 6,3 2,4 3,6 2,44 3,65 2,52 3,78
60 80 7,92 9,82 6,3 8,4 3,6 4,8 3,65 4,87 3,78 5,06
80 100 9,82 12,15 8,4 10,5 4,8 6,0 4,87 6,09 5,06 6,3
100 120 12,1514,58 10,5 12,6 6,0 7,2 6,09 7,31 6,3 7,76
II. SNH HNG CA GING ST TI H THNG IN:
- Vit Nam, trong khun kh ti cp nh nc KC - 03 - 07 lp t
cc thit b ghi st v b ghi tng hp trn cc ng dy ti in trong nhiu
nm lin tc, kt qu thu thp tnh hnh s c li in 220 KV min Bc t
nm 1987 n nm 1992 c ghi trong bng 3.- Trong tng hp s c vnh cu ca ng dy trn khng 220 KV Ph Li
- H ng, nguyn nhn do st l 8/11 chim 72,7%. Sdly kt qu s c
ca ng dy Ph Li - H ng lm kt qu chung cho s c li nh
Min Bc v y l ng dy quan trng ca Min Bc v s cng dy
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ny nh hng rt ln n tnh hnh truyn ti in nng trn li in. Ngoi
ra theo ti nghin cu khoa hc - nghin cu gii php k thut hon thin
h thng chng st bo v li in trung p ca Vin Nng Lng tp
trung thng k, phn bit tnh hnh s c li in 35 n 110 KV ca Min
Bc ni chung v li in quc gia ni ring t nm 1976 n nm 1982.
Kt qu thng k c th hin bng 1- 3.
Bng 1- 3:
NmDi 220 KV ng dy Ph Li - H ng
Tng s Vnh cu Tng s Vnh cu Do st
1987
1988
1989
1990
1991
1992
2
5
24
25
30
19
1
2
3
4
2
1
2
5
6
2
3
4
1
2
2
1
1
4
1
1
1
1
1
3
Tng s 105 16 22 11 8
- T thng k ny cho ta thy rng: Tng s c do st gy ra i vi ng
dy trung p l ln hn tng s ca ng dy cao p, nhng tc hi ca n
mang li i vi h thng in l rt ln. S ln s c li in 35KV chim
t l kh ln n 54%.
Bng 1- 4:
Nm theo di Cp in p KVTng chiu di
ng dy (Km)
Sut s c (ln /100Km)
Tng s Do st
1979110
35
1050
3400
8,5
11,5
1,85
6,00
1980110
35
1100
3400
6,8
10,7
1,46
5,40
1981 110 1150 5,6 2,50
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35 3800 7,7 1,20
1982110
35
3300
11200
6,9
10,8
2,50
5,10
*Kt lun: Qua nghin c tnh hnh ging st Vit Nam v nhng thit
hi do s gy ra cho li in l rt ln nn vic m bo chng st cho ngdy in v trm bin p l rt cn thit nhm gim n mc thp nht s c ct
in ng dy. V vy vic u t nghin cu chng st l rt quan trng
nng cao tin cy cung cp in v trong vn hnh li in quc gia.
CHNG I
------o0o-----
TNH TON TRNG ST NH TRCTIP VO TRM BIN P
rong h thng in (HT), trm bin p (TBA) ng vai tr
quan trng. N quyt nh rt ln v tin cy cung cp
in ca ton HT v mt yu t quan trng dn n s mt n nh ca
HT trong nhiu yu t l do sng st qu in p truyn vo trm tng
dy v do st nh trc tip vo trm.
- Qua thng k, ngi ta thy s ln s c ca HT do b st nh chim t
l rt ln so vi cc trng hp khc. V vy vic tnh ton bo v TBA do st
nh hn mc ti thiu l mt nhim v ht sc quan trng v cn thit.
Khi b st nh vo trm, cc thit bin s b h hng v c th dn n
vic cung cp in b ngng ton b , ng thi c th lm nh hng trc
tip n qu trnh sn xut in nng v cc ngnh kinh t khc.
- m bo chng st nh trc tip vo trm cn dng cc ct thu st vdy chng st. Cc ct thu st ny c thc t c lp hoc trong iu
kin cho php ta vn dng chiu cao ca cc ct, x ct n chiu sng. Ring
ct c lp thng tn km hn v kinh t nn ch dng khi khng th tn
dng c chiu cao ca cc ct, x...
T
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- Nu t cc ct thu st trn cc kt cu ca trm phn phi ngoi tri v
dng dy chng st bo v cho don dy dn ni t s cui cng ca trm
n ct u tin ca ng dy th chng sc ni t chung vi h thng
ni t chung ca trm v tng ln nh ny cn tu thuc vo thng s ca
dng in st v in trni t xung kch ca h thng. Khi in p ny
vt qu gii hn cho php th c th gy nn nguy him cho cc thit bin,
do vy ch trong iu kin cho php mi c t ct thu st trn cc cng
trnh trong trm hoc dy chng st trong trm.
- Khi thit k bo v chng st nh trc tip vo trm bin p, ngoi cc
yu cu v k thut cn phi ch n vn kinh t v vn m thut.
I. CC YU CU K THUT:
- Vi mc ch gim vn u t khi thit k bo v chng st nh trc tip
vo trm bin p ngoi tri, ngi ta thng b tr ct trn cc cao c sn
nh ct, x... i vi trm 110KV do c mc cch nh cao nn cc ct thu
st c tht trc tip trn cc kt cu ca trm. Cc tr ca cc kt cu trn
c t cc ct thu st phi l ngn nht sao cho dng in st (Is) khuych
tn vo t theo 3 4 thanh ci ca h thng ni t.
- i vi trm 22KV khi b tr ct thu st trc tip trn x th phi tng
cng cch in n cp 110KV, do vy s c tn tht v kinh t. Khi dng
ct thu st c lp th phi ch n khong cch gia cc ct thu st ti cc
b phn ca trm trnh kh nng phng in t ct n cc thit b bo v.
Khi dng ct n chiu sng lm gi cho ct thu st th phi cho dy dn
in n bng n vo ng ch v chn di t.
- m bo cho cgii v chng n mn kim loi cn phi theo ng quy
nh v cc loi vt liu, tit din dy khi dng trn mt t v di t.
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II. GII THIU SLC V THIT K TRM 110/22KV:
-Trm c kch thc: 60 x 100m2
-Tng din tch trm: 6000m2
- Chiu cao ln nht cn c bo v pha 110KV l 11m pha 22KV l 8m
- Trong trm c 2 my bin p 110/22KV, c 2 lng dy 110KV i vo
v 5 lng dy 22KV i ra.
- Trn cc l 110KV c dy chng st nn cc thit b ca trm nm di
on ng dy vo trm n x n dy u c bo v nn ta thit k
chng st cho pha 110KV c th khng cn tnh n phm vi ny.- Trong phn thit k ny ta a ra cc phng n b tr ct thu st bo
v trm bin p ngoi tri theo cc yu cu k thut, m thut v kinh t sau
chn phng n hp l tnh ton.
III - PHM VI BO V CA CT THU ST:
1. Phm vi bo v ca mt ct thu st (H1-1):- Bn knh bo v cho cao hx l:
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)hh(
h
h1
6,1r x
x
x +
=
Trong:
h - L cao ca ct.
hx - L c cao cn bo v.
h - hx = ha - L chiu cao hiu dng ca ct.
Nu hth3
23 th )8,0
1(5,1h
hh xx =
hhx 3
2> th
=
h
hxx 175,0
- Nu chiu cao ca ct h > 30m th ta phi kn h siu chnh nghal:
Prmhr xx .)30( => (vi P l h shiu chnh)
.5,5
hP=
2. Phm vi bo v ca hai ct v nhiu ct thu st :
- Nu hai ct c chiu cao bng nhau: gi s 2 ct cch nhau mt khong l
a.+ Nu a = 7h th bt kim no trn mt t gia hai ct sc bo
v an ton.
+ Nu a = 7h th ct bo vc cao l0
0
h
h c xc nh t h).
.77 00a
hha
hh ==
+ Ta xem h0 nh cao mt ct v tnh phm vi bo v cho cao hx
=>
=
000
0000
175,03
2
8,015,1
3
2
h
hhrhh
h
hhrhh
xoxx
xxx
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+ Khi ct thu st c h> 30 thP
ahh
70=
Hnh 1-2
-Nu hai ct c cao khc nhau:
Hnh 1 - 3- Nu c nhiu ct: Ta phi xc nh cho tng nhm ct gn nhau (3 hay 4
ct) phi hp bo v thu st sao cho nh cc ct cng nm trn mt ng
trn ngoi tip, cc ct to thnh mt tam gic hoc a gic th phn bn
trong sc bo v hon ton nu:
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ax hhhD 8)(8 =
Trong:
D - ng knh ng trn ngoi tip a gic do cc ct to thnh.
h L chiu cao ca cc ct thu st.
hx - L chiu cao ca vt cn bo v.
Hnh 1- 4
IV. KHONG CCH AN TON TRONG KHNG KH V T.
- Khong cch an ton trong khng kh SK.
- Bin in c xung (Dmax) trong trng hp chung xc nh theo cngthc:
[ ] )KV()WL(RR2
axImmaxU
22
XKXK ++=
Trong:
Imax - Dng in st tnh ton (KA)
RXK- in trxung kch ca b phn ni t (R)
L - cm ng ca dy dn t b phn ni t n im kho st )( H cn
gi l h sin cm.
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- Khi ta thay Imax = 150 KA; bn cch in ca khng kh l 500KV/m v
ax
IW
Im
2 = vi
S
KAI
60= th ta sc cng thc tnh ton dng xc nh
khong cch an ton cho php ca khng kh nh sau:
[ ]
[ ] )(8,015,0150
60.215,0
)()(2.500
150
500
max
222
2
22
mLRRLRR
KVWLRRU
S
XKXKXKXK
XKXKx
++=
++=
++==
- Khong cch an ton trong t S
- Tiu chun khong cch an ton Sc xc nh bng biu thc sau:
S= a.Rxk
Trong:
a- H s ph thuc tr s dng in st chn trong tnh ton vi Imax = 150
th a = 0,5.
Rxk- Tr sin trtn xung kch ca dng in st i qua b phn ni t
)(
xk0,5RdS =
Hnh 1- 5
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V. TRNH TTNH TON CHNG ST NH TRC TIP.
1 - B tr cc ct thu st :
2 - Xc nh chiu cao hiu dng ca ct :
- Tm ng knh D ca ng trn ngoi tip a gic i qua nh thu ct
st sao cho din tch a gic c bo v cho hx, p dng ton trm. Ta
phi tnh cao L ca ct thu st: h = hx + ha.
3- Kim tra kh nng bo vi vi vt nm ngoi phm vi ct
thu st bo v:
- Tnh bn knh bo v ca mt ct thu st.
- Tnh bn knh ca khu vc bo v gia 2 ct thu st v tnh bn knh r0x
m h0 bo vc.
- V khu vc bo v theo kch thc tnh c.
4. Kim tra li ton b:
- Kim tra li ton b bn v thit k nu c cao x no m cn bo v
cn nm ngoi bn knh bo v ro x th cn phi xem xt li thit k: tng
cao ct hoc b tr thm ct sao cho cc cao cn c bo v phi nm
trong phm vi bo v ca cc ct thu st.
