B GIO DC V O TOI HC NNGZ\]^[Y

NGUYN HU C

NGHIN CU PHNG PHP

XC NH PHN ONLI IN PHN PHI HNH TIA

LUN VN THC S K THUT

NNG - Nm 2004

B GIO DC V O TOI HC NNG

NGUYN HU C

NGHIN CU PHNG PHP

XC NH PHN ON

LI IN PHN PHI HNH TIA

Chuyn ngnh: MNG V H THNG IN

M s: 60.52.50

LUN VN THC S K THUT

Ngi hng dn khoa hc: TS. INH THNH VIT

NNG - Nm 2004

LI CAM OAN

Ti xin cam oan y l cng trnh nghin cu ca ring ti.

Cc s liu, kt qu nu trong lun vn l trung thc v cha tng c

ai cng b trong bt k cng trnh no khc.

Tc gi

Nguyn Hu c

MC LC

M U Trang 1Chng 1: TNG QUAN V VN PHN ON

LI PHN PHI 41.1. Tm quan trng ca vic xc nh phn on.

trong li phn phi. 4

1.2. Mt s phng php xc nh phn on.

li phn phi c nghin cu. 5

Chng 2: XY DNG M HNH TON HC XC NH PHN ON LI PHN PHI 9

2.1. Li phn phi-Phn on li phn phi. 9

2.2. Sut s c li phn phi. 16

2.3. Thit hi mt in khch hng. 18

2.4. Gi in. 20

2.5. Hm mc tiu tnh ton. 23

Chng 3: CHNG TRNH TNH TON PHN ON LI PHN PHI 28

3.1. M hnh li phn phi s dng tnh ton. 28

3.2. Nghin cu k thut gii quyt-Cc thut ton. 41

3.3. Chn cng c lp trnh. 68

3.4. Chng trnh tnh ton. 69

Chng 4: NG DNG XC NH PHN ONCHO LI PHN PHI TP. QUY NHN 81

4.1. Gii thiu li phn phi TP Quy nhn. 81

4.2. Tnh ton - Kt qu. 82

KT LUN KIN NGH 84 TI LIU THAM KHO. 85PH LC 86

K HIU V VIT TT

Trong ti, mt s cc t vit tt c hiu nh sau:

EVN : Tng Cng ty in lc Vit nam

TBP: Thit b phn on.

DCL: Dao cch ly.

RE.: Recloser.

MBA: My bin p.

MC: My ct

FCO: Cu ch t ri

- 86 -

PH LC

1. CHNG TRNH CHI TIT

2. CC S LIU TNH TON

3. CC BN V

Ph lc 1

CHNG TRNH CHI TIT

Ph lc 2

CC S LIU TNH TON

Ph lc 3

CC BN V

- 1 -

M U

I TNH CP THIT CA TI:

S chn la c hiu qu thit b phn on (TBP) li in phn phi

(LPP) l cng vic quan trng trong quy hoch, thit k v vn hnh.

Trong tt c cc tiu ch cn nhc v bi ton phn on, tiu ch tnh

ton tin cy, c bit l nh gi thit hi mt in ph ti ang tr nn rt

cn thit trong cc bc tnh nh gi hiu qu u t cng nh cht lng

cung cp in. y cn l vn lin quan n cht lng ci thin dch v

cung cp in cho khch hng - mt trong nhng mc tiu ln m Tng cng

ty in lc Vit nam (EVN) v ang trin khai, u t kh mnh theo

kp vi mc ci thin dch v cung cp in ca khu vc v trn th gii.

Mt khc, ti cc in lc hin rt thiu cc cng c tnh ton (cc chng

trnh tnh ton k thut) da trn cc c s l thuyt tnh ton tin cy

gii quyt bi ton nu trn mt cch thit thc.

II I TNG V PHM VI NGHIN CU:

i tng ca ti l phn on li phn phi, phm vi nghin cu bao

gm s lp t cc phn on, trong l kt qu ti u v s lng, v tr

lp t cc phn on ca chng nhm cp in hiu qu, gim thiu thit hi

mt in.

III - MC TIU V NHIM V NGHIN CU:

Mc tiu xy dng mt cng c tnh ton nhm xc nh mt cch c hiu

qu cc phn on, ti s tin hnh cc nhim v sau:

1/ Nghin cu gii php k thut p dng gii quyt bi ton xc nh

phn on .

2/ Xy dng thut ton.

- 2 -

3/ Thit lp mt chng trnh tnh ton.

IV - T TN TI:

"Nghin cu phng php xc nh phn on li in phn phi

hnh tia"

V TIN TRNH NGHIN CU:

ti nghin cu mt s phng php ph bin, tm hiu cc k thut

s dng, t xy dng mt phng php ph hp nht, n gin ng thi

d p dng cho iu kin thc t, xy dng thnh chng trnh tnh ton nhm

to cng c h tr thm cho cc vn t ra trong vn hnh li phn phi

ti in lc.

Tuy nhin do kh nng v trnh lp trnh khng chuyn, v vy chng

trnh cn nhiu hn ch. D kin chng trnh s c ci tin hn v thut

ton, cng nh thm cc phn tnh ton khc cho nhng li in phc tp,

m rng vic p dng hn gii quyt cc vn thc t trong qun l vn

hnh li in c ni dung lin quan n phn on.

Vi mong mun c c mt cng c nh nhng thun tin cho cng tc

qun l k thut li in ti cc in lc, ng thi gp phn nng cao kin

thc trong hc tp, ti chn thnh tip thu v c bit cm n nhng kin

ng gp ca cc thy c, cc ng nghip nhm to c thm mt chng

trnh hu ch. Ti chn thnh cm n s gip qu bu, y trch nhim v

tn tnh ca Gio vin hng dn: Tin s inh Thnh Vit - B mn h

thng in, Khoa in-Trng i hc Bch khoa Nng; S quan tm

gip nhit tnh ca Ph gio s, Tin s Trn Bch; Cc thy c thuc B

mn H Thng in, Trng i hc Bch Khoa H Ni; Cc thy, c trong

khoa in Trng i hc Bch khoa Nng; Cc anh, ch ng nghip

trong Cng ty in lc 3 v ti cc in lc, c bit l Ban Lnh o v cc

- 3 -

ng nghip ti in lc Bnh nh - ni ti lm vic, h tr ti rt nhiu

v iu kin thc hin ti cng nh qu trnh tm hiu, nghin cu v hon

tt ti.

VI - B CC LUN VN:

Chng 1: TNG QUAN V VN PHN ON LPP

Chng 2: XY DNG M HNH TON HC XC NH

PHN ON LPP

Chng 3: CHNG TRNH TNH TON PHN ON LPP

Chng 4: NG DNG CHNG TRNH XC NH PHN

ON CHO LPP TP. QUY NHN

- 4 -

Chng 1:

TNG QUAN V VN PHN ON

LI IN PHN PHI

1.1. TM QUAN TRNG CA VIC XC NH PHN ON

TRONG LI IN PHN PHI:

Mt vn quan trng trong vic lp quy hoch h thng nng lng l

nhn bit nhng thit b c s dng trong h thng tin cy theo nhu cu

ca dch v khch hng v cho chi ph thp nht. S lng v v tr ca cc

TBP l vn quan trng trong quy hoch v thit k li in phn phi.

Gii php phn on li phn phi lm tng ng k tin cy, gim

c tn tht kinh t do mt in, nhng cn phi u t vn. Do phn

on l mt bi ton cn phi c cn nhc, trong cn tm s lng, v tr

t v loi TBP s dng sao cho c c hiu qu kinh t cao nht, chi ph

thp nht.

Trong cc nh gi v phn tch u t vo li in, thit hi mt in

c nh gi u tin trc khi mt thit b mi c lp t, qu trnh nh

gi s c lp li cho cc phng n lp t thit b thm vo vi s lng

v v tr khc nhau. Cc phng n c quyt nh chn la sau khi so snh

tng cc chi ph (thit hi) ca chng. Cng vic ny vn phi tip tc c

thc hin mi khi c s thay i tng t pht sinh trong li in (v y l

cng vic thng xuyn phi thc hin i vi cng ty in lc, vi cc k

hoch u t v xy dng c bn v sa cha, bo dng nh k li in

hng nm).

Chng loi, s lng, v tr cc TBP khc nhau c s dng c th

c xc nh bng kinh nghim ca k s qun l, vn hnh hay c th da

- 5 -

trn kt qu ca phn tch tin cy, c thc hin bng cc tnh ton s b,

khng h thng, nhng hiu qu hn c l s dng chng trnh tnh tin

cy ph hp. Bng cch xem xt, nghin cu cc c im hin trng, cc

hng quy hoch pht trin v kt cu ca li in phn phi thc t ti

cng ty in lc, trn c s nhiu k thut x l tnh ton khc nhau, chn la

c k thut ph hp nht, tnh ton li in da trn cc nt c th nht,

hoc l cc phng php nhm n gin ha li in tnh ton, dng cc c

s l thuyt ton cng nh k thut ti u gii quyt yu cu v ti u s

lng, v tr ca TBP trn li phn phi.

1.2. MT S PHNG PHP XC NH PHN ON LI IN

PHN PHI C NGHIN CU:

1.2.1. K thut dng chng trnh nh phn [4]:

*Nhm tc gi:

- Farajollah Soudi, Kevin Tomsovic : Trng i hc Washington.

*Ni dung: Nhm tc gi s dng cng c ton hc, thit lp cc h phng

trnh v bt phng trnh tuyn tnh cho bi ton tm v tr v s lng cc

phn on ti u trn li phn phi, vi cc bin l tp hp cc TBP trn

li in, cc bin ny ly cc gi tr 0, 1 km theo tp hp cc iu kin rng

buc v v tr, gii hn vn u t Cc tham s l cc gi tr v ph ti, sut

s c, thi gian s cQua tin hnh gii bi ton bng cng c my tnh

(phn mm CPLEX-s dng cc thut ton ca l thuyt tp m) c kt

qu cc phn on cn lp t theo yu cu.

*nh gi:

- 6 -

-S dng cng c my tnh v phn mm chuyn dng rt phc tp, i

hi c trnh x l v chng trnh my tnh v kh nng x l cc

vn l thuyt ton hc v l thuyt m.

-Kha cnh gii quyt: nh gi v thit hi mt in khch hng.

-Phm vi gii quyt: rng v kt qu kh chi tit, mang tnh tng qut

cao.

1.2.2. K thut dng chng trnh nh phn [4]:

*Nhm tc gi:

- Rosawan Bupassiri v Naruemon Wattanapongsakorn: B mn my tnh

Trng i hc King Mongkut, Thailand.

- Jamnarn Hokierti: B mn k thut in, Trng i hc Kasetsart,

Thailand.

*Ni dung: Nhm tc gi s dng cng c ton hc, thit lp cc h phng

trnh v bt phng trnh tuyn tnh cho bi ton tm v tr v s lng cc

phn on ti u trn li phn phi, vi cc bin l tp hp cc TBP trn

li in, cc bin ny ly cc gi tr 0,1 km theo tp hp cc iu kin rng

buc v v tr, gii hn vn u t Cc tham s l cc gi tr v ph ti, sut

s c, thi gian s cQua tin hnh gii bi ton bng cng c my tnh

(phn mm LINDO) c kt qu cc phn on cn lp t theo yu cu.

*nh gi:

-S dng cng c my tnh v phn mm chuyn dng phc tp nh

nhm tc gi trn, i hi c trnh x l v my tnh v kh nng x

l ton hc cc bi ton ti u.

-Hng gii quyt: nh gi v thit hi mt in khch hng.

-Phm vi gii quyt: rng v kt qu chi tit, mang tnh tng qut cao.

- 7 -

1.2.3. K thut dng chng trnh my tnh, tnh ton tin cy [5]:

*Nhm tc gi:

- Ying he v Gran Andersson: Vin Cng ngh Hong gia Thy in.

- Ron N. Allan : Trung tm nng lng in Manchester, Anh.

*Ni dung: Nhm tc gi s dng cng c my tnh (phn mm RADPOW)

s dng cc thut ton nh gi tin cy cp in cho cc b TBP li

in phn phi vi cc iu kin rng buc v vn u t cho cc TBP, cc

thut ton ny c pht trin t thut ton ti thiu ha cc b cu hnh mt

in ca li in phn phi ca tc gi Billinton Roy B mn K thut

in, i hc Saskatchewan, Canada, c kt qu cc phn on cn lp t

theo yu cu.

*nh gi:

-S dng cng c my tnh v phn mm chuyn dng phc tp nh

nhm tc gi trn, i hi c trnh x l v my tnh.

-Hng gii quyt: nh gi v thit hi mt in khch hng.

-Phm vi gii quyt: rng v kt qu chi tit, mang tnh thc t cao.

1.2.4. K thut dng chng trnh my tnh, kt hp l thuyt ton quy

hoch :

*Tc gi:

- PGS. TS. Trn Bch: B mn H thng in, trng i hc Bch khoa H

ni.

*Ni dung: Tc gi s dng cng c ton hc, a bi ton v dng quy

hoch tuyn tnh vi iu kin rng buc l h bt phng trnh tuyn tnh,

cc n l cc bin ca cc TBP. Vit chng trnh my tnh (phn mm

PDH.C-Vit trn ngn ng C ) tnh ton, c kt qu cc phn on cn lp

t theo yu cu.

- 8 -

*nh gi:

-S dng cng c my tnh v xy dng phn mm chuyn dng.

-Hng gii quyt: nh gi v thit hi mt in khch hng.

-Phm vi gii quyt: rng v kt qu chi tit, mang tnh thc t cao.

Tm li, cc k thut tip cn tm li gii ca bi ton xc nh phn

on li phn phi, phn ln c nghin cu v trin khai kh cng phu

bng cng c l thuyt ton hc phc tp (nh: lp v gii quyt h bt

phng trnh tuyn tnh, phi tuyn; Dng thut ton di truyn; l thuyt

m c p dng), sau s dng cc chng trnh tnh ton ti u rt

mnh (nh RADPOW ca Thy in; LINGO ca Thi lan) tnh ton.

ti s nghin cu phng php gii quyt bi ton xc nh phn

on li in phn phi hnh tia bng k thut nh s [3], kt hp tnh u

vit ca cng c my tnh in t m xy dng thnh phng php tm kim

d tm li gii, qua thit lp mt chng trnh tnh ton nh nhng thit

thc v tin dng bng ngn ng MATLAB.

