Do an Mon Hoc NMD-OK4

Chng I

n mn hc Nguyn Dip Tng- Lp HT 4- K42

Chng I

tnh ton ph tI v cn bng cng sut

Cht lng in nng l mt yu cu quan trng ca ph ti. m bo cht lng in nng ti mi thi im , in nng do cc nh my pht in pht ra phi hon ton cn bng vi lng in nng tiu th cc h tiu th k c tn tht in nng. V in nng t c kh nng tch lu nn vic cn bng cng sut trong h thng in l rt quan trng.

Trong thc t lng in nng tiu th ti cc h dng in lun lun thay i. Vic nm c quy lut bin i ny tc l tm c th ph ti l iu rt quan trng i vi vic thit k v vn hnh. Nh vo th ph ti m ta c th la chn c cc phng n ni in hp l , m bo cc ch tiu kinh t v k thut, nng cao tin cy cung cp in. Ngoi ra da vo th ph ti cn cho php chn ng cng sut cc my bin p v phn b ti u cng sut gia cc t my pht in trong cng mt nh my v phn b cng sut gia cc nh my in vi nhau.

Theo nhim v thit k nh my in thu in c tng cng sut t l 400 MW gm c 4 my pht in kiu thy in cung cp cho ph ti 3 cp in p my pht, 110 KV v ni vi h thng cp in p 220 KV.

Ta chn my pht in loi CB-1500/170-96 c cc thng s sau:

SFm(MVA)PFm(MW)cos(mUFm(KA)IFm(KA)XdXdXd

117,6471000,8513,84,920,210,290,65

Trong nhim v thit k cho th ph ti ca nh my v th ph ti ca cc cp in p di dng bng theo phn trm cng sut tc dng (Pmax) v h s (cos(tb) ca tng ph ti tng ng t ta tnh c ph ti ca cc cp in p theo cng sut biu kin nh cng thc sau:

Trong :

S(t) : Cng sut biu kin ca ph ti ti thi im t tnh bng (MVA)

P% : Cng sut tc dng ti thi im t tnh bng phn trm cng sut cc i

Pmax : Cng sut ca ph ti cc i tnh bng (MW)

cos(tb :H s cng sut trung bnh ca tng ph ti

1-1. th ph ti ca ton nh my.

Nhim v thit k cho nh my gm 4 t my pht thy in c :

PFm = 100 MW , cos(tbm = 0,85.

Do cng sut biu kin ca mi t my l :

MVA

Tng cng sut t ca ton nh my l:

PNMm = 4PFm = 4.100= 400 MW

hay SNMm = 4SFm= 4.117,647 = 470,588 MVA

T th ph ti ca nh my in tnh c cng sut pht ra ca nh my tng thi im l:

vi

Kt qu tnh ton cho bng 1-1 v th v hnh 1-1:

Bng 1-1

t (gi)0 ( 55 ( 88 (1111(1414(1717( 2020( 2222(24

PNM(%)809010010010010010080

SNM(t) (MVA)376,47423,53470,59470,59470,59470,59470,59376,47

1-2.Ph ti t dng ca nh my

Theo nhim v thit k h s ph ti t dng ca nh my ( =0,9% cng sut nh mc ca nh my vi cos(tddm = 0,85 tc l bng h s cng sut nh mc ca nh my v c coi l hng s vi cng thc :

Std(t)=(.SNM = 0,009.470,588 = 4,235 (MVA)

1-3. th ph ti a phng cp in p UF ( 13,8 KV )Ph ti a phng ca nh my c din p 13,8 KV, cng sut cc i PUfmax = 14 MW , cos(tb = 0,85 : bao gm 2 kp*4 MW*3 Km v 2 n*3MW*3 Km. xc nh th ph ti a phng phi cn c vo s bin thin ph ti hng ngy cho v nh cng thc :

vi

Kt qu tnh c theo tng thi im t cho bng 1-3 v th ph ti a phng cho hnh 1-3. Bng 1-3t (gi)0 ( 55 ( 88 (1111(1414(1717( 2020( 2222(24

PUf(%)70809090901009080

SUf(t) (MVA)11,5313,17614,82314,82314,82316,47014,82313,176

1-4. th ph ti trung p (110 KV)

Nhim v thit k cho P110max = 280 MW v cos(tb = 0,82 :gm 2 kp*80 MW v 4 n*50 MW. xc nh th ph ti pha trung p phi cn c vo s bin thin ph ti hng ngy cho v nh cng thc :

vi

Kt qu tnh c theo tng thi im t cho bng 1-4 v th ph ti pha trung p cho hnh 1-4

Bng 1-4

t (gi)0 ( 55 ( 88 (1111(1414(1717( 2020( 2222(24

PT(%)80909090909010080

ST(t) (MVA)273,17307,32307,32307,32307,32307,32341,46273,17

1-5. th ph ti v h thng (220 KV).

Ton b cng sut tha ca nh my c pht ln h thng qua ng dy kp di 150 Km .Tng cng sut h thng SHT=2800 MVA vi in khng nh mc XHT=1,13. D tr quay ca h thng SdtHT=200 MVA . Nh vy phng trnh cn bng cng sut ton nh my l:

SNM(t) = SUf(t) + ST(t) + SVHT(t) + St d(t)

T phng trnh trn ta c ph ti v h thng theo thi gian l:

SVHT(t) = SNM(t) - {SUf(t) + ST(t) + St d(t)}

T ta lp c bng tnh ton ph ti v cn bng cng sut ton nh my nh bng 1-5 v th ph ti trn hnh 1-5.

Bng 1-5

t (gi)0 ( 55 ( 88 (1111(1414(1717( 2020( 2222(24

SNM(t)376,47423,53470,59470,59470,59470,59470,59376,47

Std(t)4,2354,2354,2354,2354,2354,2354,2354,235

SUf(t)11,52913,17614,82314,82314,82316,4714,82313,176

ST(t)273,17307,32307,32307,32307,32307,32341,46273,17

SVHT(t) (MVA)87,53598,8144,21144,21144,21142,57110,185,89

1-6. Nhn xt chung.

Ph ti nh my phn b khng u trn c ba cp in p v gi tr cng sut cc i c tr s l: SUfmax = 16,47 MVA

STmax = 341,46 MVA

SVHTmax = 144,21 MVA

Tng cng sut nh mc ca h thng l 2800 MVA, d tr quay ca h thng SdtHT = 200 MVA. Gi tr ny ln hn tr s cng sut cc i m nh my pht ln h thng SVHTmax = 144,21 MVA.

Ph ti in p trung chim phn ln cng sut nh my do vic m bo cung cp in cho ph ti ny l rt quan trng.

T cc kt qu tnh ton trn ta xy dng c th ph ti tng hp ca nh my nh sau:

Chng II

la chn phng n ni in chnh

Chn s ni in chnh l mt trong nhng nhim v ht sc quan trng trong thit k nh my in. S ni in hp l khng nhng em li nhng li ch kinh t ln lao m cn p ng c cc yu cu k thut

Theo nhim v thit k nh my c 4 t my pht, cng sut nh mc ca mi t my l 100 MW c nhim v cung cp in cho ph ti ba cp in p sau:

Ph ti a phng cp in p Uf c:

SUfmax = 16,47 MVA

SUfmin = 11,53 MVA

Ph ti trung p cp in p 110 KV c:

STmax = 341,46 MVA

STmin = 273,17 MVA

Ph ti v h thng cp in p 220 KV c:

SVHTmax = 144,21 MVA

SVHTmin = 85,89 MVA

Theo nhim v thit k th ph ti a phng pha in p my pht c cp bng cc ng cp kp m in p u cc my pht l 13,8 KV. Cng sut c ly t u cc ca hai my pht ni vi t ngu v mi my cung cp cho mt na ph ti a phng. Trong trng hp mt my b s c th my cn li vi kh nng qu ti s cung cp in cho ton b ph ti a phng.

Nh my c ba cp in p l 13,8 KV; 110KV; 220KV, trong li 110KV v 220KV u l li c trung tnh trc tip ni t v vy lin lc gia ba cp in p ta dng my bin p t ngu .

T nhng nhn xt trn y ta c th xut mt s phng n nh sau:

2-1. Phng n I (Hnh 2-1).

Do ph ti cao v trung p ln hn nhiu so vi cng sut nh mc ca my pht nn mi thanh gp 110 KV v 220 KV c ni vi mt b my pht in - my bin p ba pha hai dy qun ln lt l F3-B3 v F4-B4. cung cp in thm cho cc ph ti ny cng nh lin lc gia ba cp in p dng hai b my pht in -my bin p t ngu (F1-B1 v F2-B2).

Ph ti a phng Uf c cung cp din qua hai my bin p ni vi hai cc my pht in F1,F2.

u im ca phng n ny l b tr ngun v ti cn i. Tuy nhin phi dng n ba loi my bin p. Ngoi ra khi SVHTmin = 85,89MVA < SFm = 117,647MVA nn nu cho b F4-B4 lm vic nh mc th c th pha trung p nhn c nng lng phi qua hai ln bin p (v ph ti trung p rt ln), ln th nht qua B4, ln th hai qua B1 v B2.

2-2. Phng n II(Hnh 2-2).

khc phc nhc im phng n I, chuyn b F4-B4 t thanh gp 220 KV sang pha 110KV. Phn cn li ca phng n II ging nh phng n I.

