DATN bảo vệ trạm biến áp 220/110/35 kV ĐHBK

Thit k h thng bo v rle cho TBA

NGUYN ANH TUN KT2 K54 1

LI M U

Ngy nay, nn kinh t nc ta ang pht trin mnh m, i sng nhn

dn cng c nng cao nhanh chng. Bn cnh , qu trnh cng nghip ha

hin i ha cng ko theo nhu cu v cung cp nng lng, c bit l in

nng. Nhu cu s dng in trong cc lnh vc cng nghip, nng nghip, dch

v v sinh hot tng trng khng ngng. iu i hi mt h thng cung cp

in an ton v ng tin cy.

Trm bin p l mt mt xch rt quan trng trong h thng in, l u

mi lin kt cc ng dy, cc h thng in. Cc thit b trong trm bin p c

gi thnh ln, thng t gp s c hn cc phn t khc ca h thng in, tuy

nhin nu xy ra s c th c th gy ra thit hi nng n nu khng c x l

kp thi. V l , vic thit k mt h thng bo v cho trm bin p hot ng

chnh xc, tin cy trc cc s c l mt cng on ht sc cn thit.

ti tt nghip em c giao c tn: Thit k h thng bo v r le cho

trm bin p 220 kV. Trong thi gian lm n, em nhn c s hng dn

nhit tnh ca cc thy c trong b mn H thng in v c bit l TS. Nguyn

Xun Tng. Do nhng hn ch v kin thc v kinh nghim thc tin nn n

c th cn nhng sai st, rt mong nhn c s ch bo, gp ca cc thy c

gio trong b mn.

Em xin chn thnh cm n!

H Ni, ngy 13 thng 06 nm 2014

Sinh vin thc hin

Nguyn Anh Tun



LI M U ......................................................................................................... 1

CHNG 1 ............................................................................................................. 5

GII THIU CHUNG V VAI TR CA TRM BIN P TRONG H

THNG IN V TRM BIN P C THIT K ..................................... 5

1.1. Vai tr ca trm bin p trong h thng in .........................................................5

1.2. Tng quan chung v trm bin p cn thit k .......................................................5

1.2.1. S ni in chnh ca trm bin p ........................................................................ 5

1.2.2. Cc thng s chnh ca trm ....................................................................................... 6

CHNG 2 ............................................................................................................. 8

TNH TON NGN MCH PHC V CHNH NH H THNG BO V

R-LE ..................................................................................................................... 8

2.1. Gii thiu chung .....................................................................................................8

2.1.1. L do cn thit ............................................................................................................. 8

2.1.2. Cc gi thit tnh ngn mch ....................................................................................... 8

2.2. Cc ch tnh ngn mch v cc im ngn mch cn tnh ton ........................8

2.2.1. Cc ch tnh ngn mch ......................................................................................... 8

2.2.2. Cc im ngn mch cn tnh ton .............................................................................. 9

2.3. Quy i cc thng s phn t ...............................................................................10

2.3.1. H thng .................................................................................................................... 11

2.3.2. ng dy 110 kV .................................................................................................... 11

2.3.3. My bin p t ngu .................................................................................................. 12

2.4. Tnh ton dng in ngn mch ...........................................................................13

2.4.1. Khi trm vn hnh 1 MBA trong ch max ........................................................... 13

2.4.2. Khi trm vn hnh 2 MBA trong ch max ........................................................... 26

2.4.3. Khi trm vn hnh 1 MBA trong ch min ............................................................ 42

2.4.4. Khi trm vn hnh 2 MBA trong ch min ............................................................ 42

CHNG 3 ........................................................................................................... 51

LA CHN PHNG THC BO V ............................................................ 51

3.1. Cc dng h hng i vi my bin p ................................................................51

3.2. S phng thc bo v my bin p ...............................................................52

CHNG 4 ........................................................................................................... 56

GII THIU TNH NNG V THNG S CC RLE S DNG ............... 56

4.1. R le bo v so lch P633 ....................................................................................56

4.1.1 Gii thiu tng quan v rle P633 ............................................................................ 56



4.1.2 Mt s thng s k thut ca rle P633 .................................................................... 57

4.2. R le hp b qu dng s 7SJ64 ..........................................................................63

4.2.1 Gii thiu tng quan v rle 7SJ64 ........................................................................... 63

4.2.2 Cc chc nng ca 7SJ64 .......................................................................................... 63

4.3. R le 7RW60 ........................................................................................................65

4.3.1. Gii thiu chung ........................................................................................................ 65

4.3.2. Mt s thng s k thut chnh ................................................................................. 66

CHNG 5.................................................................................................................67

TNH TON CC GI TR CHNH NH V CI T CHO RLE .................67

5.1. Cc chc nng s dng v thng s ci t cho my bin p ca r le P633 ......67

5.1.1. Chn my bin dng in v my bin in p ........................................................ 67

5.1.2. Cc thng s k thut ca my bin p t ngu trong trm ...................................... 69

5.2. Tnh ton chnh nh cc chc nng bo v trong r le P633 ..............................69

5.2.1. Chc nng bo v so lch c hm ............................................................................. 69

5.2.2. Bo v chng chm t hn ch (REF) ..................................................................... 71

5.2.3. Chc nng bo v qu ti nhit ................................................................................. 71

5.3. Nhng chc nng bo v dng r le 7SJ64 ..........................................................72

5.3.1. Bo v qu dng ct nhanh c hng ....................................................................... 72

5.3.2. Bo v qu dng ct nhanh th t khng c hng .................................................. 73

5.3.3. Bo v qu dng c hng thi gian tr ................................................................... 74

5.3.4. Bo v qu dng TTK c hng c thi gian tr ...................................................... 75

5.3.5. Bo v chng my ct t chi tc ng 50BF ........................................................... 75

5.4. Bo v qu in p th t khng pha 35 kV (59N, U0>) .....................................80

CHNG 6 ........................................................................................................... 81

KIM TRA S LM VIC CA CC BO V .............................................. 81

6.1. Bo v so lch c hm (87T) ................................................................................81

6.1.1. Kim tra s lm vic an ton khi c ngn mch ngoi .............................................. 81

6.1.2. Kim tra nhy ca r le khi c ngn mch trong vng bo v ............................. 84

6.2. Bo v qu dng c hng c thi gian tr ..........................................................86

6.3. Bo v qu dng TTK c hng c thi gian tr .................................................88

CHUYN ........................................................................................................ 90

S DNG MY TNH GIAO TIP, CHNH NH V PHN TCH ............. 90

BN GHI S C CA RLE ............................................................................ 90

1. Gii thiu chung ......................................................................................................90



2. Xy dng d liu chnh nh cc chc nng bo v ...............................................91

2.1. Gii thiu phn mm MiCom S1 Studio ...................................................................... 91

2.2. Quy trnh to b d liu chnh nh ............................................................................. 93

2.3. Kt qu chnh nh ....................................................................................................... 94

3. Giao tip vi r le truy xut d liu ........................................................................94

3.1. Gii thiu b m phng h thng in NE9171 ........................................................... 94

3.2. Quy trnh thc hin giao tip gia b m phng v r le ............................................ 94

4. Bn ghi s c v p dng thc t phn tch bn ghi s c ......................................98

4.1. Gii thiu chung v bn ghi s c ................................................................................ 99

4.2. p dng thc t phn tch bn ghi s c Nha Trang Krng Buk ............................ 100

TI LIU THAM KHO ................................................................................... 105

PH LC ............................................................................................................ 106



CHNG 1

GII THIU CHUNG V VAI TR CA TRM BIN P TRONG H

THNG IN V TRM BIN P C THIT K

1.1. Vai tr ca trm bin p trong h thng in

Trong h thng in, trm bin p c s dng rng ri phc v cho

vic truyn ti in nng t cc li in cp in p khc nhau trong qu trnh

truyn ti v phn phi nng lng in. Thng in nng nh my khi n cc

h tiu th in th phi qua 3 n 4 ln bin p, v vy cng sut tng ca cc

trm bin p cng ln hn tng cng sut ca cc my pht t 3 n 4 ln.

Trm bin p cn thit k bo v l trm bin p cp in p 220 kV ng

vai tr quan trng trong vic lin kt h thng in, truyn ti mt lng cng

sut rt ln, l mt nt quan trng trong h thng. V vy, mi s c xy ra

trm bin p 220 kV c th gy hu qu nghim trng n cc thit b trong

trm, gy thit hi kinh t ln do ngng cung cp in v thm ch nu khng

c thit k bo v chnh xc c th gy mt n nh h thng v r li.

1.2. Tng quan chung v trm bin p cn thit k

i tng cn thit k y l trm bin p 220 kV gm 2 my bin p t

ngu lm vic song song. Trm c lin kt vi h thng in pha 220 kV v

110 kV.

1.2.1. S ni in chnh ca trm bin p



1.2.2. Cc thng s chnh ca trm

H thng:

Thng s v h thng (gi thit) da trn Scb = 100 MVA; Ucb = Utb

Ti thanh ci h thng pha 220kV:

Ch max: X1 = 0.052 X0 = 0.023

Ch min: X1 = 0.061 X0 = 0.03

Ti thanh ci h thng pha 110kV :

Ch max: X1 = 0.15 X0 = 0.07

Ch min: X1 = 0.19 X0 = 0.13

My bin p:

My bin p t ngu 125/125/63 MVA T s bin p: 230 8 1,25%/121/36,5 kV S u dy: Y0 auto / -11 in p ngn mch ( qui i v theo cng sut nh

mc ca my bin p):

UNC-T

= 10,78% ; UNC-H

= 32,72% ; UNT-H

= 20,35%

Dng in nh mc cc pha: 314/596/997 A

HT2

BI 0 230 kV 121 kV

36,5 kV

BI 2

HT1

BI 3

BI 1

S ni in chnh trm bin p



ng dy 110 kV

Loi dy AC 240, chiu di 55km Thng s x0=2,9x1



CHNG 2

TNH TON NGN MCH PHC V CHNH NH H THNG BO

V R-LE

2.1. Gii thiu chung

2.1.1. L do cn thit

Ngn mch trong h thng in l hin tng cc pha chp nhau, pha

chp t (hay chp dy trung tnh), lc xy ra ngn mch tng tr ca h thng

gim i, dng in tng ln ng k gi l dng in ngn mch. Trong thit k

bo v rle, vic tnh ton ngn mch nhm xc nh cc tr s dng in ngn

mch ln nht (INmax) v dng in ngn mch b nht (INmin) i qua i tng

bo v la chn thit b bo v r le, ci t, chnh nh cc thng s v kim

tra nhy ca r le.

2.1.2. Cc gi thit tnh ngn mch

Tn s ca h thng khng thay i.

B qua bo ha t.

B qua cc lng nh trong thng s ca mt s phn t.

H thng sc in ng ba pha ca ngun l i xng.

2.2. Cc ch tnh ngn mch v cc im ngn mch cn tnh ton

2.2.1. Cc ch tnh ngn mch

tnh c cc tr s dng in ngn mch ln nht (INmax) v dng in

ngn mch b nht (INmin) i qua i tng bo v trc tin ta cn phn tch:

- c dng INmax th khi cng sut ca ngun l ln nht, v tr s ca

dng ngn mch 3 pha lun ln hn dng in ngn mch 2 pha nn tm INmax

ta s tnh cc dng ngn mch N(3), N(1), N(1,1) trong ch max. T nhn xt ny

ta c th thy tm INmin ta s tnh cc dng ngn mch N(2)

, N(1)

, N(1,1)

trong ch

min.

