Download - Bai Tap Dai Ky Thuat Dien Cao AP
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8/3/2019 Bai Tap Dai Ky Thuat Dien Cao AP
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chng I
bo v chng st nh trc tip
1.1. cc yu cu:
Tt c cc thit b cn bo v phi c nm trn trong phm
vi bo v an ton ca h thng bo v.
Tu thuc vo c im mt bng trm v cc yu cu c
th, h thng cc ct thu st c th c t trn cc cao c
sn nh x, ct n chiu sng... hoc c t c lp.
Khi t h thng ct thu st trn kt cu ca trm s tn
dng c cao vn c ca cng trnh nn s gim c caoca ct thu st. Tuy nhin t h thng thu st trn cc thanh
x ca trm th khi c st nh s gy nn mt in p ging
trn in tr ni t v trn mt phn in cm ca ct. Phn
in p ny kh ln v c th gy phng in ngc t h thng
thu st sang cc phn t mang in khi cch in khng
ln. Do iu kin t ct thu st trn h thng cc
thanh x trm l mc cch in cao v in tr tn ca b phn
ni t nh.
i vi trm phn phi ngoi tri t 110kV tr ln do c cch
in cao nn c th t ct thu st trn cc kt cu ca trm
phn phi. Cc tr ca kt cu trn c t ct thu st th
phi ni t vo h thng ni t ca trm phn phi theo -
ng ngn nht v sao cho dng in IS khuych tn vo t
theo 3 - 4 cc ni t. Ngoi ra mi tr ca kt cu y phi c
ni t b sung ci thin tr s in tr ni t.
Ni yu nht ca trm phn phi ngi tri in p 110kV
tr ln l cun dy ca my bin p. V vy khi dng chng st
van bo v my bin p th yu cu khong cch gia hai
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im ni t vo h thng ni t ca ct thu st v v my
bin p theo ng in phi ln hn 15m.
Khi b tr ct thu st trn x ca trm phn phi ngoi tri
110kV tr ln cn ch ni t b sung ch ni cc kt cutrn c t ct thu st vo h thng ni t nhm m bo
in tr khuch tn khng c qu 4.
Khi dng ct thu st c lp phi ch n khong cch
gia ct thu st n cc b phn ca trm trnh kh nng
phng in t ct thu st n vt c bo v.
Vic lp t cc ct thu st lm tng xc sut st nh
vo din tch cng trnh cn bo v, do cn chn v trlp t cc ct thu st mt cch hp l
Tit din cc dy dn dng in st phi ln m
bo tnh n nh nhit khi c dng in st chy qua.
Khi s dng ct n chiu sng lm gi cho ct thu st
th cc dy dn in n n phi c cho vo ng ch v
chn vo.
1.2. phm vi bo v ca h thng thu st:
1.2.1. Phm vi bo v ca ct thu st.
Ct thu st lthit b khng phi trnh st m ngc li
dng thu ht phng in st v pha n bng cch s dng
cc mi nhn nhn to sau dn dng in st xung t.
S dng cc ct thu st vi mc ch l st nh
chnh xc vo mt im nh sn trn mt t ch khngphi l vo im bt k no trn cng trnh.Ct thu st to ra
mt khong khng gian gn ct thu st (trong c vt cn bo
v), t c kh nng b st nh gi l phm vi bo v.
a. Phm vi bo v ca mt ct thu st c lp.
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Phm vi bo v ca mt ct thu st c lp l min c gii
hn bi mt ngoi ca hnh chp trn xoay c ng knh xc
nh bi phng trnh.
)(1
6,1 XX
X hh
h
hr
+=
(1-1)
Trong : h: cao ct thu st.
hX: cao cn bo v.
ha=h-hX: cao hiu dng ct thu st.
rX: bn knh ca phm vi bo v.
d dng v thun tin trong tnh ton thit k thng
dng phm vi bo v dng dng n gin ho ng sinh ca
hnh chp c dng ng gy khc nh hnh sau:
Rx
0,2h
a
b
c
0,75h 1,5h
0,8hh
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Hnh 1.1: Phm vi bo v ca mt ct thu st.