Nhn xt :

tin cy cung cp in m bo, gim c vn u t do ni b cp in p thp hn thit b r tin hn. Phn cng sut lun tha bn trung c truyn qua my bin p t ngu a ln h thng (v tng cng sut cc b bn trung lun ln hn ph ti cc i bn trung). u im ca phng n ny l ch dng hai loi my bin p. Ngoi ra do STmin = 273,17MVA > 2SFm =2.117,647 =235,3MVA nn 2 b ni vi thanh gp 110KV c th lun lun lm vic ch nh mc.

2-3. Phng n III(Hnh 2-3).

Nhn xt :

- S lng my bin p nhiu i hi vn u t ln, ng thi trong qu trnh vn hnh xc sut s c my bin p tng, tn tht cng sut ln.

- Khi s c b bn trung th my bin p t ngu chu ti qua cun dy chung ln so vi cng sut ca n.Tm li: Qua nhng phn tch trn y ta li phng n I v phng n II tnh ton, so snh c th hn v kinh t v k thut nhm chn c s ni in ti u cho nh my in.

Chng III

Chn my bin p v tnh tn tht in nng

3-1.Chn my bin p - phn phi cng sut cho my bin p.

Gi thit cc my bin p c ch to ph hp vi iu kin nhit mi trng ni lp t nh my in . Do vy khng cn hiu chnh cng sut nh mc ca chng.

I.Phng n I (hnh 2-1).

1. Chn my bin p :

- Cng sut nh mc ca cc my bin p t ngu B1, B2 c chn theo iu kin sau: SB1m = SB2m ( SFm

Trong ( l h s c li ca my bin p t ngu

Do : SB1m = SB2m ( MVA

T kt qu tnh ton trn ta chn t hp ba my bin p t ngu mt pha cho mi my bin p B1,B2 loi: AOTH-120 c cc thng s k thut nh bng 3-1 (l thng s cho mt pha trong t hp 3 pha ):

Bng 3-1

Sm(MVA)Um (KV)

UN%(*)(P0(KW)(PN%

I0(%)Gi (106)

UCUTUHC-TC-HT-HC-HC-TC-HT-H

120

13,810--2103452202350,56440

Nh vy tng cng sut ca t hp 3 t my bin p t ngu mt pha l:

3.120 = 360 MVA

My bin p B3 c chn theo s b :

SB3m ( SFm = 117,647 MVA

Do ta chn my bin p tng p ba pha 2 cun dy c Sm = 125 MVA l loi : T-125 (121/13,8) c cc thng s k thut nh bng 3-2 Bng 3-2

Sm(MVA)UCm(KV)UHm(KV)(P0(KW)(PN(KV)UN%I0%Gi

(106 )

12512113,810040010,50,55120

- My bin p B4 cng c chn theo s b nh i vi B3:

SB4m ( SFm = 117,647 MVA

Do ta chn my bin p tng p ba pha 2 cun dy c Sm = 125 MVA l loi : T-125 (242/13,8) c cc thng s nh bng 3-3 .

Bng 3-3

Sm(MVA)UCm(KV)UHm(KV)(P0(KW)(PN(KV)UN%I0%Gi (106 )

12524213,8115380110,56480

2.Phn b cng sut cho cc my bin p.

- thun tin trong vn hnh, cc b my pht- my bin p hai cun dy F3-B3 v F4-B4 cho lm vic vi th bng phng sut c nm. Do cng sut ti ca mi my l:

SB3 = SB4 = SFm = 117,647MVA < SB3,B4m=125 MVA

Do i kin lm vic bnh thng B3 v B4 khng b qu ti

- Ph ti qua mi my bin p t ngu B1v B2 c tnh nh sau :

Ph ti truyn ln pha trung p ca mi my bin p t ngu l :

Ph ti truyn ln pha cao p ca mi my bin p t ngu l :

Ph ti truyn ln pha h p ca mi my bin p t ngu l :

Kt qu tnh ton cho trn bng 3-4:

Bng 3-4

t (h)0(55(88(1111(1414(1717(2020(2222(24

SB1=SB2117,647117,647117,647117,647117,647117,647117,647117,647

SCT(t)77,7694,83594,83594,83594,83594,835111,9177,76

SCC(t)-15.06-9,42413,28313,28313,28312,46-3,774-15,88

SCH(t)62,7185,412108,12108,12108,12107,3108,1361,883

Du - chng t cng sut i t pha thanh gp h thng 220KV sang thanh gp 110KV b xung lng cng sut thiu pha 110KV.

Qua bng phn b cng sut 3-5 thy rng:

SCCmax = 13,283 MVA < SB1,B2m=360 MVA

SCTmax = 111,91 MVA < SM = (.SB1m = 180 MVA

SCHmax = 108,13 MVA < SM = 180 MVA

Nh vy cc my bin p chn khng b qu ti khi lm vic bnh thng.

3. Kim tra cc my bin p khi b s c.

V cng sut nh mc ca cc my bin p hai cun dy c chn theo cng sut nh mc ca my pht in nn vic kim tra qu ti ch cn xt i vi my bin p t ngu.

Coi s c nng n nht l lc ph ti trung p cc i STmax= 341,463 MVA.

Khi SVHT =110,1 MVA ; SUf =14,823 MVA

a) Gi thit s c b F3-B3.

Kim tra iu kin : 2.Kqtsc. (.SB1m ( STmax

( 2.1,4.0,5.360 =504 > 341,463 MVA ( tho mn iu kin )

Lc ny cng sut ti ln trung p qua mi my l:

SCT-B1 = SCT-B2 = STmax/2 = 170,732 MVA

Cho cc my pht F1v F2 lm vic vi gi tr nh mc. Do cng sut qua cun h ca B1 v B2 l:

SCH-B1,B2 = SFm - SUf /2 - Std /4

= 117,647 - 14,823/2 - 4,235/4 = 109,177 MVA

Cng sut ti ln cao p ca 1 MBA:

SCC-B1,B2 = SCH-B1,B2 - SCT-B1,B2 = 109,177 - 170,732 = - 61,555 MVA

Du - chng t cng sut i t thanh gp h thng 220 kV sang thanh gp trung p 110 kV b vo phn cng sut thiu vi tr s 2.61,555 MVA. Khi lng cng sut nh my cp cho pha cao p cn thiu mt lng :

Sthiu = SVHT - SB4- 2.SCC-B1,B2

= 110,1- 117,647- (-61,555) = 155,56 MVA < SdtHT =200 MVA

Vi lng cng sut thiu ny nh hn d tr quay ca h thng.

Qua trn thy rng khi s c b F3-B3,hai my bin p t ngu B1,B2 lm vic khng b qu ti.

b) Khi s c my bin p t ngu B1(hoc B2).

Khi B1s c th F1 ngng. Trng hp ny kim tra qu ti ca B2:

Kim tra iu kin : Kqtsc. (.SB1m ( STmax SB3

( 1,4.0,5.360 =252 >341,463 - 117,647=223,816 MVA ( tho mn iu kin )

- Cng sut ti ln trung p:

SCT-B2 = STmax- SB3 = 341,463 - 117,647 = 223,816 MVA

- Cng sut qua cun h ca B2:

SCH-B2 = SFm- SUf - Stdmax/4 =

= 117,647 - 14,823 - 4,235/4 = 101,765 MVA

- Cng sut ti ln pha cao p:

SCC-T2 = SCH-B2 - SCT-B2 = 101,765 - 223,816 = - 122,051 MVA

Nh vy khi s c B1, m bo cho ph ti trung p cc i phi ly cng sut t thanh gp h thng sang thanh gp 110 kV mt lng 122,051 MVA. Khi lng cng sut nh my cp cho pha cao cn thiu l:

Sthiu=SVHT - SB 4 - SCC-B2 =

= 110,1 - 117,647 - (-122,051) = 114,504 MVA< SdtHT=200 MVA

Lng thiu ny nh hn d tr quay ca h thngnn B2 cng khng b qu ti.

II.Phng n II (hnh 2-2).1. Chn my bin p.

-Hai my bin p B3 v B4 c chn theo s b .Do hai my bin p ny cng ni vi thanh gp in p 110 KV nn c chn ging nhau v chn ging my bin p B3 phng n I l my bin p loi : T-125-121/13,8 c cc thng s k thut nh bng 3-2 (phng n I ).