- i vi cu hnh ca trm:

Khi trm vn hnh mt my th dng in tng ti im ngn mch

tng ti im ngn mch s nh hn so vi trng hp 2 my vn hnh song

song. Tuy nhin ton b dng in ny s i qua bo v.



Khi trm vn hnh 2 my bin p song song th dng in ngn mch

tng ti im ngn mch s ln hn trng hp vn hnh 1 my bin p c lp

nhng dng qua bo v ch bng mt na dng in ngn mch tng.

T nhn xt trn ta thy tm gi tr dng dng in ngn mch ln nht

(INmax) v dng in ngn mch b nht (INmin) ta phi tnh cho c 2 trng hp:

vn hnh c lp 1 MBA v vn hnh song song 2 MBA.

i vi cc dng ngn mch khng i xng, dng in cc pha

khng ging nhau, dng in m ta tnh ra l ca pha c bit (pha khng b s

c i vi ngn mch 2 pha v 2 pha chm t, cn i vi s c ngn mch 1

pha th l pha s c), v vy khi tm dng chy qua cc BI ta phi ch n

vn ny.

2.2.2. Cc im ngn mch cn tnh ton

i vi trm bin p, v tr ca cc im ngn mch cn tnh ton ph

thuc vo v tr t BI, nu s c xy ra trc BI th s khng c dng in qua

BI nn bo v s khng cm nhn c s c v ngc li. Ngoi ra i vi bo

v so lch c khi nim vng bo v c gii hn bng cc BI, cho nn hai im

ngn mch c th c cng tr s dng in ngn mch nh N1 v N1 nhng bo

v so lch ch tc ng khi xy ra ngn mch N1.

i vi my bin p t ngu ta cn phi tnh ton c cho trn hnh v

di y. Tng cng c 6 im ngn mch cn tnh ton (gm 3 im ngn mch

trong vng {N1, N2

, N3

} v 3 im ngn mch ngoi vng {N1, N2, N3}).



S cc im cn tnh ton ngn mch

2.3. Quy i cc thng s phn t

Chn cc i lng c bn:

Scb = Sm BA= 125 MVA

Ucb = Utb cc cp

T hai thng s trn ta tnh c Icb cc cp in p:

Cp in p 220 kV c Ucb1= 230 kV

cbcb1

cb1

S 125I = = = 0,314

3U 3.230kA

Cp in p 110 kV c Ucb2= 115 kV

cbcb2

cb2

S 125I 0,628

3U 3.115

kA

Cp in p 35 kV c Ucb3 = 36,5 kV

cbcb3

cb3

S 125I 1,977

3U 3.36,5

kA

HT 1

36.5kV

230 kV 121kVBI1 BI2

BI3

BI0

N1

N1'

N2

N2'

AT1

AT2

N3

N3'

HT 2



2.3.1. H thng

Pha 220 kV:

in khng ca h thng pha 220kV cho (vi Scb=100MVA, Ucb =

Utb):

Ch max: X1 = 0,052; X0 = 0,023

Ch min : X1 = 0,061; X0 = 0,03

Qui i sang h n v tng i c bn chn :

o Ch max: X0HT1 = 0,023.125

100 = 0,029

X1HT1 = X2HT1 = 0,052. 125

100 = 0,065

o Ch min: X0HT1 = 0,03.125

100 = 0,038

X1HT1 = X2HT1 = 0,061. 125

100 = 0,076

Pha 110 kV:

in khng ca h thng pha 110kV cho (vi Scb=100MVA, Ucb =

Utb):

Ch max: X1 = 0,15; X0 = 0,07

Ch min : X1 = 0,19; X0 = 0,13

Qui i sang h n v tng i c bn chn :

o Ch max: X0H2 = 0,07.125

100 = 0,088

X1H2 = X2H2 = 0,15. 125

100 = 0,188

o Ch min: X0H2 = 0,13.125

100 = 0,163

X1H2 = X2H2 = 0,19. 125

100 = 0,238

2.3.2. ng dy 110 kV

ng dy 110 kV l loi dy AC 240, chiu di 55 km, thng s X0 =

2,9 X1. Trong ch min tnh dng ngn mch nh nht ta s tnh khi vn

hnh 1 ng dy.

Tra bng ta c thng s ca ng dy nh sau:

ro = 0,12 (/km), xo = 0,401 (/km), bo = 2,84 (10-6/km)



T ta tnh c:

cb1D 0 2 2

cb

S1 1 125X .x .l .0,401.55. 0,104

2 U 2 115

X0D = 2,9. X1D = 2,9 . 0,104 = 0,302

n gin cho vic tnh ton v trnh by, ta s gp in khng ca

ng dy vi in khng ca h thng pha thanh ci 110 kV (ch min s

tnh trng hp 2 ng dy vn hnh song song v trng hp mt ng dy

vn hnh c lp), ta c:

Ch max:

X1HD = X2HD = X1H2 + X1D = 0,188 + 0,104 = 0,292

X0HD = X0H2 + X0D = 0,088 + 0,302 = 0,39

Ch min:

- Trng hp 1 ng dy vn hnh c lp:

X1HD = X2HD = X1H2 + 2.X1D = 0,238 + 2.0,104 = 0,446

X0HD = X0H2 +2.X0D = 0,163 + 2.0,302 = 0,767

- Trng hp 2 ng dy vn hnh song song:

X1HD = X2HD = X1H2 + X1D = 0,238 + 0,104 = 0,342

X0HD = X0H2 + X0D = 0,163 + 0,302 = 0,465

2.3.3. My bin p t ngu

in p ngn mch UN% ca my bin p t ngu AT1, AT2 nh sau:

C-T C-H T-HN N N10,78%; 32,72%; 20,35%;U U U

T ta rt ra:

C C-T C-H T-H

N N N N

T T-H C-T C-H

N N N N

H C-H T-H C-T

N N N N

1 1U (U U U ) (10,78 32,72 20,35) 11,575%

2 2

1 1U (U U U ) (20,35 10,78 32,72) 0,795%

2 2

1 1U (U U U ) (32,72 20,35 10,78) 21,145%

2 2

in khng thay th ca my bin p t ngu:



N

N

C

cbC

dmBA

T

H

cbH

dmBA

U % S 11,575 125X . . 0,116

100 S 100 125

X 0

U % S 21,145 125X . . 0,211

100 S 100 125

2.4. Tnh ton dng in ngn mch

2.4.1. Khi trm vn hnh 1 MBA trong ch max

a. Ngn mch pha 220 kV: N1 v N1

Ta c cc s thay th th t thun, th t nghch v th t khng ti

im ngn mch nh sau:

S thay th th t thun

S thay th th t nghch

EHT1X1HT10,065

XC0,116

X1HD0,292

EHT X

0,056

EHT2N1 N1

' XT 0

BI1 BI2

X2HT10,065

XC0,116

X2HD0,292

X

0,056

N1 N1' XT

0

BI1 BI2



S thay th th t khng

Tnh ton cc in khng tng ng:

X1 = X2 = 1HT1 C 1HD

1HT1 C 1HD

X .(X X ) 0,065.(0,116 0,292)0,056

X X X 0,065 0,116 0,292

X0 =

0HD H0HT1 C

0HD H

0HD H0HT1 C

0HD H

X .X 0,39.0,211X .(X ) 0,029.(0,116 )

X X 0,39 0,2110,026

X .X 0,39.0,2110,029 0,116X X

0,39 0,211X X

Ngn mch 3 pha

Dng in ngn mch ti im ngn mch c tr s:

I1 = HT

1

117,836

0,056

E

X

Dng in ngn mch do h thng 1 cung cp:

INHT1 = I1.C 1HD

1HT1 C 1HD

X X

X X X

= 17,836.

0,116 0,29215,385

0,065 0,116 0,292


INHT2 = I1 - INHT1 = 17,836 -15,385 = 2,451

Xc nh tr s dng in ngn mch qua cc BI

T y ta tm c dng i qua cc BI trong 2 trng hp ngn mch N1

v N1:

o Ngn mch ti N1:

IBI1 = IBI2 = IHT2 = 2,451

X0HT10,029

XC0,116

X0HD0,39

X00,026

N1 N1' XT

0

BI1 BI2XH0,211



IBI3 = IBI0 = 0

o Ngn mch ti N1:

IBI1 = IHT1 = 15,386 ; IBI2 = IHT2 = 2,451

IBI3 = IBI0 = 0

Ngn mch 1 pha

Cc dng in thnh phn i xng ti im ngn mch ca pha s c l:

HT1 2 0

1 2 0

I I IE

X X X

17,238

0,056 0,056 0,026

Phn b thnh phn dng in th t thun v th t nghch:

I1N HT2 = I2N HT2 = 1HT1

1

1HT1 C 1HD

XI .

X X X

=

0,0657,238. 0,995

0,065 0,116 0,292

I1N HT1 = I2N HT1 = 1I - I1N HT2 = 7,238 0,995 = 6,244

Phn b thnh phn dng in th t khng:

Dng in th t khng chy qua pha cao MBA:

I0C = I00HT1

0HD H0HT1 C

0HD H

X

X .XX

X XX

=0,029

7,238. 0,7450,39.0,211

0,029 0,1160,39 0,211

I0N HT1 = I0 - I0C = 7,238 0,745 = 6,494

Dng TTK chy qua pha trung ca MBA (cng l dng chy qua in

khng h thng 2):

I0N HT2 =I0T = I0C .

H

0HD H

X

X X

0,2110,745. 0,261

0,39 0,211

Dng TTK chy qua dy trung tnh ca MBA (trong h n v c tn):

I0tt = 3.(I0T.Icb(110) - I0C.Icb(220)) = 3.(0,261.0,628 0,745.0,314) = 0,209(kA)

Xc nh tr dng in ngn mch chy qua BI

o Ngn mch ti N1:

- BI1:



If(BI1) = I1N HT2 + I2N HT2 + I0C = 0,995 + 0,995 + 0,745 = 2,735

I0(BI1) = I0C = 0,745

Loi b thnh phn th t khng ta c:

IBI1(-0) = If(BI1) - I0(BI1) = 2,735 0,745 = 1,99

- BI2:

If(BI2) = I1N HT2 + I2N HT2 + I0N HT2 = 0,995 + 0,995 + 0,261 = 2,251

I0(BI2) = 0,261


IBI1(-0) = If(BI1) - I0(BI1) = 2,251 0,261 = 1,99

- Dng qua BI3 bng 0

- Dng qua BI0 l 0,209 kA.

o Ngn mch ti N1:

- BI1:

If(BI1) = I1N HT1 + I2N HT1 + I0N HT1 = 6,244+ 6,244 + 6,494 = 18,982

I0(BI1) = I0C = 6,494


IBI1(-0) = If(BI1) - I0(BI1) = 18,982 6,494 = 12,488

- BI2: Tng t nh ngn mch ti N1

- Dng qua BI3 bng 0


Ngn mch 2 pha chm t

Cc gi tr dng in thnh phn i xng ti pha khng s c (gi s l

pha A):

- Thnh phn th t thun:

I1Na = 2 0

1

2 0

X .XX

X X

E

= 1

13,5430,056.0,026

0,0560,056 0,026

- Thnh phn th t nghch:



I2Na = - I1Na 0

2 0

X

X X

= -13,543

0,0264,293

0,056 0,026

- Thnh phn th t khng:

I0Na = - I1 2

2 0

X

X X

= -13,543

0,0569,251

0,056 0,026

- Phn b dng in th t thun:

I1Na HT2 = I1Na 1HT1

1HT1 C 1HD

X.X X X

=0,065

13,543. 1,8610,065 0,116 0,292

I1Na HT1 = I1Na - I1Na HT2 = 13,543 1,861 = 11,682

- Phn b dng in th t nghch:


1HT1 C 1HD

X.X X X

= 0,065

4,293. 0,590,065 0,116 0,292

I2Na HT1 = I2Na I2Na HT2 = -4,293 (-0,59) = -3,703

- Phn b dng in th t khng


I0aC = I0Na0HT1

0HD H0HT1 C

0HD H

X

X .XX

X XX

=0,029

9,251. 0,9520,39.0,211

0,029 0,1160,39 0,211

I0N HT1 = I0 - I0C = (-9,251) (-0,952) = -8,299


khng h thng 2):

I0Na HT2 = I0aT = I0aC .

