205

Chng 5

TNH TON C HC NG DY TI IN TRN KHNG

5.1 KHI qut chung v ng dY TRN KHNG

5.1.1 ng dy trn khng ng dy trn khng bao gm dy cc ct in, trn c cc x v dy dn c treo vo cc x qua cc x cch in. Ct in c chn xung t bng cc mng vng chc, lm nhim v dy trn cao so vi mt t, do gi l ng dy trn khng. Trn ct cn c th treo dy chng st st khng nh trc tip vo dy dn. 1.Dy dn Dy dn c lm bng ng (M), nhm (A), nhm li thp (AC), thp (K, TK). C cc loi dy dn sau: - Dy n ch c mt si duy nht: thng l dy thp, dy lng kim li thp h ng ngoi; - Dy vn xon ng nht: nhiu si nh (ng, nhm hay thp) vn xon li vi nhau; - Dy vn xon nhm li thp, tng bn ngi ta lm li thp trong, cc si nhm bn ngoi; - Dy vn xon nhm li thp c thm cc si ph bng cht cch in tng bn knh dng cho in p 220 kV tr ln. 2.Ct in Ct in lm bng g, btng ct thp hay bng thp. Theo chc nng ct in gm c: - Ct no v no gc: ct no gi chc u dy ni vo ct qua chui s no; ct no gc dng khi ng dy i hng; - Ct v gc lm nhim v dy dn ni vo ct qua chui s . Ct cng chia ra ct thng v ct gc. Khi dng dy i hng, nu gc i hng t 10 n 200 th dng ct gc, nu gc ln hn th dng ct no gc. Nu dng ct gc th thng treo thm t cn bng chui s khng b lch qu; - Ct cui dng u v cui ng dy; - Ct vt l ct cao hoc rt cao s dng khi ng dy qua chng ngi cao hoc rng nh ng dy in, ng dy thng tin, sng rng,.. Ct vt c th l ct no hay ; - Cn c cc ct dng chuyn v cc dy pha (ct o pha) v ct ni cc nhnh r (ct r). Cng c cc ct c bit trn t dao cch ly, t b,... Khong cch gia hai im treo dy trn hai ct k nhau gi l khong ct. Nu hai ct k nhau l ct no th gi l khong ct no. Khong gia hai ct no gm nhiu ct lin tip gi l khong no. Khong no bao gm nhiu khong ct thng. Khi ng dy vt qua chng ngi th ta c khong vt, khong vt c th c mt hoc nhiu khong ct.

3.S cch in v ph kin

206

S cch in c th l s dng hay s treo. S ng dng cho in p trung tr xung, mi dy pha dng mt s cm trn cc cc d t trn x ct. S treo gm cc bt s treo ni tip thnh chui dng cho in p trung n siu cao. C chui s v chui s no dng cho ct v ct no. Trn chui s c th c cc kim ca khe h chng st v thit b iu ha phn b in th trn chui s. Dy dn c gn vo chui s nh cc kp dy. i vi ng dy trn khng cn c cc thit b khc nh: - Qu t chng rung tiu hao nng lng do dao ng ring ca dy dn, chng hin tng cng hng tn s dao ng ring vi tn s cng nghip, m bo dy khng b rung; - chng qu in p trn ng dy dng dy chng st, ni t cc ct in, t chng st ng, to cc khe h phng in.

Trn bng 5.1 gii thiu mt vi thng s c trng cc ng dy trn khng

Bng 5.1 Mt vi s liu c trng cc ng dy trn khng

in ap;kV Ct Khong ct; m 10 B tng ct thp 80150 35 B tng

Thp 200260 220270

110 B tng Thp

220270 250350

220 B tng Thp

220300 350450

500 B tng Thp

250300 300450

5.1.2 Cc trng thi lm vic ca ng dy trn khng

Xt v mt c hoc, ng dy trn khng s vn hnh trong cc trng thi khc nhau m mi trng thi chng chu tc ng ca cc lc tng ng. Mi trng thi c c trng bi tp hp cc thng s mi trng v tnh tng dy dn v dy chng st. Trng thi mi trng dy l tc gi v nhit khng kh. C 5 trng thi xem xt c hc cho ng dy nh sau:

1.Trng thi nhit thp nht: Khi nhit thp nht, dy dn b co li, gy ng sut trong dy ln nht. Dy b co li c th gy lc ko ngc chui s v nh ct.

2. Trng thi bo: Trng thi ny dy dn chu ti trng c hc ln nht, ng sut trong dy ln nht v dy b lch khi mt phng ng.

3. Trng thi nhit trung bnh: y l trng thi lm vic lu di ca dy dn. Dy dn chu s rung ng thng xuyn do gi gy mi dy v gy nguy c t cc si dy cc ch kp dy.

4. Trng thi nhit cao nht: Dy dn b gin ra nhiu nht lm cho khong cch t dy dn thp nht n t l nh nht. vng trong trng thi ny l vng ln nht.

207

5. Trng thi qu in p kh quyn: Trng thi ny xy ra trong cc gi ging st, cc dy dn b gi lm dao ng n gn nhau v gn ct, lm cho kh nng gy phng in rt cao. Trng thi ny dng tnh dy chng st v kim tra lch chui s.

Bng 5.2 gii thiu s liu v iu kin tnh ton c hc cho ng dy ti in trn khng.

Bng 5.2 S liu v iu kin tnh ton c hc cho ng dy trn khng

Trng thi iu kin tnh ton

Nhit ; 0C

p lc gi; daN

Tc gi; m/s

1. Nhit khng kh thp nht 5 0 0 2. Trng thi bo 25 qmax vmax 3. Nhit khng kh trung bnh 25 0 0 4. Nhit khng kh cao nht 40 0 0 5. Qu in p kh quyn 20 0,1qmax, nhng

6,25 daN/mm2 v0,3vmax

1kG lc =0,98 daN (deca Newton); qmax l p lc gi ln nht trong khu vc ng dy i qua.

5.1.3 Ti trng c hc i vi ng dy trn khng

C hai ti trng tc ng ln dy dn l: - Ti trng do trng lng dy gy ra; - Ti trng do p lc gi tc ng ln dy dn.