- 9 -

Chng 2:

XY DNG M HNH TON HC XC NH PHN ON LI PHN PHI

2.1. LI PHN PHI:

Li phn phi trung p (c cp in p t 6kV-35kV) l cc xut

tuyn t cc trm ngun 110kV; 35kV cp in cho cc trm bin p ph ti

h p ti cc khu vc thnh th, nng thn. Li phn phi trung p c u

t theo quan im chung l m bo vn hnh cp in tin cy, an ton vi

cc chi ph nm trong mt gii hn thp nht, m bo hiu qu v kinh

doanh in nng.

Phn ln thi gian mt in ca khch hng c nguyn nhn chnh t

cc s c li phn phi trung p. c im ca li in phn phi trung p

l c nhiu on, tng chiu di ln v c xc sut s c cao. Do , mt bin

php c hiu qu nng cao tin cy cung cp in cho cc h tiu th l

phn on li in. Trong li in phn phi hnh tia c mt ngun cung

cp, do khng c d phng nn s c xy ra mt ni s gy mt in ly lan

c mt vng rng nu khng dng TBP.

Cc li in phn phi hnh tia c ngun d phng (hoc li in

vng vn hnh h) th tin cy c tng ln nhiu v vic t thm cc

TBP s lm tng thm hiu qu ca ngun d phng.

Nhn chung, TBP s dng trn li in phn phi bao gm my ct

i vi cc rle bo v ti trm ngun, mt s cc TBP ng dy, vi cc

chc nng v c tnh khc nhau trn trc chnh v nhnh r ca xut tuyn

ng dy. Cc thit b c th l Re. ng dy, thit b ng ct, b lin

lc bng dao ct ti, DCL v cu ch. C th tm tt chung tng loi theo

nhm nh sau (theo chc nng):

- 10 -

*My ct: Kt hp h rle bo v, c kh nng ct s c v t ng ng

li. Vic t ng ng li nhm loi tr cc s c thong qua v khi phc

cp in cho ph ti trong thi gian rt nh. Ngy nay, kt hp cc rle dng

k thut s nn c nhiu chc nng c m rng hn: bo v, gim st, xc

nh vng s c, phi hp, ghi chp s kin

*Recloser: C kh nng ct s c v t ng ng li, lp t trn ct

ng dy hoc thnh tng cabin ngm. Re. c kh nng phi hp nhau trong

tng s c th, t ng khi phc cp in cho khch hng, khc phc mt

in do cc s c thong qua ch trong cc khong thi gian phi hp rt nh.

*Thit b ng ct, lin lc: khng c kh nng ct dng s c, c chc

nng thao tc c lp khu vc s c khi cc thit b Re. hoc my ct t pha

ngun cung cp ct in s c. Chng cng c chc nng thao tc ng

ct trong cc ng dy lin lc chuyn ngun cp, khi phc cp in cho

khch hng theo nhiu phng thc. Cc thit b ny c th l DCL, dao ct

ti iu khin t ng (khi gn thm t iu khin lin ng vi cn iu

khin) hoc bng tay (cn thao tc). Gia iu khin bng tay hoc t ng,

trong tnh ton ta xem chng ch khc nhau v thi gian thao tc.

*Cu ch t ri: c s dng ph bin nht, c gi r, c kh nng ct s

c nhng khng t ng ng li. Cu ch c s dng trong trm bin p

phn phi v cc nhnh r trung p c cng sut ph ti nh.

Khi nh gi phng n lp t TBP (v s lng v v tr lp) trn

li phn phi, phi nh gi gi tr thit hi do mt in trn li in lc

cha lp t v sau khi lp t TBP (nh gi cho nhng phng n lp

khc nhau), so snh tm phng n c thit hi nh nht. Cui cng l so

snh tng chi ph bao gm thit hi mt in vi vn u t ca cc phng

n c thit hi mt in nh nht [7].

- 11 -

Vi mt s lng TBP nht nh, phng n ti u v v tr lp t s

c chn theo thit hi nh nht khi mt in khch hng trn li phn

phi, cn vi s lng khc nhau cc TBP, phng n ti u v s lng v

v tr lp t c chn theo tng chi ph (bao gm vn u t TBP v thit

hi mt in khch hng) nh nht.

Bn cht ca vn l khi lp t TBP mi ln li in, ton b cc

khch hng t v pha ngun c bo v hoc c khi phc cp in

nhanh chng khi s c xy ra on ng dy pha sau TBP (thit hi

mt in gim i). Cn on ng dy s c s mt in cho n khi khc

phc xong h hng. Nh th, khi tin hnh xem xt phng n lp t TBP,

mt s cc iu kin cn c thng nht nh sau [7]:

- Cc ri ro do s c t l vi chiu di ng dy.

- Thit hi do mt in t l vi lng in nng mt.

- Tng trng ca ph ti nh nhau ti cc nt ph ti.

- S dng cc gi tr trung bnh: cng sut, thi gian mt in, thi gian

thao tc, gi in phc v tnh ton.

- Vn u t lp TBP c b ra thc hin trong mt nm (khng dn

tri nhiu nm). Khng xt t l lm pht ca nn kinh t trong ton b i

sng khai thc hiu qu TBP.

2.1.1. Li phn phi hnh tia- Phn on li phn phi [1].

S li phn phi khng phn on c th hin trn hnh 2.1.

Li phn phi trn hnh 2.1a l li phn phi hnh tia khng phn

on. i vi li phn phi ny, hng hc bt k ch no cng gy mt

in ton li phn phi. Khi ngng in cng tc cng vy.

- 12 -

(1) 1 (2) 2 (3) 3 (4) 4

a) P1 P2 P3 P4 TBP

b) P1 P2 P3 P4

TBP

on li I, LI on li II, LII

PI PII

c)

Hnh 2.1

S s c ton li phn phi trong 1 nm l:

100.0

LSC OO (2.1)

0O - Sut s c (v/100km.nm);

L- di li phn phi (km).

S v s c tng cng l:

CTSCND OOO (2.2)

CTO - S v ngng in cng tc.

Thi gian ngng in do s c trong mt nm l:

SCSCNDSC TT .O (2.3)

TSC- thi gian sa cha s c.

Thi gian ngng in cng tc l:

CTCTNDCT TT .O (2.4)

- 13 -

TCT- thi gian trung bnh mt ln ngng in cng tc.

Tng thi gian ngng in l:

NDCTNDSCND TTT (2.5)

in nng mt do s c l:

(2.6)tbNDSCSC PTA .

in nng mt do ngng in cng tc l:

(2.7)tbNDCTCT PTA .

tng cng tin cy, li phn phi hnh tia c phn thnh nhiu

on bng thit b ng ct phn on.

Trong trng hp phn on bng DCL, nu xy ra s c mt phn

on no my ct u ngun s nhy tm thi ct ton b li phn phi.

DCL phn on c ct ra c lp phn t b s c vi ngun. Sau ngun

c ng li tip tc cp in cho cc phn on nm trc phn on s c

v pha ngun.

Nh vy, khi xy ra s c mt phn on no th ph ti ca phn

on s c v cc phn on c cp in qua phn on s c (tc l nm

sau n tnh t ngun) b mt in trong sut thi gian sa cha phn on s

c. Cn ph ti ca cc phn on nm trc phn on s c v pha ngun

th ch mt in trong thi gian thao tc c lp phn t s c.

Trong trng hp phn on bng my ct, khi mt phn t b s c,

my ct phn on u phn t s c s t ct v c lp phn t s c. Cc

phn t trc phn t s c hon ton khng b nh hng.

Trn hnh 2.1b l li phn phi phn on gm hai on v trn hnh

2.1c l li phn phi ng tr ca n. Tnh t ngun, on li I ng trc

on li II.

Ta tnh tin cy cho tng loi li:

- 14 -

on li I: on I c th b ngng in do bn thn n b hng hoc do

nh hng ca s c trn on li sau.

- on I c s ln ngng in l v thi gian ngng in nm l T

(nu l ngng in s c hay ngng in cng tc th dng cc cng thc

tng ng tnh).

'IO

'I

- nh hng ca s c trn cc on li sau n, y l on II, nh

hng ny ph thuc TBP.

+ Nu dng my ct th on II hon ton khng nh hng n on I:

0 IIIO ; 0 IIIT (2.8)

+ Nu dng DCL, th s c on II lm ngng in on I trong thi

gian thao tc c lp s c Ttt ,do :

; T'IIIII OO ttIII T (2.9)

Tng s ln ngng in v thi gian ngng in ca on li I l:

; T (2.10)IIIII OOO'

IIIII TT '

on li II: on II c th b ngng in do chnh n b s c hoc do

nh hng ca s c trn cc on trc n v sau n. C th y ch c

on I trc n.

- S ln s c ca on li II l v thi gian ngng in nm l T'IIO .'II- nh hng ca on I n on II l ton phn khng ph thuc TBP,

ngha l on II chu s ln hng hc v thi gian ngng in ca on I:

; T (2.11)'IIII OO

'IIII T

Tng s ln ngng in v tng thi gian ngng in ca on li II l:

; T (2.12)'' IIIII OOO '' IIIII TT

Do c th rt ra kt lun chung nh sau:

- 15 -

Cc on li pha sau chu nh hng ton phn ca cc on li pha

trc, cn cc on li pha trc ch chu nh hng khng ton phn ca

cc on li pha sau. nh hng ny ph thuc TBP.

Trong tnh ton trn b qua hng hc ca TBP v s dng TBP

khng phi bo dng nh k.

2.1.2. Li phn phi kn vn hnh h.

Li phn phi kn vn hnh h gm nhiu ngun v nhiu ng dy

phn on to thnh li kn nhng khi vn hnh th mt s phn on m

to thnh li h. Khi mt on li ngng in th ch ph ti on li

mt in, cn cc on khc ch mt in trong thi gian ngn thao tc

hoc phi hp cc TBP, sau li c cp in nh thng.

Li phn phi kn vn hnh h c tin cy c nng cao rt nhiu,

c bit l khi cc thao tc thit b ng ct v phn on c iu khin t

xa hoc t ng.

Khi xy ra s c trong li phn phi, cc my ct bo v (hoc Re.) s

ct. Sau c th khi phc vic cung cp in cho mt s h ph ti bng

hai bc [7]:

1/ M thm DCL c lp vng b s c ca li phn phi, thu hp

phm vi nh hng ca s c.

2/ ng my ct (hoc Re.) hoc DCL khi phc cung cp in t

ngun c hoc t ngun d phng cho cc h tiu th ang b mt in nhng

c cch ly vi vng s c. Vic ng ngun d phng c th c t

ng ha bng cch dng thit b t ng ng ngun d phng.

2.1.3. ng tr cc on li [1].

Thc t, thun li trong cng tc qun l vn hnh li phn phi,

trong qu trnh pht trin, cc nhnh r ng dy u c lp t cu ch t

ri ti u tuyn (i vi cc nhnh r nh) hoc Re. bo v, thao tc d

- 16 -

lp tm s c. Tuy nhin, mt s nhnh r c th lp DCL hoc dao ct ti u

tuyn (trng hp ny thng c s lng nh), khi , trong vn hnh, cc

s c xy ra trn nhnh r u nh hng n trc chnh ng dy (m trc

tip l nh hng n on trc ng dy m nhnh r u ni). Nh

vy, trong tnh ton, ta ng tr cc on nhnh r theo cc biu thc nh sau:

- di ng tr ca m on li lin nhau thnh on li j l:

m

iij ll

1 (2.13)

- Xut s c ca on li ng tr l:

1000OO jj l (2.14)

- Tng ph ti trung bnh (h s ng thi bng 1):

(2.15)

m

iij pp

1

2.2. SUT S C LI PHN PHI:

C s tnh v s dng sut s c li in phn phi ti Cng ty in

lc da theo Quy nh thc hin cc ch tiu qun l k thut h thng in

ca EVN ban hnh (cng vn s 3322 EVN/VP ngy 29 thng 06 nm 2001)

c th nh sau:

2.2.1. Phn loi v cch tnh s c.

a. S c ng dy trung p t 6 n 35kV:

S c trn li in phn loi nh sau:

x S c thong qua: l s c m cc phn t s c c khi phc

trong thi gian nh hn hoc bng 20 pht.

x S c vnh cu: l s c c thi gian khi phc ln hn 20 pht.

- 17 -

x S c khch quan: l s c do cc tc nhn nm ngoi s kim

sot v trch nhim ca n v qun l vn hnh, qun l thit b,

v d nh:

+ Cc s c do bo, lt, ging, st...gy ra vi iu kin tt c cc

bin php phng chng theo qui trnh, qui phm, cc hng dn ca Tng

Cng ty v Cng ty thc hin y .

+ Cc s c lan trn t h thng nh hng ln thit b.

+ Cc s c do ph hoi hoc do cc tc nhn nm ngoi s kim

sot ca n v qun l thit b nh: Bn sng vo ng dy, nm cnh cy,

cht cy ng vo ng dy, o p knh mng st chn ct in gy

nghing, ct hoc xe c gii hc vo ct in ven ng.

+ Cc s c do cc n v ngoi ngnh thi cng cng trnh lm vt

l ri vo ng dy hoc do nhng ngi b thiu nng tm thn gy nn.

+ Cc s c do nh hng ca qu trnh qu , nh hng ca qu

ti, li ca thit b ht tui th, li ca nh ch to trong thi gian vn hnh,

hoc li ca phn thit k ca cc n v t vn.

* Lu : Cc s c v nguyn nhn khch quan khng tnh vo ch

tiu s c.

x S c ch quan: Ngoi cc nguyn nhn trong phm vi nu

trn th cc s c cn li c coi l s c ch quan.

Cc ch tiu s l:

- S v s c thong qua/100km.nm.

- S v s c vnh cu/100km.nm.

b. S c trm bin p:

S c trm bin p khng chia ra s c thong qua hay vnh cu v ch

tiu s c c tnh nh sau:

x i vi trm bin p 6-35kV:

- 18 -

+ Sut s c trm s l: S v s c/trm.nm.