H

0HD H

X

X X

0,2110,952. 0,334

0,39 0,211


I0tt = 3.(I0aT.Icb(110) - I0aC.Icb(220)) = 3.((-0,334).0,628 (-0,952).0,314) = 0,267(kA)


o Ngn mch ti N1:

Dng ngn mch m ta tnh l ca pha c bit tc pha khng b s c,

tnh dng ngn mch qua BI ta phi s dng ton t quay, gi s vit vi dng

in pha B, v ch quan tm n tr s:



INb = | a2 . I1Na + a . I2Na + I0Na |

- BI1:

If(BI1) = | a2. I1Na HT2 + a.I2Na HT2 + I0C|

21 3 1 3( j ) .1,861 ( j ).( 0,59) ( 0,952) 1,588 j2,122 2,6512 2 2 2

I0(BI1) = I0C = -0,952


2

BI1( 0)

1 3 1 3I ( j ) .1,861 ( j ).( 0,59) 2,216

2 2 2 2

- BI2:

If(BI2) = | a2. I1Na HT2 + a.I2Na HT2 + I0T|

21 3 1 3( j ) .1,861 ( j ).( 0,59) ( 0,334) 2,3332 2 2 2

I0(BI2) = -0,334


2

BI2( 0)

1 3 1 3I ( j ) .1,861 ( j ).( 0,59) 2,216

2 2 2 2

- Dng qua BI3 bng 0


o Ngn mch ti N1:

- BI1:

If(BI1) = | a2. I1Na HT1 + a.I2Na HT1 + I0NaHT1|

21 3 1 3( j ) .11,682 ( j ).( 3,703) ( 8,299) 18,1252 2 2 2

I0(BI1) = I0C = -8,299




2

BI1( 0)

1 3 1 3I ( j ) .11,682 ( j ).( 3,703) 13,908

2 2 2 2


If(BI2) = 2,333

I0(BI2) = - 0,334


IfBI2(-0) = 2,216

- Dng qua BI3 bng 0


b. Ngn mch pha 110 kV: N2 v N2


im ngn mch nh sau:

S thay th th t thun


EHT1X1HT10,065

XC0,116

X1HD0,292

EHT X

0,112

EHT2N2N2

'XT 0

BI1 BI2

X2HT10,065

XC0,116

X2HD0,292

X

0,112

XT 0

BI1 BI2

N2N2

'



S thay th th t khng


X1 = X2 = 1HT1 C 1HD

1HT1 C 1HD

(X X ).X (0,065 0,116).0,2920,112

X X X 0,065 0,116 0,292

X0 =

0HT1 C H0HD

0HT1 H C

0HT1 C H0HD

0HT1 H C

(X X ).X (0,029 0,116).0,211X . 0,39.

X X X 0,029 0,211 0,1160,070

(X X ).X (0,029 0,116).0,2110,39X

0,029 0,211 0,116X X X

Ngn mch 3 pha


I1 = HT

1

18,95

0,112

E

X


INHT1 = I1.1HD

1HT1 C 1HD

X

X X X = 8,95.

0,2925,525

0,065 0,116 0,292


INHT2 = I1 - INHT1 = 8,95 -5,525 = 3,425



v N2:

o Ngn mch ti N2:

X0HT10,029

XC0,116

X0HD0,39

X

0,070

XT 0

BI1 BI2XH0,211

N2N2

'



IBI1 = IBI2 = INHT1 = 5,525

IBI3 = IBI0 = 0

o Ngn mch ti N2:

IBI1 = INHT1 = 5,525 ; IBI2 = INHT2 = 3,425

IBI3 = IBI0 = 0

Ngn mch 1 pha


HT1 2 0

1 2 0

I I IE

X X X

13,403

0,112 0,112 0,07


I1N HT1 = I2N HT1 = 1HD

1

1HT1 C 1HD

XI .

X X X

=

0,2923,403. 2,101

0,065 0,116 0,292

I1N HT2 = I2N HT2 = 1I - I1N HT2 = 3,403 2,101 = 1,302


Dng in th t khng chy qua pha trung MBA:

I0T = I00HD

0HT1 C H0HD

0HT1 H C

X

(X X ).XX

X X X

=0,39

3,403. 2,788(0,116 0,029).0,211

0,390,116 0,029 0,211

I0N HT2 = I0 - I0T = 0,613

Dng TTK chy qua pha cao ca MBA (cng l dng chy qua in

khng h thng 1):

I0N HT1= I0C = I0T .

H

0H1 C H

X

X X X

0,2112,788. 1,653

0,029 0,116 0,211






o Ngn mch ti N2:

- BI1:

If(BI1) = I1N HT1 + I2N HT1 + I0C = 2,101 + 2,101 + 1,653 = 5,855

I0(BI1) = I0C = 1,653


IBI1(-0) = If(BI1) - I0(BI1) = 5,855 1,653 = 4,202

- BI2:

If(BI2) = I1N HT1 + I2N HT1 + I0T = 2,101 + 2,101 + 2,788 = 6,99

I0(BI2) = 2,788


IBI1(-0) = If(BI1) - I0(BI1) = 6,99 2,788 = 4,202

- Dng qua BI3 bng 0


o Ngn mch ti N2:


- BI2:

If(BI2) = I1N HT2 + I2N HT2 + I0N HT2 = 1,302 + 1,302 + 0,613 = 3,217

I0(BI2) = 0,613


IBI1(-0) = If(BI1) - I0(BI1) = 3,217 0,613 = 2,604

- Dng qua BI3 bng 0


Ngn mch 2 pha chm t


pha A):




I1Na = 2 0

1

2 0

X .XX

X X

E

= 1

6,4540,112.0,07

0,1120,112 0,07


I2Na = - I1Na 0

2 0

X

X X

= - 6,454

0,072,495

0,112 0,07


I0Na = - I1 2

2 0

X

X X

= - 6,454

0,1123,959

0,112 0,07


I1Na HT1 = I1Na 1HD

1HT1 C 1HD

X.X X X

= 0,292

6,454. 3,9840,065 0,116 0,292

I1Na HT2 = I1Na - I1Na HT1 = 6,454 3,984 = 2,47


I2Na HT1 = I2Na 1HD

1HT1 C 1HD

X.X X X

= 0,292

2,495. 1,540,065 0,116 0,292

I2Na HT2 = I2Na I2Na HT2 = -2,495 (-1,54) = -0,955



I0T = I0Na0HD

0HT1 C H0HD

0HT1 H C

X

(X X ).XX

X X X

=0,39

3,959. 3,244(0,116 0,029).0,211

0,390,116 0,029 0,211

I0N HT2 = I0 - I0T = - 0,715


khng h thng 1):


H

0H1 H C

X

X X X

0,2113,244. 1,923

0,029 0,211 0,116


I0tt = 3.(I0T.Icb(110) - I0C.Icb(220)) = 3.(-3,244.0,628 (-1,923).0,314) = 4,301(kA)




o Ngn mch ti N2:

- BI1:

If(BI1) = | a2. I1Na HT1 + a.I2Na HT1 + I0C|

21 3 1 3( j ) .3,984 ( j ).( 1,54) ( 1,923) 5,7252 2 2 2

I0(BI1) = I0C = -1,923


2

BI1( 0)

1 3 1 3I ( j ) .3,984 ( j ).( 1,54) 4,938

2 2 2 2

- BI2:


21 3 1 3( j ) .3,984 ( j ).( 1,54) ( 3,244) 6,5452 2 2 2

I0(BI2) = -3,244


2

BI2( 0)

1 3 1 3I ( j ) .3,984 ( j ).( 1,54) 4,938

2 2 2 2

- Dng qua BI3 bng 0


o Ngn mch ti N2:


- BI2:


21 3 1 3( j ) .2,47 ( j ).( 0,955) ( 0,715) 3,3122 2 2 2

I0(BI2) = -0,715




2

BI2( 0)

1 3 1 3I ( j ) .2,47 ( j ).( 0,955) 3,061

2 2 2 2

- Dng qua BI3 bng 0


c. Ngn mch pha 35 kV: N3 v N3

Pha 35 kV cun u tam gic nn khng xt s c chm t, ch xt ngn

mch ba pha.

S thay th th t thun

Ngn mch 3 pha

1HT1 C 1HD1 H

HT1 C 1HD

(X X ).X (0,065 0,116).0,292X X 0,211 0,323

X X X 0,065 0,116 0,292


I1 = 1

13,098

X 0,323

E


INHT1 = I1.1HD

1HT1 C 1HD

X

X X X = 3,098.

0,2921,913

0,065 0,116 0,292


INHT2 = I1 - INHT1 = 3,098 -1,913 = 1,186

EHT1X1HT10,065

XC0,116

X1HD0,292 EHT2

XT 0

BI1 BI2

XH0,211

BI3

N3

N3'

EHT X

0,323





v N1:

o Ngn mch ti N3:

IBI1 = 1,913; IBI2 = 1,186

IBI3 = 3,098

IBI0 = 0

o Ngn mch ti N3:

IBI1 = 1,913; IBI2 = 1,186

IBI3 = IBI0 = 0

Kt qu tnh ton c tng hp trong bng 2.1.

2.4.2. Khi trm vn hnh 2 MBA trong ch max

a. Ngn mch pha 220 kV: N1 v N1


im ngn mch nh sau:

S thay th th t thun

EHT1X1HT10,065

XC0,116

EHT X

0,055

XT 0

BI2 X1HD0,292 EHT2

BI1

N1'

N1

XC0,116

XT 0




S thay th th t khng


X1 = X2 =

C1HT1 1HD

C1HT1 1HD

X 0,116X .( X ) 0,065.( 0,292)

2 2 0,055X 0,116

0,065 0,292X X22

X2HT10,065

XC0,116

X

0,055

XT 0

BI2 X2HD0,292

BI1

N1'

N1

XC0,116

XT 0

X0HT10,029

XC0,116

X

0,024

XT 0

BI2 X0HD0,390

BI1

N1'

N1

XC0,116

XT 0 XH

0,211

XH0,211



X0 =

H0HD

C0HT1

H0HD

H0HD

C0HT1

H0HD

X 0,211X . 0,39.

X 0,1162 2X .( ) 0,029.( )X 0,2112 2

X 0,392 2 0,024

X 0,2110,39.X .

X 0,116 22 0,029X0,211X 22

0,39X22

Ngn mch 3 pha


I1 = HT

1

118,242

0,055

E

X


INHT1 = I1.

C1HD

C1HT1 1HD

XX

2X

X X2

= 18,242.