VI. LA CHN PHNG N B TR CT:
1. Phng n 1:
B tr 10 ct thu st trong c 2 ct thu st t thm 2 u ca x n dy
u trm cn 8 ct dng c lp
a. Xt cc cp ct:
- Cc cp ct 1 - 2 -10 ta c:
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)m(30l
)m(8l
101
21
=
=
Do to thnh mt tam gic vung nn:
)m(05,31830l 22102 =+=
)m(88,38
31
8
Dh
)m(31)05,3153,34)(3053,34)(853,34(53,342
05,31.30.8D
)m(05,312
05,31308P
h ===
=
=
=++
=
vi hx =11m cho nn chiu cao ca cc ct l : 11+3,88=14,88(m)
Vy ta chn chiu cao cc ct l 19 (m)
- Cc ct 2 - 3 - 10:
)m(04,413028l
)m(05,31l
)m(30l
22
103
102
32
=+=
=
=
Do cc ct ny to thnh mt tam gic thng nn.
)m(15,5
8
21,41
8
Dh
)m(21,41)04,4105,51)(05,3105,51)(3005,51(05,512
05,41.05,31.30D
)m(05,512
04,4105,3130P
a ===
=
=
=++
=
Vi hh = 11m chiu cao ca cc ct thu st l 11+5,15 = 16,15 (m)
Vy ta chn chiu cao ct l 19 (m).
- Cc ct 3 -4 -10.
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Ta c
)m(56l
)m(04,41l
)m(99,343018l
104
103
22
43
=
=
=+=
Do cc ct to thnh mt tam gic thng nn:
)m(02,78
15,56
8
Dh
)m(15,56)5602,66)(04,4102,66)(99,3402,66(02,662
56.04,41.99,34D
)m(02,662
5604,4199,34P
a ===
=
=
=++
=
Vi hx = 11m ta c chiu cao ca cc ct l : 11+7,02 = 18,02(m)
Vy ta chn chiu cao ct l 19 (m)
- Cc ct 4-5-9-10:
)m(56ll
)m(30ll
95104
10954
==
==
Do cc ct to thnh hnh ch nht nn:
)m(94,78
53,63
8
Dh
)m(53,635630D
a
22
===
=+=
Chiu cao ca cc ct thu st vi hx = 11m l 11+ 7,94 = 18,94m
Vy ta chn chiu cao ca cc ct l :19(m).
- Cc ct 5 - 6 -7:
)m(28l
)m(33l
76
65
=
=
Do cc ct to thnh mt tam gic vung nn:
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)m(41,58
28,43
8
Dh
)m(28,43)28,4314,52)(2814,52)(3314,52(14,522
28,43.28.33D
)m(14,522
28,432833P
)m(28,432833l
a
22
75
===
=
=
=++
=
=+=
Vi hx = 8m chiu cao ca cc ct l: 8+ 5,41=13,41(m)
Vy ta chn chiu cao ca cc ct l : 16 (m)
- Cc ct 5 - 7 - 9:
)m(28,433328L
)m(56l
)m(28,43l
22
97
95
75
=+=
=
=
Do cc ct to thnh mt tam gic nn ta c:
).m(09,78
76,56
8
Dh
)m(76,56
)28,4328,71)(5628,71)(28,4328,71(28,712
28,43.56.28,43D
)m(28,712
28,435628,43P
a ===
=
=
=++
=
Vi hx = 8m chiu cao ca cc ct l: 8+ 7,09 = 15,09(m)
Vy ta chn chiu co ca cc ct l: 16(m)
-Cc ct 7 - 8 -9:
)m(28,43l
)m(33l)m(28l
97
98
87
=
==
Do to thnh tam gic thng nn:
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)m(41,5828,43
8Dh
)m(28,43)28,4314,52)(3314,52)(2814,52(14,522
28,43.33.28D
)m(14,522
28,433328P
a ===
=
=
=++
=
Vi hx = 8m cho nn chiu cao ca cc ct thu st l 8+5,41=13,41(m).
Vy ta chn chiu cao ca cc ct l: 16(m).
* Vy ta xc nh c:
+Cc ct 1-2-3 -4 -10 cao l 19 m
+Cc ct 5 - 6 -7 - 8 -9 cao l: 16m.
b. Xt cc cp ct bao quanh trm:
- Cp ct 1 -2:
+ Bn knh bo v ca mi ct cao 19m l:
=
h
hhr xx 8,0
15,1 v )m(67,12h3
2m11h
1x=
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- Cp ct 2-3:
+ Bn knh bo v ca mi ct cao 19m l 7,88 (m)
+Phm vi bo v gia hai ct l:
v
)m(86,314,16
11114,16.75,0
h
h1h75,0r
)m(76,10h3
2m11h
)m(14,167
2019
7
ahh
o
x0x0
0x
o
=
=
=
=>=
===
- Cp ct 3-4:
+ Bn knh bo v ca mi ct cao 19m l 7,88m
+Phm vi bo v gia 2 ct l:
v:
)m(25,21411114.75,0
hh1h75,0r
)m(33,9h3
2m11h
)m(147
99,3419
7
ahh
0
x0x0
0x
10
= =
=
=>=
===
- Cp ct 4-5:
+ Bn knh bo v ca ct 4 l 7,88m
Bn knh bo v ca ct 5 l:
V:)m(75,3
16
11116.75,0
h
h1h75,0r
)m(67,10h3
2m11h
5
x5x
2x
=
=
=
=>=
+ Phm vi bo v gia 2 ct: v 2 ct c cao khc nhau nn
bn knh bo v ca ct 5 vi cao hx = 16(m) l:
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V:)m(25,2
19
16119.75,0
h
h1h75,0r
)m(67,12h3
2m16h
4
x44
4x
=
=
=
=>=
)m(75,2725,230raa 4 ===
7
75,2716
7
ahh 50 =
=
V:)m(06,3
04,12.8,0
8104,12.5,1
h8,0
h1h5,1r
)m(03,8h3
2m8h
0
x0x0
ox
=
=
=
==
)m(47,229,11
8129,11.75,0rox =
=
-Cp ct 6-7:
+ Bn knh bo v ca mi ct cao 16m l: 3,75 (m)
+ Phm vi bo v gia hai ct:
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=
===
o
xoox
60
h
h1h75,0r
)m(127
2816
7
ahh
V )m(8h3
2m8h 0x ===
)m(312
8112.75,0rox =
=
-Cp ct 7-8:
+ Bn knh bo v ca ct cao 16m l 3,75(m)
+ Phm vi bo v gia hai ct:
)m(127
2816
7
ahh 60 ===
=
0
x0ox
h
h1h75,0r
V: )m(8h32)m(8h 0x ===
)m(312
8112.75,0rox =
=
- Cp ct 8- 9:
+Bn knh bo v ca cc ct cao 16m l 3,75(m)
+Phm vi bo v gia hai ct l:
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)m(46,228,11
8128,11.75,0
)m(52,7h3
2m8hvi
h
h1h75,0r
)m(28,117
3316
7
ahh
ox
o
xoox
8o
=
=
=>=
=
===
-Cp ct 9 -10
+ Bn knh bo v ca ct 9 cao 16m l 3,75m.
Bn knh bo v ca 10 ct cao 19m l 7,88m
+Phm vi bo v gia 2 ct: V 2 ct c cao khc nhau nn bn knh
bo v ca ct 10 vi cao h h =16m l:
=
10
x1010
h
h1h75,0r
V: )m(67,12h3
2m16h 10x =>=
)m(25,219
16
119.75,0rox =
=
)m(75,2725,230raa 10 ===
)m(04,127
75,2716
7
ahh
90===
V: )m(03,8h3
2m8h 0x =
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+ Phm vi bo v gia 2 ct:
)m(71,147
3019
7
ahh 100 ===
V: )m(8,9h32m11h 0x =>=
)m(78,271,14
11171,14.75,0
h
h1h75,0r
o
x0ox =
=
=
*T cc s liu tnh ton trn ta c bng sau:
Bng 1-6
STT Cc cp ct hx(m) ho(m) rox(m)
1 1-2 11 17,86 6,17
2 2-3 11 16,14 3,84
3 3-4 11 14 2,25
4 4-5 8 12,04 3,06
5 5-6 8 11,29 2,47
6 6-7 8 12 3
7 7-8 8 12 3
8 8-9 8 11,28 2,46
9 9-10 8 12,04 3,06
10 10-1 11 14,71 2,78
- T bng s liu trn ta v phm vi bo v ca cc ct thu st (theo s
b tr ct thu st)
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2 - Phng n 2:B tr 12 ct thu st trong 2 ct t trn x n u trm, ct cn
li t c lp.
