Xac Dinh Tam Phu Tai

8/2/2019 Xac Dinh Tam Phu Tai

1/96

LI NI U

Nng lng in chnh c dng trong cc cng trnh, nh .N c dng

trong mi lnh vc nhm phc v ti a nhu cu ca con ngi.Vn t ra l

lm th no cung cp in cho ph ti mt cch hiu qu v tin cy. Mt

phng n cung cp in hp l l phi kt hp hi ho gia yu t kinh t - k

thut: tin cy trong cung cp in, an ton cao, tin li cho vic vn hnh,sa cha khi hng hc v phi m bo c cht lng in nng nm trong

gii hn cho php, tn tht in nng l thp nht, hn na phi m bo cho

vic m rng v pht trin trong tng lai vi chi ph u t nh nht.

Trn tinh thn vi s hng dn tn tnh ca thy Nguyn Hu Khoathng qua n ny em hiu r hn v trnh t thit k cung cp in

Do thi gian v kin thc c hn nn chng em khng trnh khi nhng

thiu st rt mong c s ch bo ca thy v cc bn.

Cui cng em xin cm n qu thy c b mn, thy Nguyn Hu Khoa


2/96

DANH SCH CC THI T B

KHMB Tn thi t b Pm (kw) K sd cos

1 My canh 1 15 0.4 0.6

2 My canh 2 15 0.4 0.7

3My canh phn

h ng10 0.4 0.6

4 My h 1 10 0.6 0.7

5 My h 2 10 0.54 0.5

6 My h 3 10 0.7 0.6

7 M d t ki 9 0 7 0 7

S 1 12


3/96


4/96

Trng C in Lc TP. HCM N CUNG CP IN

Yu cu tnh ton

Cn c vo vic b tr ca phn xng v yu cu lm vic thun tin nht, lm vic c hiu qu nht thng qua ccchc nng hot ng ca cc my mc thit b.

Ngoi cc yu cu v k thut th ta phi t yu cu v kinh t, khng nn t qu nhiu cc nhm lm vic ng thi, qunhiu cc t ng lc nh th s khng li v kinh t.Phn nhm ph ti s quyt nh t phn phi trong phn xng, s tuyn dy i ra ca t phn phi

Phn nhm ph ti cho phn xng da vo cc yu t sau:

Cc thit b trong cng mt nhm nn c cng chc nng.

Phn nhm theo khu vc: cc thit b gn nhau du cng sut cho cc nhm: tng cng sut ca cc nhm gn bng nhau. Tnh ton v d bo nhu cu ph ti in: Thit k chiu sng, xc nh ph ti,tng hp ph ti, d bo ph ti,

Tnh ton b cng sut phn khng: xc nh cng sut b, phn b cc thit b b, nh gi hiu qu, la chn s nguyn l t ng iu chnh cng sut phn khng.

Tnh ton ngn mch (1 pha, 3 pha) t ta c Dng ti ca tng nhm gn vi dng ti ca CB chun v tnh ton anton cho ngi s dng.

GVHD: Nguyn Hu Khoa 4


5/96


Nhn xtMSSV H v tn Nhn xt im

09CH1E_30 Nguyn Hong Nhi

09CH1E_32 Trn Nhn Nh

09CH1E_54 Lm Hng

Phc

MC LC

TRANG

I.XC NH TM PH TI 4II. TNH TON V D BO NHU CU PH TI 9III. TNH TON CHIU SNG 17IV. TNH TON CHIU SNG CHO PHN XNG 22V. TNH TON B CNG SUT CHO PHN XNG 22V CHN MBA PHN XNGV. TNH TON CHO T PHN PHI 24VII.CHN DY CHO T PHN PHI V T NG LC 24VIII.TON ST P 32IX.TNH NGN MCH 40X.TNH TON AN TON 61



6/96


THIT K CUNG CP IN CHO PHN XNG DT

I.XC NH TM PH TIMc ch xc nh tm ph ti: m bo v kinh t k thut, an ton cho ngi v thit b .V vy khi thit k cung

cp in cho mt phn xng th chng ta cn xc nh tm ph ti, v tr t t phn phi chnh , t ng lc, trm bin pnhm cung cp in vi tn tht in p v tn tht cng sut nh nht, chi ph kim loi mu hp l.