Bn knh c tnh ton theo cng thc sau:
Nu hhX3
2 th )
8,01(5,1
h
hhr xX =
(1.2)
Nu hhX3
2> th )1(75.0
h
hhr xX =
(1.3)
Cc cng thc trn ch ng khi ct thu st cao di 30m. Hiu
qu ca ct thu st cao trn 30m gim i do cao nh hng
ca st gi hng s. C th dng cc cng thc trn tnh
ton phm vi bo v nhng phi nhn thm h s hiu chnh
hp
5,5= v trn honh ly cc gi tr hp75,0 v hp5,1 .
b. Phm vi bo v ca hai ct thu st c cao bng nhau.Phm vi bo v ca hai hoc nhiu ct thu li th ln hn
tng phm vi bo v cc ct n cng li. Nhng cc ct thu
li c th phi hp c th khong cch a gia hai ct phi tho
mn ha 7 (trong h l cao ca ct thu st). Phn bn
ngoi khong cch gia hai ct c phm vi bo v ging nh ca
mt ct. Phn bn trong c gii hn bi vng cung i qua 3
im l hai nh ct v im c cao h0 - phm vi bo v
cao ln nht gia hai ct c xc nh theo cng thc:
70
ahh =
(1.4) Khong cch nh nht t bin ca phm vi bo v ti ng
ni hai chn ct l rx0v c xc nh nh sau:
Nu 03
2hhx th )
8,01.(.5,1
0
00h
hhr xx =
(1.5)
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Nu 03
2hhx > th )1.(.75,0
0
00h
hhr xx =
(1.6)
Khi cao ca ct thu st vt qu 30m th c cc hiu chnh
h sh
p5,5
= ; trn honh ly cc gi tr hp75,0 v hp5,1 ; khi
h0 tnh theo cng thc
p
ahh
70 = (1.7)
0,75ha
1,5h
R
rox
1 2
hx h0h
rx
Hnh 1.2: Phm vi bo v ca hai ct thu st c cao
ging nhau.
c. Phm vi bo v ca hai ct thu st c cao khc nhau.Trng hp hai ct thu st c cao h1 v h2 khc nhau th
vic xc nh phm vi bo v c xc nh nh sau:
V phm vi bo v ca ct cao (ct 1) v ct thp (ct 2)
ring r. Qua nh ct thp (ct 2) v ng thng ngang gp
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ng sinh ca phm vi bo v ct cao im 3 im ny c
xem l nh ca mt ct thu st gi nh. Ct 2 v ct 3 hnh
thnh i ct c cao bng nhau v bng h2 vi khong cch
a. Bng cch gi s v tr x c t ct thu li 3 c cao h2.im ny c xem nh nh ca mt ct thu st gi nh. Ta
xc nh c cc khong cch gia hai ct c cng cao h2 l
a' v x nh sau:
0,2h1
0,8h1h1
ho
2
1
3
0,75h2 a' x 0,75h1
1,5h2 a 1,5h1
0,8h2
0,2h2
h2
Hnh 1.3: Phm vi bo v ca hai ct thu st c cao khc
nhau.
Nu 123
2hh ta c cng thc : )
8.01(5.1
1
21
h
hhx =
Nu 123
2hh > ta c cng thc: )1(75,0
1
21
h
hhx =
(1.8)
xaa ='
(1.9)Phn cn li ging phm vi bo v ct 1.
d. Phm vi bo v ca mt nhm ct thu st (s ct >2). bo v c mt din tch gii hn bi mt a gic th
cao ca ct thu li phi tho mn:
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ahD 8
(1.10)
Trong : D l ng knh vng trn ngoi tip a gic to bi
cc chn ct.Nhm ct tam gic c ba cnh l a, b,c c: +
))()((422
cpbpapp
abcRD
==
(1.11)
vi p l na chu vi :2
cbap
++=
(1.12)
Nhm ct to thnh hnh ch nht:
22 baD +=
(1.13)
vi a, b l di hai cnh hnh ch nht.
cao tc dng ca ct thu st ha phi tho mn iu kin:
8
Dha
(1.14)
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D
a
b
c
D
a
b
Hnh1.4: Phm vi bo v ca nhm ct to thnh tam gic v
ch nht.
1.2.1. Phm vi bo v ca dy thu st.a. Phm vi bo v ca mt dy thu st.
Phm vi bo v ca dy thu st l mt di rng. Chiu
rng ca phm vi bo v ph thuc vo mc cao hx c biu
din nh sau:
0,6h
0,2h
0,8h h
a
b
a'c
1,2h
2bx
Hnh1.5: Phm vi bo v ca mt dy thu st.Mt ct thng ng theo phng vung gc vi dy thu
st tng t ct thu st ta c cc honh 0,6h v 1,2h.