a. Xt cc cp ct:- Cp ct 1 - 2 - 11
)m(8l 21 =
)m(30l 111 =
Do cc ct to thnh mt tam gic vung nn:
)m(88,38
31
8
Dh
)m(31)05,3153,34)(3053,34)(853,34(53,342
05,31.30.8D
)m(53,342
05,31308
P
)m(05,31308l
a
22
112
===
=
=
=
++
=
=+=
Vi hx = 11m chiu cao ca cc ct l: 11+ 3,88 = 14,88 (m)
Vy ta chn chiu cao ca cc ct l 18 (m)
- Cc ct 2 -3 -11:
)m(04,412830l
)m(05,31l
)m(30l
22
113
112
32
=+=
=
=
Do cc ct to thnh mt tam gic cho nn:
)m(15,58
21,41
8
Dh
)m(21,41)04,4105,51)(05,3105,51)(3005,51(05,512
04,41.05,31.30D
)m(05,51
2
04,4105,3130P
a ===
=
=
=++
=
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Vi hx = 11m chiu cao ca cc ct l: 11+ 5,15 = 16,15(m).Vy chn
chiu cao ca cc ct l 18(m)
- Cc ct 3 -11- 12:
)m(73,292810l
)m(40l
)m(04,41l
22
1211
123
113
=+=
==
Do cc ct to thnh tam gic thng nn:
)m(45,58
58,43
8
Dh
)m(58,43)73,2942,55)(4042,55)(05,4142,55(42,552
73,29.40.05,41D
)m(42,552
73,294005,41P
a ===
==
=++
=
Vi hx = 11m th chiu cao ca cc ct l: 11+ 5,45 = 16,45(m). Vy ta
chn chiu cao ca cc ct l18 (m)
- Cc ct 3 - 4 -12:
)m(83,482840l
)m(40l
)m(28l
22
124
123
43
=+=
=
=
Do cc ct to thnh tam gic thng nn:
)m(1,68
8,48
8
Dh
)m(8,48)83,4842,58)(4042,58)(2842,58(42,582
83,48.40.28D
)m(42,582
83,484028P
a ===
=
=
=++
=
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Vi hx = 11m th chiu cao ca cc ct l: 11+ 6,1 = 17,1 (m). Vy ta
chn chiu cao ca cc ct l: 18(m)
- Cc ct 4 -5 -12:
)m(73,291028l
)m(83,48l
)m(30l
22
125
124
54
=+=
==
Do cc ct to thnh tam gic thng nn ta c:
)m(48,68
86,51
8
Dh
)m(86,51)63,2928,54)(83,4828,54)(3028,54(28,542
73,29.83,48.30D
)m(28,542
75,2983,4830P
a ===
==
=++
=
Vi hx = 11m chiu cao ca cc ct l: 11+6,48 = 17,48 (m)
Vy ta chn chiu cao cc ct l 18 (m)
- Cc ct: 5 -6 -12:
)m(41,342028l
)m(73,29l
)m(30l
22
126
125
65
=+=
=
=
Do cc ct to thnh tam gic thng nn ta c:
)m(57,48
54,36
8
Dh
)m(54,36)41,3407,47)(73,2907,47)(3007,47(07,472
41,34.73,29.30D
)m(07,472
41,3473,2930P
a ===
=
=
=++
=
Vi hx = 8m chiu cao ca cc ct l: 11+4,57 =15,57 (m)
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Vy ta chn chiu cao cc ct l: 18 (m)
- Cc ct 10 -11-12:
)m(41,343028l
)m(30l
)m(73,29l
22
1210
1110
1211
=+==
=
Do cc ct to thnh tam gic cn nn:
)m(57,48
54,36
8
Dh
)m(54,36)41,3407,47)(73,2907,47)(3007,47(07,472
41,34.73,29.30D
)m(07,472
41,3473,2930P
a ===
=
=
=++
=
Vi hx = 11m chiu cao ca cc ct l: 11+4,57 =15,57(m). Vy ta
chn chiu cao ca cc ct l 18(m)
-Cc ct 6 -7 - 8:
)m(28l
)m(33l
87
76
=
=
Do cc ct to thnh tam gic vung nn:
)m(41,58
28,43
8
Dh
)m(28,43
)28,4314,52)(2814,52)(3314,52(14,522
28,43.28.33D
)m(14,522
28,432833P
)m(28,432833l
a
22
86
===
=
=
=++
=
=+=
Vi hx = 8 m chiu cao ca cc ct l: 8+5,41= 13,41m
Vy ta chn chiu cao ca cc ct l 16 (m)
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- Cc ct 6 -8 -10:
)m(56l
)m(28,432833l
)m(28,43l
106
22
108
86
=
=+=
=
Do cc ct to thnh tam gic thng nn:
09,78
76,56
8
Dh
)m(76,56)5628,71)(28,4328,71)(28,4328,71(28,712
56.28,43.28,43D
28,712
5628,4328,43P
a===
=
=
=++
=
Vi hx = 8m th chiu cao ca cc ct l: 8+7,09 = 15,09(m)
Vy ta chn chiu cao ca cc ct l 16(m)
- Cc ct 8 -9 - 10:
)m(28,43l
)m(28l
)m(33l
108
98
109
=
=
=
Do cc ct to thnh tam gic vung nn:
)m(41,58
28,43
8
Dh
)m(28,43
)28,4314,52)(2814,52)(3314,52(14,522
28,43.28.33D
)m(14,522
28,432833P
a ===
=
=
=++
=
Vi hx =8(m) th chiu cao ca cc ct l 8+5,41=13,41(m)
Vy ta chn chiu cao ca cc ct l 16(m)
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* Vy ta xc nh c:
+ cc ct 1 - 2 -3 -4 - 5 - 11 - 12 cao 18 (m)
+ cc ct 6 - 7 - 8 -9 - 10 cao16 (m)
b. Xt cc cp ct bao quanh trm:
-Cp ct 1 - 2:
+ Bn knh bo v ca mi ct cao 18 (m)
V: m12h3
2m11h 1x =
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=
===
0
x0x0
0
h
h1h75,0r
)m(14,157
2018
7
ahh
V: )m(09,10h3
2m11h 0x =>=
)m(11,314,15
11114,15.75,0rox =
=
- Cp ct 3 - 4:
+ Bn knh bo v ca ct cao 18m vi hx = 11m l 6,38 (m)
+ Phm vi bo v ca 2 ct
=
===
o
xoox
3o
h
h1h75,0r
)m(147
2818
7
ahh
V )m(33,9h3
2m11h
ox=>=
)m(25,214
11114.75,0rox =
=
- Cp ct 4 - 5:
+ Bn knh bo v ca ct 4 cao h = 18m l 6,38m
+Phm vi bo v gia 2 ct:
)m(71,137
3018
7
ahh
10===
V )m(14,9h3
2m11h 0x =>=
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)m(03,271,13
111h75,0r 0ox =
=
- Cp ct 5 - 6:
+ Bn knh bo v ca ct 5 cao 14m v hx= 8m l 6,38(m)
Bn knh bo v ca Ct 6 cao 16 m l:
V: )m(67,10h3
2m8h 0x ==
V: )m(12h3
2m16h
5x =>=
=
==
=
===
=
=
=
0
x0ox
60
5
5
x55
h
h1h75,0r
)m(93,117
5,2816
7
ahh
)m(5,285,130raa
)m(5,118
16118.75,0
h
h1h75,0r
V: )m(93,11h32m8h
0x=>=
)m(95,293,11
8193,11.75,0rox =
=
- Cp ct 6-7:
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+ Bn knh bo v ca ct cao 16m l 9 (m)
+ Phm vi bo v gia hai ct:
=
===
o
xoox
6o
h
h1h75,0r
)m(28,11
7
3316
7
ahh
V )m(52,7h3
2m8h
0x =>=
)m(46,228,11
8128,11.75,0rox =
=
-Cp ct 7- 8:
+ Bn knh bo v ca ct cao 16m l 9 (m)
+ Phm vi bo v gia hai ct.
=
===
o
x0ox
7o
h
h1h75,0r
)m(127
2816
7
ahh
V )m(8h3
2
m8h 0x ===
)m(312
8112.75,0rox =
=
-Cp ct 8 -9:
+ Bn knh bo v ca ct cao 16m l 9 (m)
+ Phm vi bo v gia hai ct.
=
===
o
x0ox
8o
h
h1h75,0r
)m(127
28167
ahh
V )m(8h3
2m8h
0x===
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)m(312
8112.75,0rox =
=
-Cp ct 9 -10:
+ Bn knh bo v ca ct cao 16m l 9m
+ Phm vi bo v ca 2 ct:
=
===
0
xoox
9o
h
h1h75,0r
)m(28,117
3316
7
ahh
V )m(52,7h3
2m8h xx =>=
)m(46,228,11
8128,11.75,0rox =
=
-Cp ct 10 -11:
+ Bn knh bo v ca ct 11 cao 18m l 6,38 (m).
Bn knh bo v ca ct 10 cao 16m l 9 (m).
+ Phm vi bo v gia 2 ct: v 2 ct c cao khc nhau cho nn
bn knh bo vct 11 vi cao hx = h10 = 16m l:
=11h
h1h75,0r x1111 v )m(12h
3
2m16h 11x =>=
)m(5,118
16118.75,0r11 =
=
=
===
===
0
x0x0
100
11
h
h1h75,0r
)m(93,117
5,2816
7
'ahh
)m(5,285,130raa
v )m(95,7h3
2m8h
0x=>=
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)m(95,293,11
8193,11.75,0rox =
=
-Cp ct 11-1:
+ Bn knh bo v ca ct cao18m l 6,38m
+Phm vi bo v gia hai ct:
=
===
o
xoox
o
h
h1h75,0r
)m(71,137
3018
7
ahh
V )m(14,9h3
2m11h ox =>=
)m(03,271,13
11171,13.75,0rox =
=
* T nhng s liu tnh ton trn ta c bng tng kt sau y:
Bng 1-8
STT Cc cp ct h0 (m) hx = (m) r0x (m)
1 1 - 2 16,86 11 4,39
2 2 - 3 15,14 11 3,11
3 3 - 4 14 11 2,25
4 4 - 5 13,71 11 2,03
5 5 - 6 11,93 8 2,95
6 6 - 7 11,28 8 2,46
7 7 - 8 12 8 3
8 8- 9 12 8 3
9 9 - 10 11,28 8 2,46
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10 10 -11 11,93 8 2,95
11 11-1 13,71 11 2,03
T bng s liu trn ta vc phm vi bo v ca cc ct thu st (theo
s b tr ct thu st)
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* So snh 2 phng n:
+ V mt k thut th c 2 phng n u m bo thit k bo v
trm
+ V mt kinh t th phng n 1 kinh t hn v phng n 1 ch
phi dng 10 ct trong c 8 ct c lp so vi phng n 2 l phi dng
ti 12 ct trong c ti 10 ct phi dng c lp.
Do vy ta chn phng n 1 thit k thi cng bo v trm bin p
110/22KV.
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CHNG II
-----o0o-----
TNH TON NI T CHO TON TRM
him v ca ni t l tn dng in xung t m bo an
ton cho vt cn ni c tr s b. Trong HT ngi ta chia lm 3
loi ni nh sau:
- Ni t lm vic: Nhim v l m bo cho s lm vic
bnh thng ca thit b theo ch c quy nh sn. Loi ni t ny
gm ni t im trung tnh MBA o lng v ca khng dng trong b
ngang trn cc ng dy ti in.
- Ni t an ton (ni t bo v): c nhim vm bo an ton cho
ngi v thit b khi cch in b h hng tc l ni t mi b phn kim loi
bnh thng khng mang in nh: v MBA, my ct, cc gi kim loi...
khi cch in h hng trn cc b phn ny s xut hin in nh th nhng
do c ni t nn gic mc in th thp m bo an ton cho ngiv thit b khi tip xc vi nhng b phn ny.
- Ni t chng st: Mc ch l nhm m bo tn dng in st vo
trong t khi c st nh xung ct thu st hoc trn ng dy. Do s hn
chc dng in st ti cc thit bin cn c bo v.
- trm bin p v nguyn tc phi thit k tch ri 2 h thng ni t
l ni t lm vic v ni t chng st. phng khi c dng ngn mchln hay dng in st i vo h thng ni t lm vic ln hay nh s khng
gy nn in th cao trn h thng ni t an ton.
- Trong thc t cng vic ny rt kh thc hin c v nhiu l do nn
ngi ta thng ch dng mt h thng ni t chung thc hin c 2 nhim
N
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v. V l nn h thng ni t chung l phi m bo yu cu ca c 2 loi
ni t, ngha l phi c in trni t nh hn hoc bng in trni t
nh nht ca mt trong hai h thng ni t k trn.
- Cc yu t cn chung cho h thng ni t.
+ B phn ni t c tr sin trni t cng nh th s thc hin
c cng tt nhim vu tn dng in st trong t v in th trn cc thit
bc ni t mc n nh. Tuy nhin vic gim in trni t s lm
cho chi ph vu t xy dng tng ln nhiu (do s lng kim loi tng).
Do cn phi c quy nh tiu chun tr s cho php ca in trni t.
* Vi cc thit bin ni t trc tip th in trni t yu cu l R)(5,0
* Vi cc thit b c im trung bnh tnh ni t trc tip th in tr
ni t yu cu l IR /250 (nu h thng ni t ch dng cho cc thit b
cao p)
* Vi h thng c im trung tnh cch in v h thng ni t cho c
thit b cao p v h p th in tryu cu l: IR /125 (nhng R chc
)(10 Dng in I tu theo mi trng hp c tr s khc nhau.
+ i vi ni t an ton th in trc chn sao cho cc gi tr
in p bc v in p tip xc trong mi trng hp u phi khng vt
qu tr s cho php.
+ Ngoi vic m bo tr sin trni t quy nh v gim in tr
ni t ca h thng ni t cn cn phi ch n vic ci thin s phn b
in th trn ton b din tch trm.