-Hai my bin p t ngu B1 v B2 c chn tng t nh phng n I

Cng sut nh mc ca cc my bin p t ngu B1, B2 c chn theo iu kin sau: SB1m = SB2m ( SFm

Do : SB1m = SB2m ( MVA

Ta chn my bin p c k hiu: ATTH-250 c cc thng s k thut nh sau :

Sm(MVA)Um (KV)

UN%(*)(P0(KW)(PN%

I0(%)Gi (106)

UCUTUHC-TC-HT-HC-HC-TC-HT-H

25023012113,8113220120520--0,510000

2. Phn phi cng sut cho cc my bin p.

m bo kinh t v thun tin trong vn hnh, cc my pht F3,F4 cho lm vic vi th ph ti bng phng sut c nm. -Do cng sut ti qua mi my bin p B3,B4 l:

SB3 = SB4 = SFm = 117,647 MVA

- Ph ti qua cc my bin p t ngu T1v T2 c tnh nh sau :

Ph ti truyn ln pha cao p ca mi my bin p t ngu l :

Ph ti truyn ln pha trung p ca mi my bin p t ngu l :

Ph ti pha h p ca mi my bin p t ngu l :

Da vo bng 1-5 tnh chng I v cc cng thc trn ta tnh c ph ti cho tng thi im , kt qu ghi trong bng 3-5Bng 3-5

t (h)0(55(88(1111(1414(1717(2020(2222(24

SB1=SB2117,647117,647117,647117,647117,647117,647117,647117,647

SCT(t)18,93836,01236,01236,01236,01236,01253,08518,938

SCC(t)43,76849,472,10672,10672,10671,28355,0542,945

SCH(t)62,7185,412108,12108,12108,12107,29108,1361,883

Du - chng t cng sut i t pha thanh gp h thng 220KV sang thanh gp 110KV b xung lng cng sut thiu pha 110KV.

Qua bng phn b cng sut 3-5 thy rng:

SCCmax = 72,106 MVA < SB1,B2m=250 MVA

SCTmax = 53,085 MVA < SM = (.SB1m = 125 MVA

SCHmax = 108,13 MVA < SM = 125 MVA

Nh vy cc my bin p chn khng b qu ti khi lm vic bnh thng.

3. Kim tra cc my bin p khi b s c.

Cng coi s c nguy him nht l xy ra khi ph ti trung p cc i. i vi cc b my pht in - my bin p hai cun dy khng cn kim tra qu ti v cng sut nh mc ca cc my bin p ny c chn theo cng sut nh mc ca my pht in. Do vic kim tra qu ti ch tin hnh vi cc my bin p t ngu.

a) Khi s c b F3-B3 (hoc F4-B4).

Kim tra iu kin : 2.Kqtsc. (.SB1m ( STmax

( 2.1,4.0,5.250 =350 > 341,463 MVA ( tho mn iu kin )

Khi cng sut ti ln cc pha qua mi my bin p t ngu c xc nh nh sau:

- Pha trung p:

SCT-B1 = SCT-B2 = ( STmax - SB4)= (341,463 - 117,647) = 111,908 MVA

- Cng sut qua cun h:

SCH-B1 = SCH-B2 = SFm - SUf/2 - Std/4 = 109,178 MVA

- Cng sut pht ln pha cao:

SCC-B1 = SCC-B2 = SCH-B1- SCT-B1 = 109,178 - 111,908 = - 2,73 MVA

Khi ph ti h thng thiu mt lng cng sut l:

Sthiu = SVHT - (SCC-B1 + SCC-B2) = 110,1 - 2.(- 2,73) = 115,562 MVA

Lng cng sut thiu ny nh hn d tr quay ca h thng =200MVA.

Qua trn thy rng khi s c b F3- B3 th cc my bin p t ngu B1,B2 khng b qu ti.

b) Khi s c t ngu B1 (hoc B2).

Khi B1 b s c th F1 ngng, ta kim tra qu ti ca B2.

Kim tra iu kin : Kqtsc. (.SB1m ( STmax - 2.SB3

(1,4.0,5.250 =175 >341,463 -2.117,647=106,169 MVA (tho mn iu kin )

Cng sut ti qua cc pha ca B2 nh sau:

Pha trung p:

SCT-B2 = STmax - (SB3 + SB4) = 341,463 - 2.117,647 = 106,169 MVA

- Pha h p:

SCH-B2 = SFm - SUf - Std/4 = 101,765 MVA

Pha cao p:

SCC-B2 = SCH-B2 - SCT-B2 = 106,169 - 101,765 = - 4,404 MVA

Ph ti h thng b thiu mt lng cng sut l:

Sthiu = SVHT - SCC-B2 = 110,1 + 4,404= 114,504 MVA< SdtHT=200MVA

Lng ny nh hn d tr quay ca h thng.

Do trong trng hp ny B2 cng khng b qu ti.

Tm li: Cc my bin p chn u tho mn cc yu cu k thut khi lm vic bnh thng v khi s c.

3-2 Tnh ton tn tht in nng.

Tnh ton tn tht in nng l mt vn khng th thiu c trong vic nh gi mt phng n v kinh t v k thut. Trong nh my in tn tht in nng ch yu gy nn bi cc my bin p tng p.

I. Phng n I (Hnh 2-1).

tnh ton tn tht in nng trong cc my bin p ta da vo bng phn b cng sut ca my bin p cho bng 3-4

1. Tn tht in nng hng nm ca my bin p B3.

Cng thc tnh ton:

Trong : T: l thi gian lm vic ca my bin p, T= 8760h.

SB3: ph ti ca my bin theo thi gian v c ly theo th ph ti hng ngy.

Ta c B3 l my bin p ba pha hai cun dy loi T-125-121/13,8 c :

(P0 = 100 kW, (PN = 400 kW, SB3 = 117,647 MVA = hng s.

Suy ra : (AB3 = 0,1. 8760 + 0,4 . = 3979,9 MWh.

2.Tn tht in nng hng nm ca my bin p B4.

Tng t nh tnh (AB3, B4 l my bin p ba pha hai cun dy loi T -125-242/13,8 c:

(P0 = 115kW; (PN = 380kW; SB4 = 117,647 MVA = hng s .

Suy ra : (AB4 = 0,115. 8760 + 0,38 . = 3956,1 MWh.

3.Tn tht in nng hng nm trong my bin p t ngu.

tnh tn tht in nng trong my bin p t ngu ta coi my bin p t ngu nh my bin p ba cun dy. Khi cun ni tip, cun chung v cun h ca my bin p t ngu tng ng vi cun cao, cun trung v cun h ca my bin p ba dy cun. Tn tht cng sut trong cc cun c tnh nh sau:

My bin p t ngu mt pha loi : AOTH-120- c

(P0=210 kW v (PNC-T =345 kW , (PNC-H = 220 MW, (PNT-H = 235 MW

T ta tnh c :

MW

MW

MW

T cc kt qu bng 3-4 v cng thc tnh trn ta c cng thc tnh tn tht in nng ca my bin p t ngu 3 pha c t hp t 3 my bin p mt pha nh sau :

(AB1=(AB2

y: SiC , SiT , SiH l ph ti pha cao p , trung p v h p ca mi my bin p t ngu ti thi im ti ghi trong bng 3-4 tnh trn .

T = 8760 h.

(PN , (Po , SBm : l ca mt my bin p mt pha.

Vit gn li:

SiC.ti= (-15,056)2.5 + (-9,423)2.3 + 13,2832.3+13,2832.3+13,2832.3+

+ 12,4592.3 +(-3,774)2.2 +(-15,879)2.2=3986,19 MVA2.h

SiT.ti=77,7622.5+94,8352.3+94,8352.3 +94,8352.3 +94,8352.3 +94,8352.3

+ 111,912. 2 + 77,7622.2 = 202281,4 MVA2.h

SiH.ti= 62,7062.5 + 85,4122.3 + 108,1182.3+ 108,1182.3+ 108,1182.3 +

+ 107,2942.3+ 108,1352.2+ 61,8832.2 = 212332,7 MVA2.h

Do :

(AB1=(AB2= =3.0,21.8760+[0,1425.3986,19+0,2025.202281,4+0,7375.212332,7]

= 5704 MWh

Nh vy tng tn tht in nng trong cc my bin p ca phng n I l:

(A( = (AB1+(AB2+(AB3+(AB4 =

= 2.5704 + 3956,1 + 3979,9 = 19345,6 MWh

II-Phng n II (hnh 2-2).

1. Tn tht in nng hng nm ca cc my bin p B3 v B4.

Theo cng thc nh phng n I :

(AB3 = (AB4=

My bin p B3 v B4 chn l my bin p kiu T-125-121/13,8 c thng s nh bng 3-2 do tn tht in nng ca my bin p B3 v B4 phng n ny bng nhau v ng bng tn tht trong my bin p B3 phng n I trn:

(AB3 = (AB4 = 3979,9 MWh

2. Tnh tn tht in nng hng nm ca my bin p t ngu B1 v B2.

Tng t ta phng n I, ta c:

My bin p t ngu 3 pha : ATTH-250-230/121/13,8 c (P0=120 kW v (PNC-T =520 kW , (PNC-H = (PNT-H =(PNC-T/2=260 MW v da vo bng 3-5

T ta tnh c :

MW

MW

MW

(AB1= (AB2

Vit gn li:

SiC.ti= 43,7682.5 + 49,42.3 + 72,1062.3+72,1062.3+72,1062.3+

+ 71,2832.3 +55,12.2 +42,92.2= 88685,9 MVA2.h

SiT.ti=18,942.5+36,0122.3+36,0122.3 +36,0122.3 +36,0122.3 +36,0122.3

+ 53,0852. 2 + 18,942.2 = 27599,53 MVA2.h

SiH.ti= 62,7062.5 + 85,4122.3 + 108,1182.3+ 108,1182.3+ 108,1182.3 +

+ 107,2942.3+ 108,1352.2+ 61,8832.2 = 212332,7 MVA2.h

Suy ra:

(AB1 = (AB2 =

=0,12.8760+[0,26.88685,9+0,26.27599,53+0,78.212332,7]

=2194,985 MWh

T cc kt qu tnh ton tn tht in nng trong cc my bin p trn ta c tng tn tht in nng trong cc my bin p phng n II l :

(A( = (AB1 + (AB2 + (AB3 + (AB4 =2(AB1 + 2(AB3 =

=2.2194,985 + 2.3979,9 = 12349,8 MWh.