Nhm 1:

KHMB Tn thit b

S

lng

Pm

(kW) Ksdcos Xi Yi

1 My canh 1 1 15 0.4 0.6 6,59 43,122 My canh 2 1 15 0.4 0.7 12,17 43,123 My canh phn 1 10 0.4 0.6 17,74 43,124 My h 1 1 10 0.6 0.7 6,59 25,25 My h 2 1 10 0.54 0.54 12,17 25,26 My h 3 1 10 0.7 0.7 17,74 25,2

Tm ph ti c tnh theo cng thc:

Xi= = = 11,76m

Yi= = = 35,44m



7/96


thun tin thao tc v v m quan ta di t ng lc v ta :X=0m Y=35m

Nhm 2:

KHMB Tn thit b S lng Pm Ksd cos Xi Yi11 My dt CTD 36 11.5 0.55 0.6 34,98 50,4


Xi= = = m

Yi= =


Nhm 3:KHMB Tn thit b S lng Pm Ksd cos Xi Yi

12 My dt CTM 36 11.5 0.55 0.6 61,6 50,4Tm ph ti c tnh theo cng thc:

Xi= = = m

Yi= =



8/96


thun tin thao tc v v m quan ta di t ng lc v ta :X=56m Y=70mNhm 4:

KHMB Tn thit b S lng Pm Ksd cos Xi Yi11 My dt CTD 36 9 0.7 0.7 76,04 50,4


Xi= = = 76,04m

Yi= =


KHMB Tn thit b S lng Pm Ksd cos Xi Yi11 My dt CTD 36 9 0.7 0.7 34,08 19,6


Xi= =34,97m

Yi= =



9/96


thun tin thao tc v v m quan ta di t ng lc v ta :X=34,08m Y=0mNhm 6:

KHMB Tn thit b S lng Pm Ksd cos Xi Yi

8 My dt kim 27 9 0.7 0.7 61,69 19,69 My dt kim 9 9 0.7 0.6


Xi= = =55,76m

Yi= =


KHMB Tn thit b S lng Pm Ksd cos Xi Yi13 My dt CTM 36 11.5 0.55 0.7 76,04 19,6


Xi= = =76,04m

Yi= =




10/96


Nhm 8:

KHMB Tn thit b S lng Pm (kW) Ksd cos Xi Yi13 My dt kimCTM 8 11,5 0,55 0.7 12,7 35,4

14 My dt kimCTM 27 11,5 0,55 0.6


Xi= = =12,7m

Yi=


XC NH TM PH TI CHO T PHN PHI CHNH

Nhm 1 2 3 4 5 6 7 8Xnhm(m) 0 35 56 77 35 56 77 110Ynhm(m) 35 70 70 50,4 0 0 0 35

Pnhm(kW70 414 414 324 324 324 414 368Ppx (kW) 2652




11/96


X= 62,86m

Y=


TNG KT TM PH TI CA T NG LC V T PHN PHI CHNH

Nhm 1 2 3 4 5 6 7 8 TPPCXnhm(m) 0 35 56 77 35 56 77 110 0Ynhm(m) 35 70 70 50,4 0 0 0 35 60Pnhm(kW70 414 414 324 324 324 414 368 2625

II. TNH TON V D BO NHU CU PH TI



12/96


13/96


Dng tnh ton ca nhm:Itt=

Dng in nh mc ca thit b: Idm=

BNG TNH TON PH TI CA NHM 1KHMB

Pm(kW) Ksd

cos Idmi

1 15 0.4 0.6 37,9842 15 0.4 0.7 32,5573 10 0.4 0.6 25,3224 10 0.6 0.7 21,7055 10 0.54 0.54 30,3866 10 0.7 0.7 25,322