- H s ma: Ta bit rng t l mt mi trng phc tp khng ng
nht v kt cu v thnh phn m ch yu l do kh hu. Do vy thit k h
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thng ni t cn tp trung v ch n tr s ln nht v tr s tnh ton in
trsut ca t c th c trong cc ma.
Kd .0 =
K l h s ma
0: in trsut c theo ma
- Phng php ni t: Theo chc nng ngi ta phn lm nhiu loi.
H thng ni t bao gm cc in cc trong t lm gim in trni t
theo tiu chun ca tng loi ni t. Cc in cc thng l cc thanh di
nm ngang hoc cc thng ng in p bc nh. Khi tnh ton ta phn ra
lm cc loi ni t t nhin v ni t nhn to.
Ni t t nhin l s dng cc ni t c sn nh dy chng st, ct
thu st, cc kt cu kim loi ca cng trnh cc ng ng nc. Cn ni t
nhn to nhm mc ch tho mn cc yu cu ni t ca cc cng trnh khi
ni t t nhin khng m bo c.
- Cc tham snh hng n ni t :
+ nh hng ca kch thc hnh hc
+ nh hng ca cc b tr in cc.
+ nh hng ca tr sin trt.
+ Hin tng phng in xung kch: Khi c dng xt i vo in cc
ni t, th gy ra mt in trng ln trn b mt in cc v trong t. in
trng t n gii hn th xy ra qu trnh phng in trong t. Cc tia
la in phng in pht trin xung quanh in cc to ra vng h quang, cc
ni t xem nh l to ra v in trni t gim, in trni t c tnh
bng cng thc:
=R
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40
Vi 1
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b- H thng ni t nhn to ca ni t an ton:
H thng ni t an ton ta s dng mch vng hnh ch nht c kch
thch nh sau:
Hnh 2-1
Mch vng c chn di phn in tch trm 110KV su 0,8m.
Phn mch vng ny cch tng ro 5m.
Ta chn loi thp dp 50 x 1mm = b = 50 mm = 0,05 m lm mch
vng.
Ta c:dt
KL
LR ttmv .
ln2
2
=
L- Chu vi mch vng L = ( 50+60).2 = 220 m
t- chn su ca mch vng t = 0,8 m.
d- ng knh quy i ca thanh dt )(025,0205,0 md ==
K- H s ph thuc t l 2,12
1 =L
Ltra bng ta c K = 5,64.
tKmdott .. = vi 6,1. =tm
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)(52,4025,0.8,0
220.64,5.ln.
220.2
10.28,1
)(10.28,1)(10.28,16,1.10.8,022
244
==
===
mvR
mcm
Ta thy )(1)(52,1 >=mvR cho nn ta ta phi ng cc dc theo mch
vng gim in trca h thng ni t. Ta ng cc cc dc theo mch
vng c 3m mt cc v mi cc di 3m c ng knh l 40mm v nh vy
s cc l 743
220= ( cc).
)(57,28
38,0.4
38,0.4ln
2
1
10.40
3.2ln
3.2
10.8,0
)4
4
2
12(
2
3
2
=
++=
+
+=
lt
ltn
d
lm
lRc
in trtn ca h thng ni t c tnh nh sau:
74..52,1.57,28
52,1.57,28
.. ctcmvtc
mvcnt nRR
RRR
+=
+=
Tra bng ta c:
=
=
4,0
22,0
c
t
)(1)(85,04,0.48,11222,0.57,28
43,43
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Hnh 2-2
Mch vng c chn di phn din tch trm 22KV c su 0,8m
phn mch vng ny c cch tng ro 5m
Ta chn loi thp dp 50 x 1mm = b = 50mm = 0,05m lm mch
vng. Ta c:dt
KL
LR ttm
=2
ln2
Trong : L - Chu vi mch vng: L= (30+50) 2 = 160(m)
t - chn su ca mch vng t = 0,8m
d - ng knh quy i ca thnh dp
)(025,02
05,0md ==
K - h s ph thuc t l 67,12
=
L
L Tra bng ta c K =
6,05
Kmttt .0 = vi Kmt = 1,6 )(10.28,1)(10.28,1116.10.8,0444 mcm ===
)(38,2025,0.8,0
160.05,6
160.2
10.28,1 22=
= LnRm
Nh vy
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II. TNH TON NI T CHNG ST CHO TRM:
1. Mu:
Khi c dng in i vo b phn ni t nu tc bin thin ca dng
in theo thi gian rt ln th trong thi gian u in cm s ngn cn khngcho dng in i ti cc phn cui ca in cc, khin cho in p phn b
khng u, sau mt thi gian nh hng in cm mt dn v in p s phn
bu hn.
Thi gian ca qu trnh qu ni trn ph thuc vo hng s thi
gian: 2LglT = (T l t trng vi in cm tng L .1 v in dn gl ca in
cc)
T biu thc trn cho thy khi dng in tn trong t l dng mt
chiu hoc xoay chiu tn s cng nghip nh hng ca L khng ng k v
bt k hnh thc ni t no (thngng hoc nm ngang) cng u biu th
biir tr sin trtn.
Khi dng in tn trong t l dng in st tham s biu th bi ni t
tu thuc vo tng quan hng s thi gian T v thi gian u sng in. Khi
T
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t an ton ca trm. Nh vy s gp trng hp ni t phn b di, tng tr
xung kch ZXK c th rt ln v ln gp nhiu ln so vi tr sin tr tn
xoy chiu l in ngc ln cc phn mang in ca trm. Do ta phi tnh
ton kim tra yu cu ca ni t chng st trng hp c dng in st i
va h thng ni t.
+ Dng sng tnh ton ca dng in st.
Trong tnh ton thit k ta chn dng sng tnh ton cu dng in st
l sng xin gc c bin khng i. Dng sng tnh ton ca dng in st
c biu din nh sau:
= .)( TdsikhiI
Tdsikhiattis
Hnh 2-3
Trong:
a - c dc dng in st quy nh a = tg = 30(KA/s).
I = 150KA - Bin dng in st i vi trm bin p
Nh vy thi gian u sng:
)(530
150s
aI
Tds ===
* Cc yu cu k thut i vi ni t chng st
t
i
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Ta s kim tra theo iu kin m bo an ton cho cch in ca my
bin p ( y ta xt in p ti thi im khi c dng in st i vo h
thng ni t).
%50),0(. UTZI dc <
Trong:
I- Bin ca dng in st ly bng 150 (KA)
Z(0,Ts)- Tng trxung kch ti thi t = Ts v ngay ti ch dng
in st i vo in cc.
U50%- Tr sin p phng in xung kch t ln cch in ca my
bin p.
- Nu iu kin trn khng tho mn th phi tin hnh tnh ton ni t b
xung ti ni i vo h thng ni t ( tc l ni thm 1 in trsong song vi
h thng nhm mc ch lm gim tng trxung knh ).
2- Tnh ton ni t chng st cho trm 110 KV:
- Do ta dng ni t an ton lm ni t chng st cho nn trc ht ta phi
hiu chnh li tr s Rnt
( in trnhn to) theo yu cu ca ni t chng st
c ngha l ta phi hiu chnh h s ma ca ni t an ton theo h s mu
ca ni t chng st ( Tra bng 19 - 2 gio trnh kthutin cao p).
Kma( NCS ) = 1,4 - i vi thanh chn su 0,8m
Theo tnh ton ca phn ni t an ton:
)1()(.
ln.2
)(.)(
2
=dt
KL
L
NDATKSNDATR
md
mvo
Cn i vi t chng st. Ta c:
)2(.
ln2
)(.)(
2
dt
KL
L
NDCSKSNDCSR mdmv
o
=
Chia c hai v ca ( 2) cho ( 1) ta c:
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)(33,152,1.6,1
4,1)(.
6,1
4,1)(
6,1
4,1
)(
)(
)(
)(
===
==
NDATRNDCSR
NDATK
NDCSK
NDATR
NDCSR
mvmv
m
m
mv
mv
Vy in trni t nhn to ng vi ni t chng st l:
)(83,074.4,0.33,122,0.57,28
33,1.57,28
.).(.
)(.)( =
+=
+=
nNDCSRR
NDCSRRNDCSR
cmvtc
mvcnt
a- Tnh ton tng trsng xung kch:
- tnh ton tng trsng xung kch ta c vi gi thit sau:
+ B qua in trni t t nhin.
+B qua cc thanh ni t cn bng p trong trm bin p+ Trong qu trnh tnh ton dn gin ho ta b qua qu trnh phng
tia la in trong t v gi thit in trsut ca t l khng i. Mt khc
in tr ca bn thn vt liu dng l ni t b hn rt nhiu so vi in
khng L ca n ( ng vi tr s dng in st) v nh hng n in dung C
ca in ni t cng rt nh hn nhiu so vi nh hng ca in dn ni t.
Nh vy ta c th coi mch vng ni t gm hai tai di L1 v L2 ghp
song song vi nhau v mi tia c chiu di l:
)(1102
220
221m
LLL ====
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Hnh2-4
S thay th ca 1 tia h thng ni t.
Hnh2-5
Da vo s thay th trn ta c cc gi trin cm Lo v in dn Go
c tnh ton nh sau:
Go - in dn ca in cc tnh trn mt n v di
=mlNDCSR
Gont
1
)(2
1
Lo- in cm ca in cc ni t tnh trn 1 n v di
=m
H
p
lLo
)31,0(ln2,0
Trong:
l = L1 = L2 = 110 (m) ( chiu di mt tia)
)(83,0)( =NDCSRnt - in trni t nhn to ng vi ni t chng
st.
P- Bn knh ca cc ni t do ta chn b = 40 mm.
)(10.20202
3 mmmb
p ===
GoGoGoGo
Lo Lo Lo Lo
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Thay cc thng s vo ta tnh c.
)(641,131,010.20
100ln2,0
110.610.6,0
100.83,0.2
1
3
32
m
HLo
mGo
=
=
===
Tnh ton phn b di khi khng xt n qu trnh phng in trong t
t s thay th ta c th lp c h phng trnh vi phn nh sau:
=
=
gUX
It
iL
X
U
.
Gii h phng trnh trn ta c in p ti thi im bt k trn in
cc:
+=
=
l
XKe
KT
lG
aU
K
T
t
txt cos)1(
121
. 121
0),(
Vi hng s thi gian:
22
200
.KlGLTK
= vi = 1K
Mt khc:
.2
12
200
1K
TT
lGLT K =
=
Tng trxung kchu vo in cc ni t (i vi 1 tia)
+=
=
0
12
1
0).0( )1(
121
1
K
Kt
tte
Kt
T
lGZ
Tng trxung kch tng ng u vo in cc ni t gm hai tia
ghp song song ti thi im: .5 sTt ds ==
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+==
=
12
1
0)5,(),0( )1(
121
1
2
1
K
T
T
dsot
K
ds
eKT
T
lGZZ
tnh )5,0(Z ta xt chui:
=
=
=
=
1 1 1222
1)1(
K K K
TT
TT
K
e
KK
e Kds
K
ds
=
=
=1
2
26433,1
6
1
K K(stay ton hc)
=
++++=1
2222 21K
T
T
T
T
T
T
T
T
K
eee
K
e Kds
K
ds
K
ds
K
ds
LL
Ta c: e-3 = 0,0497
e-4 = 0,0183
e-5 = 0,0067
e-6 = 0,0024
e-7 = 0,00091
Cho nn ta ly cp chnh xc l 0,001 th khi 7>K
ds
T
Tth ta coi:
02
K
e Kds
T
T
Do gii hn: 7K
ds
T
T
Mt khc ta c:
.2
7 12
12
1
ds
dsK T
TKK
T
T
K
TT =
Vi
)(98,91.14,3
100.641,1.10.6
. 22
23
21
2
20
1 sK
lGLT o ==
=
(v K l snguyn dng)
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74,35
98,9.7=K
Vy ta ch tnh n gi tr K < 3,74 Tnh K = 1 3
Vi K = 1 3 ta c cc gi tr 21
.KT
Tds tng ng l:
2
11 .1 KT
TK ds=
Thay cc gi tr va tnh c trn vo biu thc ca Z(0,5) ta c:
)(093,2)641,036,1(5
98,9.21
110.10.6
1.