Bng so snh tn tht in nng gia hai phng n:

Bng 3-6:

Tn tht in nng(A((MWh)

Phng n I19345,6

Phng n II12349,8

Chng IV

Tnh ton kt-kt Chn phng n ti u

Vic quyt nh bt k mt phng n no cng u phi da trn c s so snh v mt kinh t v k thut, ni khc i l da trn nguyn tc m bo cung cp in v kinh t quyt nh s ni dy chnh cho nh my in.

Trn thc t vn u t vo thit b phn phi ch yu ph thuc vo vn u t my bin p v cc mch thit b phn phi. Nhng vn u t ca cc mch thit b phn phi ch yu ph thuc vo my ct, v vy chn cc mch thit b phn phi cho tng phng n phi chn cc my ct.Trong tnh ton ch tiu kinh t-k thut ta ch cn chn s b cc my ct.

4-1. Chn s b my ct ca cc phng n.

I-Xc nh dng in lm vic cng bc ca cc mch.

1-Phng n I (Hnh 2-1).

a) Cp in p v h thng 220 KV.

- Mch ng dy ni vi h thng: Ph ti cc i ca h thng l SVHTmax = 144,21 MVA . V vy dng in lm vic cng bc ca mch ng dy c tnh vi iu kin mt ng dy b t . Khi

KA

-Mch my bin p ba pha 2 cun dy B4 : Dng in lm vic cng bc c xc nh theo dng in cng bc ca my pht in.

KA

-Mch my bin p t ngu B3(B4) :

Khi lm vic bnh thng th dng cng bc ca mch ny l :

KA

Khi s c b bn trung th dng cng bc l

KA

Khi s c mt my bin p t ngu th dng cng bc l

KA

Nh vy dng in lm vic ln nht cp in p 220 kV ca phng n I ny l :

Icbcao = 0,3785 KA

b) Cp in p trung 110 kV.

-Mch ng dy : Ph ti trung p c cp bi 2 ng dy kp *80MW ,

4 n*50MW, ta c :

Dng in lm vic cng bc l :

KA

-Mch my bin p ba pha hai cun dy :

KA

-Mch my bin p t ngu :


KA


KA


KA

Vy dng in lm vic cng bc ln nht pha 220 kV c ly l :

Icbtrung = 1,175 KA

c) Cp in p 13,8 kV.

Dng in lm vic cng bc mch ny chnh l dng in lm vic cng bc ca my pht in nn ta c :

KA

Bng kt qu tnh ton dng in lm vic cng bc cu phng n I l :

Bng 4-1

Cp in p220 kV110 kV13,8 kV

Icb (kA)0,37851,1755,168

2-Phng n II (Hnh 2-2).

a) Cp in p 220 kV.

-Mch ng dy cng nh phng n I ta c : Ilvcb = 0,3785 KA



KA


KA

Vy dng in lm vic cng bc ln nht cp in p 220 kV ca phng n II l : Icbcao = 0,3785 KA

b) Cp in p 110 kV.

-Mch ng dy tng t nh phng n I ta c : Ilvcb = 0,512 KA

-Mch my bin p ba pha hai cun dy nh phng n I ta c: Ilvcb = 0,6484 KA



KA


KA


KA

Nh vy dng in lm vic cng bc ln nht cp in p 110 kV l:

Ilvcb = 0,6484 KA

c) Cp in p 13,8 kV.

Tng t nh phng n I ta c : Ilvcb = 5,168 KA

Bng kt qu tnh ton dng in lm vic cng bc ca phng n II l :

Bng 4-2

Cp in p220 kV110 kV13,8 kV

Icb (kA)0,37850,64845,168

II-Chn my ct cho cc phng n.

Cc my ct kh SF6 vi u im gn nh, lm vic tin cy nn c dng kh ph bin. Tuy nhin cc my ct loi ny c nhc im l gi thnh cao, vic thay th sa cha thit b kh khn.

Vi nh my thit k u dng cc my ct kh SF6 c ba cp in p.Ta chn s b my ct theo iu kin sau:

UmMC ( UliImMC ( Icbmax

Cc thng s k thut ca my ct cho bng 4-3. Bng 4-3Phng nCp

in

p

(KV)Dng

Ilvcb(KA)Loi my cti lng nh mc

U

(KV)I

(KA)Ict(KA)Il(KA)

I

2200,37853AQ1245440100

1101,1753AQ1145440100

13,85,1868BK4117,512,580225

II2200,37853AQ1245440100

1100,64843AQ1145440100

13,85,1868BK4117,512,580225

4-2. So snh ch tiu kinh t gia cc phng n.

*Vn u t cho mt phng n l : V = VB + VTBPP Trong : Vn u t cho my bin p : VB = (ki. vBi

ki=1,4 : H s tnh n chuyn tr v xy lp.

vBi: Tin mua my bin p.

Vn u t cho my ct: VTBPP = ((nC.vC + nT.vT + nH.vH)

nC,nT,nH : S mch phn phi.

vC,vT,vH :Gi tin mi mch phn phi.

*Ph tn vn hnh hng nm ca mt phng n c xc nh nh sau:

P = Pkh + P(A

Trong :

Pkh = : Khu hao hng nm v vn v sa cha ln .

a=8,4: nh mc khu hao (%).

P(A = (.(A : Chi ph do tn tht hng nm gy ra.

( : l gi 1 kWh in nng (( = 400 ng /kWh)

(A : l tn tht in nng hng nm

I.Phng n I.

1. Chn s ni in.

Pha 220 kV : Dng s h thng hai thanh gp.

Pha 110 kV : Dng s h thng hai thanh gp c thanh gp ng vng v s nhnh vo ra nhiu.

Pha 13,8 kV : Khng dng thanh gp in p my pht v ph ti in p my pht chim khng qu 15% cng sut b.

S ni in phng n 1:(hnh 4-1)

220 kV

110kV

F4 F1

F2

F3

2.Tnh hm chi ph tnh ton.

a) Vn u t.

Vn u t cho my bin p : Phng n I dng 3 loi my bin p l :

- Hai t hp ca 3 my bin p t ngu mt pha kiu AOTH - 120

Vi gi tin : 161.103 R/1 pha(1R = 40.103 ng) nn gi tin ca c 3 pha l 3.161.103.40.103 = 19,32.109 ng v

- Mt my bin p 3 pha hai cun dy loi T-125 - 242/13,8

Vi gi tin : 162.103.40.103 = 6,48.109 ng v k = 1,4.

- Mt my bin p 3 pha hai cun dy loi T-125 - 121/13,8

Vi gi tin : 128.103.40.103 = 5,12.109 ng v k = 1,4.

Nh vy tng vn u t cho my bin p ca phng n I l :

VB1 = 2.1,4.19,32.109 + 1,4.6,84.109 + 1,4.5,12.109 =

= 70,336.109 ng.

Vn u t cho thit b phn phi:T s hnh 4-1 ta nhn thy :

Cp in p 220 kV gm c 4 mch my ct kiu 3AQ1 gi tin mt mch l : 71,5.103.40.103 = 2,86.109 / mch .

Vy gi 4 mch my ct kiu 3AQ1 l : 4.2,86.109 = 11,44.109 ng

Cp in p 110 kV gm c 5 mch my ct kiu 3AQ1 gi tin mt mch l 31.103.40.103 = 1,24.109 / mch .

Vy gi tin 5 mch my ct 3AQ1 l : 5.1,24.109 = 6,2.109 ng

Cp in p 13,8 kV gm c 2 mch my ct ,gi tin mt mch l 15.103.40.103= 0,6.109 / mch .

Vy gi tin ca 2 mch my ct l : 2.0,6.109 = 1,2.109 ng

Tng vn u t cho thit b phn phi l :

VTBPP1 = 11,44.109 + 6,2.109 + 1,2.109 = 18,84.109 ng

T tnh c tng vn u t ca phng n I l:

V = VB1 + VTBPP1 = 70,336.109 + 18,84.109 = 89,176.109 ng

b) Tnh ph tn vn hnh hng nm .*Khu hao hng nm v vn u t v sa cha ln:

Suy ra: ng

*Chi ph do tn tht in nng hng nm:

P(A= (.(A1 = 400.19345,6.103 = 7,738.109 ng

Nh vy ph tn vn hnh hng nm l:

P1 = Pkh + P(A = 7,49.109 + 7,738.109 =15,22.109 ngII-Phng n II.

1. Chn s ni in.

Tng t nh phng n I cp in p 220kV ta dng s h thng 2 thanh gp . Cp in p 110 kV dng h thng hai thanh gp c thanh gp vng nh s hnh 4-2 .

S ni in phng n 2:220 kV

110kV

F1 F2

F3 F4

Hnh 4-2

2. Tnh ton hm chi ph tnh ton.

a) Vn u t.