2.2.2. Ch tiu sut s c thc hin:ng dy 6-35kV. Trm bin

p 6-35kV.Li h

pVC

V/100km.nm.TQ

V/100km.nm.V/trm.nm. V/100km.

nm.3,6 12 1,8 120

Bng 2.1

Cc ch tiu s c s c tnh c h s ma. Trong ma ma bo h s

ma l 1,2; trong ma kh h s l 0,8. H s ma c tm tnh cho tng

in lc trong bng sau: Bng 2.2H s ma STT in lc

0,8 1,201 Qung Bnh Thng 1- thng 6 Thng 7- thng 12 02 Qung Tr Thng 1- thng 6 Thng 7- thng 12 03 Tha Thin Hu Thng 1- thng 6 Thng 7- thng 12 04 Nng Thng 1- thng 6 Thng 7- thng 12 05 Qung Nam Thng 1- thng 6 Thng 7- thng 12 06 Qung Ngi Thng 1- thng 6 Thng 7- thng 12 07 Bnh nh Thng 1- thng 6 Thng 7- thng 12 08 Ph Yn Thng 1- thng 6 Thng 7- thng 12 09 Khnh Ho Thng 1- thng 6 Thng 7- thng 12 10 Gia Lai Thng 1,2,3,4,11,12 Thng 5- thng 10 11 Kon Tum Thng 1,2,3,4,11,12 Thng 5- thng 10 12 k Lk Thng 1,2,3,4,11,12 Thng 5- thng 10

2.3. THIT HI MT IN KHCH HNG.

Thit hi do mt in khch hng ph thuc vo loi khch hng v c

khong thi gian mt in. Theo tiu chun phn loi cng nghip (SIC) chia

khch hng thnh cc dng: Ph ti c bit ln (cng sut ln hn 8MW),

cng nghip, thng mi, nng nghip, sinh hot, hnh chnh vn phng [6].

Thit hi v kinh t do mt in, ngoi vic Ngnh in b mt mt lng

in nng thng phm m bnh thng vn cung cp cho khch hng m

cn tnh ton ti cc thit hi tht ca ph ti (hng sn phm, thit b, dy

chuyn) trn quan im ca h thng in. N nh gi mc tha mn

nhu cu ca ph ti v tin cy h thng in m ng thi m bo hiu

- 19 -

qu kinh t ca h thng in. Thit hi c tnh cho tng loi ph ti, cho

mt ln mt in, cho 1 KWh in nng mt v cng c tnh cho di thi

gian mt in. Khi mt in, gi tr thit hi c xem xt n l [1]:

- Gi tr ca sn lng in thng phm b mt do khng cung cp c

cho vng ph ti mt in. Hin nay, cc cng ty in lc ang tnh thit hi

ny bng gi tr ca lng in nng trung bnh b mt ti cc trm bin p

ph ti. Gi tr ny c s dng trong cc thit k v quy hoch li in.

-Gi tr thit hi tnh trn quan im h thng in t cc thit hi c tht

ti ph ti s dng in. Gi tr thit hi ny rt kh xc nh do thiu thng

tin hoc thng tin thiu tin cy ( chnh xc thp). Cn thc hin cc phng

php thc nghim cng phu c biu gi hp l, hin nay c cc phng

php chnh xc nh nh sau [6]:

+Kho st khch hng: tin hnh kho st thit hi do mt in cho

tng loi khch hng nu trn vi nhng khong thi gian mt in khc

nhau, tp hp thng tin v thc hin cc iu chnh ph hp c biu gi

thit hi.

+Nghin cu c th cc trng hp mt in: tin hnh nghin cu cc

trng hp mt in c th trong phm vi ln, t c c cc d liu v

cc thit hi trc tip v c thit hi gin tip c lin quan.

+Tnh ton da trn t l cc ch s ca nn kinh t v in nng tiu

th: t l GNP, GDP, tng sn phm mt ngnh kinh tvi tng in nng

tiu th, t suy ra thit hi khi mt in theo tng thnh phn kinh t.

Gi tr thit hi do mt in khch hng v th rt phc tp xc nh.

Trong ti, biu gi tr thit hi ny c xem xt t tham kho cc biu gi

ca mt s quc gia a ra mt biu gi dng tnh ton. Biu gi tr

thit hi c s dng trong cc tnh ton thit k, quy hoch h thng in,

tm gi l biu gi n b mt in.

- 20 -

Ta c tng thit hi kinh t do mt in ti khch hng:

)( dbbq ccAICOST (2.16)

Trong :

ICOST- Thit hi do mt in khch hng (ng).

A Lng in nng khng cung cp do mt in (KWh).

cbq, cdb- Theo th t l gi bn in bnh qun v gi n b mt in

(/KWh).

2.4. GI IN:

Ngnh in thc hin vic bn in cho khch hng thng qua h thng

o m in nng lp t ti v tr s dng in, ty theo nhu cu s dng

in ca khch hng m h thng o m c th t cc cp in p khc

nhau. i vi li phn phi hin nay, h thng o m in nng ch yu

c lp t pha h p (380V v 220V) ca cc trm bin p phn phi, nh

vy gi mua bn in nng gia Ngnh in v khch hng xem xt trong

ti ny ch da trn biu gi in cp in p di 6kV theo quy nh ca

Chnh ph nc Cng Ha X Hi Ch Ngha Vit nam trong quyt nh s

124/2002/Q-TTg ngy 20/09/2002 ca Th tng Chnh ph v gi bn

in.

xc nh thit hi cho mt lng in nng b mt do s c mt in

hoc mt in do tm dng sa cha, bo dng li in theo k hoch, khi

tnh ton, cn xc nh cc gi tr thit hi sau:

- Gi tr thit hi ca lng in nng thng phm khng bn c do

mt in ti khch hng.

- Gi tr thit hi do phi n b cc h hng sn phm, ngng tr trong

dy truyn sn xut, do nguyn nhn mt in.

- 21 -

2.4.1. Gi bn in ( cp in p di 6kV):

Di y l mt s biu gi in ca mt s loi khch hng ph bin,

chim t l ln (cha c thu VAT) (cng nm trong phn loi theo SIC):

- Gi in cho sn xut cng nghip:

+Gi bnh thng: 895 /KWh

+Gi thp im: 505 /KWh

+Gi cao im: 1480 /KWh

- Gi in cho hnh chnh s nghip: 920 /KWh

- Gi in sinh hot:

+ Cho 100 KWh u tin: 550 /KWh

+ Cho 50 KWh tip theo: 900 /KWh

+ Cho 50 KWh tip theo: 1210 /KWh

+ Cho 100KWh tip theo: 1340 /KWh

+ T KWh th 301 tr ln: 1400 /KWh

- Gi sinh hot nng thn: 390 /KWh

- Gi in kinh doanh, dch v, du lch:

+Gi bnh thng: 1410 /KWh

+Gi thp im: 815 /KWh

+Gi cao im: 2300 /KWh

Mt trong cc ch tiu nh gi hiu qu kinh doanh in nng Cng ty

Phn phi in v ti cc in lc l gi bn in bnh qun (/KWh), y

cng l gi in c s dng trong cc tnh ton phn tch hiu qu kinh t

ti chnh ca mt d n Ngnh in, th hin doanh thu kinh doanh in nng,

ti cng s dng gi in bnh qun tnh gi tr thit hi mt in. Gi

bn in bnh qun l gi in trung bnh theo t l lng in nng ca cc

loi ph ti trong tng lng in nng thng phm ca mt in lc, cng

ty trong mt n v thi gian xem xt (1 qu, 1 nm). Biu thc tnh ton

- 22 -

gi bn in bnh qun cho cc loi ph ti tiu biu trn nh sau (trong 1

nm):

KDDVshntASSHhcsnsx

KDDVshntASSHhcsnsxbq AAAAA

RRRRRc

(2.14)

Trong :

- Rsx, Rhcsn, RASSH, Rshnt, RKDDV: ln lt l doanh thu trong mt nm ng vi

tng loi ph ti sn xut, hnh chnh s nghip, nh sng sinh hot, sinh

hot nng thn v kinh doanh dch v (ng).

- Asx, Ahcsn, AASSH, Ashnt, AKDDV: ln lt l tng in nng thng phm

trong mt nm ca tng loi ph ti sn xut, hnh chnh s nghip, nh

sng sinh hot, sinh hot nng thn v kinh doanh dch v (KWh).

Gi in bnh qun ca in lc Bnh nh nm 2003 l 646 /KWh.

2.4.2. Gi tr ca thit hi khi khch hng mt in:

Hin ti, cha c cc kho st quy m no cp quc gia c th ly

lm c s php l nh gi thit hi khi mt in khch hng, cng nh cha

c mt biu gi c th no p dng cho cc cng vic nh gi c lin

quan. Tuy nhin ti a ra mt biu gi tham kho v gi tr thit hi do

mt in khch hng, da trn so snh, tnh ton t cc biu gi c ca mt

s quc gia, tnh quy ra gi tr thit hi cho 1KWh in nng b mt, trn c s

quy v mt bng kinh t ca Vit nam. Xc nh nh sau (t gi ly

1AusD=11400 VN) [1]:

Gi bn in trung bnh Australia khong:

0,17 AusD/KWh =1938 /KWh.

Gi tin trung bnh 1 KWh in nng ca ph ti sn xut b ngng

cung cp in Australia:

-K hoch: 2,5 AusD/ KWh = 28500 /KWh

-S c: 7,5 AusD/ KWh = 85500 /KWh.

- 23 -

Gi bn in trung bnh nc ta l : 785 /kWh.

Chnh lch gi tr 1 KWh in nng trung bnh gia Vit nam v

Australia khong 40,4 %. Ly gi tr n b ti Vit nam bng 40% Australia.

Nh th, trung bnh gi 1 KWh in nng ngng cung cp in nc ta c

th ly gn ng nh sau:

- K hoch: 4664 /kWh.

- S c : 14000 /kWh.

2.5. HM MC TIU TNH TON:

2.5.1. Hm mc tiu kinh t [1]:

nh gi cc phng n phn on v mt kinh t, c th s dng

hm hiu qu kinh t. Khi so snh cc phng n ch cn xt n phn hiu

qu kinh t th hin s khc nhau gia cc phng n ch khng cn xt n

hiu qu tuyt i ca tng phng n. Do , i vi cc phng n phn

on li in th hiu qu kinh c th tnh l:

ICCZ mdbq '' (ng) (2.15)

Trong :

- l li ch thu c do nng cao tin cy cung cp in (gim in

nng thiu do ngng in).

mdC'

'''mdmdmd CCC ' (2.16)

- v C : ln lt l thit hi ca doanh nghip ngnh in do thiu

ht in nng cc h tiu th v ngng cung cp in tnh trc khi v sau

khi thc hin phn on li in phn phi.

'mdC

''md

- : chnh lch gia chi ph bo qun TBP trc v sau khi phn

on li in phn phi.

bqC'

.0''' ' bqbqbq CCC (2.17)

- 24 -

- I : chi ph vn ca phng n phn on.

Vy hm hiu qu kinh t c th vit l:

ICCCCZ mdmdbqbq ''''''

(2.18))()( '''''' mdbqmdbq CCICC

Ta cn chn phng n c Z max - ngha l ch cn chn phng n t

cc tiu hm chi ph

o

.C

min.o '''' mdbq CCIC

Chi ph C bng tng cc chi ph C t ca li in phn phi trong cc

nm t tnh t khi bt u b vn cho n ht thi gian s dng cc TBP (gi

s l T nm) ri tr i gi tr cn li (gi tr o thi) ca cc thit b ,

trong chi ph cc nm khc nhau cn c qui i v thi im hin ti

(nm gc 0).

T

ttt

t

qD

qCC

1

(2.19)

Trong : q l h s tnh li: q=1+r.

Vi r l li sut tnh ton khi u t vn vo ngnh in lc (khong 8 n

12%).

Nu gi V (ng) l chi ph vn ca phng n phn on, qui i v

nm gc 0 v nu b qua gi tr o thi ca cc TBP th:

T

tt

bqmd

qtCtC

VC1

)()(

(2.20)

- Cmd(t) l thit hi trong nm th t v mt in cc h tiu th ca

li in phn phi ang xt (ng).

- Cbq(t) l chi ph bo qun TBP nm th t.

- 25 -

nh gi kinh t cc cng trnh ln nh nh my thu in, cng trnh

thu li,... vi tui th ln t 50 nm n hng trm nm th thng dng

phng php ng tc l c xt n s thay i theo thi gian ca cc chi ph

v thu nhp hng nm lin quan n i tng u t.

i vi cc cng trnh c tui th ngn hn, n gin m hnh ton hc

th c th dng phng php tnh, tc l tnh ton vi gi tr trung bnh hng

nm ca cc lng thu, chi hoc s dng gi tr ca nm u tin vi chnh

xc d bo cao nht.

Cc TBP c thi gian s dng khng di qu (c 20 - 25 nm) nn trong

cc tnh ton sau s s dng phng php tnh. Khi , tnh ton cc thit hi

Cmd(t) theo gi tr ph ti trung bnh ca li in phn phi v vi gi tr c

nh ca gi mt in (sut thit hi (ng/KWh) do thiu ht in nng v

ngng cp in h tiu th) tc l c:

Cmd(t) H = const l k vng thit hi kinh t hng nm do ngng cp

in. Tng t, cng tnh C

|

bq(t)=Cbq=const.

Vy:

T

t

tbq qCHVC

1/)( (2.21)

hay l:

).()1(

1bqT

T

CHqq

qVC (2.22)

Mt cch tng ng c th xt hm mc tiu:

..1

)1(.1

)1( HCVq

qqCq

qqC bqTT

T

T

(2.23)

hay l:

- 26 -

HCVC bq .0D (2.24)

Trong :

-1

)1( T

T

qqq

0D gi l h s thu hi vn

- C : gi l chi ph qui i nm hay chi ph qui dn.

t tbpdbq CCV .0D gi l chi ph cho TBP.