0,1160,292

2 15,3850,116

0,065 0,2922


INHT2 = I1 - INHT1 = 18,242 -15,385 = 2,857



v N1:

o Ngn mch ti N1:

IBI1 = IBI2 = 0,5 . IHT2 = 0,5 .2,857 = 1,429

IBI3 = IBI0 = 0

o Ngn mch ti N1:

HT2BI1 HT1

2,857I15,385 16,814I I

2 2

IBI2 = 0,5 . IHT2 = 0,5 .2,857 = 1,429

IBI3 = IBI0 = 0

Ngn mch 1 pha




HT1 2 0

1 2 0

I I IE

X X X

17,480

0,055 0,055 0,024


I1N HT2 = I2N HT2 = 1HT1

1C

1HT1 1HD

XI .

XX X

2

=0,065

7,480. 1,1720,116

0,065 0,2922

I1N HT1 = I2N HT1 = 1I - I1N HT2 = 7,480 1,172 = 6,308



I0C = I0.0HT1

H0HD

C0HT1

H0HD

X

XX .

2XX2

X2

X

=0,029

7,480. 1,2760,211

0,39.0,116 20,029

0,21120,39

2

I0N HT1 = I0 - I0C = 7,480 1,276 = 6,204


khng h thng 2):

I0N HT2 = I0T = I0C .

H

H0HD

X

2

X2

X

0,211

21,276. 0,2720,211

0,392


I0tt = 3 . (I0T.Icb(110) - I0C.Icb(220)) = 3.(0,272.0,628 1,276.0,314) = 0,69(kA)


o Ngn mch ti N1:

- BI1:

1HT21BI1 2BI1

I 1,172I I 0,586

2 2

0C0BI1

I 1,276I 0,638

2 2



If(BI1) = I1BI1 + I2BI1 + I0BI1 = 0,586 +0,586 + 0,638 = 1,81


IBI1(-0) = If(BI1) - I0(BI1) = 1,774 0,638 = 1,172

- BI2:

1HT21BI2 2BI2

I 1,172I I 0,586

2 2

0T0BI2

I 0,272I 0,136

2 2

If(BI2) = I1BI2 + I2BI2 + I0BI2 = 0,586 +0,586 + 0,136 = 1,308


IBI2(-0) = If(BI2) - I0(BI2) = 1,272 0,136 = 1,172

- Dng qua BI3 bng 0

- Dng qua BI0: 0,5 . 0,69 = 0,345 kA.

o Ngn mch ti N1:

- BI1:

1HT21BI1 2BI1 1HT1

I 1,172I I I 6,308 6,894

2 2

0C0BI1 0HT1

I 1,276I I 6,204 6,842

2 2

If(BI1) = I1BI1 + I2BI1 + I0BI1 = 6,894 +6,894 + 6,842 = 20,63


IBI1(-0) = If(BI1) - I0(BI1) = 20,63 6,842 = 13,788


- Dng qua BI3 bng 0

- Dng qua BI0: 0,5 . 0,69 = 0,345 kA.

Ngn mch 2 pha chm t


pha A):




I1Na = 2 0

1

2 0

X .XX

X X

E

= 1

13,9790,055.0,024

0,0550,055 0,024


I2Na = - I1Na 0

2 0

X

X X

= -13,979

0,0244,263

0,055 0,024


I0Na = - I1 2

2 0

X

X X

= -13,979

0,0559,716

0,055 0,024



C1HT1 1HD

X.

XX X

2

= 0,065

13,979. 2,1890,116

0,065 0,2922

I1Na HT1 = I1Na - I1Na HT2 = 13,979 2,189 = 11,789



C1HT1 1HD

X.

XX X

2

=0,065

4,263. 0,6680,116

0,065 0,2922

I2Na HT1 = I2Na I2Na HT2 = -4,263 (-0,668) = -3,595



I0aC = I0Na0HT1

H0HD

C0HT1

H0HD

X

XX .

2XX2

X2

X

=0,029

9,716. 1,6570,211

0,39.0,116 20,029

0,21120,39

2

I0N HT1 = I0 - I0C = (-9,716) (-1,657) = -8,059


khng h thng 2):

I0Na HT2 = I0aT = I0aC .

H

H0HD

X

2

X2

X

0,211

21,657. 0,3530,211

0,392




I0tt = 3.(I0aT.Icb(110) - I0aC.Icb(220)) = 3.((-0,353).0,628 (-1,657).0,314) =

0,896(kA)


o Ngn mch ti N1:

- BI1:

1HT21aBI1

I 2,189I 1,095

2 2

1HT22aBI1

I 0,668I 0,334

2 2

0aC0aBI1

I 1,657I 0,829

2 2

If(BI1) = | a2. I1aBI1 + a.I2aBI1 + I0aBI1|

21 3 1 3( j ) .1,095 ( j ).( 0,334) ( 0,829) 1,7302 2 2 2

I0(BI1) = I0C = -0,829


2

BI1( 0)

1 3 1 3I ( j ) .1,095 ( j ).( 0,334) 1,295

2 2 2 2

- BI2:

1HT21aBI2

I 2,189I 1,095

2 2

2HT22aBI2

I 0,668I 0,334

2 2

0aT0aBI2

I 0,353I 0,177

2 2


21 3 1 3( j ) .1,095 ( j ).( 0,334) ( 0,177) 1,3572 2 2 2

I0(BI2) = -0,177




2

BI1( 0)

1 3 1 3I ( j ) .1,095 ( j ).( 0,334) 1,295

2 2 2 2

- Dng qua BI3 bng 0

- Dng qua BI0 bng 0,5 . 0,896 = 0,448 kA.

o Ngn mch ti N1:

- BI1:

1aHT21aBI1 1aHT

I 2,189I I 11,789 12,884

2 2

2HT22aBI1 2HT1

I 0,668I I 3,595 3,929

2 2

0aC0aBI1 0HT1

I 1,657I I 8,059 8,888

2 2

If(BI1) = | a2. I1Na HT1 + a.I2Na HT1 + I0NaHT1|

21 3 1 3( j ) .12,884 ( j ).( 3,929) ( 8,888) 19,7652 2 2 2

I0(BI1) = I0C = -8,888


2

BI1( 0)

1 3 1 3I ( j ) .12,884 ( j ).( 3,929) 15,233

2 2 2 2


- Dng qua BI3 bng 0

- Dng qua BI0 bng 0,5 . 0,896 = 0,448 kA.

. b. Ngn mch pha 110 kV: N2 v N2


im ngn mch nh sau:



S thay th th t thun


S thay th th t khng

EHT1X1HT10,065

XC0,116

EHT X

0,087

XT 0

BI2 X1HD0,292 EHT2

BI1

N2' N2

XC0,116

XT 0

X2HT10,065

XC0,116

X

0,087

XT 0

BI2 X2HD0,292

BI1

N2'

N2

XC0,116

XT 0

X0HT10,029

XC0,116

X

0,042

XT 0

BI2 X0HD0,390

BI1

N2' N2

XC0,116

XT 0 XH

0,211

XH0,211




X1 = X2 =

C1HT1 1HD

C1HT1 1HD

X 0,116(X ).X (0,065 ).0,292

2 2 0,087X 0,116

0,065 0,292X X22

X0 =

H0HD

C0HT1

H0HD

H0HD

C0HT1

H0HD

X 0,211X . 0,39.

X 0,1162 2(X ). (0,029 ).X 0,2112 2

X 0,392 2 0,042

X 0,2110,39.X .

X 0,116 22 0,029X0,211X 22

0,39X22

Ngn mch 3 pha


I1 = HT

1

111,555

0,087

E

X


INHT1 = I1.1HD

C1HT1 1HD

X

XX X

2

= 11,555.0,292

8,1300,116

0,065 0,2922


INHT2 = I1 - INHT1 = 11,555 8,130 = 3,425



v N2:

o Ngn mch ti N2:

IBI1 = IBI2 = 0,5 . INHT1 = 0,5 .8,130 = 4,065

IBI3 = IBI0 = 0

o Ngn mch ti N2:

HT1BI1

8,130I4,065I

2 2



HT1BI2 HT2

8,130I3,425 7,49I I

2 2

IBI3 = IBI0 = 0

Ngn mch 1 pha


HT1 2 0

1 2 0

I I IE

X X X

14,639

0,087 0,087 0,042


I1N HT1 = I2N HT1 = 1HD

1C

1HT1 1HD

XI .

XX X

2

=0,292

4,639. 3,2640,116

0,065 0,2922

I1N HT2 = I2N HT2 = 1I - I1N HT2 = 4,639 3,264 = 1,375



I0T = I00HD

C H0HT1

0HDCH

0HT1

X

X X(X ).

2 2XXX

X2 2

=0,39

4,639. 4,1330,116 0,211

( 0,029).2 20,39

0,116 0,2110,029

2 2

I0N HT2 = I0 - I0T = 0,505


khng h thng 1):


H

CH0HT1

X

2

X2 2

XX

0,211

24,133. 2,2650,211 0,116

0,0292 2




o Ngn mch ti N2:



- BI1:

1HT11BI1 2BI1

I 3,264I I 1,632

2 2

0C0BI1

I 2,265I 1,133

2 2

If(BI1) = I1BI1 + I2BI1 + I0BI1 = 1,632 + 1,632 + 1,133 = 4,397


IBI1(-0) = If(BI1) - I0(BI1) = 4,397 1,133 = 3,264

- BI2:

1HT11BI2 2BI2

I 3,264I I 1,632

2 2

0T0BI2

I 4,133I 2,067

2 2

If(BI2) = I1BI2 + I2BI2 + I0BI2 = 1,632 + 1,632 + 2,067 = 5,331


IBI2(-0) = If(BI2) - I0(BI2) = 5,331 2,067 = 3,264

- Dng qua BI3 bng 0

- Dng qua BI0: 0,5 . 5,653 = 2,827 kA.

o Ngn mch ti N2:


- BI2:

1HT11BI2 2BI2 1HT2

I 3,264I I I 1,375 3,007

2 2

0T0BI2 0HT2

I 4,133I I 0,505 2,572

2 2

If(BI2) = I1BI2 + I2BI2 + I0BI2 = 3,007 +3,007 + 2,572 = 8,586


IBI2(-0) = If(BI2) - I0(BI2) = 8,586 2,572 = 6,014

- Dng qua BI3 bng 0



- Dng qua BI0: 0,5 . 5,653 = 2,827 kA.

Ngn mch 2 pha chm t


pha A):


I1Na = 2 0

1

2 0

X .XX

X X

E

= 1

8,6930,087.0,042

0,0870,087 0,042


I2Na = - I1Na 0

2 0

X

X X

= - 8,693

0,0422,862

0,087 0,042


I0Na = - I1 2

2 0

X

X X

= - 8,693

0,0875,830

0,087 0,042


I1Na HT1 = I1Na 1HD

C1HT1 1HD

X.

XX X

2

= 0,292

8,693. 6,1160,116

0,065 0,2922

I1Na HT2 = I1Na - I1Na HT1 = 8,693 6,116 = 2,576


I2Na HT1 = I2Na 1HD

C1HT1 1HD

X.

XX X

2

=0,292

2,862. 2,0140,116

0,065 0,2922

I2Na HT2 = I2Na I2Na HT1 = -2,862 (-2,014) = -0,848



I0T = I0Na0HD

C H0HT1

0HDCH

0HT1

X

X X(X ).