Vn u t cho my bin p:

Phng n ny gm c :

- Hai t hp my bin p t ngu 3 pha kiu : ATTH-250 vi gi tin : 250.103.40.103=10.109 ng /3 pha

- Hai my bin p 3 pha hai cun dy kiu : T-125-121/13,8 gi 5,12.109 ng vi k = 1,4

Nh vy tng vn u t cho my bin p l :

VB = 2.1,4.10.109 + 2.1,4.5,12.109 = 42,336.109 ng

Vn u t cho thit b phn phi :T s 4-2 ta nhn thy :

Cp in p 220 kV gm c 3 mch my ct kiu 3AQ1 gi tin mt mch l 2,86.109 / mch

Vy gi 3 mch my ct kiu 3AQ1 l : 3.2,86.109 = 8,58.109

Cp in p 110 kV gm c 6 mch my ct kiu 3AQ1 gi tin mt mch l 1,24.109 / mch

Vy gi tin 6 mch my ct 3AQ1 l : 6.1,24.109 = 7,44.109

Cp in p 13,8 kV gm c 2 mch my ct, gi tin ca 2 mch my ct l : 1,2.109

Tng vn u t cho thit b phn phi l :

VTBPP2 = 8,58.109 + 7,44.109 + 1,2.109 = 17,22.109

T tnh c tng vn u t ca phng n I l:

V2 = VB2 + VTBPP2 = 42,336.109 + 17,22.109 = 59,556.109 ng

b) Tnh ph tn vn hnh hng nm.

* Chi ph do tn tht in nng :

T cng thc tnh nu trn v tn tht in nng (A a tnh c chng III ta c : P(A = (.(A2 = 400.12349,77.103 = 4,94.109 ng

* Khu hao hng nm :

ng

Vy ph tn vn hnh hng nm l :

P2 = Pkh + P(A = 5.109 + 4,94.109 = 9,94.109 ng

Bng kt qu tnh ton kinh t ca hai phng n bng 4-4:

Bng 4-4

Phng nVn u t V

( x109 )Ph tn vn hnh P

( x109 )

I89,17615,22

II59,5569,94

Qua trn ta chn phng n II l phng n ti u cho bn n thit k ny do VII < VI v PII < PIChng V

TNH TON DNG IN NGN MCH

Mc ch ca vic tnh ton ngn mch l chn cc kh c in v dy dn, thanh dn ca nh my in theo cc iu kin m bo v n nh ng v n nh nhit khi c ngn mch. Dng in ngn mch tnh ton l dng in ngn mch ba pha.

tnh ton dng in ngn mch ta dng phng php gn ng vi khi nim in p trung bnh v chn in p c bn bng in p nh mc trung bnh ca mng.

Chn cc lng c bn:

Cng sut c bn: Scb =100MVA;

Cc in p c bn: Ucb1 = 230 kV; Ucb2 =115 kV; Ucb3 =10,5 kV

5-1. Tnh cc in khng trong h n v tng i c bn.

1. in khng ca h thng in .

Nhim v thit k cho in khng tng i ca h thng th t thun ca h thng l XHT = 1,13 v cng sut nh mc ca h thng

SHTm =2800 MVA. Do in khng ca h thng qui i v c bn l:

XHT = XHT.

2. in khng ca nh my pht in.

Cc my pht in c in khng siu qu dc trc l Xd =0,21. Do in khng qui i v lng c bn l:

XF = Xd.

3. in khng ca ng dy 220kV

Theo nhim v thit k, nh my c ni vi h thng qua hai ng dy cao p 220kV c chiu di 150km. C X0 = 0,4 (/km

Tr s in khng qui i v lng c bn l:

XD = X0 .l.

4. in khng ca my bin p 3 pha 2 cun dy.

Loi - 125-121/13,8

XB3 = XB4 =

5. in khng ca my bin p t ngu.

Nh ch to cho in p ngn mch gia cc pha in p ca my bin p t ngu. T ta c:

UNC% = 0,5.( UNC-T + UNC-H - UNT-H ) = 0,5.( 11 + 32 - 20 ) = 11,5

UNT% = 0,5.( UNC-T + UNT-H - UNC-H ) = 0,5.( 11 + 20 - 32 ) = -0,5 ( 0

UNH% = 0,5.( - UNC-T + UNC-H + UNT-H ) = 0,5.( -11 + 32 + 20 ) = 20,5T y tnh c in khng qui i ca my bin p t ngu ba pha v lng c bn:

XC = = = 0,046

XT = 0

XH == = 0,082

5-2. Tnh ton dng in ngn mch.

1.S ni in (Hnh 5-1).

2.S thay th (Hnh 5-2)

3.Tnh ton ngn mch

chn kh c in cho mch 220 kV,ta chn dim ngn mch N1 vi ngun cung cp l ton b h thng v cc my pht in. i vi mch 110kV,im ngn mch tnh ton l N2 vi ngun cung cp gm ton b cc my pht v h thng.Tuy nhin vi mch my pht in cn tnh ton hai im ngn mch l N3 v N3. im ngn mch N3 c ngun cung cp l ton b cc my pht(tr my pht F2) v h thng . im ngn mch N3 c ngun cung cp ch c my pht F2. So snh tr s ca dng in ngn mch ti hai im ny v chn kh c in theo dng in c tr s ln hn. chn thit b cho mach t dng ta c im ngn mch tnh ton N4. Ngun cung cp cho im ngn mch N4 gm ton b cc my pht v h thng in. Dng ngn mch ti N4 c th xc nh theo dng ngnmch ti N3 v N3

a) Ngn mch im N1

S tnh ton im ngn mch N1(Hnh 5-3):

T s hnh 5-2 ta c s thay th tnh ton im ngn mch N1 nh hnh 5-3 c cc thng s nh sau :

X1 = XHT + XD = 0,04 + 0,057 = 0,097

X2 = X5 = XC = 0,046

X3 = X6 = XH = 0,082

X4 = X7 = X9 = X11 = XF = 0,179

X8 = X10 = XB3,B4 = 0,084

Bng cch ghp ni tip v song song cc in khng ta c hnh 5-4:

X12 =

X13 =

EMBED Equation.3

X14 =

Ghp song song F1,F2 vi F3,F4 ri ni tip vi X13 ta c

X15 =

X16 = X13 + X15 = 0,023 + 0,0655 = 0,0885

S hnh 5-5 l s ti gin c hai u cung cp in cho N1

in khng tnh ton t pha h thng n im ngn mch N1 l :

Tra ng cong tnh ton ca nh my thu in ti t= 0 ses v t= (I*" = 0,37 ; I*( = 0,41

i ra h n v c tn ta c :

I" = KA

I( = KA

in khng tnh ton t pha nh my n im ngn mch N1 l :

Vi (SFm = 4SFm = 4.117,647 = 470,588 MVA

Tra ng cong tnh ton ca thu in ta c :

I* = 2,5 ;I*( = 2,7

i ra h n v c tn ta c

I" = KA

I( = KA

Nh vy tr s dng in ngn mch tng ti im N1 l :

- Dng ngn mch siu qu : IN1" = 2,6 + 2,95 = 5,55 KA

- Dng ngn mch duy tr: I(N1 = 2,88 + 3,19 = 6,07 KA

- Dng in xung kch : ixkN1 = .kxk.IN1" = .1,8.5,55 = 14,12 KA

b) im ngn mch N2

tnh ton im ngn mch N2 c th li dng kt qu khi tnh ton ,bin i s ca im N1 trn. T hnh 5-3 ta c:

Cng nh i vi im N1 ta cng ghp F1,F2 v F3,F4 ta c s hnh 5-7

X15 = X1 + X13 = 0,097 + 0,023 = 0,12

X16 =


V XttHT > 3 nn p dng cng thc tnh : I*" = I*( =


I" = I( = KA


Tra ng cong tnh ton ca thu in ta c : I* = 3,8 ;I*( = 3,2


I" = KA

I( = KA


- Dng ngn mch siu qu : IN2" = 4,18 + 9 = 13,18 KA



c) im ngn mch N3

Ta bit im ngn mch N3 c cung cp bi h thng v nh my (tr my pht F2) . Tng cng sut ca nh my cung cp cho im ngn mch N3 l

(SFm = 3SFm = 3.117,647 = 352,94 MVA

T s hnh 5-3 ta c s thay th hnh 5-8

Bin i tng ng ta c s hnh 5-9 nh sau :

Ta c:

X12 =

EMBED Equation.3 X13 =

X14=

Ghp F1 vi F3,F4 ta c s hnh 5-9.