T bng trn ta suy ra dng nh mc ln nht ca nhm 1:

Idmmax=37,984A Dng nh nhn ca nhm thit b

Ikdmax=Kmm.Idmmax=5.37,984=189,22A(chn Kmax=5 v Pmax


14/96


Da vo bng 2-2 trang 112 sch mch cung cp v phn phi in ta s chn h s Kmax theo h s s dng Ksd v sch sthit b hiu qu nhq:

nhq= =

=>Kmax=1.133 Cng sut tc dng trung bnh :Ptb=Ksd.Pdm=0,55.36.11,5=227,4KW

Cng sut tc dng tnh ton:Ptt=Kmax.Ksd. Pdm=0,55.36.11,5.1,133=257,98KW

Cng sut phn khng tnh ton:Qtt=Ptb.tan =227,4.tan(arccos0,55)=302,84KVAr

Cng sut biu kin tnh ton: S tt= =397,82KVA


Dng in nh mc ca thit b: Idm= = =29,12A

Idmmax=29,12A

Dng nh nhn ca nhm thit bIkdmax=Kmm.Idmmax=5.29,12=145,6A(chn Kmax=5 v Pmax


15/96


cos= =

Da vo bng2-2 trang 112 sch mch cung cp v phn phi in ta s chn h s Kmax theo h s s dng Ksd v sch sthit b hiu qu nhq:

nhq= =

=>Kmax=1.133

Cng sut tc dng trung bnh :Ptb=Ksd.Pdm=0,55.36.11,5=227,4KW

Cng sut tc dng tnh ton:Ptt=Kmax.Ksd. Pdm=0,55.36.11,5.1,133=257,98KW





Idmmax=29,12A



16/96


cos= = =0,7

Da vo bng2-2 trang 112 sch mch cung cp v phn phi in ta s chn h s Kmax theo h s s dng Ksd v sch s

thit b hiu qu nhq:

nhq= =

=>Kmax=1.09

Cng sut tc dng trung bnh :Ptb=Ksd.Pdm=0,7.36.9=226,8KW

Cng sut tc dng tnh ton:Ptt=Kmax.Ksd. Pdm=0,7.36.9.1,09=247,212KW





Idmmax=19,534A



17/96


Ksd= = =0,7

cos= = =0,7


nhq= =

=>Kmax=1.09

Cng sut tc dng trung bnh :Ptb=Ksd.Pdm=0,7.36.9=226,8KW

Cng sut tc dng tnh ton:Ptt=Kmax.Ksd. Pdm=0,7.36.9.1,09=247,212KW





Idmmax=19,534A



18/96


Ksd= = =0,7

cos= = =0,675


nhq= =

=>Kmax=1.09

Cng sut tc dng trung bnh :Ptb=Ksd.Pdm=0,7.(27.9+9.9)=226,8KW

Cng sut tc dng tnh ton:Ptt=Kmax.Ksd. Pdm=0,7.(27.9+9.9).1,09=247,212KW

Cng sut phn khng tnh ton:Qtt=Ptb.tan =226,8.tan(arccos0,675)=247,906KVAr Cng sut biu kin tnh ton: Stt= Dng tnh ton ca nhm: Itt=


Suy ra :Idmmax=22,790A

Dng nh nhn ca nhm thit b

Ikdmax=Kmm.Idmmax=5. 22,790 =113,95A(chn Kmax=5 v Pmax


19/96


20/96


Ikdmax=Kmm.Idmmax=5.24,916=124,58A(chn Kmax=5 v PmaxKmax=1.09

Cng sut tc dng trung bnh : Ptb=Ksd.Pdm=0,55.( )=202,4KW

Cng sut tc dng tnh ton:Ptt=Kmax.Ksd. Pdm=0,625.( ).1,09=230,94KW

Cng sut phn khng tnh ton:Qtt=Ptb.tan =202,4.tan(arccos0,625)=252,798KV Ar Cng sut biu kin tnh ton: S tt= =342,403KVA