2
13)5,0(
=
+=Z
Vi )(93,2)5,0( =Z ta kim tra iu kin ni t chng st xem c tho
mn khng nu khng tho mn th ta phi tin hnh ni t b sung.
)460)110(43993,2.150. %50)5,0( KVKVUZIs =>
Trong tnh ton ly dng in st Is = 150KA; a = 30KA/ ,S in p
trn b phn ni t c xc nh.
1=
3
1=2
54
2
250
22
22
13
22
12
2
11
6410=3
+2
+==
54=39895
=3=
2=29895
=2=
50=19895
=1=
K K
KT
TT
T
ds
ds
ds
eee
K
e
K
e
KT
TK
KT
TK
KT
TK
K
ds
K
ds
,
,.,
.
.,.
,.,
.
,,
.
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dd
PdKPddsd
KKPd
KPddSQ
SEURIU
SEUdt
dLRIU
..
..
.
= vi L- chiu di cc = 2,5m.
)(105,0
55,0
5,0
)(83,203,0
5,2.1,50,6
==
=
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Hnh 2-6
Trong:
L- Chiu di cc = 2,5 m v cc c lm bng thp gc 4040.4mm.
mhl
t 05,28,02
5,2
2/ =+=+=
d- ng knh ca cc = 0,95.40.10-3= 0,038 m
)(71,405,205,2.4
5,205,2.4ln
2
1
038,0
5,2.2ln
5,2.14,3.2
45,1.10.8,0 2=
+
+= cR
Tnh in trthanh nm ngang bng thp dt :
dt
Ka
aR ttt .
ln2
2
=
Trong:
t = 0,8 m - L chn su .
K = 1 L h s hnh dng ( tia ngang).
D - L ng knh quy i ca thanh thp dp vi b = 40 mm
mmmb
d 02,0202
===
a = 10 m - L di ca thanh .
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Hnh 2-8
Tr s xung kch ca tia XKT . Tra bng 10 (trang 87 - sch hng dn thit k
tt nghip) ta c cc dng st vc th quan h XKC v d . T th ta
c:
IS = 5KA 10KA 20KA 40KA
Th XKT = 1,05 0,93 0,83 0,74
Hnh2-9
0,8 2 4 6 8 10
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1
1,1
Is = 40(KA)
Is = 20(KA)
Is = 10(KA)
Is = 5(KA)
(10 .cm)4
kxc
0,8 2 4 6 8 10
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1
1,1
Is = 40(KA)
Is = 20(KA)
Is = 10(KA)
Is = 5(KA)
(10 .cm)4
kxt
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* Tr s XK t bng 10 (trang 87 - sch hng dn thit k tt nghip)
ta tra c h s s dng vi cc dng in xung kch l 0,8. T cc gi tr
IS(KA) ta tnh c cc ga tr Ut tng ng vi chng nh bng 2-11
Bng 2-10
Is (KA) 5 10 20 40
XKC 0,87 0,8 0,171 0,57
RXKC )( 32,174 47,976 42,579 34,183
Uc(KV) 326,087 399,70 1064,467 1709,145
XKC 1,05 0,93 0,85 0,74
RXKC )( 19,221 25,882 23,655 20,594
Ut(KV) 182,634 323,524 59,387 1029,71
Vi cc gi tr IC = 5;10; 20;40 KA ta ging ln ng cong Uc ri x
ngang ct ng Ut v h vung gc vi trc I ta c cc gi tr I v tnh
c IS = 3IC + 2I
Hnh 2-11
1200
600
I(KA)
U(KV)
1800
5 10 25 40
Ut = f(It)
Uc = f(Ic)
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Uc(KV) 326,087 599,70 1064,467 1709,146
IC(KA) 5 10 20 40
It (KA) 11 19 42 67
IS (KA) 37 68 144 254
Bng 2-13
T cc gi tr Uc v IS ta s thu c th quan h gia U v IS v ta
xc nh U bng cch t I = 150KA dng ln th v dng sang tr U ta c
U = 1200KV
)(8150
1200 == XKR
Phng n ni t m bo yu cu k thut vi ct thu st c lp
pha trm 22KV. T cc phn tnh ton trn, ta vc s ni t cho ton
trm
Hnh 2-14
Phng n ni t m bo yu cu k thut vi ct thu st c lp
pha trm 22KV. T cc phn tnh ton trn ta sc s ni cho ton
trm
Bng tng kt s lng kim loi dng cho ton trm
I(KA)
U(KV)
60 120 180 240
600
1200
1800
Ud = f(Is)
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Thp dp 50 x 1mm 300m
Cc 40 x 40 x 4mm 24 cc = 60m
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Hnh 2.12. S nit ton trm 110/22kV
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III. TNH TON NI T CT NG DY 110KV:
1- Nhim v:
Nhim v ch yu ca ni ct ng dy l thot dng in st khic st nh vo ng dy chng st. Ngoi ra i vi cc ng dy 3
20KV khi i qua khu vc ng dn c th ni t cn c nhim v bo v an
ton cn trong h thng ng dy c dng ngn mch ln ( KVUdm 110 ) do
trong h thng in c rle bo v tc ng nhanh do y nhim v bo
v an ton khng t ra. Nh ta bit khi c st nh dng st rt ln c th
lm hng ng dy v trm, do vy ni t ct in trng dy l ht sc
cn thit trong xy dng v vn hnh h thng in.
Gi R l RXK l in trni t n nh v in trni t xung kch,
ta c:
RRXK .= vi R l h s xung kch
Thng i vi ng dy ngi ta quy nh in trni t ng vi
tn s cng nghip.
i vi in trni t ca ng dy cng cn phi ch n qu
trnh sng in p v sng in lan truyn n cc cc. Do tc dng ca
in cm ca cc trong vic khuych tn dng in st nn i vi ni t ca
ct ng dy cng chia ra l ni t tp trung v ni t ko di.
in trni t ca ng dy ph thuc vo in trxut ca t ,
iu xut pht t ch nu in trsut cao th dng in s b v cc vng
y vic thit k xy dng v lp t h thng t s gp rt nhiu kh khn v
gi thnh xy dng rt cao.
Tr s quy nh ca in trni t tn s cng nghip cho bng di y:
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+ Vi loi mng cc th c th chn cc ngy vo l mng hay dng
ni t dng tia - cc.
i vi tia th di gii hn l:
Chiu di gii hn ca tia in trsut ca t )( m
40m 410.5
60m 410.25
100m 410.100
Bng 2-16
- Ta chn mng l mc lm mng cho ct ng dy c treo mt
dy chng st 110KV v ta tin hnh tnh ton ni t vi dng l tia - cc.
Ta c in tr )(10.8.04
cm= nn ta tnh ton ni t sao cho in trni tca ct ng dy l R
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- Tia di 6m bng thp c b = 40mm = 0,04m tia c chn su 1,5m- Cc dng loi cc bng thp gc 40 x 40 x4mm
)(038,0)(38940.040.95,0 mmmdd ====
Khong cch cc cc l 3m, chiu di cc l 3m
- H s ma: Kmt = 1,6 ; Kmc = 1,45
* in trca thanh ni:bt
l
lR ttt .
2ln
.2
2
=
Trong:
)(10.28,16,1.10.8.0 22 mtt ==
l = 6m, t = 1,5m ; b = 0,04m
)(09,2404,0.5,1
6.2ln
6.14,3.2
10.28,1 22== tR
* in trca cc thp gc 40 x 40 x 4mm
+
+
=14
14ln
2
12ln
2 t
t
d
l
lR ttc
Trong:
)(10.16,145,1.10.8.0 22 mtt ==
l = 3m
t/ = 1/2 = 1,5m
d = 0,04m
)(23,3435,1.4
35,1.4ln
04,0
3.2ln
3.14,3.2
10.16,1 2=
+
+= cR
* in trca h thng ni t tia - cc:
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ctc
ctnt nRR
RRR
..
.
+=
Trong:
n - L s cc cn s dng l 2 cc
c v t - L h s s dng c xc nh trong bng sch hng dn
thit k tt nghip cao p vi a/l =1
Ta c:85,0
87,0
=
=
c
t
)(61,112.87,0.09,2485,0.23,34
23,34.09,24=
+= ntR
Ta thy rng
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Vi: )(10.16,145,1.10.8,0 22 ==tt
L = 3 ( m)
D = 0,038( m)
T/ = 1,5 m
)(55,3435,1.4
35,1.4ln
2
1
038,0
3.2ln
3.14,3.2
10.16,1 2=
+
+= cR
in trca h thng ni t tia cc:
nRR
RRR
cttc
tcHT ...
.
+=
Trong:
n = 4 ( l slng cc sdng)
t v c tra trong bng 3,5 sch hng dn thit k tt nghip k thut
in cao p vi 2=l
a .
)(5,84.74,0.09,2477,0.55,34
09,24.55,34
74,0
77,0
=+
=
=
=
HT
c
t
R
Ta thy rng
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ng dy ti in l phn t di nht ca HT nn thng b st nh
v chu tc ng ca in p kh quyn, c th dn n ct my, ct ng
dy nh hng ti s cung cp in ca li ng thi ph hoi thit b trong
trm. Do ta phi nghin cu chng st cho ng dy ti in, c bit l
nhng on gn trm, v khi st nh c thr to nn sng truyn vo trm
gy nguy him cho ngi v thit b trong trm.
Qua in p kh quyn xut hin trn ng dy l do st nh trc tip
vo ng dy dn, dy chng st, ct hoc xung t gn ng dy. Khi
xt n ch tiu kinh t ta ch c th chn theo mc hp l v kinh t v k
thut, tan khng th chn c mc cch in theo yu cu ca qu in p
kh quyn v tr s ca n l ln.
Vi mc treo cao trung bnh ca dy trn cng ( dy chng st) l h,
ng dy s thu v cc phng in st trn di t rng 6h v chiu di bng
chiu di ng dy. Tng s ln st nh thng ln ng dy hng nm
c tnh l:
ngsTB nLhN ..10.)9,06,0(3=
Trong : hTB- cao trung bnh ca dy cao nht ( m)
Nngs- s ngy st nh hng nm trong khu vc dy i qua
10070 =ngsN ngy.