Nh vy: .HCC tbpd (2.25)

Nu trong li in phn phi t nhiu TBP th:

.)()( i

itbpdtbpd CC (2.26)

v iu kin ti u l:

.min)( o HCCi

itbpd (2.27)

Cc cng thc (2.23 - 2.25) c s dng khi cn tnh dng gi tr thit hi

hng nm cha quy i v nm hin ti.

Vy iu kin la chn phng n ti u l cc tiu hm chi ph qui

dn ca phng n phn on li in phn phi bao gm tng chi ph cho

TBP v thit hi do ngng in - ngha l ta phi i xc nh s lng v v

tr lp t TBP cc tiu hm chi ph qui dn ca phng n phn on

li in phn phi (2.22).

Sau khi chn c phng n phn on li in phn phi, c th s

dng cc ch tiu ca phng php ng nh li rng NPV, li sut ni ti

FIRR,... nh gi tnh kh thi ca phng n.

- 27 -

2.5.2. Cc gi thit tnh ton.

1/ Cc iu kin rng buc v k thut c xem nh c m bo, c

th l:

+Li phn phi c thit k sao cho trong cc ch lm vic sau s

c th ngun c cng sut v li in kh nng ti trong mi ch

linh hot cp in.

+Do kh nng v vn u t, do c im pht trin ph ti, c bit l

mt ph ti cha cao, nn u t lp t phn on li phn phi mc

va phi tc l phn on cho tng cm trm bin p ph ti (ch yu cc

li in mch vng vn hnh h khu vc cc khu th), khng xt cc s

c trang b b thit b cp vng ring-main units cho trm bin p ph ti,

s trang b b khi ng (iu khin bng vi x l) TBP kt hp vi thit

b iu khin t xa ch th vng s c (RFSI)).

2/ V cc trm bin p phn phi u c my ct hoc cu ch nn s c

my bin p khng nh hng n cc TBP khc. Do , vic phn on

li in phn phi khng lm nh hng n thit hi kinh t do s c trong

trm bin p phn phi. V vy, khi tnh hm mc tiu, khng xt n phn

ca k vng thit hi do ngng in gy ra bi s c hay bo qun nh k

trong ni b trm bin p phn phi.

3/ Gi thit khng c 2 s c xy ra ng thi trong li in phn phi

ang xt.

4/ V cng hng hc ca cc TBP rt nh so vi ng dy. V

nhiu TBP hin i nh my ct SF6 khng phi sa cha nh k cho nn

khi tnh ton H khng k n s c hay bo qun nh k cc TBP.

- 28 -

Chng 3:

CHNG TRNH TNH TON PHN ON LI PHN PHI

3.1. M HNH LI IN PHN PHI S DNG:

nh hng gii quyt vn nm trong khun kh cc iu kin v

cc c im c th, thc t ca li in phn phi trung p trong khu vc.

3.1.1. Cc c im thc t c bn ca li phn phi cn quan tm

nh hng gii quyt trong chng trnh:

3.1.1.1. Cc c im thc t in hnh:

Li in phn phi tnh ton c cp in p 22kV, 35kV, 15kV,

10kV, 6kV. Li in phn phi trung p ch yu lm c s tnh ton trong

ti l li in 22kV. Tuy nhin, chng trnh vn c th p ng cho cc

tnh ton trn li in ca cc cp in p cn li.

Theo iu kin v tnh cht pht trin thc t, hin nay ph bin hai cu

hnh li in trung p ch yu hnh thnh v pht trin, l li in

trung p nng thn v li in trung p khu vc thnh ph, th trn. Hai cu

hnh ny do iu kin u t cng nh lch s pht trin khc nhau do c

nhng nt khc nhau c bn (ch yu v ng dy) nh sau:

a) Li in trung p nng thn:

Hnh 3.1

Ngu?n

MC R

DCLDCL

FCO

- 29 -

-Cu hnh chung dng hnh tia, trc ng dy xut tuyn ko di, mt

ph ti thp, tha tht. Cc trm bin p phn phi c cp in t mt

ngun n.

-Trc kia, cc TBP trc hoc nhnh r trn ng dy, thm ch

ngay c thit b ng ct u xut tuyn u l FCO vi chc nng va thao

tc ng ct phn on, va bo v ngn mch ng dy, sut mt in ca

cc h tiu th do vy rt cao, ch bo v thiu tin cy, tin cy li in

thp.

-Hin nay, cc xut tuyn ng dy trung p khu vc nng thn

c u t a vo s dng cc DCL trung p, cc Re. trung p c th lp

t phi hp vi FCO bo v ng dy, iu ny ci thin r rt cc s

c, c bit l cc s c thong qua, hn na khi s dng cc thit b ny,

cng vic phn on, tm v c lp s c trn ng dy cng thun li hn,

khu vc mt in v th c th c gim thiu hn.

-Tuy nhin do mt ph ti thp, ri rc, mc s dng in khng

cao, hiu qu kinh doanh bn in khu vc nng thn hin cn rt thp, do

kh nng u t hon chnh TBP cho li in khu vc ny cn hn ch.

b) Li in trung p thnh ph, th trn:

Hnh 3.2

- 30 -

-Mt ph ti cao, tp trung, mc s dng in rt cao, kt cu

li i dc theo cc ng ph, c nh hnh r rt. Cu hnh li in khu

vc ny ch yu dng vng kn, vn hnh h, cc trm bin p phn phi

thng c cp in t 2 ngun n.

-Li in thnh ph v ang c u t lp t cc TBP nh

Re., dao ct ti, DCL ct c ti, FCO vi mt dy hn, hin nay, vic lp

t cc thit b ny c xem xt, lp t cho tng on ng dy trn xut

tuyn, mi on khong vi trm bin p phn phi gim thiu khu vc

mt in do s c, tng cng s phi hp bo v nhm gim thi gian mt

in do c lp s c, tin cy cung cp in ngy mt tng.

3.1.1.2. nh hng gii quyt trong chng trnh:

-Cu hnh li in phn phi c nghin cu trong ti: hnh tia v m

rng cho s vng vn hnh h.

-TBP xc nh: DCL, dao ct c ti (LBS), DCL iu khin t ng v

Re. ng dy. Ph hp tnh hnh u t hin ti cho phn on ng dy

ch yu l cc loi thit b ny.

-Vic xc nh ti u s lng v v tr cc TBP c tnh ton s dng cho

cc trng hp:

+Xut tuyn ng dy hon ton cha lp t TBP no

+Xut tuyn ng dy c mt s phn on l DCL, Re. ng

dy.

- ph hp vi kh nng vn, s cho trc s lng ti a cc Re. ng

dy s lp t trn xut tuyn, t xc nh v tr ti u ca chng cng cc

DCL lp t thm.

-Cn c vo cc c im thc t ca li in phn phi nh sau:

- 31 -

+Pht trin dn tng on ng dy ni tip nhau ty thuc nhu cu

pht trin ph ti (khng c xy dng hon chnh mt ln-chiu di

khng n nh).

+Cc nhnh r trn trc xut tuyn u c lp t cu ch t ri u

nhnh r - nu l nhnh r nh, hoc l Re. - nu l cc nhnh r

c pht trin ln v mc ti v chiu di. c im ny ch yu

xut pht t khu qun l li in, va m bo chc nng bo v

nhnh r, ng thi to thun li cho cng tc thao tc c lp cc nhnh

r phc v cng vic ci to, bo dng li in cng nh cc cng

tc nghim thu ng in cc cng trnh mi pht trin trn on ng

dy. Do vy trc xut tuyn hoc cc ng trc ca cc nhnh r

hon ton khng b nh hng mt in do cc s c xy ra trn cc

nhnh r.

Do , vi mt li in phc tp v cu trc: nhiu nhnh r vi nhiu

cp, vic xc nh cc phn on ca ton b li in c gii quyt nh

sau:

+Tnh ton xc nh cc phn on trn ng trc xut tuyn, vi lu

l cc nhnh r c thay th bng cc nt ph ti tng ng

(trc tin, thc hin ng tr cc nhnh r khng c TBP ri nhp s

sau ng tr cho file d liu tnh ton).

+Tnh ton xc nh cc phn on trn cc trc nhnh r nh mt bi

ton ring v hon ton tng t nh khi tnh ton ng trc bc

u tin, vi lu ngun cp cho nhnh r xem nh ti cu ch hoc

Re. u nhnh r.

+Kt qu cui cng l s tng hp kt qu cc bc tnh nu.

- 32 -

-i vi li in trong khu th thng l mch vng (mt ngun hoc

nhiu ngun) vn hnh h. im ct ca cc mch vng li in vn

hnh h thng c tnh ton v thay i nh k hng nm c s vn

hnh ti u nht. Khi li phn phi vng chia thnh hai on li phn

phi con hai bn im ct. im ct do vy ng vai tr l mt ngun d

phng . V vy c th gii bi ton phn on ti u cho tng li phn phi

con ring r.

3.1.2. M hnh ton hc:

Xt mt xut tuyn ng dy trung p vi cc thng s v d liu ban

u nh sau:

-ng dy c n nt ph ti.

-Cng sut trung bnh mi nt ti: pi (kW).

-Chiu di mi nhnh ng dy : li (km).

-Sut s c vnh cu ca ng dy : O (ln/100km.nm).

-Sut s c thong qua : J (ln/100km.nm).

-Thi gian ti a cho mt s c thong qua: ttq (gi).

-Thi gian thao tc : ttt (gi) Thi gian thao tc ng ct DCL L thi gian

thao tc s c (gi), c tnh t khi xy ra s c (my ct, Re. ct) cho n

khi c lp c phn on b s c v ng my ct tr li tip tc cung cp

in cho cc phn on pha trc.

-Thi gian s c (hay thi gian khc phc s c): tsc (gi)- tnh t khi nhy

my ct u ngun n khi khi phc cp in nhnh s c li theo s vn

hnh bnh thng.

-Tng thi gian sa cha k hoch nm ca ng dy : tkh (gi).

-Thi gian phi hp Re. ng dy: tph (gi) - Khi xut tuyn c lp t t

2 Re., vic chnh nh thi gian phi hp gia chng c tnh n tng

- 33 -

hiu qu cc bo v, tng tin cy ca h thng, thi gian phi hp ny hin

ang th nghim ti mt s in lc, ly t 55 90 s.

-Gi tin bn in bnh qun ca in lc (hoc cng ty): cbq (ng/kWh).

-Gi tin n b thit hi do mt in s c (tham kho): cdb(ng/kWh).

-Gi tin n b thit hi do mt in k hoch (tham kho): ckh(ng/kWh).

-Gi tr thit hi do mt in trong thi gian s c: W1 (ng).

-Gi tr thit hi do mt in trong thi gian thao tc: W2 (ng).

-Gi tr thit hi do mt in trong thi gian phi hp: W3 (ng).

-Gi tr thit hi do mt in tng cng: W (ng).

3.1.2.1. Li in hnh tia:

Cc k hiu trong hnh v minh ha:

N Ngun. li- chiu di on ng dy

i- S th t nt pi- Cng sut ti

1 2 li i nNMC

p1 p2 pi pn

Hnh 3.3

a) Khng phn on:

Theo nh s li nh trn, s nhnh bng s nt.

Khi mt in (do s c hoc sa cha k hoch) ti nhnh i , ton b

xut tuyn s mt in trong khong thi gian mt in nhnh i, khi hm

thit hi (W) c tnh nh sau (xt cng thc tnh cho trng hp s c,

trng hp sa cha theo k hoch xt tng t ):

- 34 -

LPttcc

ptltlcc

WW tqscdbbq

n n

itqiscidbbq )(

100)(

)(100

)(1

1 1JOJO

(3-1)

Vi - Chiu di xut tuyn. n

ilL1

Vi - cng sut trung bnh ca xut tuyn. n

ipP1

b) Lp t phn on DCL:

1 2 li i nNMC

DCL

p1 p2 pi pn

Hnh 3.4

*Trng hp s c ti cc im nm trc DCL:

Trong trng hp ny, ton b cc ph ti ca xut tuyn mt in

trong sut thi gian sa cha s c, hm thit hi s l:

Ptltlcc

ptltlcc

WWi

ktqksck

dbbqn

jj

i

ktqksck

dbbq

1

11

1

1)(

100)(

)(100

)(1 JOJO (3-2)

*Trng hp s c ti cc im nm sau phn on DCL:

Trong trng hp ny, my ct u ngun ct trong thi gian thao tc

c lp s c, sau ng in li cho ph ti trc phn on v pha ngun,

cn cc ph ti t phn on v sau mt in trong sut thi gian sa cha

(cng tng t i vi sa cha k hoch). Gi s DCL lp trn nhnh i, nh

vy hm thit hi s l:

- 35 -

1

122

1

1222

11

111 )(100

)()(

100)(

21i

jj

i

kttkttk

dbbqn

ijj

n

iktqksck

dbbq ptltlcc

ptltlcc

WWW JOJO

(3-3)

c) Lp t phn on bng Re. v DCL:

1 lR li i nNMC R

DCL

p1 p2 pi pn

Hnh 3.5

n gin, xem R va l k hiu Re., va l k hiu v tr lp ca Re.

trn nhnh R.