2 2XXX

X2 2

=0,39

5,83. 5,1950,116 0,211

( 0,029).2 20,39

0,116 0,2110,029

2 2

I0N HT2 = I0 - I0T = -0,635




khng h thng 1):


H

CH0H1

X

2

X2 2

XX

0,211

25,195. 2,8470,211 0,116

0,0292 2


I0tt = 3 . (I0T.Icb(110) - I0C.Icb(220)) = 3.(-5,195.0,628 (-2,847).0,314) = 7,106(kA)


o Ngn mch ti N2:

- BI1:

1HT11aBI1

I 6,116I 3,058

2 2

1HT22aBI1

I 2,014I 1,007

2 2

0aC0aBI1

I 2,847I 1,424

2 2


21 3 1 3( j ) .3,058 ( j ).( 1,007) ( 1,424) 4,2892 2 2 2

I0(BI1) = -1,424


2

BI1( 0)

1 3 1 3I ( j ) .3,058 ( j ).( 1,007) 3,667

2 2 2 2

- BI2:

1HT11aBI2

I 6,116I 3,058

2 2

1HT22aBI2

I 2,014I 1,007

2 2

0aT0aBI2

I 5,195I 2,598

2 2




21 3 1 3( j ) .3,058 ( j ).( 1,007) ( 2,598) 5,0522 2 2 2

I0(BI2) = -2,598


2

BI2( 0)

1 3 1 3I ( j ) .3,058 ( j ).( 1,007) 3,667

2 2 2 2

- Dng qua BI3 bng 0

- Dng qua BI0 bng 0,5 . 7,106 = 3,553 kA.

o Ngn mch ti N2:


- BI2:

1aHT11aBI2 1aHT2

I 6,116I I 2,576 5,634

2 2

2aHT12aBI2 2aHT2

I 2,014I I 0,848 1,855

2 2

0aT0aBI2 0aHT2

I 5,195I I 0,635 3,233

2 2


21 3 1 3( j ) .5,634 ( j ).( 1,855) ( 3,233) 8,2652 2 2 2

I0(BI2) = -3,233


2

BI1( 0)

1 3 1 3I ( j ) .5,634 ( j ).( 1,855) 6,755

2 2 2 2

- Dng qua BI3 bng 0

- Dng qua BI0 bng 0,5 . 0,896 = 3,553 kA.

c. Ngn mch pha 35 kV: N3 v N3



Pha 35 kV cun u tam gic nn khng xt s c chm t, ch xt ngn

mch ba pha.

S thay th th t thun

Ngn mch 3 pha

C1HT1 1HD

H1

CHT1 1HD

X 0,116(X ).X (0,065 ).0,292

X 0,2112 2X 0,192X 0,1162 2

0,065 0,292X X22


I1 = 1

15,207

X 0,192

E


INHT1 = I1.(1HD

C1HT1 1HD

X

XX X

2

) = 5,207.0,292

3,6640,116

0,065 0,2922


INHT2 = I1 - INHT1 = 5,207 3,664 = 1,543



v N1:

EHT1X1HT10,065

XC0,116

EHT X

0,192

XT 0

BI2

X1HD0,292 EHT2

BI1

XC0,116

XT 0

XH0,211

XH0,211

N3'

N3

BI3



o Ngn mch ti N3:

IBI1 = 0,5 . 3,664 = 1,832

IBI2 = 0,5 . 1,543 = 0,772

IBI3 = 0,5 . 5,207 = 2,604

IBI0 = 0

o Ngn mch ti N3:

IBI1 = 0,5 . 3,664 = 1,832

IBI2 = 0,5 . 1,543 = 0,772

IBI3 = 0

IBI0 = 0

Kt qu tnh ton c tng hp trong bng 2.2.

2.4.3. Khi trm vn hnh 1 MBA trong ch min

Sau khi tnh ton dng ngn mch trong ch max, trong gii hn n

em s khng trnh by c th cc bc tnh ton v iu ny tng t nh vi ch

max v kt s c tng hp trong bng kt qu cui mi phn.

Kt qu tnh ton ch min vn hnh 1 my bin p v 1 ng dy

c tng hp trong bng 2.3, kt qu tnh ton ch min vn hnh 1 my bin

p v 2 ng dy vn hnh song song c tng hp trong bng 2.5.

2.4.4. Khi trm vn hnh 2 MBA trong ch min

Tnh ton hon ton tng t nh, ta thu c kt qu tnh ton ch

min vn hnh 2 my bin p v 1 ng dy trong bng 2.4, kt qu tnh ton ch

min vn hnh 2 my bin p v 2 ng dy trong bng 2.6.



Bng 2.1. Bng kt qu tnh ton ngn mch trong ch cc i vn hnh 1 MBA

im ngn mch

Dng ngn mch

Dng in chy qua cc BI

BI1 BI2 BI3 BI0

(kA)

N1

N(3)

If 2,451 2,451

N(1)

If 2,735 2,251 0,209

I0 0,745 0,261

If I0 1,990 1,990

N(1,1)

If 2,651 2,333 0,267

I0 -0,952 -0,334

If I0 2,216 2,216

N1

N(3)

If 15,386 2,451

N(1)

If 18,982 2,251 0,209

I0 6,494 0,261

If I0 12,488 1,990

N(1,1)

If 18,125 2,333 0,267

I0 -8,299 -0,334

If I0 13,908 2,216

N2

N(3)

If 5,525 5,525

N(1)

If 5,855 6,990 3,696

I0 1,653 2,788

If I0 4,202 4,202

N(1,1)

If 5,725 6,545 4,301

I0 -1,923 -3,244

If I0 4,938 4,938

N2

N(3)

If 5,525 3,425

N(1)

If 5,855 3,217 3,696

I0 1,653 0,613

If I0 4,202 2,604

N(1,1)

If 5,725 3,312 4,301

I0 -1,923 -0,715

If I0 4,938 3,061

N3

N(3)

If 1,913 1,186 3,098

N3

N(3)

If 1,913 1,186



Bng 2.2. Bng kt qu tnh ton ngn mch trong ch cc i vn hnh 2 MBA

im ngn mch

Dng ngn mch


BI1 BI2 BI3 BI0

(kA)

N1

N(3)

If 1,429 1,429

N(1)

If 1,810 1,308 0,345

I0 0,638 0,136

If I0 1,172 1,172

N(1,1)

If 1,730 1,357 0,448

I0 -0,829 -0,177

If I0 1,295 1,295

N1

N(3)

If 16,814 1,429

N(1)

If 20,63 1,308 0,345

I0 6,842 0,136

If I0 13,788 1,172

N(1,1)

If 19,765 1,357 0,448

I0 -8,888 -0,177

If I0 15,233 1,295

N2

N(3)

If 4,065 4,065

N(1)

If 4,397 5,331 2,827

I0 1,133 2,067

If I0 3,264 3,264

N(1,1)

If 4,289 5,052 3,553

I0 -1,424 -2,598

If I0 3,667 3,667

N2

N(3)

If 4,065 7,490

N(1)

If 4,397 8,586 2,827

I0 1,133 2,572

If I0 3,264 6,014

N(1,1)

If 4,289 8,265 3,553

I0 -1,424 -3,233

If I0 3,667 6,755

N3

N(3)

If 1,832 0,772 2,604

N3

N(3)

If 1,832 0,772 2,604



Bng 2.3. Bng kt qu tnh ton ngn mch trong ch cc tiu vn hnh 1 MBA v 1 ng

dy

im ngn mch

Dng ngn mch


BI1 BI2 BI3 BI0

(kA)

N1

N(2)

If 1,542 1,542

N(1)

If 2,135 1,577 0,381

I0 0,711 0,153

If I0 1,424 1,424

N(1,1)

If 2,037 1,667 0,476

I0 -0,888 -0,192

If I0 1,603 1,603

N1

N(2)

If 11,395 1,891

N(1)

If 15,790 1,577 0,381

I0 5,264 0,153

If I0 10,526 1,424

N(1,1)

If 15,075 1,667 0,476

I0 -6,580 -0,192

If I0 11,860 1,603

N2

N(2)

If 4,510 4,510

N(1)

If 5,504 6,589 3,447

I0 1,488 2,573

If I0 4,016 4,016

N(1,1)

If 5,391 6,190 4,086

I0 -1,763 -3,050

If I0 4,665 4,665

N2

N(2)

If 4,510 1,942

N(1)

If 5,504 2,027 3,447

I0 1,488 0,299

If I0 4,016 1,728

N(1,1)

If 5,391 2,126 4,086

I0 -1,763 -0,354

If I0 4,665 2,008

N3

N(2)

If 1,753 0,755 2,508

N3

N(2)

If 1,753 0,755



Bng 2.4. Bng kt qu tnh ton ngn mch trong ch cc tiu vn hnh 2 MBA v 1 ng

dy

im ngn mch

Dng ngn mch


BI1 BI2 BI3 BI0

(kA)

N1

N(2)

If 0,859 0,859

N(1)

If 1,427 0,882 0,443

I0 0,620 0,075

If I0 0,807 0,807

N(1,1)

If 1,360 0,930 0,568

I0 -0,795 -0,096

If I0 0,898 0,898

N1

N(2)

If 12,254 0,859

N(1)

If 17,045 0,882 0,443

I0 5,537 0,075

If I0 11,508 0,807

N(1,1)

If 16,324 0,930 0,568

I0 -7,097 -0,096

If I0 12,798 0,898

N2

N(2)

If 3,231 3,231

N(1)

If 4,006 4,889 2,577

I0 0,970 1,853

If I0 3,036 3,036

N(1,1)

If 3,920 4,656 3,307

I0 -1,245 -2,378

If I0 3,375 3,376

N2

N(2)

If 3,231 5,173

N(1)

If 4,006 6,956 2,577

I0 0,970 2,096

If I0 3,036 4,860

N(1,1)

If 3,920 6,695 3,307

I0 -1,245 -2,690

If I0 3,375 5,403

N3

N(2)

If 1,597 0,480 2,077

N3

N(2)

If 1,597 0,480 2,077



Bng 2.5.Bng kt qu tnh ton ngn mch trong ch cc tiu vn hnh 1 MBA v 2 ng

dy

im ngn mch

Dng ngn mch


BI1 BI2 BI3 BI0

(kA)

N1

N(2)

If 1,891 1,891

N(1)

If 2,517 1,982 0,275

I0 0,777 0,242

If I0 1,740 1,740

N(1,1)

If 2,418 2,070 0,342

I0 -0,966 -0,301

If I0 1,967 1,967

N1

N(2)

If 11,395 1,891

N(1)

If 15,826 1,982 0,275

I0 5,338 0,242

If I0 10,488 1,740

N(1,1)

If 15,093 2,070 0,342

I0 -6,637 -0,301

If I0 11,852 1,967

N2

N(2)

If 4,510 4,510

N(1)

If 5,507 6,612 3,506

I0 1,513 2,618

If I0 3,994 3,994

N(1,1)

If 5,394 6,202 4,127

I0 -1,781 -3,081

If I0 4,662 4,662

N2

N(2)

If 4,510 2,532

N(1)

If 5,507 2,743 3,506

I0 1,513 0,501

If I0 3,994 2,242

N(1,1)

If 5,394 2,824 4,127

I0 -1,781 -0,59

If I0 4,662 2,617

N3

N(2)

If 1,661 0,932 2,593

N3

N(2)