1. Ti trng c hc do trng lng dy

Trng lng 1m dy l G [kg/m], hoc thnh n v [kg/m] = 9,81[daN//m]. Khi t ti g1 do trng lng bn thn dy dn c xc nh theo cng thc:

201201 mm.m/daN;FG

81,9ghay;]mm.m/kg[;F

Gg (5.1)

trong F l tit din dy dn;[mm2]

Do cu to ca dy vn xon gm nhiu si vn xon vi nhau nn chiu di thc t ln hn, khi t ti g1 vn tnh theo cng thc (5.1) nhng nhn thm vi h s khong 1,021,03.

2.Ti trng do gi

Gi thit gi thi ngang vung gc vi chiu di dy dn, gy ra p lc (sc p) i vi dy dn. p lc gi Pv trn 1m dy dn c tnh theo cng thc sau:

m/kg;S.q.k.aPv

trong : a- h s khng u ca gi; k- h s ng lc ca khng kh; S- b mt cn gi; m2.

208

q- ng nng ca gi; [kg/m2]

Nu gi thi chch i th phi nhn thm vi sin ( l gc hp bi tuyn dy v chiu gi). H s khng u ca gi dc theo khong vt ph thuc vo tc v ca gi nh sau:

Tc gi; m/s 20 25 30 40 H s a 1 0,85 0,75 0,7

H s ng lc k ca khng kh ph thuc vo hnh dng, b mt m gi tc dng vo, ng knh dy dn, c th nh sau:

- Dy dn c ng knh d20 mm th k=1,1 - Dy dn c ng knh d

209

Ti ton (x,y) c tung cch mt t l y, cch trc tung l x chiu di ca ng dy tnh t im O l Lxy. ng cong ca ng dy c dng dy xch [1], c phng trnh biu din:

ooxy

oox

yxshyL

yxchyy

(5.5)

trong xo v yo l ta ca im O.

Trin khai cc cc hm hypebolic thnh chui

!5!3sh

!4!21ch

53

42

p dng cho (5.5) ta c

5o

5

3o

3

oo

ooxy

4o

4

2o

2

oo

ox

y!5x

y!3x

yxy

yxshyL

y!4x

y!2x1y

yxchyy

(5.6)

Tnh gn ng, b qua cc thnh phn bc cao, ch gi li hai phn t u ca chui th cc phng trnh ca (5.6) s l :

2o

2

xy

o

2

ox

y6xxL

y2xyy

(5.6a)

T (5.6a) ta thy rng phng trnh yx ca ng dy tr thnh dng parabol. iu ny cng r nt hn nu ly trc honh i qua im dy thp nht v trc tung vn qua im gia ng cng dy (hnh 5.2), khi phng trnh yx s l:

210

o

2

x y2xy (5.7)

2.ng sut, vng v chiu di dy dn

a) ng sut Dy dn b vng xung vi gi thit: - Dy vng l tng, ngha l khng bi ko cng; - Trng lng dy dn phn b u dc theo chiu di dy; - Ti ta im bt k (x,y) u c mt lc ko Txy tc ng theo phng tip tuyn ti im (hnh 5.1). Lc ko chnh bng trng lng Gy ca on ng dy t im qua rng rc l tng (khng c ma st) n trc honh: y.F.gTxy (5.8)

trong : g - t ti ca dy dn; [kg/m.mm2] y khong cch t im ta (x,y) n trc honh; [m]

Sc cng ti im thp nht bng: oo y.F.gT (5.8a) Sc cng ti im treo dy A bng: f.F.gTfy.F.gy.F.gT ooAA (5.8b) Vi f l vng ca dy dn (xem hnh 5.1)

ng sut trong vt liu dy dn bng sc cng trn mt n v tit din dy dn. Vy ta c:

f.gfy.gF

Tg

y;y.gFT

;F

T

o0A

A

ooo

oo

xyxy

(5.9)

trong Aoxy ,, l ng sut day dn ti cc ta (x,y), im O v im A.

T cng thc (5.8) ta thy rng ng sut ca dy dn ti im treo dy dn ln hn ng sut ti im thp nht. Ti cc ng dy c chiu di khong vt trung bnh, th s khc

211

nhau gia A v o rt t (khong 0,3%) nn c th b qua v ly o tnh ton. Nu khong vt ln (trn 700m) th phi dng cng thc tnh ng nh nu trn.

b) vng f

T phng trnh (5.5) ti ta treo dy im A (hnh 4.1) vi x=2 , ta c

ooA y2chyy , m yA=yo+f, suy ra :

1

2g.ch

g1

y2chyyyf

o

o

oooA

.

Trin khai hm

1

2g.ch

o

thnh chui c:

116!.4g

4.2g11

2g.ch 4

o

44

2o

22

o

Khi vng f s l :

4

o

44

2o

22o

16.24g

4.2g

gf

Gn ng, gi li hai s hng u ca chui ta c cng thc tnh gn ng vng nh sau:

0

2

8gf

(5.10)

Vi cc khong vt di 700m, vng c xc nh gn ng theo cng thc (5.10) ch t sai s rt nh (chng t 0,1% n 0,3%).

c) Chiu di dy dn L

T phng trnh (4.5) ti ta treo dy im A (hnh 4.1) vi x=2 , ta c

o

o

ooA 2

g.shg

2y2

shy2y2L

Trin khai hm o2

g.sh thnh chui c:

5o

55

3o

33

oo 32!.5g

8!.3g

2g

2g.sh

Khi di L s l :

5

o

55

3o

33

o

o

32!.5g

8!.3g

2g

g2L

212

Gn ng, gi li hai s hng u ca chui ta c cng thc tnh gn ng vng nh sau:

3f8

24gL

2

2o

23

(5.11)

Vi cc khong vt di 700m, vng c xc nh gn ng theo cng thc (5.11) ch t sai s rt nh (chng t 0,1% n 0,3%).

Cc cng thc (5.10) v (5.11) l nhng cng thc c bn tnh ton c kh ca ng dy trn khng. C hai cng thc u c dng phng trnh parabol. Vi php tnh gn ng ta a cng thc dng dy xch v dng parabol.

4.2.2 Dy dn treo trn hai im khng cng cao 1. vng Ta vit li phng trnh (5.7) ca trng hp dy dn treo trn hai im c cng cao, c

xt n (5.9) c:

0

2

x 2x.gy

(5.12)

y l phng trnh parabol c im thp C trng vi im gc ta 0 (xem hnh 5.2). Khi hai im treo dy A v B khng cng cao, ta c th khng nh rng: Dy dn vn to thnh ng parabol duy nht i qua hai im treo dy A,B (hnh 5.3). Tm ca trc ta s gn v pha im treo dy thp hn. Phng trnh (5.12) vn p dng c cho trng hp ny. T hnh 5.3 thay a v b vo 5.12 ta c:

0

2

B0

2

A 2b.gh;

2a.gh

(5.13)

hA v hB l cao ca hai im treo dy.