Ta c: X15 =X1 +X12 =0,097 + ,0,023 = 0,12

Bin i s sao X6 , X14 , X15 thnh s tam gic thiu X16 , X17 :

X16 = X6 + X15 + = 0,082 + 0,12 + = 0,315

X17= X6 + X14 + = 0,082 + 0,0874 + = 0,229


V XttHT > 3 nn p dng cng thc tnh : I*" = I*( =


I" = I( = KA


Tra ng cong tnh ton ta c : I* = 1,4

;I*( = 1,6


I" = KA

I( = KA


- Dng ngn mch siu qu : IN3" = 17,46 + 27,17 = 44,63 KA



d) im ngn mch N3'

im ngn mch N3 chnh l ngn mch u cc my pht in F2 nn ngun cung cp ch gm c mt my pht F2 v c s thay th nh hnh 5-11

in khng tnh ton: Xtt = X7 = Xd" = 0,179


- Dng ngn mch siu qu v duy tr:

IN3" = I(N3 = KA

- Dng in xung kch :

ixkN3 = .kxk.IN3" = .1,9.36,14 = 97,11 KA

(Ngn mch u cc ly kxk =1,9)

e) im ngn mch N4

T s thay th hnh 5-2 ta thy :

IN4 = IN3 + IN3' t ta c :

-Dng ngn mch siu qu :I"N4 = I"N3 + I"N3' = 44,63 + 36,14 = 80,77 KA

-Dng ngn mch duy tr :I(N4 = I(N3 + I(N3' = 48,51 + 36,14 = 84,65 KA

-Dng in xung kch : ixkN4 = 205,6 kA

Kt qu tnh ton ngn mch ca phng n :

Bng 5-1

Cp in p

( kV )im ngn mchI"

(KA)I((KA)ixk(KA)

220N15,556,0714,12

110N213,1811,7434,1

13,8N344,6348,51113,6

N3'36,1436,1497,11

N480,7784,65205,6

4.Chn my ct v dao cch ly

Ta c tiu chun chn my ct l :

UmMC ( UmliImMC ( IcbmaxIctm ( I

im ( ixk

Ta c iu kin chn dao cch ly l :

UmCL ( UmliImCL ( Icbmaxim ( ixk

T ta c bng chn my ct v dao cch ly nh sau:

MchThng s tnh tonThng s my ct, dao cch ly

Um( KV)Icb

(KA)I(KA)Ixk

(KA)LoiK.hiuUm(KV)Im

(KA)Ictm

(KA)Im

(KA)

Cao p2200,38755,5514,12MC3AQ1245440100

DCLSGC

245/8002450,8--80

Trung p1100,648413,1834,1MC3AQ1145440100

DCLSGCP

123/8001230,8--80

H p13,85,18644,63113,6MC8BK4117,512,580225

DCLSGCP

36/12503612,5--125

Chng VI

Chn dy dn v kh c in

6-1.Chn thanh dn cng, thanh dn mm , thanh gp.

1. Chn thanh dn cng u cc my pht.

Thanh dn cng dng ni t my pht ti cun h ca my bin p t ngu v my bin p ba pha hai cun dy. Tit din ca thanh dn c chn theo iu kin pht nng lu di.

a) Chn tit din thanh dn .

Gi thit nhit lu di cho php ca thanh dn bng ng l (cp = 70oC, nhit mi trng xung quanh l (0 = 35oC v nhit tnh ton nh mc l (m = 200C. T ta c h s hiu chnh theo nhit l :

Tit din ca thanh dn cng c chn theo dng in lu di cho php : Ilvcb ( Icp.Khc

Do ta c : Icp ( kA

Nh vy ta chn thanh dn cng bng ng , c tit din hnh mng nh hnh 6-1, qut sn v c cc thng s nh bng 6-1:

Bng 6-1

Kch thc (mm)Tit din

1cc

mm2 M men tr khng (cm3)M men qun tnh (cm4)Icp

c 2 thanh A

hbcr1 thanh2 thanh1 thanh2 thanh

WxxWyyWyoyo JxxJyyJyoyo

1506571017857414,71675606812607000

b) Kim tra n nh nhit khi nhn mch.

Thanh dn chn c dng in cho php Icp > 1000 A nn khng cn kim tra iu kin n nh nhit khi ngn mch.

c) Kim tra n nh ng.

Ly khong cch gia cc pha l a = 60 cm, khong cch gia hai s l l = 200 cm.

* Tnh ng sut gia cc pha:

Lc tnh ton tc dng ln thanh dn pha gia trn chiu di khong vt l:

Ftt = 1,76.10-2..ixk2 = 1,76.10-2..113,62 = 757,1 kG. ( khd = 1 )

M men un tc dng ln chiu di nhp :

Mtt = = = 15142 kG.cm

V ng sut do lc ng in gia cc pha l :

(tt = = = 90,67 kG/cm2

vi Wyoyo =167 cm3 l m men chng un ca tit din ngang thanh dn.

* Xc nh khong cch gia 2 s :

Lc tc dng ln 1 cm chiu di thanh dn do dng ngn mch trong cng pha gy ra: f2 = 0,26.10-2. .ixk2 = 0,26.10-2. .113,62 = 5,16 kG/cm

ng sut do dng in trong cng pha gy ra :

(2 = = kG/cm2

iu kin n nh ng ca thanh dn khi khng xt n dao ng l :

(cpCu ( (tt + (2 hay (2 ( (cpCu - (ttl2 (

Vi thanh dn ng (cpCu = 1400 kG/cm2. Vy khong cch ln nht gia cc s m thanh dn vn m bo n nh ng l :

l2max = = 211,57 cm

Gi tr ny ln hn khong cch ca khong vt l = 200 cm. Do khng cn t ming m ti hai u s m thanh dn vn m bo n nh ng.

2. Chn s thanh dn cng.

S thanh dn cng c chn theo iu kin sau:

Loi s: S t trong nh.

in p: USm ( Ummg = 13,8 kV

iu kin n nh ng.Ta chn s O(- 20-2000KB.Y3 c: Um = 20 kV ; Fcp = 2000 kG ; HS = 315 mmKim tra n nh ng:S c chn cn tho mn iu kin : Ftt ( 0.6 Fph

trong : Fph- Lc ph hoi cho php ca s.

Ftt- Lc ng in t trn u s khi c ngn mch.

Ftt = Ftt

Vi : Ftt Lc ng in tc ng ln thanh dn khi c ngn mch

Hs Chiu cao ca s

H Chiu cao t y s n trng tm tit din thanh dn

Thanh dn chn c chiu cao h = 150 mm.

Do : H= Hs + 0,5.h =315 + 0,5.150 = 390 mm.

Lc ph hoi tnh ton ca s :

kG

Lc ny nh hn lc ph hoi cho php ca s. Vy s chn hon ton tho mn

3. Chn dy dn mm.

Thanh dn mm c dng t u cc pha cao, pha trung ca my bin p t ngu v cun cao ca my bin p hai cun dy ln cc thanh gp 220 kV v 110 kV. Tit din ca thanh gp v thanh dn mm c chn theo iu kin nhit lu di cho php. Khi dng in cho php hiu chnh theo nhit l:

Ihccp ( Ilvcb/Khc

Trong : Ihccp l dng in lm vic lu di cho php ca dy dn c hiu chnh theo nhit ti ni lp t.

Ilvcb : dng in lm vic cng bc.

Khc: H s hiu chnh,Khc = 0,837

Cc dy dn mm ny treo ngoi tri, c n nh nhit tng i ln nn ta khng cn kim tra n nh nhit khi ngn mch.

a) Chn tit din.

T kt qu tnh ton dng in lm vic cng bc chng trc tnh c dng cho php ( hiu chnh theo nhit ) ca cc cp in p.

Mch in p 220 kV:- Dng lm vic cng bc ca dy dn trong mch ny l: Ilvcb = 0,3785 kA

Ta phi chn dy dn c :

Icp ( kA

Nh vy ta chn loi dy dn AC-400 c S =400 mm2 v Icp = 835 A.

Mch in p 110 kV: Dng in lm vic cng bc ca mch: Ilvcb = 0,6484 kA

Ta phi chn dy dn c :

Icp ( kA

Nh vy chn dy AC-400 c tit din S = 400 mm2, Icp =0,835 kA.

b) Kim tra n nh nhit khi ngn mch.

Tit din nh nht dy dn n nh nhit l : Smin =

BN : Xung lng nhit ca dng in ngn mch ( A2.s ) .

C : Hng s ph thuc vo vt liu dy dn ( ) .

Vi dy dn AC c C = 70 .

Tnh xung lng nhit :

BN = BNck + BNkck

Gi thit thi gian tn ti ngn mch l 0.5 sec. Khi c th tnh gn ng xung lng nhit ca thnh phn dng in ngn mch khng chu k:

BNkck1 = IN12.Ta = ( 5,55.103 )2. 0,05 = 1,54.106 A2.s

BNkck2 = IN22.Ta = ( 13,18.103 )2. 0,05 = 8,68.106 A2.s

Thnh phn xung lng nhit ca thnh phn dng in ngn mch chu k c xc nh theo phng php gii tch th :

BNCK =

T s thay th tnh ton ngn mch im N1 v N2 ( a v hai bin i ) ca phng n ti u ( phng n 2) ta tnh c gi tr dng in ngn mch ti N1 v N2 theo thi gian nh sau :

t ( sec )

Dng in 00,10,20,5

IN1 ( kA )5,555,815,555,59

IN2 ( kA )13,1812,4511,9811,74

im N1:

I2tb1 = = = 32,31 kA2

I2tb2 = = = 32,31 kA2I2tb3 = = = 31,04 kA2

T : BNck1 = 32,31.0,1 + 32,31.0,1 + 31,04.0,3 = 15,77 kA2.s = 15,77.106 A2.s

Vy xung lng nhit ca dng ngn mch ti im N1 :

BN1 = BNck1 + BNkck1 = 1,54.106 + 15,77.106 = 17,31.106 A2.s

im N2:

I2tb1 = = = 164,14 kA2

I2tb2 = = = 149,29 kA2I2tb3 = = = 140,7 kA2

T : BNck1 = 164,14.0,1 + 149,29.0,1 + 140,7.0,3= 73,56 kA2.s= 73,56.106 A2.s

Vy xung lng nhit ca dng ngn mch ti im N2 :

BN2 = BNck2 + BNkck2 = 8,68.106 + 73,56.106 = 82,24.106 A2.s

Tit din dy dn nh nht m bo n nh nhit cc cp in p 220 kV v 110 kV l :

Smin1 = = = 59,4 mm2 .