If 1,661 0,932



Bng 2.6.Bng kt qu tnh ton ngn mch trong ch cc tiu vn hnh 2 MBA v 2 ng

dy

im ngn mch

Dng ngn mch


BI1 BI2 BI3 BI0

(kA)

N1

N(2)

If 1,083 1,083

N(1)

If 1,674 1,135 0,393

I0 0,662 0,123

If I0 1,012 1,012

N(1,1)

If 1,590 1,182 0,500

I0 -0,842 -0,156

If I0 1,129 1,129

N1

N(2)

If 12,478 1,083

N(1)

If 17,338 1,135 0,393

I0 5,676 0,123

If I0 11,662 1,012

N(1,1)

If 16,890 1,182 0,500

I0 -7,673 -0,156

If I0 13,017 1,129

N2

N(2)

If 3,231 3,231

N(1)

If 4,013 4,917 2,638

I0 0,993 1,897

If I0 3,020 3,020

N(1,1)

If 3,925 4,672 3,356

I0 -1,264 -2,413

If I0 3,374 3,374

N2

N(2)

If 3,231 5,763

N(1)

If 4,013 7,693 2,638

I0 0,993 2,307

If I0 3,020 5,386

N(1,1)

If 3,925 7,404 3,356

I0 -1,264 -2,935

If I0 3,374 6,013

N3

N(2)

If 1,542 0,604 2,146

N3

N(2)

If 1,542 0,604 2,146



Bng 2.7. Tng hp tr s dng ngn mch ln nht qua cc BI

im ngn mch

Loi dng ngn mch

Dng ngn mch qua cc BI

BI1 BI2 BI3 BI0 (kA)

N1 If 2,735 2,451

I0 0,966 0,334 0,568

N1

If 20,630 2,451

I0 8,888 0,334 0,568

N2 If 5,855 6,990

I0 1,923 3,244 4,301

N2

If 5,855 8,586

I0 1,923 3,233 4,301

N3 If 1,913 1,186 3,098

N3

If 1,913 1,186 2,604

Bng 2.8. Tr s dng ngn mch nh nht qua cc BI

im ngn mch

Loi dng ngn mch

Dng ngn mch qua cc BI

BI1 BI2 BI3 BI0 (kA)

N1

If 0,859 0,859

I0 0,620 0,075 0,209

I2 0,238 0,238

N1

If 11,395 0,859

I0 5,264 0,075 0,209

I2 3,29 0,238

N2

If 3,231 3,231

I0 0,970 1,853 2,577

I2 0,892 0,892

N2

If 3,231 1,942

I0 0,970 0,299 2,577

I2 0,892 0,609

N3 If 1,542 0,48 2,077

I2 0,890 0,277 1,199



N3

If 1,542 0,48 2,077

I2 0,890 0,277 1,199



CHNG 3

LA CHN PHNG THC BO V

3.1. Cc dng h hng i vi my bin p

i vi my bin p, chng ta c th chia cc dng h hng thng xy ra

lm 2 nhm: h hng bn trong v h hng bn ngoi.

Cc s c i vi my bin p:

- Phng in s xuyn;

- S c pha-pha, pha t vi cun dy cao v h p;

- S xm m ca hi nc vo du cch in;

- St nh lan truyn vo trm lm hng cch in cun dy;

- S c gia cc vng dy trn cng mt cun dy.

Nhng ch lm vic bt thng ca my bin p bao gm:

- Qu ti: Dng in hoc cng sut vt qu gi tr nh mc, my bin

p ch cho php lm vic trong mt thi gian nht nh gi l thi gian cho php;

- Mc du tng cao hoc gim thp;

- Hng b chuyn i u phn p;

- Li t b qu t thng.

i vi cc loi s c trn, ta thng c cc loi bo v nh sau:

Bng 3.1. Cc loi bo v thng dng cho my bin p

Dng s c Loi bo v

S c pha-pha v pha-t trong

cun dy

- Bo v so lch

- Bo v qu dng

- Bo v chng chm t hn ch

S c gia cc vng dy - Bo v so lch

- R le kh (Buchholz)

S c li t - Bo v so lch


S c thng du my bin p - Bo v so lch




- Bo v chng chm t thng du

MBA

Qu t thng - Bo v chng qu t thng

Qu nhit - Bo v chng qu ti

3.2. S phng thc bo v my bin p

Ty theo cng sut, v tr v vai tr ca my bin p trong h thng m

ngi ta la chn cc phng thc bo v thch hp cho my bin p. y, da

vo Quy nh v cu hnh h thng bo v, quy cch k thut ca r le bo v

cho ng dy v TBA ca EVN, Mc 2, iu 9, ta c s phng thc bo

v cho my bin p t ngu 220/110/35 nh sau:

Bo v chnh 1: c tch hp cc chc nng bo v 87T, 49, 64, 50/51,

50/51N tn hiu dng in cc pha c ly t my bin dng chn s

MBA

Bo v chnh 2: c tch hp cc chc nng bo v 87T, 49, 64, 50/51,

50/51N tn hiu dng in cc pha c ly t my bin dng ngn my

ct u vo cc pha MBA

Bo v d phng cho cun dy 220 kV: c tch hp cc chc nng

bo v 67/67N, 50/51, 50/51N, 27/59, 74, 50BF tn hiu dng in c

ly t my bin dng ngn my ct u vo pha 220 kV ca MBA, tn

hiu in p c ly t my bin in p thanh ci 220 kV

Bo v d phng cho cun dy 110 kV: tng t nh pha 220 kV

Bo v d phng cho cun dy trung p: c tch hp cc chc nng

bo v 50/51, 50/51N, 74, 50BF tn hiu dng in c ly t my bin

dng chn s cun trung p MBA

Chc nng rle bo v nhit du/ cun dy MBA (26), rle p lc MBA

(63), rle gas cho bnh du chnh v ngn iu p di ti (96), rle bo mc

du tng cao (71) c trang b ng b vi MBA, c gi i ct trc tip

my ct 3 pha thng qua rle ch huy ct hoc c gi i ct ng thi

thng qua hai b bo v chnh v d phng ca MBA



S phng thc bo v MBA

67/67N

50BF

27/59

74

96F 26Q

63

26W

71Q1 71Q2 FRD

67/67N

50BF

27/59

74

96B46

87T

49

51 50BF

74

87T

49

64

59N

220kV110 kV

35 kV

46

64

7SJ64

7SJ64 7SJ64

P633

7UT613

7RW60



Cc chc nng c s dng bo v cho trm bin p trn s phng thc

c tng hp trong bng

Bng 3.2. Cc chc nng c s dng trong s phng thc

STT K hiu Chc nng

1 87T Bo v so lch MBA

2 87N Bo v so lch TTK (REF)

3 49 Bo v chng qu ti

4 67 Bo v qu dng c hng 2 cp tc ng (ct

nhanh v c thi gian tr)

5 67N Bo v qu dng TTK c hng 2 cp tc

ng (ct nhanh v c thi gian tr)

6 51 Bo v qu dng c thi gian tr

7 46 Bo v qu dng TTN

8 27 Bo v km p

9 59/59N Bo v qu p/ Bo v qu p TTK

10 50BF Bo v trong trng hp my ct t chi tc

ng

11 74 Chc nng gim st mch ct

12 96B Rle kh cho bnh du MBA (Buchholz)

13 96F Rle kh cho thng du ngn iu p di ti

14 26Q Bo v theo nhit du

15 26W Bo v theo nhit cun dy

16 63 Rle p lc MBA

17 71 Rle bo mc du tng cao

18 PRD Thit b x p lc

Vi cc chc nng trn, ta s s dng cc r le sau:

Dng 2 b rle bo v so lch c hm l 7UT613 ca SIEMENS v

P663 ca AREVA. Rle 7UT613 c ly tn hiu t BI ngay u chn

s my bin p. Cc chc nng bo v xut s dng: chc nng bo

v so lch c hm (87T), bo v chng chm t hn ch (87N) v bo

v chng qu ti nhit (49). Cn rle P633 ly tn hiu dng in t cc

BI pha thanh gp.

Rle 7SJ64 l rle a chc nng ca hng SIEMENS vi cc chc nng

c bn: 67, 67N v 50BF.



- Chc nng chng h hng my ct (50BF) tc ng khi c s

c hng my ct nhn c lnh ct.

- Chc nng bo v qu dng c hng (67) v bo v qu dng

in th t khng c hng (67N) l chc nng bo v d

phng ng dy c hng v pha ng dy.

Bo v chng chm t pha 35kV: dng r le 7RW60 ca SIEMENS

bo v tc ng vi thnh phn 3U0, tn hiu in p 3U0 do cun tam

gic h ca bin in p loi 3 pha 3 tr cung cp. Khi c s c chm

t, nu tr s 3U0 vt qu ngng Ut th bo v s gi tn hiu cnh

bo c s c chm t pha 35kV.



CHNG 4

GII THIU TNH NNG V THNG S CC RLE S DNG

4.1. R le bo v so lch P633

R le P633

4.1.1 Gii thiu tng quan v rle P633

Rle s P633 do hng Alstom ch to, c s dng bo v chnh cho

MBA 3 cun dy hoc my bin p t ngu tt c cc cp in p. Rle ny

cng c th dng bo v cho cc loi my in quay nh my pht in, ng

c. Cc chc nng khc c tch hp trong rle P633 lm nhim v d phng

nh bo v qu dng, qu ti nhit, bo v qu kch thch, chng h hng my

ct. Bng cch phi hp cc chc nng tch hp trong P633 ta c th a ra

phng thc bo v ph hp v kinh t cho i tng cn bo v ch cn s dng

mt rle. y l quan im chung ch to cc rle s hin i ngy nay.

Rle s P633 l thit b vi chc nng chnh l bo v so lch tch hp

thm cc chc nng sau y:

Thc hin x l hon ton tn hiu s t o lng, ly mu, s ho cc i

lng u vo tng t n vic x l tnh ton v to cc lnh, cc tn

hiu u ra.

Cch li hon ton v in gia mch x l bn trong ca P633 vi cc

mch o lng iu khin v ngun in do cc cch sp xp u vo

tng t ca cc b chuyn i, cc u vo, u ra nh phn, cc b

chuyn i DC/AC hoc AC/DC.



Gii thiu cc chc nng bo v c tch hp trong rle P633.

- H thng bo v so lch 3 pha, bo v cho i tng 3 cun dy.

- T cn bng pha v cn bng t u dy.

- Lc dng in th t khng cho mi cun dy, c th khng kch hot.

- c tnh hm 3 on.

- Hm b sung phn ng theo thnh phn sng hi bc hai (2f0), tu chn

cho cc ng dng, c th khng kch hot.

- Hm qu t thng phn ng theo thnh phn sng hi bc nm (5f0), c

th khng kch hot.

- Tng tnh n nh vi b pht hin bo ho.

- Bo v so lch chng chm t hn ch.

- Bo v qu dng c tnh thi gian c lp (3 cp tc ng, tc ng

theo tng pha, h thng o lng ring cho tng pha, phn ng theo thnh phn

th t nghch v thnh phn th t khng (TTK)).

- Bo v qu dng c tnh thi gian ph thuc (3 cp tc ng, tc ng

theo tng pha, h thng o lng ring cho tng pha, phn ng theo thnh phn

th t nghch v thnh phn TTK).

- Bo v qu ti nhit.

- Bo v tn s f< >.

- Bo v in p V< >.

- Gim st gi tr ti hn.

- Lp trnh lgc.

Ngi s dng c th la chn hay hu b cc chc nng trn tu theo yu

cu s dng.

