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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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 3 l- Trm 110 kV c 5 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
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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.
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
- 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
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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:
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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.
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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.
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
- 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:
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
- 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.
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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:
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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 = 16,5mi 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:
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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 = 16,5 + 10 = 26,5 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 + 9 = 20 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=16,5(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)
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
Vy bn knh ca khu vc bo v gia hai ct thu li l:
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)
Vy bn knh ca khu vc bo v gia hai ct thu li l:
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)
Vy bn knh ca khu vc bo v gia hai ct thu li l:
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:
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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
ng knh D ca ng trn ngoi tip i qua cc nh ct l:
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
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
+Cnh (17-18)=50m=k
+Cnh (18-21)=40m=l
ng knh D ca ng trn ngoi tip i qua cc nh ct 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)
Vy bn knh ca khu vc bo v gia hai ct thu li l:
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)
Vy bn knh ca khu vc bo v gia hai ct thu li l:
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
+Xt 2 ct (20-21): khong cch gia 2 ct l a=50 m
Ta c: h0=h-a
7=19-
50
7= 11,857 (m)
hx=11(m) > 23
h0= 23
.11,857=7,905(m)
Vy bn knh ca khu vc bo v gia hai ct thu li l:
rx20-21=0,75.h0(1-x
0
h
h)=0,75.11,857.(1-
11
11,857)=0,643 (m)
+Xt 2 ct (18-21): khong cch gia 2 ct l a=40 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)
Vy bn knh ca khu vc bo v gia hai ct thu li l:
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rx18-21=0,75.h0(1-x
0
h
h)=0,75.13,286.(1-
11
13,286)=1,714 (m)
+Xt 2 ct (15-18): khong cch gia 2 ct l a=30 m
Ta c: h0=h- a7
=19- 307
= 14,714 (m)
hx=11(m) >2
3h0=
2
3.14,714=9,809(m)
Vy bn knh ca khu vc bo v gia hai ct thu li l:
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
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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
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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.
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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.
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
*)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
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+ 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:
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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.
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- 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
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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
5,535,53
5,81
6,42
8,17
10,4
7,144
l1/l2
K
1 1,5 2 2,444 3 4
T th ng vi t s l1/l2 =220/90=2,444 ta ni suy c K=7,144
Nh vy in tr mch vng l :
( )2 2
ttMV 2
K.L 160 7,144.620R .ln .ln 0,779
2. .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
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Ta c in tr ni t ca h thng:
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 yucu 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.
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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.
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 .k
k=
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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:
setset set setmuaNT MV mua
atmua
R 0,779.1,25R R .k 0,608
k 1,6= = = =
Thay s vo ta c in dn ca in cc l:
3
0
1 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 .= = = =
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
Thay s:( )
2
3
2
1 2 2
6202,145.2,653.10 .
L.G.l 2T 55,41 s
3,14
= = =
Tng tr tin ti tr s n nh khi:k
t 4T
12
54 k 6,658
Tk
ly k = 6
k k
t t2 6T T
2 2
k 1 k 1
1 11 e e
k 6 k
= =
=
5 5.4 5.9 5.16 5.25 5.362
55,41 55,41 55,41 55,41 55,41 55,411 1 1 1 11. . . . . .6 4 9 16 25 36
0,488
= + + + + +
=
e e e e e e
Do tng tr sng u vo s c tnh bng:
(0,t) 3
1 2.55,41Z .(1 .0,488) 7,184
2,653.10 .620 5= + =
Tng tr sng xung kch Zxk =1
Z(0, t) 3,5922 =
iu kin ni t chng st tha mn khi Uxk= IS. Zxk< U50%
Trong , dng in st IS = 150kA, v in p 50% ca trm U50% = 460kV
Ta c in p xung kch : Uxk= 3,592. 150 = 538,8kV > U50%.
Tc l gi tr in tr ni t ny cha t nn ta cn tin hnh tnh ton ni t b
sung
Ni t b sung:Vic tin hnh ni t b sung nhm m bo cho gi tr in tr ni t t tr s yu
cu v m bo an ton. Chng ta thc hin vic ny bng cch t thm mt in tr
ni t ngay chn ct thu li, chn my bin p. V hnh thc th ni t b sung l loi
ni t tp trung (gm thanh hay thanh v cc).
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
* 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.
SVTH:Nguyn Tin Hng Lp:1-H2 33 Khoa: H Thng in
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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
SVTH:Nguyn Tin Hng Lp:1-H2 34 Khoa: H Thng in
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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
SVTH:Nguyn Tin Hng Lp:1-H2 35 Khoa: H Thng in
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
- 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:
SVTH:Nguyn Tin Hng Lp:1-H2 36 Khoa: H Thng in
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BI TP DI MN KT CAO P GVHD:Th.S Trn Hong Hip
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.