*Trng hp s c ti cc im t ngun n trc phn on lp Re.:

Trong trng hp ny, ton b cc ph ti ca xut tuyn mt in

trong sut thi gian sa cha s c, hm thit hi s l:

Ptltlcc

ptltlcc

WWR

ktqksck

dbbqn

jj

R

ktqksck

dbbq

1

11

1

1)(

100)(

)(100

)(1 JOJO (3-4)

*Trng hp s c ti cc im t nhnh R n trc phn on lp DCL:

Trong trng hp ny, cc ph ti t trc R v ngun khng mt in,

cc ph ti t R n cui ngun mt in trong thi gian sa cha s c, hm

thit hi s l:

n

Rjj

n

Rktqksck

dbbq ptltlcc

WW )(100

)(1 JO (3-5)

*Trng hp s c ti cc im t nhnh DCL n cui xut tuyn:

Trong trng hp ny, khi s c, Re. R ct trong thi gian thao tc c

lp s c, sau ng in li cho ph ti trc phn on DCL (gi s DCL

- 36 -

t on i ) v n phn on R, cn cc ph ti t phn on v sau mt

in trong sut thi gian sa cha, cn cc ph ti t u ngun n trc

phn on lp R khng mt in. Nh vy hm thit hi s l:

1

22

222

11

111 )(100

)()(

100)(

21i

Rjj

n

ikttkttk

dbbqn

ijj

n

iktqksck

dbbq ptltlcc

ptltlcc

WWW JOJO

(3-6)

3.1.2.2. Li in vng vn hnh h:

a) Trn xut tuyn lp t cc DCL phn on:

i lq nN2

MC2N1 lilmMC1

DCL1 L1 DCL2Thng m

p1 pm pi pq pn

Hnh 3.6

Gi s: DCL1 t on m v DCL2 t on q sau DCL1 tnh t N1

i, ch vn hnh bnh thng l ngun N1 cp in, my ct MC2 thng

m, cc DCL1, DCL2 ng, ta xt cc trng hp xy ra nh sau:

*Trng hp s c xy ra trong on t N1 n trc DCL1:

Khi , MC1 ct, DCL1 c m ra, N2 cp in cho cc ph ti t

sau DCL1 n N2, cc ph ti ny mt in trong thi gian thao tc, hm thit

hi s l:

n

mjj

m

kttkttk

dbbqm

jj

m

ktqksck

dbbq ptltlcc

ptltlcc

WWW2

2

1

1222

1

111

1

1111 )(100

)()(

100)(

21 JOJO

(3-7)

*Trng hp s c xy ra trong on t DCL1 n trc DCL2:

- 37 -

Khi , MC1 ct, 2 DCL DCL1, DCL2 c m ra, ngun N1, N2 cp

in tr li qua vic ng MC1, MC2. on t DCL1 n trc DCL2 mt

in n khi sa cha s c xong, cc on cn li mt in trong thi gian

thao tc, hm thit hi s l:

n

qjj

q

mkttkttk

dbbq

m

jj

q

mkttkttk

dbbqq

mjj

q

mktqksck

dbbq

ptltlcc

ptltlcc

ptltlcc

WWW

33

1

333

1

122

1

222

1

11

1

111

)(100

)(

)(100

)()(

100)(

21

JO

JOJO

(3-8)

*Trng hp s c xy ra trong on t DCL2 n trc N2:

Khi , MC1 ct, DCL2 c m ra, N1 cp in tr li cho cc ph

ti t N1 n trc DCL2, cc ph ti ny mt in trong thi gian thao tc,

cc ph ti t DCL2 n N2 mt in n khi sa cha xong s c, hm thit

hi s l:

1

122

222

11

111 )(100

)()(

100)(

21q

jj

n

qkttkttk

dbbqn

qjj

n

qktqksck

dbbq ptltlcc

ptltlcc

WWW JOJO

(3-9)

b) Trn xut tuyn lp t cc Re. phn on:

n MC2N2N1

lilR1 i lR2MC1R1 R2

Thng m

p1 pR1 pi pR2

Hnh 3.7

Cng k hiu tng t, gi s Re. R1 lp t trn on lR1, Re. R2 lp

t trn on lR2 pha sau Re. R1 tnh t ngun i. Gi s ch vn hnh

- 38 -

bnh thng l ngun N1 cp in, my ct MC2 thng m, cc Re. R1, R2

ng. Ta xt cc trng hp xy ra nh sau:

*Trng hp s c xy ra trong on t ngun N1 n trc on R1:

Trong trng hp ny, my ct MC1 tc ng ct, Re. R1 do mt ngun

s ct ra, R2 sau thi gian phi hp vi R1 t trc cng ct ra, my ct MC2

ng, ngun N2 bt u cp in, R2 ng khi phc cp in cho ph ti t

R1 n ngun N2, vng s c t MC1 n trc R1 c c lp, mt in

trong thi gian s c. Hm thit hi khi :

n

Rjj

R

kphkphk

dbbqR

jj

R

ktqksck

dbbq ptltlcc

ptltlcc

WWW12

2

11

1222

11

111

11

1111 )(100

)()(

100)(

31 JOJO

(3-10)

*Trng hp s c xy ra trong on t R1 n trc on R2:

Trong trng hp ny, ph ti trong on t N1 n trc R1 khng

mt in, Re. R1 ct, sau thi gian phi hp, R2 ct, my ct MC2 ng cp

ngun t N2 cho cc ph ti t N2 n R2, khi hm thit hi c tnh:

n

Rjj

R

Rkphkphk

dbbqR

Rjj

R

Rktqksck

dbbq ptltlcc

ptltlcc

WWW22

2

12

1222

12

111

12

1111 )(100

)()(

100)(

31 JOJO

(3-11)

*Trng hp s c xy ra trong on t R2 n N2:

Trong trng hp ny, Re. R2 ct, ton b ph ti t R2 n N2 mt in

trong thi gian sa cha s c, cc ph ti cn li khng mt in. Hm thit

hi s l:

n

Rjj

n

Rktqksck

dbbq ptltlcc

WW22

)(100

)(1 JO (3-12)

c) Trn xut tuyn lp t cc Re. v DCL phn on:

c.1) Lp t cc Re. R1, R2 v mt DCL:

- 39 -

N1i lq n MC2

N2lilmMC1

R1DCL

R2

Thng m

p1 pm pi pq pn

Hnh 3.8

Gi s DCL lp t trn on i nm trong on R1-R2 v trong ch

vn hnh bnh thng l ng.

*Trng hp s c trong on t N1 n R1:

Xt hon ton tng t mc b) ta c cng thc (3-10).

*Trng hp s c trong on t N2 n R2:

Xt hon ton tng t mc b) ta c cng thc (3-12).

*Trng hp s c trong on t R1 n DCL:

Khi , R1 ct ra, sau thi gian phi hp ci t, R2 ct, DCL c ct

ra, my ct MC2 ng, R2 ng li cp in t N2 cho cc ph ti n DCL.

Cc ph ti trong on DCL n R2 mt in trong thi gian thao tc, ph ti

trong on R2 n N2 mt in trong thi gian phi hp Re.. Hm thit hi s

l:

n

Rjj

i

Rkphkphk

dbbqR

ijj

i

Rkttkttk

dbbq

i

Rjj

i

Rktqksck

dbbq

ptltlcc

ptltlcc

ptltlcc

WWWW

222

1

1323

12

22

1

1222

1

111

1

1111

)(100

)()(

100)(

)(100

)(321

JOJO

JO (3-13)

*Trng hp s c trong on t DCL n R2:

Khi , R1 ct ra, sau thi gian phi hp ci t, R2 ct, DCL c ct

ra, my ct MC2 ng, N2 cp in n R2, ng R1 cp in t N1 cho cc

ph ti n trc DCL. Cc ph ti trong on DCL n R2 mt in trong

- 40 -

thi gian sa cha s c, ph ti trong on R2 n N2 mt in trong thi

gian phi hp Re., ph ti trong on R1 n trc DCL mt in trong thi

gian thao tc. Hm thit hi s l:

n

Rjj

R

ikphkphk

dbbqi

Rjj

R

ikttkttk

dbbq

R

ijj

R

iktqksck

dbbq

ptltlcc

ptltlcc

ptltlcc

WWWW

222

12

323

1

122

12

222

12

11

12

111

)(100

)()(

100)(

)(100

)(321

JOJO

JO (3-14)

c.2) Lp t cc Re. R1, R2 v DCL DCL1, DCL2:

N1 N2MC1 MC2lR1 lm m lq lR2q

R1 R2DCL1 DCL2 Thng m

p1 pm pq pR2

Hnh 3.9

Gi s cc DCL lp t trn on m, q u nm trong on R1-R2 v

DCL1 nm trc DCL2 tnh t N1 i, trong ch vn hnh bnh thng cc

DCL ng.

*Trng hp s c trong on t N1 n trc R1:

Xt hon ton tng t mc b) ta c cng thc (3-10).

*Trng hp s c trong on t N2 n R2:

Xt hon ton tng t mc b) ta c cng thc (3-12).

*Trng hp s c trong on t R1 n trc DCL1:

Khi , R1 ct, sau thi gian phi hp R2 ct, ct DCL1, on R1-trc

DCL1 mt in trong thi gian sa cha s c. ng N2 qua MC2 cp in

cho ph ti on N2-R2, ng li R2 khi phc cp in cho cc ph ti t

- 41 -

trc R2 n DCL1 (cc ph ti ny mt in trong thi gian thao tc DCL1).

Hm thit hi s l:

n

mjj

m

Rkttkttk

dbbqm

Rjj

m

Rktqksck

dbbq ptltlcc

ptltlcc

WWW2

2

1

1222

1

111

1

1111 )(100

)()(

100)(

21 JOJO

(3-15)

*Trng hp s c trong on t DCL1 n DCL2:

Khi , Re. R1 ct, sau thi gian tph, Re. R2 ct, ct cc DCL DCL1 v

DCL2. ng ngun N2 cp in cho ph ti n Re. R2, ng R1 v R2 khi

phc cp in cho cc ph ti cn li n 2 DCL. Hm thit hi s l:

n

qjj

q

mkttkttk

dbbq

m

Rjj

q

mkttkttk

dbbqq

mjj

q

mktqksck

dbbq

ptltlcc

ptltlcc

ptltlcc

WWWW

33

1

333

1

122

1

222

1

11

1

111

)(100

)(100

)(100

321

JO

JOJO

(3-16)

*Trng hp s c trong on t DCL2 n trc R2:

Khi , R1 ct, sau thi gian phi hp R2 ct, ct DCL2, on DCL2

n trc R2 mt in trong thi gian sa cha s c. ng R1 khi ph cp

in cho ph ti t R1 n trc DCL2, ng N2 qua MC2 cp in cho ph

ti on N2-R2 (cc ph ti ny mt in trong thi gian thao tc DCL2).

Hm thit hi s l:

n

Rjj

m

Rkttkttk

dbbq

m

Rjj

R

qkttkttk

dbbqR

qjj

R

qktqksck

dbbq

ptltlcc

ptltlcc

ptltlcc

WWW

233

1

1333

1

122

12

222

12

11

12

111

)(100

)(

)(100

)()(

100)(

21

JO

JOJO

(3-17)

3.2. NGHIN CU K THUT GII QUYT CC THUT TON :

3.2.1. K thut nh s - Phng php gii quyt:

Mt k thut tng i gn gi vi thc t v thun li khi s dng hn

so vi cc k thut nu trn l K thut nh s [3] , c xy dng bi tc

- 42 -

gi Billinton Roy B mn K thut in, i hc Saskatchewan, Canada.

Trong ti ny s nghin cu v pht trin thnh phng php gii quyt

vn t ra, s dng cng c my tnh lp chng trnh tnh bng ngn

ng Matlab.

Di y quy c cc tn bin cng nh cc gi tr thng s dng

trong cc thut ton v mt s cc t vit tt c hiu nh sau (Ti liu tham

kho [3]):

ICOST : Tn bin, th hin gi tr thit hi tnh ca mt phng

n.

MICOST: Tn ma trn, cc phn t cha gi tr thit hi mt in nh

nht v vn u t ca mt phng n TBP.

MMICOST: Tn ma trn, cc phn t l gi tr thit hi mt in nh nht

v vn u t ca cc b phng n TBP khc nhau.

Lkl : tn phng n, b phng n v tr th k ca l TBP.

N: S v tr c th lp TBP trn li in.

NSNl: s b phng n xy ra cho l TBP lp trn N v tr.

3.2.1.1. K thut nh s:

Thit hi do mt in trong li in kho st l mt hm ph thuc

vo v tr lp t TBP. Vic xc nh v tr lp t TBP ti u vi tng

thit hi do mt in nh nht l bi ton ti u a mc tiu, thng rt phc

tp. K thut nh s a ra mt s hu hn cc gii php kh thi tnh ton

v so snh tt c cc b TBP Lkl c th c v trnh c vic nhm ln v tr

ti u cc b vi li gii v tr ti u ton cc ca bi ton.

Xt mt li in phn phi vi N v tr c th lp t TBP. Cho mt

s c nh cc TBP, s c nhiu b v tr lp t c th xy ra. Gi Lkl l b

v tr th k lp t l TBP. Vi l TBP, s b v tr c kh nng lp t

NSNl l:

- 43 -

),...,2,1,0()!(!

! NllNl

NNS lN

Tng s cc b v tr c th lp t TBP l:

N

l

NlNNSNT

02

p dng k thut nh s cho bi ton ny, TBP ti mi v tr i

c i din bi mt bin TBP Si, Si = 1 nu c TBP ti v tr i v Si = 0

nu khng c TBP ti v tr i. Cc b VTr/TBP c th c xc nh

bng vic dng tp hp cc bin phn on ca tt c cc v tr c th lp

phn on. Mi b VTr/TBP c i din bi mt s nh phn (N bt)

(S={Si}; i=1,N) m c th chuyn thnh mt s thp phn tng ng. Mi b

VTr/TBP v th tng ng vi mt s thp phn duy nht. Trnh t nh

s xc nh v tr ti u ca mt s c nh cc TBP theo cc bc sau:

(i) Xc nh s cc b VTr/TBP theo s lng c nh cc TBP

cho NSNl.

(ii) Chn mt b v tr theo th t ca s thp phn v chuyn s thp

phn ny qua s nh phn tng ng, s nh phn ny xc nh trng

thi (c/khng) TBP ti mi v tr lp phn on.

(iii) Thc hin phn tch v nh gi tng thit hi mt in ICOSTkl

cho b v tr Lkl v so snh ICOSTkl vi mt thit hi nh nht tnh

MICOST (gi tr u tin ca MICOST l thit hi mt in khi li

in khng lp TBP no).

(iv) Thay MICOST bng ICOSTkl nu ICOSTkl nh hn MICOST.

(v) Lp li bc (ii)-(iv) cho n khi tt c cc b v tr u c nh

gi v so snh, cui cng b v tr vi tng thit hi nh nht s tm

c.

- 44 -

Vi s lng l TBP c nh, khi hon i cc v tr lp t chng trn

li nh gi thit hi mt in, phn gi tr vn u t khng thay i

theo tng v tr .

3.2.1.2. Phng php tm kim li gii :

Cho mt s c nh TBP, b v tr ti u TBP c xc nh nh k

thut nh s, k thut nh s nu trn.