213

lch gia hai im treo dy l:

22o

AB ab2gtg.hhh

y l gc to bi ng thng AB vi trc honh; harctg

Thay b= - a vo phng trnh trn ta c:

a22

gab2

gho

22

o

Gii phng trnh ta c:

.Gh.T

21

.gh.

21b

.Gh.T

21

.gh.

21a

oo

oo

(5.14)

Tnh c a,b ta xc nh c h ta ca ng cng dy. C th xy ra ba trng hp i vi im O, im thp nht ca ng dy cng, l: - im thp nht nm gn khong ct (hnh 5.3): AB hhh;ba - im thp nht trng im treo dy (hnh 5.4a): BA hh;0h;b - im thp nht nm ngoi khong ct (hnh 5.4b): AB hhh;ab a)

b)

Trng hp hnh 5.4b khng c im thp nht trn thc t (khng tn ti), ch l im gi tng. Cc phng trnh (5.14) ng cho mi trng hp. Nu hA>hB th im thp nht gn B, do du ca thnh phn th hai trong (5.14) s thay i v a>b. vng gia khong ct fC (hnh 5.3) l :

CAC y2hhf (5.15)

214

yC l cao ca dy ti xC:

.g

h.a

2x oC

(5.16)

Thay xC vo (5.12) ta c:

2

oC

hg2

y

(5.17)

Thay

.g

h.2

x2

b oC

vo (5.13) ta c:

2

2o

o

2

2

22

22oo

2o

2oo

A

.g2h.

2h

8.g

.gh.

.gh..

22s2.g2

h.2g2

h

(5.18)

Thay (5.16), (5.17) v (5.18) vo (5.15) ta c:

o

2

C 8.gf

(5.19)

Ta thy vng im gia khong ct fC cng tnh ging nh hai im treo dy ngang nhau.

2. Khong ct tng ng

Ta c ng dy nh trn hnh 5.5, y ko di ng parabol A n im B ngang vi im B v c im A ngang vi im A. t trc ta vo im thp nht C v c c trng hp hai im treo cng cao tng ng. ng BAB to ra ng cng dy tng ng ln, cn AA to ra ng cng dy tng ng b. Khong ct tng ng l:

.g

h.2a2;

.gh.2

b2 otdo

td

(5.20a)

Cc thng s ca khong ct tng ng cng tun theo mi quy lut ca khong ct tht. Khong ct tng ng dng tnh ton dy dn trong cc trng thi khc nhau.

215

Ta c th dng khong ct tng ng tnh vng ca ng cng dy tng ng ln f, chnh l hB tnh theo (5.13):

o

td

o

2

B 8.g

2b.gh'f

(5.21b)

Tng t ta c vng cho ng cng dy tng ng b, chnh l hA :

o

td

o

2

A 8.g

2a..ghf

(5.21c)

Cng thc trn dng tnh ton cc khong ct c chnh lch ln. Vic xy dng cng thc tnh di dy L ( di on ACAB) trong trng hp ny c da theo cng thc (5.11), ta c:

3td3td2

o

2tdtd

2o

3td

td2o

3td

tdACB'OA

24g

2

24g.

21

24g.

21LLL

Thay tdtd , t cng thc (5.20a) vo cng thc trn, sau vi php bin i ta c di ca dy dn :

.2h

24g.L 2

o

23

(5.21d)

Nu chch lch gia A v B khng ln th t s h/2 nh, c th b qua; Khi cng thc (5.21d) tr v cng thc (5.11)

5.3 PHNG TRNH TRNG THI CA DY DN

5.3.1 t vn

i vi mt dy dn treo ln hai ct vi khong ct l , mt i kin no ( nhit , tc gi,..) th s tn ti cc thng s ng sut , vng f, di ng dy L, t ti g tng ng. Trong qa trnh lm vic iu kin s thay i th cc thng s ny s b thay i theo. Ngi thit k phi tnh ton la chn sao cho : - Trong mi bin i ca thi tit ng sut khng c vt qu gi tr cho php CP v nh vy s lm hng dy dn; - vng khng c ln qu, v s lm cho khong cch an ton ca dy b vi phm. lm c vic ny ngi thit k phi bit c quy lut bin i ca ng sut , vng f theo nhit v tc gi th hin qua t ti g. Quy lut bin i ny chnh l phng trnh trng thi ca dy dn.

5.3.2 Phng trnh trng thi ca dy dn

Gi thit trng thi m ta bit : nhit m , dy dn c t ti gm, ng sut m th di ca dy dn trong khong ct theo cng thc (5.11) l :

216

2m

2m

3

m 24g.

L

.

Khi nhit thay i t m sang n th chiu di dy dn s thay i bi mt lng bng :

mnm1 LL ,

trong l h s gin n theo nhit ca nguyn liu cu to dy dn, n v l 1/.

Tng ng vi trng thi m sang n th ng sut dy dn thay i t m sang n v do yu t ny chiu di cng b thay i thm mt lng na:

mnm2 EL

L ,

trong E l mun n hi ca dy dn, n v l kg/mm2 hay daN/mm2.

trng thi n, chiu di ca dy dn trong khong ct theo cng thc (5.11) l:

2n

2n

3

n 24g.

L

,

di Ln bng tng di Lm cng vi cc on di thm 1L v 2L . Vy ta c :

mnmmnm2m

2m

3

2n

2n

3

EL

.L24

g.24

g.

Trong tnh gn ng mL , tc l coi chiu di ca dy dn trong khong ct bng chiu di khong ct (v d =400m, f=12 m th L=400,96 m); khi c th vit:

mnmn2m

2m

2

2n

2n

2

mnmn2m

2m

3

2n

2n

3

E1.

24g.

24g.

E.

24g.

24g.

t E1

, gi l h s ko di n hi.

Chia hai v ca phng trnh trn cho , ng thi chuyn v ta c:

mn2m

2m

2

m2n

2n

2

n 24g.

24g.

(5.22a)

Phng trnh (5.22a) gi l phng trnh trng thi ca dy dn. l phng trnh c bn quan trng trong tnh ton c kh dy dn. Vi phng trnh trn, khi bit cc thng s trng thi m l mmm ,g, th c th tnh ra c n trong iu kin mi vi cc thng s

nn ,g

Nu hai im treo dy khng cng cao th phng trnh trng thi s l:

217

mn2m

22m

2

m2n

22n

2

n 24cosg.

