do an mon hoc nmd-ok4
DESCRIPTION
Đồ án môn học "Phần điện nhà máy điện và trạm biến áp" - Đại học Bách Khoa Hà Nội.Bản đồ án 63 trang gồm 7 chương cho người đọc cái nhìn khái quátTRANSCRIPT
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 :
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
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
-Mch my bin p t ngu :
Khi lm vic bnh thng th dng cng bc ca mch ny l :
KA
Khi s c mt my bin p t ngu th dng cng bc l
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
-Mch my bin p t ngu :
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 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 =
in khng tnh ton t pha h thng n im ngn mch N2 l :
V XttHT > 3 nn p dng cng thc tnh : I*" = I*( =
i ra h n v c tn ta c :
I" = I( = KA
in khng tnh ton t pha nh my n im ngn mch N2 l :
Tra ng cong tnh ton ca thu in ta c : I* = 3,8 ;I*( = 3,2
i ra h n v c tn ta c
I" = KA
I( = KA
Nh vy tr s dng in ngn mch tng ti im N2 l :
- Dng ngn mch siu qu : IN2" = 4,18 + 9 = 13,18 KA
- Dng ngn mch duy tr: I(N2 = 4,18 + 7,56 = 11,74 KA
- Dng in xung kch : ixkN2 = .kxk.IN2" = .1,8.13,39 = 34,1 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
in khng tnh ton t pha h thng n im ngn mch N3 l :
V XttHT > 3 nn p dng cng thc tnh : I*" = I*( =
i ra h n v c tn ta c :
I" = I( = KA
in khng tnh ton t pha nh my n im ngn mch N3 l :
Tra ng cong tnh ton ta c : I* = 1,4
;I*( = 1,6
i ra h n v c tn ta c
I" = KA
I( = KA
Nh vy tr s dng in ngn mch tng ti im N3 l :
- Dng ngn mch siu qu : IN3" = 17,46 + 27,17 = 44,63 KA
- Dng ngn mch duy tr: I(N3 = 17,46 + 31,05 = 48,51 KA
- Dng in xung kch : ixkN3 = .kxk.IN3" = .1,8.41,718 = 113,6 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
i ra h n v c tn ta c :
- 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
PAGE \# "'Page: '#''"
PAGE \# "'Page: '#''"
PAGE \# "'Page: '#''"
Thit k phn in nh my in v trm bin p
1
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