Bai Tap Dai Ky Thuat Dien Cao AP

8/3/2019 Bai Tap Dai Ky Thuat Dien Cao AP

1/30

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


2/30

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)


8/30

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.


10/30

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.


11/30

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


12/30

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


13/30

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 = .


14/30

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


15/30

)(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


16/30

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:


17/30

=

=

)(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:


18/30

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


19/30

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


20/30

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)


21/30

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


22/30

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 ==


23/30

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:

>=


24/30

=

==++++=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


25/30

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:


26/30

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


27/30

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:


28/30

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

=+

=


29/30

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


30/30

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 =

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