24cosg.

(5.22b)

trong harctg , iu kin

218

96,2378,847

78,84724

450.8900.10.36,3B

96,2315408900.10.3,1875,5.24

450.8900.10.36,375,5A

mm/daN75,58,14.8450.10.36,3

f8.g

2

223

62

223

2232

oo

Gii ra ta c =5,376 daN/mm2.

Tnh vng: m82,15376,5.8

450.10.36,38.gf

232

y l vng ln nht ca dy dn. b) Trng hp b) tnh tng t trng hp a) ch c khc tr s v nhit c) Trng thi bo =2 5oC, v =40 m/s

Trc tin tnh gc khi c xt theo tc gi (xem mc 5.1.3):

2323232221c

232

2

22

v2

mm.m/daN10.697,510.6,410.36,3ggg

mm/daN10.6,41000.5,453.16

40.7,27.1,1.7,0g

mm.m/kg;1000.F.16

v.d.k.aFP

g

A = -21,52 B = 2437,23

52,2123,24372c

c

Gii phng trnh trn ta c c = 8,944 daN/mm2

m123,16944,8.8

450.10.697,5.8.g

f23

c

2c

Trn l gi tr vng theo ng cho, cn tnh vng theo thnh phn dc fd v ngang fn th s l:

m027,13808,0.123,16sin.ffm496,9589,0.123,16cos.ff

808,0sin589,010.697,5

10.36,3gg

cos

n

d

3

3

2

1

5.4 KHONG CT TI HN CA DY DN

5.4.1 Khi nim chung

219

ng sut ca dy dn thay i ty theo iu kin kh hu v iu kin ph ti c gii tc ng ln n. Vy trong mt khong ct, khi no th dy dn c ng sut ln nht hay ni cch khc lc no th dy dn d t hn c. dy dn c th lm vic c th ng sut trong dy dn trong mi trng thi phi nh hn ng sut cho php cp.

Vic xc nh ng sut cho mi trng thi ca dy dn phi tin hnh trn c s phng trnh trng thi nu mc 5.3, trong cn chn mt trng thi xut pht ban u, m thng thng chn trng thi ng sut c th ln nht. Trng thi c th a li ng sut ln nht l:

(i) Nhit mi trng thp nht, v lc dy dn c di ngn nht trong mt khong ct nht nh;

(ii) Ph ti tc dng ln dy dn ln nht (tng ng vi lc gi to nht: bo);

Hai trng thi trn c th xy ra ng sut ln nht. Ngoi ra cn trng thi th ba: trng thi nhit trung bnh, y c ng sut khng ln nh hai trng thi trn nhng trong trng thi ny ng sut cho php thp hn nn cng c nguy c vt ng sut cho php ca n.

Ngi ta quy nh ng sut cho php cc ch theo cng thc sau [1a] :

at

bencp K

(5.24)

trong , cp - ng sut cho php cc ch ; [daN/mm2]

ben - gii hn bn ko ca dy; [daN/mm2]

Kat- h s an ton; Kat = 2,5 ch cc i; Kat = 4,0 ch nhit trung bnh

i vi trng thi xut pht ny ta coi = cp th khi ng xut tnh c ca tt c cc trng thi khc s thon mn iu kin nh hn ng sut cho php.

S din bin ca ng sut trong mt trng thi nht nh ph thuc vo khong ct. Do chnh khong ct l im xut pht xc nh xem ng sut vt khung s xy ra trng thi no?

Khi thit k ng dy ta chn c khong ct (m). Ta cn bit vi khong ct ny phi chn trng thi no lm trng thi xut pht? Mun vy trc ht ta phi xc nh c khong ct ti hn th

Khong ct ti hn th xc nh cho tng cp trng thi trong ba trng thi ng sut, v d cp trng thi (i) v (ii), l khong ct m mun ly trng thi no lm trng thi xut pht cng c; Cn th th phi ly trng thi (i) hoc (ii) v nu th th phi ly trng thi (ii) lm trng thi xut pht.

Ta c ba khong ct ti hn:

th1 - khong ct ti hn gia trng thi nhit trung bnh v trng thi nhit thp nht;

220

th2 - khong ct ti hn gia trng thi nhit thp nht v trng thi bo;

th3 - khong ct ti hn gia trng thi nhit trung bnh v trng thi bo;

5.4.2 Khong ct ti hn th2

Khong ct ti hn th2 l khong ct gia hai trng thi nhit mi trng thp nht v trng thi bo, y l hai trng thi c th xy ra ng sut ln nht. m bo ng sut cho php trong hai trng thi ny ta c th thc hin mt trong hai cch sau y:

1- Ly trng thi bo lm trng thi xut pht, ly ng sut trng thi ny bng cp, ngha l ng sut trong trng thi bo c m bo. Ta tnh xem trong trng thi nhit nh nht ng sut din bin th no khi khong ct thay i 1min f .

2- Ly trng thi nhit thp nht lm trng thi xut pht, ly ng sut trng thi ny bng cp, ngha l ng sut trong trng thi nhit thp nht c m bo. Ta tnh xem trong trng thi bo ng sut din bin th no khi khong ct thay i 2bao f .

Cch tnh l gii phng trnh trng thi sau khi cho bin i t 0 n , c th l:

- Cho 1min f :

baomin2CP

2bao

2

CP2min

2min

2

min 24g.

24g.

(5.25)

- Cho 2bao f

minbao2CP

2min

2

CP2bao

2bao

2

bao 24g.

24g.

(5.26)

trong cc thng s minmin g, l ng sut v t ti ca trng thi nhit thp nht,

cn bbao g, l ng sut v t ti ca trng thi bo, cp l ng sut cho php.

Sau khi tnh ta lp c th nh trn hnh 5.6

221

T hnh 5.6 thy rng hai ng cong 1min f , 2bao f ct nhau im cp.

Vi ng cong 1min f , ta thy khi th2 th min c gi tr thp hn gi tr cho php, cn khi th2 th min ln hn gi tr cho php. Vi ng cong 2bao f , ta thy khi th2 th bao c gi tr ln hn gi tr cho php, cn khi th2 th bao thp hn gi tr cho php.