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Nu hhx3
2 th )
8,01.(.2,1
h
hhb xx =
(1.15)
Nu hhx 3
2
> th )1.(.6,0 hh
hbx
x =
(1.16)
Khi cao ct ln hn 30m th iu kin bo v cn c hiu
chnh theo p.
b. Phm vi bo v ca hai dy thu st. phi hp bo v bng hai dy thu st th khong cch
gia hai dy thu st phi tho mn iu kin hS 4
Vi khong cch trn th dy c th bo v c cc imc cao h0.
40
Shh =
(1.17)
Phn ngoi ca phm vi bo v ging phm vi bo v ca
mt dy, cn phn bn trong c gii hn bi vng cung i qua
ba im l hai im treo dy thu st v im c cao h0.
0,2h
0,8hh
ho
0,6h 1,2hS
bx
1 2
Hnh 1.6: Phm vi bo v ca hai dy thu st.
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chng ii
Tnh ton ni t cho trm2.1. 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
m 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.Tu theo nhim v v hiu qu m h thng ni t c chia
lm ba loi.
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Ni t lm vic.
Nhim v chnh l m bo s lm 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 tcc 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 ho
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.
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 qu 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 c 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 hu 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
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cu ca cc 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 hn. 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 yu
t l m bo v k thut v hp l v kinh t.
Mt s yu cu v 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 tho mn: 5,0R
+ i vi cc thit b c im trung tnh cch in th:
ttI
R250
.
+ 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: ttI
R125
.
+ Khi dng ni t t nhin nu in tr ni t t nhin
tho 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 tho 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 t nhn to l: 1R . Thc
t d RTN 0,5 th ta vn phi ni t nhn to v RTN c th
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xy ra bin ng nh t y 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 cacng trnh chn trong t...Vic tnh ton in tr tn ca cc
ng ng chn trong t hon ton ging vi in cc hnh tia.
+ V t l mi 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, a xt ... cha 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 trsut cu t cng bin i trong phm vi rng. Do vy trong
tnh ton thit k v ni t th tr s in tr sut 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 hiu chnh nh sau:
mdoTT K. =
(2.1)
Trong : tt: l in tr sut tnh ton ca t.
o: in tr sut o c ca t.
Km: h s ma ca t.
H s Km 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: md =105 .
in tr ni t ct ng dy: =10cR .
Dy chng st s dng loi C- 70 c in tr n v l:
kmr /38,20 = .
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Chiu di khong vt ng dy l: ml 300= . Pha 220 kv
ml 200= . Pha 110 kv
in tr tc dng ca dy chng st trong mt khong vt
l:)(714,010.300.38,2. 30220 ===
lrRcs .
)(476,010.200.38,2. 30110 ===lrRcs .
S l trong trm: Pha 110 kv 5=n .
Pha35 kv 2=n .
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:
+
== 5,0.
//TNNT
TNNTTNNTHT
RR
RRRRR .
Trong : RTN: in tr ni t t nhin.
RNT: in tr ni t nhn to 1NTR .
a) in tr ni t t nhin.Ni t t nhin ca trm l h thng chng st ng dy
v ct in 35kV v 35kV ti trm.
Ta c cng thc sau:4
1
2
1.
1
++
=
cs
c
cTN
R
R
R
nR
.
(2.3)
Trong : n: s l dy.
Rcs: in tr tc dng ca dy chng st trong
mt khong vt.
Rc: in tr ni t ca ct in.
-Pha 110 kv
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)(468,0
4
1
714,0
10
2
1
10.
5
1220 =
++=TNR
.
-Pha 110 kv
)(978,0
4
1
476,0
10
2
1
10.2
1220 =
++=TNR
.
RTN = RTN220//RTN110 = 0,317 )( .
Ta thy RTN
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CTTC
CTHT
RnR
RRR
...
.
+= .
(2.5)
Trong :
RC: in tr tn ca mt cc.
RT: in tr tn ca tia hoc ca mch vng.
n: s cc.
T: h s s dng ca tia di hoc ca mch vng.C: h s s dng ca cc.
i vi 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. Mch vng cch mng tng bao quanh trm mi chiu
1m.
in tr mch vng ca trm l:
td
LK
LRMV
.