L- chiu di ng dy ( km)
Do tham s dng st nh l khc nhau, khng phi ln no cng dn
n
phng in trn cch in. xy ra phng in khi st nh th tr s qu
in
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p kh quyn phi ln hn mc cch in xung kch ca ng dy. Do vy
s
ln phng in trn trn cch in ph thuc vo gi tr xc sut phng in
rp
v s ln phng in c xc nh:
pdngsTbpdpd rnLhNrN ....),,(3109060==
Phng in trn cnh in ch gy ct in khi phng in xung kch
trn cch in chuyn thnh ho quang duy tr bi in p lm vic xc sut
trthnh h quang ph thuc vo graien in p lm vic dc theo ng
dy phng in:
).( LoEf = vi )(.m
KVL
ULoE
pd
LV=
Trong : ULV- in p lm vic ca ng dy.
Lp- chiu di ng phng in.
Vy s ln ct in hng nm do st nh vo ng dy ti in l:
rnLhrNN pdngstbpdc .....),,(..3109060==
so snh kh nng chu st ca ng dy ti in c tham s khc
nhau ngi ta dng tr s suts ct ng dy tc l s ln ct khi ng dy
c chiu di L = 100km c tnh l:
rhn pdtbcd ..).,,( 9060=
Do gim s ln ct din ta phi gim rp hoc .
+ Gim rp: bng cch tng cng cch in ng dy v treo dy
chng st. Treo dy chng st l c hiu qu nht. Vi vng t c cm 510>
th dy chng st s khng pht huy tc dng.
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Gim : gim c khi gim c cng in trng dc theo
ng phng in hay phii tng Lp( tng s bt s).
Ta thy rng ng dy 110KV i qua vng c st nh hot ng
mnh cn c bo v bng dy chng st. ng dy 22KV c trung tnhcch in v cch in ca ng dy l rt yu nn khng treo dy chng
st.
I.NG DY 110KV.
1.Tham s ct ng dy 110KV
- Dy dn dng loi AC - 185 c ng knh d = 18,8 mm = 18,8.10-3m pha
A1, A2 treo cao 20,8m, B1 v B2 treo cao 17,8 m, C1 v C2 treo cao 14,8m.Ct ng dy 110kv - 2l - 1 dy chng st.
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Hnh 3-1:Ct ng dy 110KV- 2 l- 1 dy chng st
- Dy chng st l dy thp C- 70 c ng knh d = 11,4 mm, r = 5,7
mm treo cao h = 25 m.
- vng ca dy chng st l fcs= 6 m. vng ca dy dn in lfdd= 8,5 m.
- Chui cch in dng loi s treo 54 , c 7 bt s.
- Khong vt ng dy l lKV= 200m.- U50% cch in 110Kv l 650 Kv/7 bt s.- H s hiu chnh vng quang cp in p 110Kv l 31= ,
2. Cc s liu tnh ton:
- treo trung bnh ca dy chng st.
Hcstb= hcs- 2/3 fcs= 25 - 6(2/3)o= 21 m.
- treo cao trung bnh ca dy dn pha A1, A2.
mfhh ddAtbAA 1315=5832820=3
2=21
,,).(,)(
- treo cao trung bnh ca dy dn B1, B2.
)(,,)(,)( mfhh ddBtbBB 1312=5832817=3
2=21
- treo cao trung bnh ca dy dn C1, C2.
)(,,).(,)( mfhh ddctbCC 139=5832
814=32
=21
- Tng trsng ca dy dn c tnh theo cng thc:
r
hZdd
260= ln.
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Trong : h - treo cao trung bnh ca dy dn.
+ r l bn knh ca dy dn: r=1049=210818 3
3
.,.,
- Tng trsng ca dy dn pha A1, A2
)(,.,
,.ln.
..ln. 612484=
1049
1315260=
260=
321
r
hZ
tbAA
dd
- Tng trsng ca dy dn B1, B2)(,
.,
,.lnln 352471=
1049
1312260=
260=
321
r
hZ
tbBB
dd
- Tng trsng ca dy dn pha C1, C2.)(,
.,
,..ln
.ln 305454=
1049
139260=
260
321
r
hZ
tbCC
dd
- Tng trsng ca dy chng st:)(,
.,
.ln
.ln 297534=
1075
21260=
260=
3r
hZ cstbcs
- Tng trsng ca dy chng st khi c vng quang:)(,
,
,998410=
31297534
==
ZZ csvqcs
- Gc bo v pha A1, A2:h
lxarctgA
=
Trong : lx- chiu di x treo dy pha tnh t tm ct.
Ac hhh = l k hiu gia chiu cao ct v chiu cao
dy dn cc pha.
o
A arctg 4625=820252
=21 ,,
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- Gc bo v pha B1, B2:o
B arctg 2924=81725253
=21 ,,
,
- Gc bo v pha C1, C2:o
C arctg 3916=814253
=21 ,,
Ta thy rng cc gc bo v cc pha n < 30o. Vy m bo yu cu
v chng st nn ta chn loi ct st v treo dy dn, dy chng st nh trn
tnh ton.
* H s ngu hp gia dy dn v dy chng st.- H s ngu hp pha A1-2 v dy chng st:
2
2
12
12
2=
r
h
d
D
Koln
ln
D12- khong cch gia dy chng st v nh hng cu pha A1-2
d12- khong cch gia dy chng st v dy dn pha A1-2
h2- bn knh dy chng st.
( ) ( )
( ) ( ) mlhhd
mlhhD
x
x
26=2+131521=+=
18536=2+21+1315=++=
22221212
22222112
,,
,,
Ta c:
1980=
1075
21226
18536
=2
=
32
2
12
12
,
.,
.ln
,
,ln
ln
ln
r
h
d
D
Ko
Khi c nh hng ca vng quang th:
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2576=311980== ,,.,.KK oorq
- H s ngu hp gia pha B1-2 v dy chng st.( ) ( )
( ) ( ) mlhhd
mlhhD
x
x
459=253+131221=+=
2933=253+1312+21=++=
22221212
22222112
,,,
,,,
Ta c:
1410=
1075
2124592933
=2
=
32
2
12
12
,
.,
.ln
,
,ln
ln
ln
r
h
d
D
Ko
Khi c nh hng ca vng quang th:
1830=311410==0 ,,.,KK orq
- H s ngu hp gia pha C1-2 v dy chng st.( ) ( )
( ) ( ) )(,,
,,
mlhhd
mlhhD
x
x
2412=3+13921=+=
330=3+139+21=++=
22221212
22222112
Ta c: 1010=
1075
2122412 330=
2=
32
2
12
12
,
.,
.ln
,,ln
ln
ln
r
h
dD
Ko
Khi c nh hng ca vng quang th:
1310=311010== 00 ,,.,.KK rq
Qua tnh ton ta thy rng h s ngu hp gia dy chng st v pha
A1-2 l ln nht do :
+ tnh sut ct do st nh vng qua dy chng st vody dn ta
xt cho pha A1-2 c gc bo v ln v treo cao hn.
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+ tnh st nh vo khong vt ca dy chng st ta tnh cho pha
c in p kh quyn t ln ln hn tc l pha c h s ngu hp nh hn.
+ tnh sut ct do st nh vo nh ct ta phi xc nh qu in p
kh quyn t ln cch in cc pha v ta tnh vi trng hp nguy him tcl c Uc(a,t) ln hn.
II.TNH TON THAM S KHI ST NH LN NG DY 110KV
1.S ln st nh vo ng dy:
Cng thc tnh: LnhN ngscstb ...),,(3109060=
Trong : + hcstb=21 m
+ L - chiu di ng dy ti in ly 110 km
+ nngs- s ngy st nh trong nm ly 100ngy
Ta c: 189126=10010010219060= 3 ...),,(N ln / nm.
M: N = N+ Nc +NKV
Trong : N- s ln st nh vng qua dy chng st vo dy dn .
Nc- s ln st nh vo nh ct v ln cn nh ct.
NKV- s ln st nh vo khong vt.
2. S ln st nh vng qua dy chng st no dy dn.
c trng cho s xc sut 2Vkhng nhng ch ph thuc vo gc bo
vma cn tng theo chiu cao ct c tnh l:
490
=hc
U
lg
l gc bo v tnh vi pha A1 - 2 = 25,460
hc l chiu cao ca ct = 25m
aUU 32 1062=5852=49025
= .,,lg25,46
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Vy s ln st nh vng qua dng dy chng st vo pha A12 l:
mnkmlnNUN //(,,.. 100490=6210189== 322
Ta ly N = 189 l gi tr ln nht tnh.
3. S ln st nh vo nh ct hoc ln cn v khong trt
)/(, mKmnlnN
NN KVdc 100594=2189
=2
==
4. Tnh sut ct do st nh vng qua dy chng st vo dy dn.
Theo cng thc: UNn Pddd ..2=
Vp l xc sut phng in do qu in p ng dy khi c st nh
vo dy dn.
l xc sut hnh thnh h quang.
* Tnh Vp : Khi st nh vo dy dn mch ca khe st ghp ni tip
vi tng tr sng ny c tr s l Zdd/2 (ta gi s st nh vo chnh gia
ng dy).
Ta c:2
=2
2+
= 00
0 Scdd
ddcc
IIZZvi
ZZ
ZII .
in p trn dy khi st nh vo l:
4= ddSdd ZIU .
Khi in p trn dy dn Udd vt qua mc in xung kch ca chui
cch in %50 UUdd th dy n phng in trn chui cch in.
%504U
ZI ddS nn
dd
SZ
UI %
. 504
Khi xc sut phng in khi qu p trn ng dy chnh l xc sut
cho:
8650===
4=
4
612484126
504
126
504
50
50
,,.,%
,
%
%
%
U
Z
U
dd
Spd
dd
S
dd
Z
UIfV
Z
UI
ll
Vi Zdd l tng trsng pha A1-2
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* Tnh : tr s ph thuc vo t scs
LVLV
l
UE =
+ ULV- in p lm vic
+ lcs- chiu di chui s.
KvU
U dddmLV 2157=110143
2
3
2=
2
3
2= ,.
,...
cp in p 110 Kv ta chn loi s cch in cho ng dy l loi
s 54 , c 7 bt s.
mlcs 21= ,
Ta c: )/(,,
,kmkv
l
UE
cs
lvLV 6847=21
2157==
T bng xc sut hnh thnh h quang ( trang 205 - sch k thut in
cao p) ta c
Bng 3-2:
Elv 50 30 20 10
0,6 0,45 0,25 0,01
Theo bng s liu trong bng ta vc th biu din quan h ELV v .
Sau ELV ta xc nh c gi tr tng ng l ELV= 47,68(kv/km) =0,58.
Thay vo cng thc:
260=58049== ,,.,.. oVNn pddd (ln/100 km.nm)
Ta thy rng gim ndd c th gim bng cch tng cng chui cch
in ... hoc gim N bng cch gim gc bo v tng cao dy chng st.
Hnh 3-3
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5. Tnh sut ct do st nh vo khong vt:
Khi c st nh vo dy chng st trong khong th ti v tr st nh
coi nh dng in st s chia u cho 2 phn ca dy chng st.
Gi thit l sng st c dng xin gc, phng trnh dng in do st l:
Is= at nu t < Ts
Is= I nu dcTt
Khi tnh ton ta phi tnh vi cc gi tr khc nhau ca dng st vi
a = 10,...100 )(s
KA v t = 1,2 ...10 )( s v gi tr 11=c
R .