Nh vy nu khong ct cho trc, khi thit k th2 m bo ng sut trong mi trng thi khng vt gi tr cho php ta phi chn trng thi nhit min lm trng thi xut pht tnh ng sut v t tnh vng thi cng v kh ng sut trong trng thi bo s nh hn ng sut cho php. Cn khi th2 th ta chn trng thi bo lm trng thi xut

pht v khi ng sut trong trng thi nhit min s c bo m nh hn gii hn.

Khong ct ti hn th2 l gii hn la chn trng thi xut pht. Khi th2 , c th chn ty mt trong hai trng thi lm trng thi xut pht v chng c ng sut bng nhau.

tnh khong ct ti hn th2 ta c th dng mt trong hai cng thc (5.25) hoc (5.26),

trong min v bao ly bng cp, kt qu c:

2min

2bao

minbaocpth2 gg

24

(5.27a)

Theo s liu v iu kin tnh ton c hc cho ng dy trn khng (bng 5.2), ta c 21

2min

22

21

2bao gg;ggg , trong g1-t ti do trng lng dy v g2 t ti do gi, thay vo

cng thc (5.27) ta c:

minbao2

cpth2 24g

(5.27b)

Khong ct ti hn th2 l khong ct ng sut trong hai trng thi nhit thp nht v bo bng nhau v bng ng sut cho php cp

Sau khi tnh c khong ct gii hn th2 ta ly khong cch thc t so vi khong ct ti hn.

- Nu th2 , ng sut ln nht xy ra trong trng bo. Vy ta phi ly trng thi bo lm trng thi xut pht, trong ng sut bng ng sut cho php tnh ton.

- Nu th2 th ng sut ln nht xy ra trong trng thi nhit thp nht v ta phi chn trng thi ny lm trng thi xut pht tnh. ng sut xy ra trong trng thi ny l ng sut cho php.

- Khi th2 th xut pht t trng thi no cng c.

5.4.2 Khong ct ti hn th1 v th3

222

Nh trnh by mc 5.4.2, khi th2 ta ly trng thi nhit nh nht lm trng

thi xut pht, cn khi th2 ta ly trng thi bo lm trng thi xut pht. Nh vy ng sut trong dy lun nh hn ng sut cho php (hnh 5.7)

T hnh 5.7 thy rng khi th2 th ng sut lc bo bng ng sut cho php, cn ng sut

khi nhit min th nh hn; khi th2 th ng sut lc nhit min bng ng sut cho php, cn ng sut khi bo nh hn.

Nu khng hn ch v ng sut khi trng thi nhit trung bnh th ch cn tnh dy dn theo th2 l c v ng sut khi trng thi trung bnh nh hn cc trng thi bo v nhit min.

Tuy nhin do ng sut cho php lc nhit trung bnh, t l cptb nh hn trng thi bo

v trng thi nhit min (xem cng thc 5.24).

xt n kh nng ny ta tnh ng sut khi trng thi nhit trung bnh theo khong ct ).(f: tb i vi mi khong ct ti hn ta s ln lt xy dng hai quan h:

31tb f v 42tb f , c th nh trnh by tip sau y.

1) th1 - khong ct ti hn gia trng thi nhit trung bnh v trng thi nhit thp nht:

- Cho 31tb f , trng thi xut pht l trng thi nhit min, ta c:

mintb2CP

2min

2

CP2tb

2tb

2

tb 24g.

24g.

(5.28a)

- Cho 42tb f , trng thi sut pht l trng thi nhit trung bnh, ta c:

tbmin2CPtb

2tb

2

CPtb2min

2min

2

min 24g.

24g.

(5.28b)

223

trong cc thng s minmin g, l ng sut v t ti ca trng thi nhit thp nht,

cn tbtb g, l ng sut v t ti ca trng thi nhit trung bnh, cp l ng sut cho php

khi nhit min, cptb l ng sut cho php khi nhit trung bnh.

Hai ng cong (5.28a) v (5.28b) ct nhau ti mt im, m chiu chng trn trc honh ta c khong ct ti hn th1 . Biu thc xc nh th1 bng cch thay gi tr im ct

vo mt trong hai phng trnh trn, v d phng trnh (5.28a) vi th1cptbtb , ; ta

c:

mintb2CP

2min

2th1

CP2CPtb

2tb

2th1

CPtb 24g.

24g.

Gii phng trnh trn ta c:

2

CPtb

tb

2

CP

min

CPtbCPtbmin

2

CPtb

tb

2

CP

min

CPtbCPtbmin

th1

gg24E

E

gg241

(5.29)

1) th3 - khong ct ti hn gia trng thi nhit trung bnh v trng thi bo:

- Cho 31tb f , trng thi xut pht l trng thi bo, ta c:

baotb2CP

2bao

2

CP2tb

2tb

2

tb 24g.

24g.

(5.30a)

- Cho 42tb f , trng thi sut pht l trng thi nhit trung bnh, ta c:

tbbao2CPtb

2tb

2

CPtb2bao

2bao

2

bao 24g.

24g.

(5.30b)

trong cc thng s tbtb g, l ng sut v t ti ca trng thi nhit trung bnh, cn

baobao g, l ng sut v t ti ca trng thi bo, cp l ng sut cho php khi bo, cptb l ng sut cho php khi nhit trung bnh.

Hai ng cong (5.30a) v (5.30b) ct nhau ti mt im, m chiu chng trn trc honh ta c khong ct ti hn th3 . Biu thc xc nh th3 bng cch thay gi tr im ct

vo mt trong hai phng trnh trn, v d phng trnh (5.30a) vi th1cptbtb , ; ta

c:

224

baotb2CP

2bao

2th3

CP2CPtb

2tb

2th3

CPtb 24g.

24g.

Gii phng trnh trn ta c:

2

CPtb

tb2

CP

bao

CPtbCPtbbao

2

CPtb

tb

2

CP

bao

CPtbCPtbbao

th1

gg24E

E

gg24

1

(5.31)

Khong ct ti hn th1 c xc nh t phng trnh (5.28a) vi trng thi xut pht l trng thi nhit min v ng cong ca phng trnh ny gi l ng cong (c). Khong ct ti hn th3 c xc nh t phng trnh (5.30a) vi trng thi xut pht l trng thi bo v ng cong ca phng trnh ny gi l ng cong (d). Ta v ng cong (c) v (d) ln cng th th2 v c th hin trn hnh 5.8a Trn trc tung t ng sut cho php

CPtb i vi trng thi nhit trung bnh v k ng nm ngang, ng ny ct ng (c) ti C v ng (d) ti D. Chiu hai im C v D ln trc honh s c cc gi tr khong ct

th3th1 , nh trn hnh 5.8a.