.ln
..2
2
=
(2.6)
Trong :
L: chu vi mch vng
Vi s mt bng trm cho ta cn quy i mch
vng trm v mch vng hnh ch nht c cc cnh l l5, l6.l1, l2 c xc nh nh h phng trnh:
Thay vo h phng trnh trn ta gii c:
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=
=
)(9 5.2 2 7
)(5 5.8 4
5
4
ml
ml
t: chn su ca thanh ly t= 0,8m.
tt: in tr xut tnh ton ca t i vi thanh lm mch
vng chn su t.
muadott k. =
Tra bng vi thanh ngang chn su t=0,8m ta c kma=1,6.
).(1686,1.105 mtt ==
d: ng knh thanh lm mch vng. Chn thanh c b rng lb=4cm.
)(10.22
10.4
2
22
mb
d
===
K: h s hnh dng ph thuc hnh dng ca h thng ni
t.
Gi tr ca )(5
4
l
lfK = c cho bng sau:
l1/l2 1 1.5 2 3 4
K 5.53 5.81 6.42 8.17 10.4
Bng 2.1: Bng )(5
4
l
lfK =
Ta c th sau:
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H s hnh dng
0
2
4
6
8
10
12
0 1 2 3 4 5
l1/l2
K
Hnh 2.1: H s hnh dng )(5
4
l
lfK =
7.255.84
95.227
5
4 ==l
lTra th ta c: 82,6=K
Vy in tr mch vng l:
)(81,010.2.8,0
625.82,6ln
625..2
1682
2
==
MVR
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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 qutrnh c nh hng khc nhau n hiu qu ni t.
in tr tn xung kch ca ni t tp trung:
in tr tn xung kch khng ph thuc vo kch thc hnh
hc ca in cc m n c quy nh bi bin dng in I,
in tr sut v c tnh xung kch ca t. V tr s in
tr tn xoay chiu ni t t l vi nn h s xung kch c gi
tr :
.
1
IR
Rxkxk ==
(2.7)
hoc dng tng qut: ),( Ifxk =
(2.8)
Tnh ton ni t phn b di khng xt n qu
trnh phng in trong t:
S ng tr ca ni t c th hin nh sau:
L R
G C
L R L R L R
G C G C G C
Hnh 2.2: S ng tr ca h thng ni t.
Trong mi trng hp u c th b qua in tr tc dng R
v n b so vi tr s in tr tn, ng thi cng khng cn
xt ti phn in dung C v ngay c trong trng hp sng xung
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kch, dng in dung cng rt nh so vi dng in qua in
tr tn.
L
G
L
G
L
G
L
G
Hnh 2.3: S ng tr rt gn
Trong :
L: in cm ca in cc trn mt n v di.
G: in dn ca in cc trn mt n v di.
( )[ ]
=
mH
rlL
31,0ln.2,0
(2.9)
Vi:
l: chiu di cc.
r: bn knh cc phn trc nu cc l thp dt c b rng
b(m).
Do 4br=
Gi Z(x,t) l in trng xung kch ca ni t ko di, n l
hm s ca khng gian v thi gian.
),(
),(),(
txI
txUtxZ =
(2.10)
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Trong U(x,t); I(x,t) l dng in v in p xc nh t h
phng trnh vi phn:
=
=
UGx
I
t
ILx
U
.
.
(2.11)
Gii h phng trnh ny ta c in p ti im bt k v ti
thi im t trn in cc:
+=
=
121
..cos.1
1..2
.),(
k
T
t
l
xke
kTt
lG
atxU K
(2.12)
Tng tr xung kch u vo ca ni t:
+=
=
12
1 11..2.
),0(k
T
t
Kekt
TtlG
atZ
(2.13)
Vi:22
2
.
..
k
lGLTK = ; 2
2
1
..
lGLT = ;
2
1
K
TTK =
Tnh ton ni t phn b di khi c xt qu trnh
phng in trong t.
Vic gim in p v mt dng in cc phn xa cain cc lm cho qu trnh phng in trong t cc ni
ny c yu hn so vi u vo ca ni t. Do in dn
ca ni t (trong s ng tr) khng nhng ch ph
thuc vo I, m cn ph thuc vo to . Tuy nhin vic tnh
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ton tng tr s rt phc tp v vy ta c th b qua trong phm
vi n ny.