Phng php tm kim li gii c bt u vi 01 TBP. B v tr ti

u tng ng ca chng xc nh c bng k thut nh s. Tip tc vi 02

TBP v cng tm b v tr ti u ca chng bng cng k thut nh s. So

snh tng thit hi (bao gm thit hi mt in ca phng n v tng vn

u t quy i ca b TBP) trong h thng tnh ca phng n 01

TBP v 02 TBP chn phng n c tng thit hi b hn. Th tc tip

tc c thc hin cho 03 TBP, 04 TBP cho n khi s lng ti u cc

TBP v b v tr ca chng c tm ra.

3.2.2. S khi tm tt ton b chng trnh:

S khi tnh ton chnh ca chng trnh ch yu bao gm nh sau:

- 45 -

Bt uKhi D liu

ban u

Nhp s liu, thng s li

Xc nh cc b Vtr/TBPc th

-Tnh vn u t b TBPTnh cc thit hi do mtin ca b TBP

Khi chngtrnh chnh Khi tnh

hm thithi mt in

nh gi tm tng chi ph nh nht (bao gmtng vn u t v gi tr thit hi caphng n). Kt lun phng n ti u

Hnh 3.10 Kt qu

-Khi d liu ban u: c tp trung v cung cp y cc d liu cn

thit tnh ton, nm trn trong file d liu, bao gm:

+Thng s k thut li in tnh ton: thng s nt, nhnh

+Thng s phc v tnh ton thit hi mt in khch hng: sut s c,

cc thi gian thao tc, khi phc s c...

+Cc thng s phc v tnh ton v ti chnh ca hm mc tiu: li

sut, thi gian vay, vn u t, gi tr thit hi...

+Cc d liu v phn on c sn (nu c), cc s liu v Re..

-Khi chng trnh chnh, x l bng phng php tm kim li gii, trn

k thut nh s, khi ny bao gm cc ni dung:

+c d liu t khi d liu cung cp.

+Xy dng v thit lp cc phng n TBP trn li in phc v

cho khi tnh ton thit hi.

+So snh cc gi tr thit hi c tnh (t khi tnh ton thit hi) cho

tng phng n a ra c phng n c thit hi nh nht.

- 46 -

+Lu tr d liu phng n chn sau phn tch thit hi mt in.

+Tnh ton tng vn u t cho phng n chn.

+Tnh tng chi ph ca phng n: gm thit hi mt in v vn u

t.

+nh gi cui cng chn phng n c tng chi ph nh nht.

+Xut kt qu phng n ti u: s lng v v tr c th TBP trn

li in.

-Khi chng trnh tnh ton thit hi ca mt phng n: do khi chng

trnh chnh gi thc hin sau khi khi chng trnh chnh xut mt

phng n lp t vi s lng TBP v v tr lp t c th xc nh trn

li in, bao gm:

+Tnh ton cc thit hi mt in gy ra do im s c n cc vng

ln cn trong cc thi gian s c v thao tc, thi gian phi hp cc

Re..

+Tnh ton cc thit hi trn c xut tuyn.

+Tng hp thnh gi tr tng thit hi mt in ca phng n.

+Chuyn kt qu cho khi chng trnh chnh x l.

3.2.3. Cc ni dung chng trnh gii quyt:

Mt chng trnh tnh ton bao gm y cc khi ch yu nu trn,

bao gm mt s cc chng trnh m nhim ton b cc tnh ton cc trng

hp nh sau:

a) Trng hp li in hnh tia (Ngun cp 1 u):

-Tnh cho li in lp mi ton b TBP:

+ Ch lp DCL : Tm ti u s lng v v tr.

+ Lp Re. & DCL: Tm ti u s lng v v tr.

+ Lp mt s lng DCL, Re. c nh: Tm ti u v tr.

-Tnh cho li in hin lp sn mt s TBP nh DCL, Re.:

- 47 -

+Lp thm mt s DCL: tm ti u s lng, v tr.

+Lp thm mt s Re. & DCL: tm ti u s lng, v tr.

+ Lp mt s lng DCL, Re. c nh: Tm ti u v tr.

b) Trng hp li in vng vn hnh h (Ngun cp 2 u):

-Tnh cho li in lp mi ton b TBP:

+ Ch lp DCL : Tm ti u s lng v v tr.

+ Lp Re. & DCL: Tm ti u s lng v v tr.

+ Lp mt s lng DCL, Re. c nh: Tm ti u v tr.

-Tnh cho li in hin lp sn mt s TBP nh DCL, Re.:

+Lp thm mt s DCL: tm ti u s lng, v tr.

+Lp thm mt s Re. & DCL: tm ti u s lng, v tr.

+Lp mt s lng DCL, Re. c nh: Tm ti u v tr.

3.2.4. Thut ton:

3.2.4.1. Thut ton chng trnh xc nh cc phn on l DCL trong

li in phn phi hnh tia:

-D liu nhp:

+Cu trc li in, bao gm nt, nhnh, chiu di, cng sut trung

bnh ti...

+Cc thng s v gi in, sut s c, cc thng s v li sut....

-Chng trnh chnh:

+Tnh thit hi mt in khi khng lp DCL (l=0) MICOSTK.

+Ln lt lp l=1 DCL, thit lp b v tr lp DCL tng ng Si-l mt

s nh phn N bt. Xc nh s nh phn ln nht v nh nht c th c i vi

l DCL lp trn N v tr. Chuyn 2 s nh phn ny qua s thp phn amin,

aMax tng ng to nn gii hn cc phng n c th c ca cc b Si.

- 48 -

+Xc nh mt s thp phn trong [amin,aMax] theo th t chuyn

qua s nh phn tng ng. Kim tra s nh phn ny ph hp s DCL lp t

l (s bt c gi tr 1 trong s nh phn Si ng bng s DCL lp l).

+nh gi thit hi mt in ca b phn on/vtr Si tm c bng

khi chng trnh Tnh thit hi mt in c thut ton tng ng (trnh by

sau). C c thit hi ICOST ca phng n.

+nh gi ICOST vi cc MICOST, rt ra gi tr nh nht, lu li trong

ma trn MICOST (l hng, 2 ct), (Ct 1 cha tng vn u t DCL, ct 2

cha gi tr thit hi nh nht).

+Mt gi tr l (mt s lng DCL lp t nht nh) c mt di cc

phng n t amin->aMax. Phng n ti u v tr c lu trong MICOST.

+Tng dn s lng DCL n N, thc hin tng t trn, ta c ma trn

MICOST cha y cc gi tr thit hi nh nht ca tng l.

+nh gi gi tr nh nht ca tng cc s hng trong ct 1 v 2 tng

ng tng hng ca MICOST ta c c b v tr, s lng ti u cc DCL.

-Thut ton tnh thit hi mt in:

+D liu u vo: b v tr chn c Si.

+Cho im s c i chy t nhnh 1 n nhnh N.

+Xy dng ma trn m xc nh xem t im s c i v ngun c

DCL khng. Ma trn ny c cc c im sau:

Mt hng, c s s hng bng s nhnh ng dy (cng s

hng vi Si). Gi tr ban u cc s hng bng nhau v bng N.

Cho gi tr j chy ngc t i v ngun, nhnh no c DCL (Si=1)

ti , s hng tng ng ca m gn gi tr (i-j).

+Nh vy, khi j chy t nhnh i v n nhnh 1, tin hnh nh gi ma

trn m, nu cc s hng ca m t j = i n j = 1 vn khng i (bng N) th

khng c TBP trc i. Nu c cc gi tr khc N, xc nh mt gi tr nh

- 49 -

nht (l duy nht) v v tr ca gi tr nh nht trong m, v tr chnh l v

tr DCL gn i nht. T tnh c cc thit hi mt in ng vi tng on.

+Thc hin tnh ton gi tr thit hi ca xut tuyn tng ng vi tng

im s c i.

+Tng hp li c thit hi mt in cn tnh, tr kt qu v chng

trnh chnh nh gi.

- 50 -

Dng chngtrnh

S

S

S

S

Kt qu bTBP ti u

MICOST(l,2)= MICOSTK

1=l+1

TmMin(MICOST(l,1)+MICOST(l,2)),suy ra l ti u

i aMax

i = i+1

Kim tra s TBPng bng l

Tnh ICOST(S)

ICOST

- 51 -

S(i)-B P/VTs1, l

i1=1

m = s1*ones(1, i1)Lp ma trn m c s hng c gi tr s1

j1=1

SS(1,j1)=1

m(1,j1) = i1-j1

j1=j1+1

Sj1 > i1

S

A = s1

i1=i1+1

i1 > s1min(m)=AS

W1(1,i1)

W2(1,i1) = 0

W1(1,i1)W2(1,i1)

[A, n] = min(m)

Kt quW = W1+W2

Hnh 3.12: Thut ton tnh thit hi khi ch c DCL

- 52 -

3.2.4.2. Thut ton chng trnh xc nh cc phn on l Re. v DCL

trong li in hnh tia:

-D liu nhp: Ging thut ton ch xc nh DCL, c thm thng tin:

+S Re. s lp t (R).

+n gi b Re..

-Chng trnh chnh:

+Tnh thit hi mt in khi khng lp TBP (l=0) MICOSTK.

+Cho s lng cc Re. s lp t bt u t Sr=1, thit lp ma trn SR

m t cc b v tr lp Re., tp hp cc s hng thnh mt s nh phn N bt,

trong c Sr bt c gi tr l 1 th hin v tr lp Re..

+Cng xc nh s nh phn nh nht v ln nht tng t nh DCL,

sau chuyn qua s thp phn tng ng aminR v aMaxR . Mi s thp

phn trong [aminR, aMaxR] chuyn qua s nh phn, xc nh c mt phng

n lp Re. (sau khi kim tra s bt ng bng R). Vi mi phng n lp

Re. ny, ta ln lt xt n lp s dao cach ly t 1 n (N-R) DCL.

+Ln lt lp l=1 DCL, thit lp b v tr lp DCL Si-l mt s nh

phn N bt. Xc nh s nh phn ln nht v nh nht c th c i vi l DCL

lp trn N v tr. Chuyn 2 s nh phn ny qua s thp phn amin, aMax

tng ng to nn gii hn cc phng n c th c ca cc b Si DCL.

+Xc nh mt s thp phn trong [amin,aMax] theo th t chuyn

qua s nh phn tng ng. Kim tra s nh phn ny ph hp s DCL lp t

l (s bt c gi tr 1 trong s nh phn Si ng bng s DCL lp l) ng thi

kim tra loi b cc b DCL c v tr trng vi R Re. ca phng n Re. ang

xt .

+nh gi thit hi mt in ca b phn on/vtr Si v SR tm c

bng khi chng trnh Tnh thit hi mt in c thut ton tng ng (trnh

by sau). C c thit hi ICOST ca phng n.

- 53 -

+nh gi ICOST vi cc MICOST, rt ra gi tr nh nht, lu li trong

ma trn MICOST (l hng, 2 ct), (Ct 1 cha tng vn u t DCL, ct 2

cha gi tr thit hi nh nht).

+Mt gi tr l (mt s lng DCL lp t nht nh) c mt di cc

phng n t amin->aMax. Phng n ti u v tr c lu trong MICOST.

+Tng dn s lng DCL n N, thc hin tng t trn, ta c ma trn

MICOST cha y cc gi tr thit hi nh nht ca tng l.

+Vi mi phng n Re., c l (l=1,(N-R)) phng n lp DCL, chn ra

c mt phng n c thit hi mt in nh nht (bng cch xt

min(MICOST)) gi tr phng n ny (bao gm thit hi v vn u t c Re.

v DCL) c lu vo ma trn MMICOST.

+nh gi gi tr nh nht ca tng cc s hng trong ct 1 v 2 tng

ng tng hng ca MMICOST ta c c b v tr, s lng ti u cc b Re.

v DCL lp t.

-Thut ton tnh thit hi mt in:

+D liu u vo: b v tr chn c SR ca Re. v Si ca DCL.

+Cho im s c i chy t nhnh 1 n nhnh N.

+Xy dng 2 ma trn mR xc nh v tr Re. t im s c v ngun

v m xc nh xem t im s c i v ngun c DCL khng. 2 Ma trn ny

c cc c im sau:

Mt hng, c s s hng bng s nhnh ng dy (cng s

hng vi SR, Si). Gi tr ban u cc s hng bng nhau v bng

N. Cho gi tr j chy ngc t i v ngun, nhnh no c Re.

(SR=1) ti , s hng tng ng ca mR gn bng (i-j) v

nhnh no c DCL (Si=1) ti , s hng tng ng ca m gn

gi tr (i-j).

- 54 -

+Nh vy, khi j chy t nhnh i v n nhnh 1, tin hnh nh gi 2

ma trn mR, m, nu cc s hng ca mR, m t j = i n j = 1 vn khng i

(bng N) th khng c TBP trc i.

+Nu c cc gi tr khc N, xc nh mt gi tr nh nht (l duy nht)

v v tr ca gi tr nh nht trong mR, m, v tr chnh l v tr Re., DCL

gn i nht, vic xc nh Re. gn im s c hn hoc DCL gn hn thng

qua so snh v tr ca 2 gi tr nh nht ca mR v m. T tnh c cc

thit hi mt in ng vi tng on.

+Thc hin tnh ton gi tr thit hi ca xut tuyn tng ng vi tng

im s c i.

+Tng hp li c thit hi mt in cn tnh, tr kt qu v chng

trnh chnh nh gi.