Ti im C, nu ly trng thi nhit min lm trng thi xut pht th ng sut trong trng thi nhit min v trng thi nhit trung bnh bng ng sut cho php ca hai trng thi ny. Nu th1 th ng sut trong trng thi nhit trung bnh s nh hn cho php.

Nh vy khi th1 ta ly trng thi nhit min lm trng thi xut pht.

Ti im D, nu ly trng bo lm trng thi xut pht th ng sut trong trng thi bo v trng thi nhit trung bnh bng ng sut cho php ca hai trng thi ny. Nu th3 th

ng sut trong trng thi nhit trung bnh s nh hn cho php. Nh vy khi th3 ta ly trng thi bo lm trng thi xut pht.

Nu th3 nhng ln hn th2 th ng sut trong trng thi nhit trung bnh s ln hn ng sut cho php, do dod phi ly trng thi nhit trung bnh lm trng thi xut pht. Nh vy khi th3th1 th ly trng thi nhit trung bnh lm trng thi xut pht.

Tm li, trong trng hp th3th2th1 nh trn hnh 5.8 ta phi p dng trng thi

xut pht v dng cng thc tnh ng sut x ca trng thi x c gx v x nh sau:

(i) Khi th1 , trng thi xut pht l trng thi nhit min v cng thc tnh :

225

minx2CP

2min

2

CP2x

2x

2

x 24g.

24g.

(5.32)

(ii) Khi th3th1 , trng thi xut pht l trng thi nhit trung bnh v cng thc tnh:

tbx2CPtb

2tb

2

CPtb2x

2x

2

x 24g.

24g.

(5.33)

(iii) Khi th3 , trng thi xut pht l trng thi bo v cng thc tnh :

baox2CP

2bao

2

CP2x

2x

2

x 24g.

24g.

(5.34)

Ngoi trng hp va xt trn, c th xy ra cc trng hp khc nh di y .

Trng hp th1th3 (hnh 5.8b):

Trong trng hp ny, khi th2 th ng sut trang trng thi nhit trung bnh lun

thp hn CPtb , cn khi th2 th ng sut trong trng thi nhit trung bnh cng lun

thp hn CPtb . Do trong trng hp ny khng phi tnh n th1 v th3 . Dy dn ch

cn tnh theo th2 , tc l khi th2 th ly trng thi nhit min lm trng thi xut pht,

cn khi th2 th ly trng thi bo lm trng thi xut pht.

226

Trng hp khng c th1 , ch c th2 v th3 (hnh 5.8c)

Trong trng hp ny khi th3 phi ly trng thi nhit trung bnh lm trng thi

xut pht, khi th3 phi ly trng thi bo lm trng thi xut pht, cn th2 khng c vai tr g.

Trng hp khng c th3 , ch c th1 v th2 (hnh 5.8d)

Trong trng hp ny th2 cng khng dng n, khi th1 th phi ly trng thi nhit

min lm trng thi xut pht, cn khi th1 th ly trng thi nhit trung bnh lm trng thi xut pht.

5.5 TNH TON DY NHM LI THP AC

5.5.1 Khi nim chung

Cc cng thc trn dng cho dy ng nht mt kim loi, cn i vi dy lng kim m thng dng nht l dy nhm li thp AC th cn phi c cc phn tch hiu chnh.

i vi dy AC, khi nhit thay i s gin n chung phi l mt, nhng bn trong cc si nhm v thp xut hin lc tng tc gia chng vi nhau. Thc th v d nh nhit tng th phn nhm gin ra nhiu hn, phn thp t hn, nhng do chng c bn cht vo nhau do ch to nn phn thp b phn nhm ko di thm, cn phn nhm b phn thp co ngn li, kt qu l chng c cng mc co gin. Nh vy c hai phn u chu mt ng sut ph do nhit gy ra, ng sut ny cng vi ng sut do ti trng c gy ra to thnh ng sut thc trong mi phn ca dy.

Bi ton l i vi dy AC l phi xc nh c cc thng chung cho chng : mun n hi E0, ng sut gi tng gt , h s gin n nhit ng tr o , h s ko di n hi ng

tr o . Sau y s phn tch vic xc nh cc thng s chung ny.

5.5.2 Mun n hi E0 v ng sut gi thit gt

Khi dy AC chu lc ko T, th lc ny c phn cho c hai phn nhm TAL v thp TFe : T = TAL+ TFe . Khi ng sut gi thit s l :

FeAL

gt FFT

FT

(5.35)

trong FAL,FFe l thit din phn nhm v phn thp.

Khi thay i trng thi, dy b thay i di v khi ta c:

Fe

Fe

AL

AL

o

gt

EEE

(5.36)

Ta rt ra:

227

Fe

oFe

AL

oALgt

o

FegtFe

o

ALgtAL

EE

.EE

.hay

E.

E.

(5.37)

Theo (5.35) ta c : T = TAL+ TFe, m T=.F nn FeFeALALgt F.F.F

Chia c hai v biu thc trn cho c:

FeFeALALFeALgt F.F.FF

,

m E

, nn FeFeALALFeALo F.EF.EFFE , rt ra :

1FF

F

F.E

1FF

F

F.EFF

F.EF.EE

Fe

ALFe

FeFe

Fe

ALFe

ALAL

FeAL

FeFeALALo

t Fe

AL

FE

a , ta c : a1

EE.aE FeALo

(5.38)

Cng thc (5.38) cho php tnh mun n hi ca ton b dy AC, v nu thay chng vo (5.37) ta c:

ALFe

FEgt

o

FegtFe

ALFe

ALgt

o

ALgtAL

aEE)a1(E

EE

.

aEE)a1(E

EE

(5.39)

Cng thc (5.39) cho php tnh ng sut ca thp v nhm theo ng sut gi tng v ngc li.

5.5.3 H s ko di n hi ng tr o v h s gin n nhit ng tr o

H s ko di n hi ng tr o c xc nh theo biu thc:

FeALo

o EaEa1

E1

(5.40)

H s gin n nhit ng tr o cn c xy dng cng thc tnh.

Gi thit cc phn nhm v thp c t do dn n, khng nh hng ln nhau. Lc u, nhit bng nhit lc ch to, hai b phn nhm v thp c di bng nhau (vch AB trn hnh 5.9). Khi nhit tng ln th phn nhm dn n nhiu hn phn thp, di ca phn nhm s chim v tr mi l EF, cn thp th dn n t hn v chim v tr mi l IK. Trn thc t phn nhm v phn thp c gn cht vi nhau, cho nn nhm v thp cng dn n v chim ti v tr CD (nhm b nn li t EF ti CD, cn thp b ko di thm t IK n CD).