Smin2 = = = 129,55 mm2 .

Vy cc dy dn v thanh gp mm chn u m bo n nh nhit .

c) Kim tra iu kin vng quang.

Kim tra iu kin vng quang theo cng thc :

Uvq ( Um vi Uvq = 84.m.Rdt.lg() kV

Trong : Uvq l in p ti hn pht sinh vng quang

m l h s c xt n x x ca b mt dy dn, ly m = 0,87

a l khong cch gia cc pha ca dy dn, ly a =500 cm (vi cp 220 kV) v a =300 cm (vi cp 110 kV)

R l bn knh ngoi ca dy dn.

in p 220 kV:

Dy AC- 400 c : Icp = 835 A, d = 26,6 mm t trn mt phng nm ngang.

Khong cch gia cc pha l a = 500 cm. Khong cch trung bnh hnh hc atb = 1,26.a = 630 cm.

Uvq = 84.0,87.1,33.lg= 260 kV > Um = 220 kV

Nh vy dy dn chn tho mn iu kin vng quang.

Tng t i vi in p 110 kV ta cng thy tho mn.

6-2.Chn cp v khng ng dy.

1.Chn cp cho ph ti 13,8 KV

- Ph ti cp in p 13,8 KV c Pmax =14 MW gm :

+ 2 ng dy cp kp P = 4 MW * 3 km , Cos( = 0,85

S = =

+ 2 ng dy cp n P = 3 MVA *3 km , Cos( = 0,85

S = =

- Tit din cp c chn theo tiu chun mt dng in kinh t Jkt

Scp = vi Ilvbt : Dng in lm vic bnh thng

Chn tit din cp n:

Cc ng cp n c S = 3,53 MVA nn dng in lm vic bnh thng l:

Ilvbt =

T th ph ti a phng ta tnh thi gian s dng cng sut cc i:Tmax= =

Tra bng c Tmax = 6329,3 h ta chn cp cch in bng giy c Jkt = 2A/mm2 Tit din cp trong trng hp ny: Scp = = 74 mm2 Tra bng chn loi cp ba li bng ng v nhm cch in bng giy tm du nha thng v cht do khng chy, v bng ch

S = 95 mm2 ; Um = 13,8 KV ; Icp = 265 A

- Kim tra cp theo iu kin pht nng lu di

K1.K2.Icp Ilvbt

K1 : H s hiu chnh theo nhit

K1 =

K2 : H s hiu chnh theo s cp t song song, vi cp n c K2=1

0,837.1.265 = 221,8 A > Ilvbt = 148 A

Vy cp chn tho mn iu kin pht nng lu di cho php

Chn tit din cp kp c S= 4,7 MVA

( Dng in lm vic bnh thng qua mi cp :

Ilvbt =

C Jkt = 2A/mm2

Tit din cp c chn :

Scp = =

( Tra bng chn loi cp ba pha li ng cch in bng giy tm du nha thng v cht do khng chy v bng ch t trong t ,ta chn cp c Um = 13,8 KV

S = 70 mm2, Icp = 215A

Tng t nh trn ta kim tra cp theo iu kin pht nng lu di

Kqt.K1.K2.Icp ( Icb = 2.Ilvbt

K1 = 0,837 ; K2 = 0,9 ( Vi 2 cp t song song ) ; Kqt = 1,3

1,3.0,837.0,9.215 = 210 A > 196 A = 2.Ilvbt

Vy iu kin pht nng lu di tho mn

Kt lun : Cp chn m bo yu cu k thut

2.Chn khng ng dy.

* Xc nh dng cng bc qua khng:

Dng cng bc qua khng c gi thit khi s c 1 khng in. Lc ny cng sut qua khng cn li l:

Squa K = S -

EMBED Equation.3 n = 16,47 - 3,53 = 12,94 MVA

Ta c:

Cng sut qua khng

MVAKhng 1Khng 2

Bnh thng8,2358,235

S c khng 1012,94

S c khng 212,940

Dng cng bc qua khng l:

IcbK = =

Tra bng ta chn khng dn PbA-10-600-10 c dng in ImK = 600A

Xc nh XK% ca khng

Trong chng tnh ngn mch ta tnh c dng ngn mch ti im N4

IN4 = 80,77 KA

Vy in khng ca h thng tnh n im ngn mch N4 l:

XHT = =

in khng ca cp 1 l: XC1 = X0.l.= = 0,218

Dng n nh nhit ca cp 1 l : InhC1 =

S1: Tit din cp 70 mm2

C1: H s cp nhm C1 = 90 As1/2/s

t1 : Thi gian ct ca my ct 1 :tct MC1 = tct MC2 + t = 0,7 + 0,3 = 1 sec

InhC1 =

Dng n nh nhit cp 2 : InhC2 = =

Ta phi chn c khng c XK% sao cho hn ch c dng ngn mch nh hn hay bng dng ct nh mc ca my ct chn ng thi m bo n nh nhit cho cp c tit din chn:

IN5 ( ICm1, InhC1 ) v IN6 ( ICm2, InhC2 )

Chn my ct u ng dy MC1: Cc my ct u ng dy c chn cng loi. Dng cng bc qua my ct c tnh ton cho ng dy kp khi 1 ng dy b s c

Icb = =

Tra bng chn my ct 8DA10 c:

Um = 15KV

Im = 2500 A Ict m = 40 KA

Mc ch ca vic chn khng in ng dy l hn ch dng ngn mch ti h tiu th ti mc c th t c my ct 8DA10 v cp ca li in phn phi c tit din nh nht l 70 mm2 theo yu cu ca u bi.

IN5 ( 40KA; 6,3KA )

IN6 ( 20KA; 7,529KA )

Vy ta chn khng c XK% sao cho ngn mch ti N6 th c dng ngn mch

IN6 7,529 KA

Khi ngn mch ti N6 th in khng tnh n im ngn mch l:

=

M ta c = XHT + XK + XC1

XK = - XHT XC1 = 0,6888 0,0642 0,218 = 0,4066

Nn XK% = XK.

Vy ta chn khng n dy nhm PbA-10-600-10

XK% = 10% ; Im = 600A ; Um= 13,8 KV

Kim tra khng va chn

in khng tng i ca khng in va chn

XK = XK%.

Dng ngn mch ti N5

IN5 =

EMBED Equation.3 Tho mn iu kin:

IN5 Ict m1= 40 KA

IN5 InhC1 = 6,3 KA

Dng ngn mch ti N6

IN6 =

Tho mn iu kin:

IN6 < ICt m2= 20 KA

IN6 < InhS2 = 7,529 KA

Kt lun: Vy khng chn m bo yu cu

6-3.Chn my bin in p v my bin dng in.

Vic chn my bin in p v my bin dng in ph thuc vo ti ca n. in p nh mc ca chng phi ph hp vi in p nh mc ca mng.

1. Cp in p 220 kV.

a) My bin in p:

kim tra cch in v cung cp tn hiu cho h thng bo v rle, o lng t cc my bin in p trn thanh gp 220 kV. Thng chn my bin in p mt pha kiu HK -220 - 58Y1 ni dy theo s Yo/Yo// c cc thng s sau:

in p s cp: Usm = 150000/ V; in p th cp 1: Ut1m = 100/ V

in p th cp 2: Ut2m = 100/3 V ; Cp chnh xc: 0,5

Cng sut nh mc: STUm = 400 VA

b) My bin dng in.

Cc my bin dng in c i km vi cc mch my ct c nhim v cung cp tn hiu cho h thng bo v rle. Vi mc ch d chn my bin in kiu TH - 220-3T c cc thng s sau:

Dng in s cp: Ism =1200 A ; Dng in th cp: Itm = 5 A

Cp chnh xc : 0,5 ; Ph ti nh mc: 2(

Dng in n nh ng : ildd = 108 kA

My bin dng chn c dng in s cp nh mc ln hn 1000A nn khng cn kim tra iu kin n nh nhit.

2. Cp in p 110 kV.

a) My bin in p: kim tra cch in v cung cp tn hiu cho h thng bo v rle, o lng t cc my bin in p trn thanh gp 110 kV. Thng chn my bin in p mt pha kiu HK -110 57,Y1 ni dy theo s Yo/Yo// c cc thng s sau:

in p s cp: Usm = 66000/ V ; in p th cp 1: Ut1m = 100/ V

in p th cp 2: Ut2m = 100/3 V ; Cp chnh xc: 0,5

Cng sut nh mc: STUm = 400 VA

b) My bin dng in.Cc my bin dng in c i km vi cc mch my ct c nhim v cung cp tn hiu cho h thng bo v rle. Vi mc ch d chn my bin in kiu TH-110M c cc thng s sau:

Dng in s cp: Ism =2000 A ; Dng in th cp: Itm = 5 A

Cp chnh xc : 0,5 ; Ph ti nh mc: 0,8(

Dng in n nh ng : ildd = 108 kA

My bin dng chn c dng in s cp nh mc ln hn 1000A nn khng cn kim tra iu kin n nh nhit.