Khi c st nh vo khong vt ca dy chng st s sinh ra cc in
p l in p tc dng ln cch in khng kh gia dy dn v dy chng st,
in p tc dng ln cch in chui s. Thnh phn in p tc dng ln
cch in khng kh ph thuc vo dc dng in st, v khong cch gia
dy dn v dy chng st. Trong thit k v trong xy dng ngi ta t
khong cch gia dy dn v dy chng st kh ln trnh chm dy nn
kh nng phng in trong khong vt do thnh phn in p tc dng ln
cch in khng kh ln hn mc cho php, ng thi nu c th xc sut hnh
thnh h quang l rt nh, do kh nng ct in ng dy l khng ng
k, v vy trong tnh ton ta b qua thnh phn ny.
ELV40302010 50
0,1
0,25
0,45
0,57
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a. in p tc dng ln cch in ca chui s.in p tc dng ln chui s khi st nh vo khong vt ca dy
chng st l:
LVCSC
ctcd Uk
dttdisLatRU +1
2+
2= )()(.)(
- Dng in st c dng is(t) = at nn c:LV
cs
cc
cd Uka
LatR
tU +1
2
+2
= )(..
)(
Trong cc thnh phn ca Ucc xc nh nhphn sau.
b. in p lm vic ( tnh phn trc)
)(,.,
. kvULV 2157=1101432
3
2=
c. in p tinh ct khi stnh vo khong vt khi c nh hng cah quang.
2
+2
1=a
LatR
kU cscc ..
)(
Trong : a- dc ca dng in st.
Lccs- in cm ca thn ct in tnh t mt t ln n im
treo dychng st v Lccs= Lo.hc. Vi Lo l in cm n v ca thn ct:
)(,m
HL 60=0 ;
hc- chiu cao ct in.
HLcsc 15=2560= .,
Rc- in trni t ca ct ng dy 11=cR
k- h s ngu hp ca dy chng st i vi dy dn c nh hng ca
h quang ( k ca dy dn pha A K pha B).
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Trong cng thc trn ta thy t l nghch vi K, vy ta s ly gi tr K
nh hn tnh vi U ln, ta c:
KA1-2= 0,275.
KB1-2= 0,183.KC1-2=0,131.
V ta ly KC1-2 l h s ngu hp tnh ton. T ta c:
2157+15+1143450=
2157+15+43450=
2157+
2
15+2
13101=
+1
2
+2
=
,)(.,
,).(.,
,.
),(
)(.
)(
ta
tRa
aatR
Uka
LatR
atU
c
c
LVcsc
ccd
Nh vy khi thay 11=R vo ta c biu thc tnh Uc theo a v t ta
c kt qu sau:
at
1)( s
2)( s
3)( s
4)( s
5)( s
6)( s
7)( s
8)( s
9)( s
10)( s
10 170,18 217,98 265,77 313,57 361,36 490,16 456,96 504,75 552,54 600,34
20 283,15 378,74 474,33 569,92 665,51 761,10 866,69 952,28 1077,87 1143,36
30 396,12 539,51 862,99 826,28 969,66 1113,05 1256,43 1399,82 1543,20 1686,59
40 509,09 700,27 891,45 1082,63 1273,81 1464,99 1656,17 1847,35 2038,53 2229,71
50 622,06 861,04 1100,01 1338,99 1577,96 1816,94 2055,91 2294,89 2533,86 2772,84
60 735,03 1021,80 1308,57 1595,34 1822,11 2168,88 2455,65 2742,42 3029,19 3315,96
70 848,00 1182,57 1517,13 1851,70 2186,26 2520,83 2855,39 3189,96 3524,52 3859,09
80 960,97 1343,33 1725,69 2108,05 2490,41 2872,77 3255,13 3637,49 4019,82 4402,21
90 1073,94 1504,10 1934,25 2364,41 2794,56 3224,72 3654,87 4085,03 4515,18 4945,34
100 1186,91 1664,86 2142,81 2620,76 3098,71 3576,66 4054,61 4532,56 5010,51 5488,46
T gi tr 11=cR ta c bng c tnh Vc(a,t) ph thuc vo thi gian
v dc u sng st a. T bng ny ta vc th Vc theo a,t v v
c c tnh phng in ca chui s.
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th gi tr ca Uc theo a v t vi 11=cR vi st nh vo khong
vt.
Ta c bng sau:
t(MS) 1 2 3 4 5 6 7 8
Ucs(KV
)
1020 930 860 818 790 780 770 770
Bng gi trin p phng in ca chui s Ucs(KV)
T th ta xc nh c cc gi tr ai v ti l giao im ca Uc vi
c tnh V - S ca chui s, do ta s xc dnh c cc cp thng s nguyhim (Ii;ai). Sau ta xc nh c ng cong nguy him I = f(a).
d. Xc sut phng in Vp: l xc sut cp thng s (I,a) ca
phng in st thuc min nguy him.
Vp= f{I,a} thuc min nguy him 3 hay
dVp=f{a = ai}* { }iIIf m
{ } 11261 == rIIf
I
i,
l ( r1 l xc sut dng in st ln hn
dc i Ii no )
aaiaii drdaadafaaf =+== )()( ( dra l xc sut dng in st
ln hn dc ai no )
9101
== ,)(I
ia aafr l
T ta xc nh c dVp= r1dra hay Vp= Str1dra, ta chia on t0
lm thnh thnh phn nh khi :
= aipd rrV vi 1+= aiai VVV
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T th Uc vi 11=cR ta xc nh c cc cp gi tr (a,t) t xc
nh c I, Vi, Va, aV nh trong bng sau:
a(KA/ )s 10 20 30 40 50 60 70 80 90 100
T( )s 11,8 6,7 3,7 2,8 2,3 1,8 1,5 1,2 0 0
I(KA) 118 134 111 112 115 108 105 96 0 0
V1(10-1 0,11 0,05 0,14 0,14 0,12 0,16 0,18 0,25 0 0
Va 0,0019 0,0005 0,0037 0,0034 0,0026 0,0049 0,065 0,0149 0 0
)( 210aV 1,0014 0,9968 1,0003 1,0008 0,9977 0,9984 0,9976 1,0149 0 0
)( 610aVV1
110,15 49,84 140,04 140,11 119,72 178,49 178,49 253,73 0 0
Vi 11=cR ta c: 61 10821151== .,.. apd VVV
Sut ct do st nh vo khong vt:
06530=6010821151594== 6 ,,..,.,.. VNN pddcKV (ln/100km.nm)
Do ta c th c th I = f(a) l gii hn ca min nguy him:
a )( sKV
Min nguy him
I(kA)
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6. Tnh ton sut ct do st nh vo nh ct v ln cn nh ct.
Ta xt trng hp st nh vo nh ct: khi dng in si vo b
phn ni t ca ct in, ng thi mt phn nh si vo b phn ni t
ca cc ct ln cn. Khi c qu p kh quyn tc dng ln cch in ng
dy th i vi cc pha khc l khng ging nhau, ta phi tnh ton in p t
ln cch in tng pha v xt trng hp c gi tr ln nht.
a. in p t ln cch in cc pha.Gi s ta tnh Uc (a;t) vi 1 gi tr a = 30 )( s
KV V t = 5 )( s vi
11=cR , khi cch in ln mi pha c tnh bi cng thc sau:
[ ] [ ]t
ticcs
c
dd
ctcstddcc
d
culvtacd d
dKLLKMMakRiUUU x
)(
)()();( )(. ++1++=
Trong : + ULV- thnh phn in p lm vic ca ng dy 57,12Kv.
+ K- h s ngu hp c nh hng ca h quang vi pha.
KA1-2= 0,257; KB1-2= 0,183; KC1-2= 0,131.
+ ddcL - tr sin cm ca phn ct in tnh t mt t ti
cao ca ng dy: ddcL = L0.hdd.
+ L0- in cm n v ca thn ct in ta ly
mHL 60=0 ,
+ hdd- treo cao ca dy dn pha:
A1-2: hdd= 20,8 ddcL = 0,6.20,8 = 12,48 H
+ B1-2: HLmh ddcdd 6810=81760=817= ,,.,,
C1-2: hdd= 14,8 m ddcL =0,6.14,8 = 8,88 H
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+ cscL - tr sin cm phn ct in tnh t mt t ti treo cao
dy chng st.
HhLL cscsc 15=2560== 0 ,. + Mdd(t)- h cm gia khe phng in st
vi mch vng dy dn v t c tnh:
1+
2
+1
+20=
h
H
h
h
H
HVhM
dd
tddtdd
ln
)(ln,)(
. V- vn tc pht trin vi phng in ngc khe vi Vt= .C
- tc tng i phng in ngc ca st (= 0,3)
C- tc truyn sng trong khng kh ( C = 300s
m )
Vt= 0,3.300.5 = 450 ( sm ) ( ly t = 5)
. H = hc+ hdd:
Pha A1-2: H = 25 + 20,8 = 45,8 (m)
B1-2: H = 25 + 17,8 = 42,8 (m)
C1-2: H = 25 + 14,8 = 39,8 (m)
. h = hc- hdd:
Pha A1-2: h =25 - 20,8 = 4,2 (m)
B1-2: h = 25 - 17,8 = 7,2 (m)
C1-2: h = 25 - 14,8 = 10,2 (m)
Vi pha A1-2:
HMAdd 9811=
1+
24845
820224
84530+1
845+4582020=21 ,
,
,ln
,.
,
,),(
,ln,.,
Vi pha B1-2:
HMBdd 6410=
1+
27
842
8172
27
84230+1
842+45081720=21 ,
,
,ln
,.
,
,),(
,ln,.,
Vi pha C1-2:
HMCdd 765=
1+
210839
8142210
83930+1
839+45081450=21 ,
,
,ln
,.
,
,),(
,ln,.,
+ Mcs(t)- trcm gia mch khe st v mnh dy chng st c tnh:
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H
hVthcM ctcs 215=
1+
2530+12252+510030
2520=
1+
+12
2+10= ,
),(
...,ln.,
)(ln,)(
+ Udc- thnh phn in cm ng ln ng dy v cc pha gy ra bi
din trng ca cc in tch khe st v trong ct.
Hhh
HvthVthvt
ha
h
hkU
c
cdd
dd
csdcu
.)(
))(()(ln
..,).(
2+1
+++101=
Vi pha A1-2:
KvUAdcu 655855=845422530+1
845+53003024+53003025+530030
308203010
82025
25701=2
21 ,,...),(
),..,)(,..,()..,(ln
,
,..,)
,.,(
Vi pha B1-2:
KvUBdcu 096756=842272530+1
842+59027+59025+590
308173010
81725
18301=2
21 ,,.,..),(
),.)(,.().(ln
,
,..,)
,.,(
Vi pha C1-2:
KvUCdcu 214643=8392102530+1
839+590210+59025+590
3081410300
14825
13101=2
21 ,,.,..),(
),.)(,.().(ln
,
,.,).,(
+ is lthnh phn dngin st i vo trong thn ct c xc nh bi
cng thc:
t
c
tcscs
tsR
MLai 21
2
2= l(. )()( )
Vi: + Lcs- in cm ca dy chng stC
LZL cscs
.=
+ L- l chiu di khong vt L = 200 m.
+ Zcs- tng trsng ca dy chng st Zcs= 410,998 )(
+ C- tc truyn sng C = 300s
m
HLcs 247=300200998410
=.,
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+ 0720=152+274
112=
2+
2=2 ,
.
.csccs
c
LL
R
Vy ta c: KAi tc 12112=1112215227430
= 50720 ,).(.
,.. .,)( l
+ bin thin dng in st : tc
tcscs
t
tic
R
MLa
d
d.)()(
...