228

Theo cc mi lin h c bn cc i lng ,E,,T, l .,E.,FT

ta c F.E..F.E.F.T ; p dng biu thc ny cho tng phn cho dy AC ta c:

ALALooALAL

FeFeooFeFe

F.E.TF.E.T

trong o l nhit ban u, cn l nhit ti trng thi xem xt.

Qui c lc ko c du cng , cn lc nn c du tr, th y lc TAL c du cng, cn lc TFe c du tr. Theo lut cn bng lc ta c th vit:

Fe

ALALo

Fe

ALALALFeoFeFe

ALALoALFeFeoFe

ALALooALFeFeooFeALFe

FF

.EFF

.EE..E

0F.EF.E.0F.E.F.E.0TT

t Fe

AL

FF

a v thay vo phng trnh trn, gii ta c:

ALFe

ALALFeFeo E.aE

E..aE.

(5.41)

5.5.4 Phng trnh trng thi i v tnh ton c hc vi dy AC Cng ging nh dy n cht, i vi dy AC nhng iu kin kh hu khc nhau th ng sut v vng cng thay i khc nhau. xc nh c cc i lng ny cho mt trng thi iu kin kh hu ta vn c th s dng phng trnh trng thi c bn (5.22a) nhng phi thay cc thng s ooogt ,,E, , c th l:

mno

o2gtm

2m

2

gtm2gtn

2n

2

gtn 24g.

24g.

(5.42)

T phng trnh trng thi (5.42) ty theo trng thi kh hu xc nh c ng sut gt

theo phng php lun trnh by trc y, v t xc nh c vng theo cng

thc quen thuc gt

2

8g.f

.

229

5.6 CHNG RUNG CHO DY

5.6.1 Hin tng rung dy

Khi gi thi l dy dn trong khong ct c th to nn mt dng xoy gn dy, dn n tc gi pha trn dy ln hn pha di lamg cho p lc gi tc ng ln dy tr thnh khng ng u.

S xut hin thnh phn dc ca p lc gi lm cho dy chuyn ng ln, sao c xung lc tc ng lin tc theo chiu ln xung. Nu tn s xung lc gn bng tn s dao ng ring ca dy th dy s dao ng mnh v duy tr. l hin tng rung ca dy dn trong khong ct.

Dy dn thng b rung duy tr tc gi iu ha, cc xung lc tc ng mt cch chu k lin tc. S rung ca dy dn ph thuc vo tc gi, thng s rung hay xy ra tc gi v = 5 8 m/s v thng c nhng khong trng (cnh ng chng hn) hoc nhng khong vt sng ln. Nu gi qu nh, xung lc s khng gy dao ng.

Tn s rung ca dy t l vi tc gi v ph thuc vo ng dy c th c xc nh theo biu thc sau:

]Hz[;dv.200f d (5.43)

trong : fd - tn s ring ca dy dn; [Hz] v - vn tc gi; [m/s] d - ng knh dy dn; mm

S rung ca dy ch xy ra khi hng gi to vi trc dy dn mt gc chng = 450 900. Khi gc ny trong khong 300 450, s rung khng duy tr. Khi < 300 th khng c hin tng rung.

5.6.2 nh hng ca rung dy

Khi b rung dy dn rt chng b h hng, c bit l cc ch kp ni dy. Khi b rung s xut hin cc ng lc tc ng chu k. Ngoi ng lc tnh cn c nhng ng lc ng do un gy ra v s ko di ca dy dn khi b rung. Bn cnh , nhng ch kp dy, cc si cn phi chu thm nhng ng lc ring gy ra bi nhng thit b kp dy. Nhng ng lc tahy i chu k lm vt liu b mi, khi vt qu gii hn mi ( gii hn bi 20 25 % ng lc ph hoi vt liu) dy s b t nhng si nh, khi t mt si cc si cn li s b t rt nhanh. Phn dy ch kp b h hi mau hn phn xa kp.

Mun ng dy n nh khng rung m khng c bin php chng rung no th ng sut trung bnh hng nm ca vt liu khng c vt qu 4 5 kg/mm2 vi dy AC v 8 10 kg/mm2 i vi dy ng.

Khi b rung, dy dn dng kp c nh b h hng nhanh hn khi dng kp ng (dy x dch c) v ch kp c nh lc tp trung gy un i vi dy dn v xut hin ng lc ph ch kp. Nu dng kp ng, tuy s h hng ca dy gim i nhiu so vi kp c nh nhng khi rung cc si nh vn b t do nhng kp nng khi lng phn ng ca kp

230

ln hn nhiu so vi nng lng dao ng. Ngay c khi dng kp quay, ta vn khng th loi tr hon ton s h hng dy dn khi rung do chu k v gc pha dao ng hai pha dao ng hai khong vt k nhau khng ph hp vi nhau nn ch kp xut hin un gy nn ng lc ng ph.

Tm li d kp loi g, vi bt c loi dy dn no u phi c bin php gim ti a s rung ca dy dn.

5.6.3 Cc bin php chng rung

Bin php chng rung c hiu qu nht v thng c p dng rng ri l dng t chng rung (hnh 5.10). T chng rung gm hai i trng (qu t) c ni vi nhau bng cn thp v c treo dy nh kp chuyn dng. Khi c rung, thit b ny s trit tiu nng lng dao ng bi ma st gia cc si ca cp treo hai i trng.

Theo quy phm trang b in th cn phi t t chng rung trong trng hp [1]: ng dy i qua khu vc trng, bng phng, khong ct trn 110m, ng lc vn hnh lc bnh thng hng nm vt qu cc tr s sau: 11,8 daN/mm2 i vi dy ng; 3,9 daN/mm2 i vi dy nhm; 5,9 daN/mm2 i vi dy AC-35, 50, 70, 95; 20 daN/mm2 i vi dy thp, 4,9 da N/mm2 i vi dy AC-120 tr ln [1]; Khi khong ct trn 500 m th bt buc phi dng t chng rung.