- 55 -

S

Sr > R

Sr=Sr+1

Sr=1-S Re. lp dt

S

S

S

S

S

S

Xut kt qu

i1 > aMaxR

i1 = i1 + 1

MMICOST(i1,2) = A

A = min(MICOST(:,2))

l > (s1-R)

l = l + 1

i2 > aMax

Sum(S)=l-Ktra ng s DCL

i 2 = i2 + 1

i 2 = amin

MICOST (l,2) = ICOST MICOST (l,1) = V

ICOST < MICOST

Tnh ton, nh gi ICOST ca cc b SMICOST (l,2) = MICOSTK

Kim tra S { SR

numbin-s nh phn s1 bt chuyn t i2Xc nh cc b VT/DCL S(1,s1)

Xc nh aMax; amin ca b DCL.

l = 1, X vn DCL (V)

Sum(SR)=R-Ktra ng s Re.

numbinR-s nh phn s1 bt chuyn t i1Xc nh cc b VT/Re. SR(1,s1)

i1 = aminR

Xc nh aMaxR; aminR ca s Re. Sr

Tnh thit hi khi li khng lp P-MICOSTKXc nh vn u t Re. VR

-S liu li s1 -S Re. (R), n gi

Hnh 3.13: Thut ton xc nh c Re. v Dao cach ly trong li hnh tia

- 56 -

m1 < m2 SS

S

S

S

S

V Chng trnh chnh W

[m1,n] = min(m)[m2,nR] = min(mR)

W1 (n ->s1)W2 (n->nR)

W = W1 + W2

W1-(nR->s1)W2 = 0

[m2,nR] = min(mR)W1-thit hi TscW2 = 0

m1 > m2 m1 = m2 = s1

min(m) = m1 min(mR) = m2

i > s1

i = i + 1

j = 0

j = i - 1

mR(1,j) = i - j m(1,j) = i - j

SR(1,j) = 1 S(1,j) = 1

j = i

i = 1 - im s c

Xy dng cc ma trntm v tr DCL v Re. gn im s c nht

m(1.s1)=mR(1,s1)=s1*ones(1,s1)

2 ma trn xc nh vtr DCL S(1, s1) v

Re. SR(1, s1)

Hnh 3.14: Thut ton tnh thit hi cho bi ton tm cc b Re. v DCL-Li hnh tia

- 57 -

3.2.4.3. Thut ton chng trnh xc nh cc phn on l Re. v DCL

trong li in hnh tia c sn mt s Re. v DCL:

-D liu nhp: Ging thut ton xc nh DCL, c thm thng tin:

+S DCL, Re. lp t.

+S Re. s lp t (R).

+n gi b Re..

-Chng trnh chnh:

Ging chng trnh xc nh Re. v dao cach ly trong li hnh tia mi.

C thm ni dung:

+Kim tra loi b cc b Re. c cng v tr cc Re. lp,

+Kim tra loi b cc b DCL c v tr trng vi cc Re. ca b

phng n Re. ang xt .

Vi lu l cc b phng n chn la tnh thit hi mt in u

c b sung cc Re. v DCL hin c.

-Thut ton tnh thit hi mt in: s dng chung vi chng trnh xc

nh Re. v DCL trong li hnh tia mi.

- 58 -

R

Sr > R

Sr=Sr+1

Sr=1-S Re. lp t

Thm cc v tr DCLhin c vo s

S

Thm cc v tr Re.hin c vo sr

sr{ (S or SR)

Xc nh cc b S-DCL hin c Xc nh cc b SR-Re. hin c

S

S

S

S

S

S

Xut kt qu

i1 > aMaxR

i1 = i1 + 1

MMICOST(i1,2) = A

A = min(MICOST(:,2))

l > (s1-R-nDCL)

l = l + 1

i2 > aMax

Sum(s)=l-Ktra ng s DCL

i 2 = i2 + 1

i 2 = amin

MICOST (l,2) = ICOST MICOST (l,1) = V

ICOST < MICOST

Tnh ton, nh gi ICOST ca cc b SMICOST (l,2) = MICOSTK

Kim tra s { (sr or S)

numbin-s nh phn s1 bt chuyn t i2Xc nh cc b VT/DCL s(1,s1)

Xc nh aMax; amin ca b DCL.

l = 1, X vn DCL (V)

Sum(sr)=R-Ktra ng s Re.

numbinR-s nh phn s1 bt chuyn t i1 Xc nh cc b VT/Re. sr(1,s1)

i1 = aminR

Xc nh aMaxR; aminR ca b Re.

Tnh thit hi khi li khnglp P-MICOSTKXc nh vn u t Re. VR

-S liu li s1; S Re.(R) cn lp mi , n gi; B phn on hinc DCL: nDCL & Re. :

Hnh 3.15: Thut ton k thut tm trong li c sn Re.& DCL li hnh tia

- 59 -

3.2.4.4. Thut ton chng trnh xc nh cc phn on l DCL trong

li in kn vn hnh h:

-D liu nhp: ging trong li hnh tia.

-Chng trnh chnh: ging trong li hnh tia.

-Thut ton tnh thit hi mt in:

+D liu u vo: b v tr chn c Si ca DCL.

+Cho im s c i chy t nhnh 1 n nhnh N.

+Xy dng 2 ma trn m1, m2 xc nh v tr DCL t im s c v

2 ngun. 2 Ma trn ny c cc c im sau:

*Ma trn m1 xc nh c DCL gia im i v ngun 1, m2 xc nh c

DCL gia ngun 2 v i.

*Hai ma trn c mt hng, c s s hng bng s nhnh ng dy

(cng s hng vi Si). Gi tr ban u cc s hng bng nhau v bng N. Cho

gi tr j chy ngc t i v ngun 1, nhnh no c DCL (Si=1) ti , s hng

tng ng ca m1 gn bng (i-j). Cho j chy t (i+1) v ngun 2, nu nhnh

no c DCL (Si=1) ti , s hng tng ng ca m2 gn gi tr (j-i).

+Nh vy, tng 2 trng hp trn, khi j chy t nhnh s1 v n nhnh

1, tin hnh nh gi 2 ma trn m1, m2, nu cc s hng ca chng vn khng

i (bng N) th khng c TBP trc v sau i.

+Nu c cc gi tr khc N, xc nh mt gi tr nh nht (l duy nht)

v v tr ca gi tr nh nht trong m1, m2, v tr chnh l v tr DCL gn

i nht tng ng v tng ngun, khi xc nh c v tr ca cc DCL ny,

do vy tnh c cc thit hi mt in ng vi tng on.

+Thc hin tnh ton gi tr thit hi ca xut tuyn tng ng vi tng

im s c i.

+Tng hp li c thit hi mt in cn tnh, tr kt qu v chng

trnh chnh nh gi.

- 60 -

V chng trnhchnh

W1(1,i1):n1->(n2-1)W2(1,i1): 1->(n1-1)&

n2->s1

W1(1,i1):n1->s1W2(1,i1): 1->(n1-1)

W1(1,i1):1->(n2-1)W2(1,i1): n2->s1

[A1, n1] =min(m1)

[A2, n2] =min(m2)

[A1, n1] =min(m1)

[A2, n2] =min(m2)

A1s1; A2=s1 A1s1; A2s1S S

A1= s1; A2s1

S

j1 < i1+1

j1=j1-1

m2(1,j1) = j1-i1

S(1,j2)=1

j2=s1

S

Kt quW = W1+W2

i1 > s1

i1=i1+1

A1=A2 = s1

W1(1,i1)W2(1,i1) = 0

min(m1)=A1min(m2)=A2

S

S

j1 > i1

j1=j1+1

m1(1,j1) = i1-j1

S(1,j1)=1

j1=1

S

m1 = s1*ones(1, s1)m 2= s1*ones(1, s1)

i1=1

S(i)-B P/VTs1, l

Hnh 3.16: Thut ton tnh thit hi khi xc nh DCL trong mch vng

- 61 -

3.2.4.5. Thut ton chng trnh xc nh cc phn on l Re., DCL trong

li in kn vn hnh h:

-D liu nhp: ging trong li hnh tia.

-Chng trnh chnh: ging trong li hnh tia.

-Thut ton tnh thit hi mt in:

+D liu u vo: b v tr chn c Si ca DCL tng ng theo b

SR ca Re..

+Cho im s c i chy t nhnh 1 n nhnh N.

+Xy dng 2 ma trn m1, m2 xc nh v tr TBP t im s c v

2 ngun. 2 Ma trn ny c cc c im sau:

*Ma trn m1 xc nh c DCL gia im i v ngun 1, m2 xc

nh c DCL gia ngun 2 v i.

*Hai ma trn c mt hng, c s s hng bng s nhnh ng

dy (cng s hng vi Si, SR). Gi tr ban u cc s hng bng

nhau v bng N. Cho gi tr j chy ngc t i v ngun 1, nhnh

no c TBP (SR=1 hoc Si=1) ti , s hng tng ng ca

m1 gn bng (i-j). Cho j chy t (i+1) v ngun 2, nu nhnh

no c TBP (SR=1 hoc Si=1) ti , s hng tng ng ca

m2 gn gi tr (j-i).

+Nh vy, tng 2 trng hp trn, khi j chy t nhnh s1 v n nhnh

1, tin hnh nh gi 2 ma trn m1, m2, tm gi tr min cc ma trn (gi

s tng ng l n v m), s c 4 trng hp xy ra nh sau:

Trng hp 1: Nu cc s hng ca chng vn khng i (n=m= N) th

khng c TBP trc v sau i. Tnh c thit hi mt in.

Trng hp 2: n = N & m N - t i v ngun 1 khng c TBP, t i

v ngun 2 c TBP. Khi , c cc trng hp nh sau:

-SR(m) = 1: C Re. 2 gn i nht. Tnh c thit hi.

- 62 -

-S(m) = 1: C DCL 2 gn i nht. Lc ny, cn tm tip sau DCL 2 c

Re. 2 khng? Xt ma trn vt(1,N) c cc s hng l N, cho j chy t N

v (m+1), nu SR(j) = 1, gn s hng tng ng vt(j) = (j-m). nh gi

tm gi tr min(vt)=A. Nu A = N, khng c Re. 2 sau cch ly 2. Tnh

c thit hi mt in. Nu khng, xc nh v tr ca A, chnh l v

tr Re. 2, t tnh c thit hi mt in.

Trng hp 3: n N & m = N: t i v ngun 1 c TBP, t i v

ngun 2 khng c TBP. C cc trng hp nh sau:

-SR(n) = 1: C Re. 1 gn i nht. T tnh c thit hi mt in.

-S(n) = 1: C cch ly 1 gn i nht. Lc ny, cn tm tip trc DCL 1

c Re. 1 khng? Xt ma trn vt(1,N) c cc s hng l N, cho j chy t

n v 1, nu SR(j) = 1, gn s hng tng ng vt(j) = (n-j). nh gi tm

gi tr min(vt)=A. Nu A = N, khng c Re. 1 trc cch ly 1. Tnh

c thit hi mt in. Nu khng, xc nh v tr ca A, chnh l v

tr Re. 1, t cng tnh c thit hi mt in.

Trng hp 4: n N & m N: t i v ngun 1 c TBP, t i v

ngun 2 cng c TBP. C cc trng hp nh sau:

-SR(n) = 1: C Re. 1 gn i nht v pha ngun 1. Khi :

+SR(m)=1: c Re. 2 gn i nht v pha ngun 2, t tnh c

thit hi mt in.

+S(m)=1: c cch ly 2 gn i nht v pha ngun 2, cn tm tip

sau DCL 2 c Re. 2 khng? Xt ma trn vt(1,N) c cc s hng l N,

cho j chy t N v (m+1), nu SR(j) = 1, gn s hng tng ng vt(j) =

(j-m). nh gi tm gi tr min(vt)=A. Nu A = N, khng c Re. 2 sau

cch ly 2. Tnh c thit hi mt in. Nu khng, xc nh v tr ca

A, chnh l v tr Re. 2, t tnh c thit hi mt in.

- 63 -

-S(n)=1: C cch ly 1 gn i nht v pha ngun 1. Lc ny, cn tm tip

trc DCL 1 c Re. 1 khng? Xt ma trn vt1(1,N) c cc s hng l

N, cho j chy t n v 1, nu SR(j) = 1, gn s hng tng ng vt1(j) =

(n-j). nh gi tm gi tr min(vt1) = A. Nu A = N, khng c Re. 1

trc cch ly 1. Khi :

+SR(m)=1: c Re. 2 gn i nht v pha ngun 2, t tnh c

thit hi mt in.

+S(m)=1: c cch ly 2 gn i nht v pha ngun 2, cn tm tip

sau DCL 2 c Re. 2 khng? Xt ma trn vt2(1,N) c cc s hng l N,

cho j chy t N v (m+1), nu SR(j) = 1, gn s hng tng ng vt2(j)

= (j-m). nh gi tm gi tr min(vt2) = C. Nu C = N, khng c Re. 2

sau cch ly 2. Tnh c thit hi mt in. Nu khng, xc nh v tr

ca C, chnh l v tr Re. 2, t tnh c thit hi mt in.

Ngc li, A N- c Re. 1 trc cch ly 1, xc nh v tr ca A (l

nR1), khi :

+SR(m)=1: c Re. 2 gn i nht v pha ngun 2, t tnh c

thit hi mt in.

+S(m)=1: c cch ly 2 gn i nht v pha ngun 2, cn tm tip

sau DCL 2 c Re. 2 khng? Xt ma trn vt3(1,N) c cc s hng l N,

cho j chy t N v (m+1), nu SR(j) = 1, gn s hng tng ng vt3(j)

= (j-m). nh gi tm gi tr min(vt3) = B. Nu B = N, khng c Re. 2

sau cch ly 2. Tnh c thit hi mt in. Nu khng, xc nh v tr

ca B, chnh l v tr Re. 2, t tnh c thit hi mt in.

+Thc hin tnh ton gi tr thit hi ca xut tuyn tng ng vi tng

im s c i.

+Tng hp li c thit hi mt in cn tnh, tr kt qu v chng

trnh chnh nh gi.