4.1.2 Mt s thng s k thut ca rle P633

Thng s dng in u vo:

Tn s nh mc fN 50Hz/60Hz/ (C th

la chn)

Dng in nh mc Inom 1A/5A/0,1A(C th

thay i)



Cng sut tiu th mi

u vo

IN = 1A Xp x 0,05VA

IN = 5A Xp x 0,3VA

IN = 0,1A Xp x 1mVA

u vo nhy cao

1A Xp x 0,05VA

Dung lng qu ti dng

in

Nhit

100IN trong 1s

30IN trong 10s

4IN lin tc

Xung 250IN (Na chu k)

Dung lng qu ti dng

in u vo nhy

cao

Nhit

300A trong 1s

100A trong 10s

15A lin tc

Xung 750A (Na chu k)

Thng s in p u vo nh mc:

in p xoay chiu 50V (f=50/60Hz)

130V

Cng sut tiu th 0,3 VA < vi U = 130V

Thng s u vo nh phn:

S lng 5

in p nh mc 24 n 250V (DC)

Dng tiu th 1,8mA

in p ln nht cho php 300V (DC)

Thng s u ra nh phn:

Kh nng ng ct

ng: 1000W/VA

Ct: 30W/VA

Ct vi ti l in tr: 40W



Ct vi ti l L/R: 250W(50ms)

in p ng ct 230V

Dng ng ct cho php 30A cho 0,5s

5A khng hn ch thi gian

c tnh tc ng

Sau khi dng u vo thch ng vi t s bin dng, t u dy, x l

dng th t khng, cc i lng cn thit cho bo v so lch c tnh ton t

dng trong cc pha IA, IB v IC, b vi x l s so snh v mt tr s theo cc cng

thc sau:

Dng in so lch : 321d

IIII

Dng in hm : )III(2

1I

321R

Trong : 1, 2, 3 ln lt l dng in chy qua cc bo v BI1, BI2, BI3

t cc pha cao, trung v h ca MBA.



c tnh vng tc ng ca bo v so lch

c tnh vng tc ng c 2 im gp. im gp u tin ph thuc vo

tr s chnh nh bo v so lch ngng thp DIFF: Idiff > . im gp th hai

c xc nh bng h s chnh nh dng in hm DIFF: IR,m2.

Ta c phng trnh c tnh ca 3 on tc ng l:

Phng trnh c tnh cho di: 0 IR 0,5Idiff

Id = Idiff >

Phng trnh c tnh cho di: 0,5Idiff > < IR IR,m2

Id = m1IR + Idiff >(1 - 0,5m1)

Phng trnh c tnh cho di: IR,m2 < IR

Id = m2IR + Idiff >(1 - 0,5m1) + 4(m1 m2)

Trong :

m1: l h s gc ca c tnh trong di 0,5Idiff > < IR IR,m2

m2: l h s gc ca c tnh trong di IR,m2 < IR

Ngng iu chnh xc nh DIFF: Idiff > xc nh theo dng qua MBA

trong ch lm vic bnh thng nhm trnh tc ng nhm do dng khng cn

bng sinh ra bi sai s ca thit b o lng.

Trn ngng iu chnh DIFF: Idiff >> rle s tc ng khng cn tnh ton

n hm hi bc cao.

Trn ngng iu chnh DIFF: Idiff >>> dng hm v b pht hin bo ho

khng cn c tnh n, lc rle s tc ng m khng cn quan tm n

bin hm v b pht hin bo ho.

Hm b sung phn ng theo thnh phn sng hi bc hai (2f0)



Khi ng my bin p khng ti hay ct dng in ngn mch ngoi s

xut hin dng in chy qua mch t MBA c gi l dng in t ho. Dng

in ny c th c gi tr ln gp nhiu ln dng in danh nh ca MBA.

Trng hp xu nht (tng ng vi dng t ho ln nht) s xy ra khi ng

my ct in vo thi im in p ngun c gi tr tc thi qua im 0.

Khi qu trnh qu chm dt, dng in t ho tr li tr s xc lp

chng vi phn trm dng in danh nh. V dng in t ho ch chy pha

cun dy MBA ni vi ngun v bin p ang ch khng ti, nn dng in

cun dy bn kia bng khng. Bo v so lch MBA trong trng hp ny c

th cm nhn vic ng MBA khng ti nh khi c ngn mch bn trong MBA,

vi ngun cung cp t mt pha v nu khng c gii php ngn chn bo v, bo

v c th tc ng nhm ct MBA.

phn bit trng hp ng MBA khng ti vi trng hp ngn mch

trong MBA, ngi ta cn c vo tnh cht ca dng in t ho xung kch v

dng in ngn mch trong MBA.

Phn tch thnh phn sng hi ca hai dng in ny ta thy, dng in t

ho xung kch c cha mt phn lng rt ln hi bc hai (khong 70% so vi

hi c s) v c th t ti gi tr cc i n khong 30% tr s dng in s c.

Nu thnh phn hi bc hai trong dng in t ho c tch ra v a

vo tng cng cho dng in hm ca bo v so lch th s ngn chn c tc

ng nhm khi ng MBA khng ti.

Rle P633 c trang b b lc dng in so lch v xc nh c thnh

phn sng hi c bn v thnh phn sng hi bc hai. Khi t l gia sng hi bc

hai v sng hi c bn I(2f0)/I(f0) vt qu mt gi tr t trc mt trong 3 pha

th tn hiu kho tc ng c thc hin theo mt trong 2 cch sau:

+) Kho c 3 pha.

+) Chn lc cho mt pha.



Tn hiu tc ng s khng b kho nu dng so lch vt qu ngng Idiff

>>.

B pht hin bo ho

Khi xy ra ngn mch ngoi vng bo v, thi im ban u, dng ngn

mch ln lm cho BI b bo ho dn n dng in khng cn bng chy qua bo

v ln, c th cao hn ngng tc ng, lm cho bo v tc ng nhm. loi

tr hin tng tc ng nhm ny rle s P633 c trang b b pht hin hin

tng bo ho.

Mi khi dng hm vt qua gi tr khng, b pht hin bo ho s gim

st s bin thin ca dng so lch trong mt khong thi gian. Vi s c trong

vng bo v, dng in so lch xut hin sau khi qua khng cng vi dng hm.

Trong trng hp dng in ln chy qua gy nn bo ho BI dng so lch s

khng xut hin cho ti khi hin tng bo ho ca BI xy ra. V vy, mt tn

hiu kho c gi i da trn ln ca dng so lch c so snh vi dng

hm. Do vic tng tnh n nh ca bo v c m bo.

Hm qu t thng

Khi MBA truyn ti cng sut phn khng gy hin tng qu in p gy

nn qu kch t xy ra trong MBA. Nu qu trnh ny khng n nh th bo v

so lch s tc ng nhm. Thc t phn tch cho thy khi xut hin qu kch t

MBA thnh phn dng in ch yu l sng hi bc 5 (5f0). V vy, rle s da

vo hin tng ny dng cho mc ch n nh. Rle P633 s lc dng in so

lch v xc nh cc thnh phn sng hi c bn I(f0) v hi bc 5 I(5f0). Nu t l

I(5f0)/I(f0) tng vt qu gi tr t trc th bo v tc ng. Nu dng hm nh

hn 4Iref tn hiu ct pha b s c s b kho mt cch c chn lc.



4.2. R le hp b qu dng s 7SJ64

4.2.1 Gii thiu tng quan v rle 7SJ64

R le 7SJ64

SIPROTEC4 7SJ64 l loi r le c dng bo v v kim sot cc l

ng dy phn phi v ng dy truyn ti vi mi cp in p, mng trung

tnh ni t, ni t qua in tr thp, ni t b in dung. R le cng ph hp

dng cho mch vng kn, mng hnh tia, ng dy mt hoc nhiu ngun cung

cp. 7SJ64 l loi r le duy nht ca h r le 7SJ6 c c im chc nng bo v

linh hot, c th ln ti 20 chc nng bo v tng ng vi cc yu cu ring.

Cc chc nng d s dng, t ng ho.

Rle ny c nhng chc nng iu khin n gin cho my ct v cc thit

b t ng. Logic tch hp lp trnh c (CFC) cho php ngi dng thc hin

c tt c cc chc nng sn c, v d nh chuyn mch t ng (kho lin

ng).

4.2.2 Cc chc nng ca 7SJ64

Bo v qu dng c thi gian (51, 51N)

Bo v qu dng ct nhanh (50, 50N)

Bo v qu dng c thi gian c hng (67, 67N)

Bo v chng chm t nhy cao

Bo v thay i in p (59N/64)

Bo v chng chm t chp chn

Bo v chng chm t tng tr cao (87N)

Hm dng xung kch



Bo v ng c (14)

Bo v qu ti (49)

Kim sot nhit (38)

Bo v tn s (81O/U)

Bo v cng sut (32)

Bo v chng h hng my ct (50BF Bo v dng th t nghch (46)

Kim sot thnh phn pha

ng b ho (25)

T ng ng li

nh v s c (21FL)

Lockout (86).

Chc nng iu khin / logic lp trnh c.

o iu khin my ct v dao cch li.

o iu khin qua bn phm, u vo nh phn, h thng DIGSI 4

hoc SCADA.

o Ngi s dng ci t logic tch hp lp trnh c (ci t kho

lin ng).

Chc nng gim st.

o o gi tr dng lm vic

o Ch th lin tc.

o ng h thi gian.

o Gim st ng ngt mch.

o 8 biu dao ng ghi li.

Cc cng giao tip

o Giao din h thng:

- Giao thc IEC 60870 5 103.

- PROFIBUS FMS/ - DP.

- DNP 3.0 / MODBUS RTU

o Cung cp giao din cho DIGSI 4 (modem) / o nhit (RTD

box)

o Giao din mt trc rle cho DIGSI 4.



o ng b thi gian thng qua IRIG B / DCF 77.

Biu cc chc nng ca r le c ch ra nh sau:

Biu chc nng ca 7SJ64

4.3. R le 7RW60

4.3.1. Gii thiu chung

Rle s 7RW60

Rle 7RW60 l rle s a chc nng c sn xut bi cng ty Siemens,

c ni vi my bin in p, tc ng khi pht hin bin ng cc i lng

tn s, in p hoc b qu kch thch, c th s dng bo v cho ng dy,

my pht, my bin p, ng c,...



Cc chc nng chnh ca rle 7RW60:

- Bo v in p (km p v qu p)

- Bo v tn s (tn s tng hoc gim qu ngng cho php, tc thay

i ca tn s vt qu gii hn)

- Bo v chng qu kch thch

4.3.2. Mt s thng s k thut chnh

Thng s mch o u vo:

Tn s nh mc fN 50Hz/60Hz/ (C th

la chn)

in p nh mc VN 100-125 V

Cng sut tiu th mi u vo < 0,2 VA

Dung lng qu ti Lin tc 200 V

Trong vng 10s 230 V

Thng s u vo nh phn:

S lng 5

in p nh mc 24 n 250V (DC)

Dng tiu th 2,5 mA



CHNG 5

TNH TON CC GI TR CHNH NH V CI T CHO RLE

5.1. Cc chc nng s dng v thng s ci t cho my bin p ca r le

P633

5.1.1. Chn my bin dng in v my bin in p

a) My bin dng in

iu kin chn:

Dng in: IdBI Ilvcp

in p: UdBI Udl

Trong :

IdBI: Dng in nh mc ca my bin dng in.