Tnh ton cho trm thit k:
y l trm 220/110kV nn cho php ni t chng st nichung vo vi ni t an ton. Do ni t chng st l ni
t phn b di dng mch vng. Lc ny mch vng c xem
nh hai tia ghp song song. Chiu di mi tia l:
)(5.3122
625
2m
Ll ===
in cm ca in cc trn mt n v di l:
( )[ ] = mH
rlL
31,0ln.2,0
C bn knh in cc )(10.14
10.4
42
22
mbd
r
====
Vy:
=
=
mHL
07.231,010.1
5.312ln.2,0
2
in dn ca in cc trn mt n v di:
MVSRlG
..2
1=
(2.14)
Vi setmuaATmua
MVATMVS k
k
RR .
.
.=
(2.15)
kmua.AT=1,6 v kmua.set=1,2
)(608,02,1.6,1
81,0==MVSR
).1(10.63,2608,0.5,312.2
1 3 mG ==
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Trong thit k tnh ton ta chn dng sng ca dng in st l
dng sng xin gc c bin khng i.
Phng trnh sng c dng sau:
>=
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=
==++++=1
2
2222645,1
6...
1...
2
1
1
11
k kk
(2.19)
=
++++=1
2222......
21.
1 21
k
TTTT
k
eeee
k
K
dsdsds
K
ds
(2.20)
Trong dy s ny ta ch xt n s hng cha e -4, t s e-5 tr i
c gi tr rt nh so vi cc s hng trc nn ta c th b qua. ta
tnh n k sao cho:
ds
ds
K
ds Tkhay
kTT
12
2
1
.44 =
; ds
Tk
1
.2
(2.21)
)(87.53277.10.14,3.98,1..
2
23
2
2
1 slGL
T
===
56,65
87.53.2.2 1 ==
ds
Tk
Ta ly gi tr k t 1-7:K 1 2 3 4 5 6 7 8
K2 1 4 9 16 25 36 49 64
TK
48.38325 12.09581 5.37592 3.02395 1.93533 1.34398 0.98741 0.75599
K
ds
T
0.10334 0.41337 0.93007 1.65346 2.58354 3.72030 5.06374 6.61386
Kds
Te
0.90182 0.66142 0.39452 0.19139 0.07551 0.02423 0.00632 0.00134
0.90182 0.16536 0.04384 0.01196 0.00302 0.00067 0.00013 0.00002
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2K
e KdsT
Bng 2.2: Bng tnh ton chui s
=
1
2.
1
k
TK
ds
e
k
Ta c
=
=1
213,1.
1
k
TK
ds
ek
Vy:
)(16,6)13,1645,1.(5
87,53.21
5.312.10.14,3.2
1),0(
3=
+= MVdsZ
Do my bin p l phn t yu nht nn ta ch cn kim tra
vi my bin p. Khi c dng in st i vo ni t m
bo an ton phi tho mn iu kin:
MBAdsXKd UZIU %50),0(. 110
%50U =460 kV.
C MBAd UU %50> vy cn phi tin hnh ni t b sung.
2.2.3. Ni t b sung.
Trong ni t b sung ta s dng dng ni t tp trung
gm thanh v cc ti chn cc ct thu st. do vic xc nh zbs
bng l thuyt li rt kh khn nn ta chn hnh thc ni t b
sung nh sau:
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Chn thanh ni t b sung l loi thp dt c:
Chiu di lt = 12(m).
B rng bt = 0,04(m).
Dc theo chiu di thanh c chn 3 cc trn c:Chiu di cc lcc = 3(m).
ng knh d = 0,04(m).
Khong cch gia hai cc a = 6(m).
chn su t = 8(m).
Ni t c tnh ton cho chng st nn ta ly h s kma nh sau:
i vi thanh ngang chn su t = 0,8(m); kma = 1,2.
i vi cc di 3m chn su t = 0,8(m); kma = 1,15.
S ni t ca h thng khi c ni t b sung nh sau:
Hnh 2.4: S ni t b sung.
in tr thanh
Cng thc s dng tnh ton:
T
T
T
TttT
dt
lk
lR
.
.ln.
..2
2
.
=
(2.23)
Trong :
l: Chiu di ca thanh l = 12(m).
t=0,8mlT
lcc
a
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t: chn su ca thanh lm tia t = 0,8(m)
tt.t: in tr sut tnh ton ca t i vi thanh lm tia
chn su t
).(1262,1.105..mk
muaoTtt ===
d: ng knh thanh lm tia. chn thanh dt c b rng b
=0,04(m) nn:
)(10.2210.42 22 mbd ===
k: H s hnh dng ly 1=k do ni t l tia ngang.