3. Cp in p my pht 15,75 kV.

Mch my pht in cc bin in p v bin dng in nhm cung cp cho ccdng c o lng. Theo quy nh bt buc mch my pht phi c cc phn t o lng sau: ampe k, vn k, tn s k, cos( k, ot k tc dng, ot k phn khng, ot k tc dng t ghi, cng t tc dng, cng t phn khng. Cc dng c o c mc nh hnh 6-3.

S ni cc dng c o vo BU v BI

a) Chn my bin in p.

My bin in p c chn phi tho mn iu kin sau :

Sdc STUm vi Sdc =

V ph ti ca bin in p l cc dng c o lng nn dng hai bin in p mt pha ni dy kiu V/V v c ni vo u cc ly cc in p dy AB v BC.

Cc dng c o lng s dng qua my bin in p c ghi bng 6-5.

Bng 6-5

Th

TTn ng hK hiuPh ti abPh ti bc

P(W)Q(Var)P(W)Q(Var)

1Vn kB27,2

2Tn s k3446,5

3Ot k tc dng 3411,81,8

4Ot k phn khng 342/11,81,8

5Ot k t ghi

\\ 338,38,3

6Cng t tc dngT0,660,661,62

7Cng t phn khngTP0,661,620,661,62

Tng cng20,403,2419,723,24

Ph ti ca bin in p ab:

Sab = = 20,7 VA ; cos(ab =

Ph ti bin in p bc:

Sbc = = 19,9 VA ; cos(bc =

V ph ti ca my bin in p l dng c o lng nn ta chn my bin in p kiu mt pha HOM-15 c cc thng s sau :

in p s: 15750 /V ; in p cun th 1: 100/ V

in p cun th 2: 100/3 V ; Cp chnh xc : 0,5

Ph ti nh mc: Sm = 75 VA

* Chn dy dn ni t bin in p n dng c o :

Gi s di t my bin in p n cc ng h o lng l l = 60 m

Dng in trong cc pha a, b, c :

Ia = A ; Ic = A

n gin trong tnh ton coi Ia = Ic= 0,2 A v coi cos(ab = cos(bc = 1

Khi Ib =.Ia = 0,34 A

Tr s in p ging trn dy dn pha a v b:

(U = (Ia + Ib) ly (Cu = 0,0175 (V mch in c cng t nn (U ( 0,5 %. Vy tit din dy dn l:

Fdd = mm2

m bo bn c ta chn dy dn ng c bc cch in c tit din l:

Fdd = 1,5 mm2b) Chn my bin dng in .

Cc bin dng c t trn c ba pha v c ni theo s sao. V cc cng t c cp chnh xc 0,5 nn cc my bin dng c chn phi c cng cp chnh xc. Ngoi ra cn phi m bo cc iu kin sau:

in p nh mc: UTIm ( UFm =13,8 kV

Dng in nh mc s: ITIm ( Ilvcb = 5,186 kA

Vy chn my bin dng in kiu T-20-1 c cc thng s k thut nh sau:

Um = 20 kV , Ism = 10000 A , Itm = 5 A , ph ti nh mc Zm = 1,2 ( , cp chnh xc 0,5.

Bng dng c o lng ni vo TI c ghi trong bng 6-6:

Bng 6-6

Th

tTn dng cK hiuPh ti (VA)

ABC

1Am pe k-302111

2Ot k tc dng-34155

3Ot k t ghi 331010

4Ot k phn khng -342/155

5Cng t tc dng 6702,52,5

6Cng t phn khngT-6722,552,5

Tng cng26626

Pha a v c ca bin dng mang ti nhiu nht Smax =26 VA

Tng tr dng c o mc vo cc pha ny l:

Z(d = ( tho mn cp chnh xc 0,5 ca my bin dng in ta cn chn dy dn n cc dng c o lng c ln cn thit. Gi s khong cch t my bin dng in n cc dng c o lng l l = 50 m.

Chn dy dn bng ng c tit din tho mn:

Fdd ( mm2

Theo iu kin v bn c ta chn dy dn ng c bc cch in c tit din F = 6 mm2.

Kim tra n nh ng my bin dng in:

My bin dng kiu T(-20-1 c s cp l thanh dn ca tht b phn phi nn n nh ng ca n quyt nh bi n nh ng ca thanh dn mch my pht. Do vy khng cn kim tra n nh ng ca my bin dng in ny.

Kim tra n nh nhit khi ngn mch:

V dng nh mc s cp ca my bin dng in ln hn 1000A nn khng cn kim tra n nh nhit.

Vy my bin dng in chn hon ton tho mn yu cu.

Chng VII

Chn s v thit b t dng

Lng in t dng ca nh my thu in l rt nh so vi nh my nhit in cng cng sut. Mt khc theo bi thit k th y l nh my thu in c cng sut trung bnh ( ( 1000 MVA ) nn s t dng ca nh my thit k c nhng c im sau :

Mt my pht hay mt nhm my pht ghp chung 1 MBA, c mt MBA t dng h t 13,8/0,4 KV. in ly ngay t cc my pht.

D phng nng cho nhau thng qua Aptomt pha h p. Khi mt MBA b s c, cc my cn li s tng cng sut thay th MBA b s c.

Pha cao ca MBA t dngkhng cn dng my ct m ch cn dng cch ly (v l MBA trong nh, thng bo qun rt tt nn hu nh khng bao gi c s c sy ra).

Pha h ca MBA t dng l Aptmt v dao cch ly phc v sa cha 380/220V ,do phi ni t an ton v c dy trung tnh ly in p pha.

1. Chn my bin p t dng.

My bin p t dng c chn theo iu kin sau:

v (n-1).SmB.Kqtsc( Smaxtddo ta chn MBA t dng TM-1,6 c thng s nh sau :

LoiSm

(MVA)UCm

(kV)UHm(kV)(P0

(kW)(PN

(kW)Un%I0%

TM1,613,80,42,816,55,51,3

Ta c s t dng ca thu in nh sau:

0

5

8

14

17

20

24

0

11

100

22

200

423,53

300

400

500

t (h)

376,47

470,59

Hnh 1-3

376,47

SNM(MVA)

Hnh 1-1

11,53

22

11

SUf(MVA)

13,176

14,823

13,176

t (h)

100

80

60

40

20

0

24

20

17

14

8

5

0

B3

(

B1

(

B2

(

B6

(

HT

SUF

ST

B5

B4

SVHT(MVA)

Hnh 1-4

16,470

307,32

22

11

ST(MVA)

273,17

341,46

273,17

t (h)

500

400

300

200

100

0

24

20

17

14

8

5

0

98,8

144,21

22

11

85,89

110,1

87,535

t (h)

250

200

150

100

50

0

24

20

17

14

142,57

8

5

0

Hnh 1-5

14,823

B3

85,89

341,463

273,17

110,1

142,57

144,21

14,823

87,535

423,53

470,59

273,17

98,8

13,176

16,47

307,32

22

11

SNM(MVA)

376,47

11,53

376,47

t (h)

500

400

300

200

100

0

24

20

17

14

8

5

0

F1

B1

B2

F4

F3

B4

F2

B3

HT.

ST

220 KV

110 KV

Hnh 2-1

f1

b1

f3

f2

f4

b2

b4

b3

HT

St

220 KV

110 KV

Hnh 2-2

(

(

(

(

B4

B2

B1

B1

(

(

(

W

A

A

A

0,4 KV

F4

F3

F2

EHT

XHT

XK

Hnh 6-3

c

b

a

G

2.HOM-15

f

V

VARh

Wh

W

VAR

XC1

B4

B3

B2

TM-1,6

Hnh 6-2

Ftt

F1

S

Thanh dn

H'=390

Hs=315 mm

Hnh 6-1

c

MC2

N6

MC1

F1

B1

N5

N4

1 n

3 MVA

1 n

3 MVA

2 Kp

8 MVA

y

~

y

y

y

h

h

x

c

b y y0 y

Hnh 6-1

A-400 AOTH-267 A-400

3AF2

~ ~ ~ ~ G4 G1 G2Hnh 5-1 G3

Hnh 5-11

N3

X7

G2

Hnh 5-10

X17

X16

NM

HT

Hnh 5-9

N3

NM

X6

X14

X12

X1

HT

Hnh 5-8

HT

F4

F3

F1

N3

X11

X9

X10

X6

X5

X4

X3

X2

X1

X8

Hnh 5-7

NM

HT

X16

X15

N2

Hnh 5-6

F12

N2

X12

X14

X13

X1

F34

HT

Hnh 5-5

NM

HT

X16

X1

N1

HT

Hnh 5-4

F34

F12

X12

X14

X13

N1

X1

Hnh 5-3

HT

F4

F3

F2

F1

N1

X11

X9

X10

X7

X6

X5

X4

X3

X2

X1

X8

XF

XF

XB3

Hnh 5-2

HT

F4

F3

F2

F1

N4

N3

N3

N2

N1

XF

XB4

XH

XC

XF

XH

XC

XD

XHT

Hnh 5-1

n2

b3

b4

n1

n3'

b2

f4

f2

f3

n3

n4

ht

b1

f1

(

B3

B2

B1

B4

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