Ilvcp: Dng in ph ti lm vic lu di cho php.

UdBI: in p nh mc ca my bin dng.

Udl: in p li in.

Ta thng chn dng nh mc s cp ca my bin dng ln hn dng

lm vic ln nht 10 n 40%, y ta chn K=1,2.

La chn h s gii hn dng in:

Pha 220kV:

Ilvmax = 1,2. 125000/(3.230) = 376,53 A

Do ta chn BI c cc thng s sau: Im BI s cp = 400A, ImBItc = 1A

INM max = 20,63

Vy INM max = 20,63 0,314 = 6,478 (kA)

T s INM max/ImBI1= 16,195

Ta chn cp chnh xc l 5P20

Tng t vi cc cp 110kV v 35kV.

Da vo cc iu kin ta chn my bin dng vi cc thng s nh sau:



Bng 5.1. Thng s ca cc BI c chn

Cp in p, kV 220 110 35

Dng in lm vic MBA, A 314 596,4 997

Dng in nh mc s cp, A 400 800 1250

Dng in nh mc th cp, A 1 1 1

T s my bin dng 400/1 800/1 1250/1

in p nh mc, kV 220 110 35

Cp chnh xc 5P20 5P20 5P20

Vi BI trung tnh th iu kin chn l BI khng b bo ha khi c ngn

mch m bo chc nng so lch chng chm t hn ch khng tc ng

nhm.

Do ta chn BI c cc thng s sau: Im BI s cp = 400A, ImBItc = 1A

INM max = 4,301 kA

T s INM max/ImBI = 10,753

Ta chn cp chnh xc l 5P20

b) My bin in p

iu kin chn:

o in p: UdBU Ung.

o Cp chnh xc ph hp vi yu cu ca dng c o.

Da vo cc iu kin ta chn my bin p vi cc thng s nh sau:

Bng 5.2. Thng s ca cc BU c chn

Cp in p, kV 220 110 35

in p nh mc, kV 220 110 35

in p s cp, V 220000 / 3 110000 / 3 36000 / 3

in p th cp, V 100 / 3 100 / 3 100 / 3

Cp chnh xc 3P 3P 3P



5.1.2. Cc thng s k thut ca my bin p t ngu trong trm

Bng 5.3. Thng s k thut ca MBA

Cp in p

Cc thng s 220 kV 110 kV 35 kV

Cng sut inh mc, Sm, MVA 125 125 63

in p inh mc, Um, kV 230 121 36,5

Dng in inh mc, Im, A 314 596 997

T u dy Y0 Y0 -11

T s bin dng 400/1 800/1 1250/1

S nc iu chnh in p 16 nc: 8 x 1,25%

5.2. Tnh ton chnh nh cc chc nng bo v trong r le P633

5.2.1. Chc nng bo v so lch c hm

- on 1: Dng so lch mc thp IDIFF > l gi tr khi ng ca dng so

lch on a, gi tr ny biu th nhy ca bo v khi xt n dng khng cn

bng c nh qua rle, trong ch lm vic bnh thng th:

IDIFF > = Kat.IKCB

Kat: h s an ton; Kat=1,2 1,3

IKCB l dng in khng cn bng, trong trng hp bnh thng, theo

nguyn l o lng ca rle 7UT613 th dng so lch bng khng, tuy nhin

trong thc t n o c dng khng cn bng bao gm nhng thnh phn sau:

IKCB= (Kn.KKCK.fi + U).IdB

- Kn l h s ng nht ca cc my bin dng, Kn=1.

- KKCK l h s k n nh hng ca thnh phn khng chu k ca dng

ngn mch trong qu trnh qu , KKCK= 1.



- fi : sai s tng i cho php ca BI, fi =0,1.

- U l phm vi iu chnh in p ca u phn p, vi phm vi iu

chnh l 8 1,25 % = 10 %, U= 0,1.

- IdB: dng in nh mc ca my bin p (ly pha 220kV lm pha c

bn).

Do IDIFF > = (1,2 1,3)(1.1.0,1 + 0,1).IdBA

IDIFF > = (0,2 0,3). ImBA

Thng chn IDIFF > =DIFF>

ddB

I

I= 0,2

- on 2: Di 0,5Idiff > < IR IR,m2

Vi IR,m2 = 4, h s gc m1 = 0,3

- on 3: IR,m2 < IR

H s gc m2 = 0,7

- Ngng tc ng cp 2: IDIFF>> = C-T

NU % =10,78% 1,2*N

1

U %= 11,132

Chn IDIFF>> = 11.ImBA

Ti ngng ny s lp tc tc ng ct 3 pha my bin p khng quan

tm n vng hm b sung.

T l thnh phn hi bc hai t n ngng chnh nh, tn hiu ct s

b kho, trnh cho rle khi tc ng nhm (20%, theo mc nh ca nh sn

xut)

T l thnh phn hi bc nm t n ngng chnh nh, tn hiu ct s

b kho, trnh cho rle khi tc ng nhm (20%, theo mc nh ca nh sn

xut).

Thi gian tr ca cp IDIFF > l 0s.

Thi gian tr ca cp IDIFF >> l 0s.



5.2.2. Bo v chng chm t hn ch (REF)

Bo v chng chm t REF dng bo v s c trong MBA lc c

trung tnh ni t. Vng bo v l vng gia my bin dng t dy trung tnh

v t my bin dng ni theo s b lc dng th t khng t pha u ra

cun dy ni hnh sao ca MBA. Dng in bnh thng qua trung tnh xp x 0,

tuy nhin nu c xt n sai s ca bin dng v dng khng cn bng th ta s

t:

IREF = kat . fi% .IdBA

kat = 1,2 : h s an ton.

fi% = 0,1 : sai s ln nht ca my bin dng.

IREF =1,2.0,1.ImBA = 0,12. ImBA

Chn dng in khi ng : IREF> = 0,2. ImBA

dc ca ng c tnh : m = 1,005

Thi gian tr tc ng : T IREF> = 0 s

5.2.3. Chc nng bo v qu ti nhit

Ngng cnh bo alarm:

- Ci t ngng cnh bo nm di ngng tc ng ct my ct gip

trnh phi ct my ct thng qua vic sm gim ti cho my bin p. Ngng

phn trm ci t c th ln n tng nhit ti hn dng in ln nht cho

php.

Ta ci t gi tr alarm = 95%.

- H s k = 1,15. Tn hiu cnh bo nn c a ra khi tng nhit

t n tng nhit ti hn dng in nh mc my bin p.

Vic chnh nh rle P633 c cho trong ph lc 1.

T y ta c c tnh bo v so lch c hm ca r le P633:



c tnh ca bo v so lch co ham

5.3. Nhng chc nng bo v dng r le 7SJ64

5.3.1. Bo v qu dng ct nhanh c hng

Dng in khi ng c xc nh theo iu kin:

Ik = Kat.INng.max, kA

Trong : Kat: H s an ton, chn Kat = 1,2

INng.max: Dng ngn mch ngoi ln nht qua BI khi c

ngn mch ngoi

Dng khi ng pha th cp ca BI c xc nh theo iu kin:

Ik>> = 3kd

I

I.10

n, (A)

Trong : nI: t s bin tng ng

Thi gian t cho cc bo v qu dng ct nhanh c hng ta chn l 0,05s.

Bo v qu dng ct nhanh pha 220 kV

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

IDIFF>= 0,2

IR,m2

= 4

m1= 0,3

m2= 0,7

I*DIFF

I*REST



INng.max = max {IN2max; IN3max } qua BI1 = 5,855

INng.max = INng.max .Icb1 = 5,855.0,314 =1,838 ( kA)

Ik220>> = 1,2 . 1,838 = 2,206 (kA)

Dng khi ng pha th cp (ca r le):

Ik220>> = 32,206 .10 = 5,515(A)

400 /1

Bo v qu dng ct nhanh pha 110 kV

INng.max = max {IN1max; IN3max } qua BI2 = 2,451

Dng khi ng pha s cp:

Ik = 1,2. INng.max . Icb2 = 1,2.2,451.0,628 = 1,847( kA)


Ik110>> = 31,847 .10 2,309(A)

800/1

5.3.2. Bo v qu dng ct nhanh th t khng c hng

Dng khi ng ca bo v:

Ik = 3.Kat.I0Nng.max, kA

Trong :

Kat: H s an ton, Kat=1,2.

I0Nng.max: Dng ngn mch ngoi th t khng ln nht qua bo v.

Dng khi ng pha th cp ca BI c xc nh theo iu kin:

3k0k

I

II 10 (A)

n

Thi gian t cho cc bo v qu dng TTK ct nhanh c hng ta chn l

0,05s.

Bo v qu dng TTK pha 220 kV:

I0Nng.max = max(I0N2) qua BI1 = 1,923

Dng khi ng pha s cp:

Ik = 1,2. I0Nng.max . Icb1 = 1,2.3.1,923.0,314 = 2,174( kA)




30k220

2,174I 10 5,434 A

400 /1

Bo v qu dng TTK pha 110 kV:

I0Nng.max = max(I0N1) qua BI2= 0,334

T ta tnh c: 0k220I 1,2.3. 0,334 . 0,628 0,755 kA

30k1100,755

I .10 0,944 A800 /1

5.3.3. Bo v qu dng c hng thi gian tr

Hai thng s cn chn l: Ik v t (chn c tnh thi gian c lp)

Ik = K ImBA

Trong : K: H s chnh nh, ly K=1,6

IdBA: Dng nh mc ca MBA

Bo v qu dng t pha 220 kV:

Ik220> = 1,6 . 314 = 502,4 (A)

Dng khi ng pha th cp BI c xc nh:

k220

502,4I 1,256(A)

400 /1


Ik110>= 1,6 . 596 = 953,6 (A)


k110

953,6I 1,192(A)

800 /1


Ik35>=1,6 . 997 = 1595,2 (A)


k35

1,5952I 1,276(A)

1250 /1

Thi gian t cho cc r le:

Gi thit thi gian ct ca cc xut tuyn ng dy cc pha l 1s v cp

thi gian chn lc l 0,5s, ta c:



tRL(35) = tD35 + t = 1 + 0,5 = 1,5s

tRL(220) = max{tD110; tRL(35)} + t = max {1; 1,5} + 0,5 = 2s

tRL(110) = max{tD220; tRL(35)} + t = max {1; 1,5} + 0,5 = 2s

5.3.4. Bo v qu dng TTK c hng c thi gian tr

Dng in khi ng ca bo v ny c chn theo iu kin sau:

I0k = (0,2 0,3).ISCBI

Trong : ISCBI l dng inh mc s cp cua bin dong

Bo v ny ch t cho 2 pha 220kV v 110kV.

- Dng khi ng pha s cp:

Ik220> = 0,2 . 400 = 80 A

Ik110> = 0,2 . 800 = 160 A

- Dng khi ng pha th cp:

Ik220> = 0,2A

Ik110> = 0,2 A

Bo v qu dng TTK s dng c tnh thi gian c lp. Thi gian lm

vic tng t nh bo v qu dng c hng c tr.

5.3.5. Bo v chng my ct t chi tc ng 50BF

Chc nng chng my ct t chi tc ng s dng gim st mch my

ct v tc ng gi tn hiu i ct cc my ct cp cao hn khi my ct t chi tc

ng. Mi bo v tc ng u gi tn hiu n cc my ct tng ng, b m

thi gian ca chc n

Documents

DATN bảo vệ trạm biến áp 220/110/35 kV ĐHBK