Vy in tr ca thanh b sung l:
)(22,1502,0.8,0
12.1
ln.12..2
126 2
== TR
in tr cc
+
+=coc
coccoc
coc
ttC
lt
lt
d
l
l '.4
'.4ln.
2
1.2ln.
..2RC
(2.24)
Trong :
ttc: in tr sut ca t i vi cc chn su
t=0,8(m).
).(75,12015,1.105. mkmuaottC ===
d: ng knh ca cc: md 04,0= .
)(3,28,02
3
2' mt
lt coc =+=+=
in tr b sung ca cc l:
)(27,3433,2.4
33,2.4ln.
2
1
04,0
3.2ln
3..2
75,120=
++=
C
R
in tr b sung
Cng thc s dng tnh ton:
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TCCT
CT
bsRnR
RRR
...
.
+=
(2.25)
Trong :n: s cc
t, c: h s s dng ca thanh v cc.
Vi n=3; lcc = 3(m); a=6(m); a/l=2.
Tra bng 3 phn ph lc(trang 82) sch hng dn thit k tt
nghip ta c:c=0,87.
Tra bng 5 phn ph lc(trang 84) sch hng dn thit k tt
nghip ta c: t=0,89.
in tr b sung l:
)(43,789,0.27,343.87,0.22,15
27,34.22,15=
+=
bsR
Tng tr ca h thng khi c ni t b sung
Ta c cng thc tnh tng tr xung kch khi c ni t b sung
nh sau
BA
e
XR
R
R
RR
RRZ
T
X
k
Kbs
setNT
setNT
setNTbs
setNTbs
dsXHbs
dsK
+=
++
+=
=
.
cos
1
.2.),0( 1
2
2
.
1
2
)(
)(
)(
)(
(2.26)
Trong :
)(56,0608,043,7
608,0.43,7.
)(
)(=
+=
+=
setNTbs
setNTbs
RR
RRA
12
2
.
1
2
)(
)( .
cos
1
.2T
X
k
Kbs
setNT
setNT
dsK
e
XR
R
R
B
=+
=
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Xt chui s1
2
2
.
1
2
)(
)(.
cos
1
.2T
X
k
Kbs
setNT
setNT
dsK
e
XR
R
RB
=
+
=
Tng t nh trn ta ch xt n s hng e -4 viT1 = 53.87s v ds = 5s
Ta tnh n Xk sao cho: 4.1
2
2
T
X dsk
62,205
87,53..2..2 1 ==
ds
K
TX
Trong Xk l nghim ca phng trnh:
KK
K
bs
setNT
K
XX
XR
RtgX
.08,0.43,7
608,0
.)(
==
=
Gii phng trnh trn bng phng php th v xc nh c
nghim nh sau :
X1=2,91; X2=5,85; X3= 8,81; X4= 11,81; X5 = 14,84; X6 =
17,89X7 = 20,96; X8 = 24,04; X9 = 27,14
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Hnh 2 7: th xc nh nghim phng trnh tgXk = -
0,08.X
Ta thy X7 = 20,96 >20,62 ln ta ch xt n X6
Ta c T1 = 53,87(s); Rbs = 7,43 (); RMVS = 0,608 ().k 1 2 3 4 5 6
Xk2,91 5,85 8,81 11,81 14,84 17,89
cos(Xk)-0,9733 0,907633 -0,8169 0,727332 -0,64638 0,573884
kX2
cos
1
1,055613 1,21389 1,498513 1,890319 2,393435 3,036353
kbx
MVS
XR
R2cos
1
+ 1,137613 1,29589 1,580513 1,972319 2,475435 3,118353
1
2
.T
X dsK
e
0,923466 0,724858 0,48201 0,269434 0,126099 0,049326
BK0,987097 0,680172 0,370844 0,166115 0,061943 0,019235
Bng 2.3: Bng tnh ton chui s
1
2
2
.
1
2
)(
)( .
cos
1
.2T
X
k
Kbs
setNT
setNT
dsK
e
XR
R
R
B
=
+
= =2,28
Tng tr xung kch l :
)(84,228,256,0),0( =+=+= BAZ dsXK
in p khi c dng in i vo ni t ti thi im dng
in st t gi tr cc i l:
)(42684,2.150),0(.max kVZIU dsXK ===
Ta thy kVUkVU MBA 460426 %50max =