Khong cch t t chng rung tnh t gia v tr t t (v tr kp) v im treo dy dn c xc nh theo cng thc sau [1]:

1

tbta gF.75,1 (5.44)

hay 1

tbta gF.d0013,0 (5.44a)

trong : ta - khong cch t t; [mm]

tb - ng sut trong trng thi nhit trung bnh trong nm; [daN/mm2]

d - ng knh dy dn; [mm] F - tit din dy dn; [mm2] g1 t ti do trng lng bn thn ca dy dn; [daN/m.mm2]

231

Ni chung khong cch t t nm trong khong 0,51,2 m tnh t kp t n im treo dy dn. Loi t chng rung ph thuc vo tit din dy dn v dy chng st (xem bng 5.3) Bng 5.3 Khong cch t t v khi lng t

Loi dy

Kch thc t chng rung

di;

mm

Trng lng;kg

Qu t Ton b

Dy thp 50 mm2 300 0,8 2,30

A-70-95; AC-70-95 350 0,8 2,35

Dy thp 70 mm2 400 1,6 4,05

A-120-185; AC-120-150; ACO-150; ACY-120-150

450 1,6 4,13

A-240; AC-185-240; ACO-185-240; ACY-185-240

500 2,4 5,60

A-300; ACO-300 550 3,2 8,10

A-400-500; AC-300-400; ACO-400-500; ACY-300-400

550 3,2 8,20

A-600; AC0-600 600 3,2 8,30

Trn y gii thiu cc kin thc c bn v cch tnh cc thng s ng sut, vng, khong ct ti hn v t chng rung ch cha th hin tnh ton chi tit cc vn v thit k c kh ng dy. Cc vn v thit k c kh ng dy c th tham kho cc ti liu [1],[1a].

232

TI LIU THAM KHO

[1]Trn Bch, Li in &H thng in- Tp 1,2&3, NXB Khoa hc&K thut- 2004

[1a] Hong Hu Thn, Hng dn thit k ng dy ti in, NXB Khoa hc&K thut- 2005

[2] Narain G.Hingorani & Laszio Gyugyi, Understanding FACTS, Concepts and Technology of Flexible AC Transmission Systems, IEEE Press, 1999.

[3] C.A Canizares, Power flow and transient stability models of FACTS controllers for voltage and angle stability studies, IEEE/PES WM on Modeling, Simulation and Application of FACTS Controllers in Angle and Voltage Stability Studies, Singapore, January 2000. [4] N. Martins, H.Pinto, J. Paserba, Using a TCSC for Power System Scheduling and System Oscillation Damping-Small Signal and Transient Stability Studies. In Proc. IEEE/PES Winter Meeting, Singapore, January. [5] R.Rajaraman, F.Alvarado, A.Maniaci, R. Camfield, S. Jalali, Determination of location and amount of series compensation to increase power transfer capability, IEEE Trans, on Power Systems, Vol.13, No 2, May 1998, pp.294-299

[6] R.J Piwko, C.A. Wegner, B.L. Damsky, B.C Furumasu, J.D. Eden, The Slatt Thyristor Controlled Series Capacitor Project-Dsign, Installation, Commissoning, and System Testing, CIGRE paper 14-104. pans. 1991

[7]S.Nyati, C.A Wegner, R.W Delmerico, D.H. Baker, R.J. Piwko, A. Edns, Effectiveness of Thyristor Controlled Series Capacitor in Enhancing Power System Dynamics: An Analog Simulator Study. IEEE Transactions on Power Delivery, April, 1994, pp 1018-1027.

MC LC

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Li ni u ............................................................................................ 4

Danh mc cc ch vit tt .................................................................... 5

Chng 1. Phn tch tnh ton thit k li in ................................. 7 1.1 Cc ni dung chnh ca thit k li in ...................................... 7 1.2 Tnh ton cn bng cng sut trong h thng in ......................... 7 1.3 Chn thit din dy dn v dy cp in ...................................... 14 1.4 Tnh ton kinh t-k thut chn phng n ti u ....................... 20

Chng 2.Tnh ton ch xc lp h thng in phc tp .............. 22

233

2.1 Tng qut chung v tnh ton ch xc lp h thng in .......... 22 2.2 H phng trnh m t ch xc lp h thng in bng ma trn tng dn Y, tng tr Z ................................................................. 26 2.3 H phng trnh m t h thng in bng cng sut nt ............. 32 2.4 Mt s m t h thng in bng ma trn graph ........................... 37 2.5 H phng trnh m t ch xc lp h thng in khi c nhiu cp in p .................................................................................. 40 2.6 Phng php nghch o ma trn tng dn Y ............................... 43 2.7 Phng php tnh trc tip ma trn tng tr Z.............................. 48 2.8 Phng php kh Gauss .............................................................. 58 2.9 Phng php lp Gauss-Seidel..................................................... 61 2.10 Phng php lp Newton-Raphson .............................................. 76 2.11 Tnh ton cc thng s ch ...................................................... 92

Chng 3.Phn tch ch lm vic ca ng dy siu cao p ..... 102 3.1 Tng qut chung v ng dy di siu cao p v h thng ti in ........................................................................................ 102 3.2 Cc phng trnh m t ch lm vic ca ng dy di thun nht ................................................................................... 105 3.3 Phn tch ch lm vic ca ng dy di thun nht .............. 121 3.4 Tnh ton ch ng dy di theo mng bn cc..................... 131 3.5 B trn ng dy di ................................................................. 150

Chng 4. H thng ti in xoay chiu linh hot FACTS .............. 163 4.1 Thyristor v khng in iu chnh thyristor ................................ 163 4.2 My b tnh iu khin bng thyristor ......................................... 169 4.3 My b ng b tnh STATCOM ................................................ 173 4.4 T b dc iu khin bng thyristor TCSC .................................. 176 4.5 Thit b b dc ng b tnh SSSC .............................................. 186 4.6 Thit b iu chnh gc pha iu khin bng thyristor TCPAR ..... 187 4.7 Cc thit b iu khin khc ......................................................... 198 4.8 Kt lun chung v FACTS ........................................................... 201

Chng 5 Tnh ton c hc ng dy ti in trn khng ............ 205 5.1 Khi qut chung v ng dy trn khng ................................... 205 5.2 Phng trnh v cc thng s c bn ca ng dy trn khng . 210 5.3 Phng trnh trng thi ca ng dy dn .................................. 219 5.4 Khong ct ti hn ca dy dn ................................................... 223 5.5 Tnh ton dy nhm li thp AC .................................................. 234 5.6 Chng rung cho dy..................................................................... 238 Ti liu tham kho ............................................................................ 242 Mc lc ............................................................................................. 243