- 64 -

Kt qu Tr.hp 1

Kt qu Tr.hp 4

Kt qu Tr.hp 3

Kt qu Tr.hp 2

Xt c th tngtrng hp 2,3,4

V Chng trnh chnhW

Trng hp 1 Trng hp 4 Trng hp 3Trng hp 2

SS

Sn s1 & m s1

S i > s1

i = i + 1

j2 < i+1 j1 > i

j2 = j2 - 1

s(1,j2) or sr(1,j2) = 1

j2 = s1

SS

S

S

W = W1 + W2 + W3

W1-thit hi TscW2 = 0;W3 = 0

n s1 & m = s1n = s1 & m s1n = m = s1

min(m1) = n min(m2) = m

j1 = j1 + 1 m2(1,j2) = j2 - i m1(1,j1) = i j1

s(1,j1) or sr(1,j1) = 1

j1 = 1

i = 1-im s c

Xy dng cc ma trntm v tr TBP gn im s c nht

v pha 2 ngunm1(1.s1)=m2(1,s1)=s1*ones(1,s1)

2 ma trn xc nh vtr DCL s(1, s1) v

Re. sr(1, s1)

Hnh 3.17: Thut ton tnh thit hi vi Re.& DCL trong mch vng

- 65 -

- 66 -

Hnh 3.18: on thut ton trng hp 2

S

S

Tr v (2)

W1: 1->(m-1)W2: m->(nR-1)

W3: nR->s1

[A, nR] = min(vt)

W1: 1->(m-1)W3 = 0

W2: m->s1

A = s1

A = min(vt)

i2 < m+1 i2 = i2 - 1

vt(1, i2) = i2 - m

sr (1, i2 ) = 1

i2 = s1

vt = s1*ones(1,s1)ma trn X sau DCL 2

c Re.2 ?

W1: 1->(m-1)W2 = 0

W3: m->s1

Trng hp 2

Ssr(1,m) 1 & s (1,m) = 1

S

sr(1,m) = 1 & s (1,m) 1

n = s1 & m s1

- 67 -

A = s1

Hnh 3.19: on thut ton trng hp 3

S

S

Tr v (3)

W1: n->s1 W2: nR->(n-1)

W3= 0

[A, nR] = min(vt)

W1: n ->s1 W3 = 0

W2: 1->(n-1)

A = min(vt)

i2 > n-1 i2 = i2 + 1

vt(1, i2) = n- i2

sr (1, i2 ) = 1

i2 = 1

vt = s1*ones(1,s1)ma trn X trc DCL 1

c Re.1 ?

W1: n->s1 W2 = 0 W3 = 0

Trng hp 3

Ssr(1, n) 1 & s (1, n) = 1

S

sr(1, n) = 1 & s(1, n) 1

n s1 & m = s1

- 68 -

S

S

S

S

W1: n->(m-1)W2: m->(nR3-1)

1->(n-1)W3:nR3 ->s1

W1: n ->(m-1)W3 = 0

W2: 1->(n-1)m->s1

[C, nR3] = min(vt)

S

C = s1

C = min(vt2)

i2 > n-1

i2 = i2 + 1

vt2(1, i2) = n- i2

sr (1, i2 ) = 1

i2 = 1

vt2 = s1*ones(1,s1)ma trn X sau DCL 2

c Re.2 ?

S

W1: n->(m-1)W2: 1->(n-1)W3: m -> s1

Xt tip trng hpring ca Tr.hp 4

s(1, m) = 1

S

sr(1, m) = 1

W1: n->(m-1)W2: m->(nR-1)

W3:nR ->s1

[A, nR] = min(vt)

W1: n ->(m-1)W3 = 0

W2: m->s1

A = s1

A = min(vt)

i2 < m+1

S

i2 = i2 - 1

vt(1, i2) = n- i2

sr (1, i2 ) = 1

i2 = s1

vt = s1*ones(1,s1)ma trn X sau DCL 2

c Re.2 ?

s(1, m) = 1 sr(1, m) = 1

Hnh 3.20: on thut ton trng hp 4

S

S

Tr v (4)

A = s1

A = min(vt1)

i2 > n-1

i2 = i2 + 1

vt1(1, i2) = n- i2

sr (1, i2 ) = 1

i2 = 1

vt1 = s1*ones(1,s1) ma trn X trc DCL 1

c Re.1 ?

W1: n->s1 W2 = 0 W3 = 0

Trng hp 4

Ss(1, n) = 1

S

sr(1, n) = 1

n s1 & m s1

- 69 -

- 70 -

[A, nR1] = min(vt1)

S

Hnh 3.21: on thut ton trng hp ring ca trng hp 4

S

S

Tr v (4)

W1: n->(m-1)W2: nR1->(n-1)

m->(nR2-1)W3: nR2->s1

[B, nR2] = min(vt3)

W1: n ->(m-1)W3 = 0

W2: nR1->(n-1)m->s1

B = s1

B = min(vt3)

i2 < m+1 i2 = i2 - 1

vt3(1, i2) = n- i2

sr (1, i2 ) = 1

i2 = s1

vt3 = s1*ones(1,s1)ma trn X sau DCL 2

c Re.2 ?

W1: n->(m-1)W2 : nR->(n-1)

W3: m->s1

Ss (1, m) = 1

S

sr(1, m) = 1

A = s1

- 71 -

3.2.4.6. Thut ton chng trnh xc nh cc phn on l Re. v DCL

trong li in vng c sn mt s Re. v DCL:

-D liu nhp: Ging thut ton xc nh Re., DCL trong li hnh tia c sn

cc phn on Re., DCL.

-Chng trnh chnh:

Ging chng trnh xc nh Re. v dao cach ly trong li mch vng

-Thut ton tnh thit hi mt in: s dng chung vi chng trnh xc

nh Re. v DCL trong li mch vng.

3.3. CHN CNG C LP TRNH:

Matlab (Matrix Laboratory) l mt ngn ng lp trnh cp cao phc v

cc tnh ton k thut ca hng Mathworks. Matlab tch hp cng c tnh

ton, m phng, lp trnh trong mt mi trng d dng, , cc vn ,

gii php dc biu din bng tp hp cc k hiu v biu thc ton hc.

Matlab l mt h thng c tnh tng tc cao, thnh phn d liu c bn ca

n l cc mng khng yu cu kch thc, iu ny cho php gii quyt rt

nhiu cc vn k thut, c bit i vi ma trn v h vc t.

Trong cc trng i hc, Matlab l cng c dy hc tiu chun

gii thiu v h tr cc bi ging v ton hc, cng ngh v khoa hc. Trong

cng nghip, Matlab c s dng nh mt cng c h tr cho nghin cu

nh gi, phn tch cc vn mang tnh k thut cao.

Vi mi trng lm vic v c php cu lnh n gin, hiu qu, kh

nng vn dng cho lp trnh mang tnh thc dng cao, d dng p dng trong

lp trnh tnh ton cc bi ton k thut in m thc t t ra.

Vi cc yu t trn, ti chn ngn ng Matlab (version 6.5) lm

cng c lp trnh gii quyt vn t ra.

- 72 -

3.4. CHNG TRNH TNH TON:

3.4.1. Cu trc li phn phi m t trn my tnh :

Cu trc ca li in c m t cho my tnh bng cc nhnh v cc nt.

-Nt c nh s t 1 n N. Nt ngun c nh s 0, s nh gn ngun

hn s ln.

-Nhnh l on li hay phn t ni gia hai nt k nhau, nhnh c nh

s trng vi nt cui ca n.

Cu trc ca li phn phi c nhn dng y nu cho bit nhnh v nt

u, nt cui ca mi nhnh.

Cp thng s nt u N(i) v nt cui NC(i) ca nhnh i c cho nh sau:

Trc ht nh s cc nt ca li in t nt ngun n nt ti cui

cng, nt ngun nh s 0, s nh gn ngun hn s ln. Li phn phi

hnh tia c s nt bng s nhnh v bng N.

Sau nh s nhnh theo quy tc s nhnh trng vi nt cui ca n.

Cch nh s ny cho php my tnh hiu bit d dng mi quan h gia cc

nt v nhnh. V d khi bit khi bit mt nhnh l i c nt u l N(i) v nt

cui NC(i) th ta bit ngay nhnh cp in cho nhnh ny l j = N(i), cn n

cp in cho nt k c nt u l N(k) = NC(i) = i.

gii tch tin cy mi nhnh v nt v ng dy, cn phi c cc

thng s s c m t c th trong phn gii thiu v file d liu.

- 73

-

Bng

3.1

3.

4.2.

S

dan

h mc

tn

cc

khi

chn

g tr

nh:

ST

TT

rn

g h

p x

t Fi

led

liu

File

ch

ng tr

nh

chn

h Fi

le t

nh th

it hi

mtin

IT

rn

g h

p liin

hn

h tia

(Ngun

cp

1

u):

I.1T

nhch

oliin

lp

mi

ton

b

TBP

:

1

+ C

h l

p D

CL

: Tm

ti

u s

ln

g v

v t

r.

Tinh

moi

.mda

nhgi

a1.m

2

+ L

p R

e. &

DC

L: T

m t

iu

s ln

g v

v t

r.da

ta1.

mTi

nmoi

RE.

mD

anhg

iaR

E.m

I.2T

nhch

oliin

hi

n

lp

sn

mt

s T

BP

nh

DC

L, R

e.:

1

+Lp

thm

mt

s D

CL:

tm

ti

u s

ln

g, v t

r.Ti

nhco

san.

mda

nhgi

a1.m

2

+Lp

thm

mt

s R

e. &

DC

L: t

m t

iu

s ln

g, v t

r.da

ta1.

mTi

nhm

oiR

Ecos

an.m

Dan

hgia

RE.

m

IIT

rn

g h

p liin

vn

gv

n h

nh h

(Ngun

cp

2

u):

II.1

Tnh

cho

liin

lp

mi

ton

b

TBP

:

1+

Ch

lp

DC

L : T

m t

iu

s ln

g v

v tr

. Ti

nhm

oi_v

ong.

mD

anhg

ia_p

dmoi

_von

g.m

2+

Lp

Re.

& D

CL:

Tm

ti

u s

ln

g v

v t

r.da

ta1.

mTi

nhm

oiR

E_vo

ng.m

Dan

hgia

RE_

vong

.m

II.2

Tnh

cho

liin

hi

n

lp

sn

mt

s T

BP

nh

DC

L, R

e.:

1+Lp

thm

mt

s D

CL:

tm

ti

u s

ln

g, v t

r.Ti

nhco

san_

vong

.mD

anhg

ia_p

dmoi

_von

g.m

2+Lp

thm

mt

s R

e. &

DC

L: t

m t

iu

s ln

g, v t

r.

data

1.m

Tinh

moi

RE_

vong

_cos

an.m

Dan

hgia

RE_

vong

.m

- 74 -

3.4.3. Cu trc chnh cc file chng trnh:

3.4.3.1. File d liu:

-Tn file: data1.m

-Dng u tin: Tn file.

-Dng th 4: Dng trong trng hp tm v tr lp t ti u cho mt s c

nh cho trc cc DCL L.

-Phn cc thng s tnh ton: ma trn ths(2,20).

+th(1,1): gi in bnh qun (/KWh), l gi kinh doanh in bnh

qun ca in lc, hoc Cng ty in lc. Gi bn in bnh qun c tng

hp hng nm t doanh thu bn in (tnh trung bnh t tng doanh thu trn

sn lng ca tng loi ph ti tng mc gi: ASSH, dch v, cng

nghip, nng nghip...).

+ths(1,2): Thi gian phi hp Re. (giy), khi xut tuyn c lp t

t 2 Re., vic chnh nh thi gian phi hp c tnh n tng hiu qu

trong ng ct, phn on, tng tin cy ca h thng, thi gian phi hp

ny hin ang th nghim ti mt s in lc, ly t 55 90 s.

+ths(1,3): Thi gian thao tc s c (gi), c tnh t khi xy ra s c

(my ct, Re. nhy) cho n khi c lp c phn on b s c v ng

my ct tr li tip tc cung cp in cho cc phn on pha trc.

+ths(1,4): n gi b DCL (triu ng), gi tin 01 b DCL trung p

3 pha bao gm h thng truyn ng c kh. Nu c t DTU, trong n gi

cng c bao gm.

+ths(1,5): i sng b TBP (nm) , thi gian s dng, khai tc hiu

qu TBP.

+ths(1,6): Li sut vay (%), li sut vn vay ca d n u t lp t

TBP.

- 75 -

+ths(1,7): Sut s c vnh cu ng dy trung p (v/100km.nm).

+ths(1,8): Thi gian vay vn (nm), thi gian hon tr vn u t (li,

gc) theo hp ng vay vn vi ngn hng.

+ths(1,9): Gi n b thit hi do s c mt in (khng thng bo)

(/KWh), n gi gi tr thit hi khi ph ti mt in do s c, tham kho

cc nc.

+ths(1,10): Gi n b mt in c k hoch (c thng bo trc cho

khch hng)(/KWh), tham kho cc nc.

+ths(2,1): Sut s c thong qua ng dy trung p (v/100km.nm).

+ths(2,2): thi gian s c thong qua (pht), ly theo quy nh phn

loi s c thong qua ca EVN.

+Cc s hng t ths(2,3)-ths(2,10): d phng khi m rng cc tnh

ton.

-Phn d liu nhnh: ma trn ldt(N,7), N s nhnh ng dy xt.

+ldt(:,1): s th t nhnh, phc v vic in n tn nt.

+ldt(:,2): Nt u nhnh.

+ldt(:,3): Nt cui nhnh.

+ldt(:,4): Chiu di (km) ca nhnh.

+ldt(:,5): thi gian khc phc s c (gi), tnh t khi nhy my ct

u ngun n khi khi phc cp in nhnh s c li theo s vn hnh

bnh thng.

+ldt(:,6): Bin TBP, s dng khi tnh ton li in c sn mt s

phn on. Nhp 0 khi trn nhnh khng c TBP, nhp 1 cho nhnh c lp

DCL, dao ct ti, 2 cho Re..

+ldt(:,7): Thi gian sa cha k hoch (gi). Trn c s k hoch sa

cha, bo dng nh k li in t u nm c duyt, tng hp thi

gian mt in d kin theo khi lng thc hin ca nhnh v nhp vo.

- 76 -

-Phn d liu nt: ma trn ndt(N,2), N s nt ng dy xt.

+ndt(:,1): s th t nt, phc v cc tnh ton trong chng trnh.

+ndt(:,2): cng sut (Kw), cng sut trung bnh tng cng ca ph ti

trn nhnh.

-Phn tn nt: ma trn tennut(N,1), N s nt ng dy. Nhp tn nt theo

thc t li in.

-Phn d liu Re.: ma trn re(1,2) cung cp thng s Re. lp mi.

+re(1,1): s Re. s u t lp mi trn xut tuyn tnh.

+re(1,2): vn u t 01 b Re. trung p (trn b) (triu n