222
2= l
Thay vo ta c: 718=0720112
215227430= 50720 ,.,.
.
,.. .,)(l
t
tic
d
d
By gita thay cc tham s tnh c v cc tham s cho trc
sts
KAaRc 5=30=11= ;; vo cng thc tnh Uc ta tnh c in p t
ln cch in cc pha:
in p t ln cch in pha A1-2
),( taUAcd21 = 57,12 + 855,655 + 102,68.11(1 - 0,257) +30
[11,98 - 0,257.15,2] + (12,48 - 0,257.15)18,7 = 2155,474 Kv
in p t ln cch in pha B1-2
),( taUBcd21 = 57,12 + 756,096 + 102,68.11( 1- 0,183) + 30
[10,64 - 0,183.152] + (10,68 - 0,183.15) 18,7 = 2120,138 Kv
in p t ln cch in pha C1-2
),( taUCcd21 = 57,12 + 643,214 + 102,68.11 (1- 0,131) + 30
[5,76 - 0,131.15,2] + (8,88 - 0,131.15).18,7 = 1924,227 Kv
Qua tnh ton trn ta thy in p t ln cchin pha A1-2 l ln
nht so vi cc pha khc nn ta ly pha A1-2tnh ton in p tc dng ln
chui s khi st nh vo nh ct ng dy in.
b. in p t ln cch in pha A1-2 trong trng hp vo nh ct.in p t ln cch in pha A1-2c tnh bi cng thc:
cscctc
dcLVtcd KVRiUUUU +++= )(
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Trong : ULV= 57,21 KV ( tnh)
d
cU- thnh phn in p cm ng in.
982585
845+9024+9025+90794=
845242530+1
845+9024+9025+90
3082010
82025
25701=
+1
+++101=
2
2
,
),)(,()(ln,
,.,..),(
),)(,()(ln
,
,..,)
,.,(
..)(
))(()(ln
..,).(
ttta
tttaHhh
Hvthvthvt
ha
h
hkU
c
cdd
dd
csdc
Vi cc gi tr ca a = 10, 20, 30.100s
KA v t = 1, 2, 3.10 (s)
Thay Uc ta c kt qu bng sau:
A t 1 2 3 4 5 6 7 8 9 10
10 148,49 204,41 239,1 263,75 285,12 301,75 315,92 328,26 339,19 349,00
20 296,97 408,83 478,24 527,49 570,24 603,51 631,85 656,53 678,39 698,00
30 445,46 613,25 957,65 791,25 855,36 905,26 947,77 984,79 1017,58 1047,00
40 593,95 817,66 1197,06 1054,99 1140,48 1207,02 1263,69 1313,05 1356,77 1396,00
50 742,44 1022,08 1436,47 1318,75 1425,59 1508,77 1579,62 1641,31 1695,96 1745,00
60 890,92 1226,49 1675,88 1582,49 1710,71 1810,53 1895,54 1969,58 2035,16 2094,01
70 1039,41 1430,91 1915,29 1846,25 1995,83 2112,28 2211,48 2297,84 2374,35 2443,01
80 1187,89 1635,33 2154,71 2109,99 2280,95 2414,04 2527,40 2626,10 2713,54 2792,01
90 1336,39 1839,74 2154,71 2373,75 2566,07 2715,79 2843,33 2954,37 3052,73 3141,01
100 1484,87 2044,16 2394,12 2637,49 2851,19 3017,55 3159,25 3282,63 3391,93 3494,02
t
cU l thnh phn in cm t.
Ta c: )( .. tddt
icddc
tc Ma
d
dLU +=
Trong : ddcU - in cm ct tnh t mt t n x treo dy dn.
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t
ic
d
d- tc bin thin ca dng in st nh vo nh ct, gi tr
ny ph thuc vo s bin thin l c phn x hay cha c t ct ln cn
trv.
1+2405459
845+906164= ,
,
,.(ln,)(
tM tdd
Thay cc gi tr t = 0, 1, 2, 3.10 s vo cng thc Mdd(t) ta c kt qu
bng sau:
T 1 2 3 4 5 6 7 8 9 10
Mdd(t) 6,592 8,707 10,102 11,145 11,98 12,673 12,267 13,783 14,25 14,666
Thnh phn in p ic Rc:
- Khi cha c phn x t ct ln cn trvc
Lt 2
Vi L l chiu di khong vt = 200 m
C l vn tc truyn sng = 300 s
st 331=3002002
,.
+ Dng in trong ct ic(t): ta c s thay th nh hnh v:
2
2+=
l
ZMZ
RZ
ai cstcstcs
ccs
taC )()(),(
i 4314=152
112+998410=
2
2+=1 ,
.
.,csc
ccs
L
RZ
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Thay vo biu thc tnh ic(a,t) ta c bng kt qu
Tc bin thin:
t
aa
t
aZtRZ
a
d
dcs
ccst
ic 94920=998432
998410=112+998410=2+=
,
,
,
).,(.)(
Gi trt
ic
d
dc tnh trong bng kt qu .
Ta c s thay th:
i:C
LZL cscs
.=
Zcs- tng trsng ca dy chng st khng c nh hng ca h quang
HLcs 198356=300200297534
= ,.,
Is
Rc
i
Zcs
2
crq
c2i
Is
Rc
i
Zcs
2
c
rq
c2i
Lcs
2Lc
cs
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+ Dng in i trong ct ic(a,t) l:
( )( )
0570=
152+198356
112=
2+
2=
122
=1
2
2
,
.,
.
)()
CS
CCS
c
t
cscs
c
c
LL
Rvi
etMLR
atai
Gi tr MCS(t) bng sau:
1+65
50+905=
ttMCS
ln)(
Bng 3.7
T 1 2 3 4 5 6 7 8 9 10
MCS(t)
9,19 9,24 9,28 9,29 9,32 9,33 9,34 9,35 9,36 9,37
Bng 3.8
a(KA/ )s 10 20 30 40 50 60 70 80 90 100
T( )s 11,3 4,9 3,5 2,4 1,6 1,3 0 0 0 0
I(KA) 113 98 105 9,6 80 78 0 0 0 0
V1(10-1) 0,132 0,234 0,179 0,253 0,466 0,504 0 0 0 0
Va .10-3 0,031 0,125 0,066 0,149 0,649 0,780 0 0 0 0
)( 210aV 0,906 1,059 0,917 0,5 0,869 1,78 0 0 0 0
)( 610aVV1 11,96 24,78 16,41 12,65 40,49 89,71 0 0 0 0
Ta c: 02220=6010196189==10196== 66 ,,..... VNnVVV pdcaiPd(ln/100Km.nm)
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7. Tnh xc sut phng in khi st nh vo nh ct v ln cnnh ct.
34150=2540+06530+02220=++= ,,,,ddKVc nnnn (ln/100km.nm)
ch tiu chng st ng dy ti in
932=341501
=1
= ,,n
m (ln /100km.nm)
* Kt lun : ta thy rng sut ct in trn ng dy 110KV l c th
chp nhn c (khng cao qu), tham s ca ct l m bo ch tiu v sut
ct cho ng dy khi c st nh vo n.
Vy gim s ln ct in do st c th dng cc bin php sau y.
- Gim xc sut phng in VP bng cch gim in trni t ctin v tng cng cch in ng dy.
- Gim tr s gc bo v gim kh nng st nh vo dy dn. Trng
hp st nh vo dy dn c xem l nguy him nht. V rt d gy phng
in, nh ng dy 110KV xc sut phng in c th t ti 80%,
ng dy in p cao hn xc sut ny vn cn gi cc tr s kh ln. Tuy
nhin cn phi bo v se yu cu ct phi cao v nh th s lm tng gi
thnh.
Bng gi tr ca d-it/ dt =0,949 a/t
t
a
10 20 30 40 50 60 70 80 90 100
1 9,492 18,98
4
28,47
6
37,96
8
47,46
0
56,95
2
66,44
4
75,93
6
85,42
8
94,92
0
2 4,746 9,492 14,23
8
18,98
4
23,73
0
28,47
6
33,22
2
37,96
8
42,71
4
47,46
0
3 3,164 6,328 9,494 12,65
6
15,82
0
18,98
4
22,14
8
25,31
2
18,47
6
31,64
0
4 2,373 4,746 7,121 9,492 11,86
5
14,23
8
16,61
1
18,98
4
21,35
7
23,73
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5 1,898 3,797 5,696 7,594 9,492 11,39
0
13,28
9
15,18
7
17,08
6
18,98
4
6 1,582 3,164 4,747 6,328 7,910 9,492 11,07
4
12,65
6
14,23
8
15,82
0
7 1,356 2,712 4,069 5,424 6,780 8,136 9,492 10,84
8
12,20
4
13,56
0
8 1,187 2,373 3,560 4,746 5,933 7,119 8,306 9,492 10,67
9
11,86
5
9 1,055 2,109 3,165 4,219 5,273 6,328 7,383 8,437 9,492 10,54
7
10 0,949 1,898 2,848 3,797 4,746 5,695 6,644 7,594 8,543 9,492
BNG GI TR CA [ ]142,0M2t998,410998,432
a)ait(i )t(csc =
t
a1 2 3 4 5 6 7 8 9 10
10 9,064 18,554 28,044 37,535 47,026 56,517 66,009 75,500 84,992 94,483
20 18,128 37,108 56,088 75,071 94,052 113,035
132,017
151,000
169,983
188,966
30 12,192 55,661 84,131 112,60
5
141,07
7
169,55
2
198,02
6
226,50
0
254,97
5
283,44
9
40 36,257 74,115 112,17
5
150,14
1
188,10
3
226,06
9
264,03
5
302,00
0
339,96
6
377,93
2
50 45,321 92,769 140,21
9
187,67
6
235,12
9
282,58
6
330,04
4
377,50
1
424,95
8
472,41
5
60 54,385 111,32
3
168,26
3
225,21
2
282,15
5
339,10
4
396,05
2
453,00
0
509,94
9
566,89
8
70 63,449 129,87
6
196,30
7
262,74
7
329,18
1
395,62
1
462,06
1
528,50
1
594,94
1
661,38
1
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80 72,513 148,43
0
224,31
5
300,28
2
376,20
6
452,13
8
528,06
9
604,00
1
679,93
3
755,86
5
90 81,577 166,98
4
252,39
4
337,81
7
423,23
2
508,65
5
594,07
8
679,50
1
764,92
4
850,34
8
100 90,642 185,53
8
280,43
8
375,55
3
470,25
8
565,17
3
660,08
7
755,00
2
849,91
6
944,83
1
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TI LIU THAM KHO
1. K thut in cao p ca V VIT N
2. Hung dn k thut in cao p ca NGUYN TH MINH CHC
3. Nh my in v trm bin p ca NGUYN HU KHI
4. Chng st cho nh v cng trnh ca VIM SUM
B GIO DC V O TOTRNG H BCH KHOA H NI
CNG HO X HI CH NGHA VIT NAMc lp - Tdo - Hnh phc
----------*******-------------
NHIM V THIT K K THUT
Ging vin hng dn : TS Nguyn nh Thng
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H v tn sinh vin :on Vn Minh
Ngnh hc : H thng in
I/ - u thit k:
TNH TON THIT K V CHNG V CHNG ST CHO NG DY V
TRM BIN P 110/22KV.
II/- Dliu ban
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