21/96



Idmmax=24,96A

Dng nh nhn ca nhm thit bIkdmax=Kmm.Idmmax=5. 24,96=145,6A(chn Kmax=5 v Pmax


22/96


23/96


8. Phn b cc n:Cch trn: h=0 (m). B mt lm vic: 0.8 (m)Chiu cao treo n so vi b mt lm vic: htt = 3.2 (m)

9. Ch s a im K = = 4,875 5

10. H s b:

Chn h s suy gim quang thng: 1 =0,8

Chn h s suy gim do bm bi: 2 =0.9

d=

11. T s treo: j= (v h=0)

12. H s s dng: U= d .ud +i .ui =0,58.1,02+0,31.0,75=0,824

13. Quang thng tng: tng = = =487090,413(lm)

14. Xc nh s b n:



24/96


Nbn =97,418(b)

Chn s b n: Nbn = 10015. Kim tra sai s quang thng:

%= = =2,65%

16. Kim tra ri trung bnh trn b mt lm vic:

Etb= = =307,951(lx)

17. Phn b cc n:



25/96

10 bLdc=3,5m


2.Tnh chiu sng cho nhm 21. Kch thc: Chiu di: a = 35 m; Chiu rng: b = 27,5 m

Chiu cao: H = 4 m; Din tch: S = 962,5 m2

2. Mu sn: Trn: trng H s phn x trn: tr = 0.75

Tng: vng nht H s phn xa tng: tg= 0.50Sn: xanh sm H s phn x sn: lv= 0.20

3. ri yu cu: Etc = 300(lx)


10 b ( Lngang =2,757m)


26/96


4. Chn h chiu sng: chung u.

5. Chn khong nhit mu: Tm = 2900 - 4200 K0

6. Chn bng n: Loi: n hunh quang mu trng universel Tm = 4000 K0

Ra = 76 P = 36 (W) d = 2500 (lm)7. Chn b n: Loi: Aresa 202

Cp b n: 2x36 (W) Hiu sut: 0.58H + 0.31TS n / 1b: 2 Quang thng cc bng/ 1b: 5000 (lm)Ldcmax =1.6htt = 5.12(m) Lngangmax =2htt = 6.4 (m)

8. Phn b cc n:Cch trn: h=0 (m). B mt lm vic: 0.8 (m)Chiu cao treo n so vi b mt lm vic: htt = 3.2 (m)

9. Ch s a im: K = =4,875 5

10. H s b:Chn h s suy gim quang thng: 1 =0,8

Chn h s suy gim do bm bi: 2 =0.9



27/96


d=

11. T s treo: j= (v h

=0)12. H s s dng: U= d .ud +i .ui =0,58.1,02+0,31.0,75=0,824

13. Quang thng tng: tng = = =487090,413(lm)

14. Xc nh s b n:

Nbn =97,418(b)

Chn s b n: Nbn = 10015. Kim tra sai s quang thng:

%= = =2,65%



28/96


16. Kim tra ri trung bnh trn b mt lm vic:

Etb= = =307,951(lx)

17. Phn b cc n:

4. Tnh chiu sng cho nhm 4 :Chn s b n: Nb n = 1005. Tnh chiu sng cho nhm 5 : Chn s b n: Nb n = 1006. Tnh chiu sng cho nhm 6 : Chn s b n: Nb n = 1007. Tnh chiu sng cho nhm 7 :Chn s b n: Nb n = 100

8. Tnh chiu sng cho nhm 8 : Chn s b n: Nb n


10 b ( Lngang =2,757m)

10 b Ldc=3,5m


29/96


30/96


Qpp = Qcspx + Qtt = Pcstng .tg(arccos 0.96) + Qtt= 20,16 + 1844,429 = 1860,709 (kVar)

Qttpx = Kt . Qpp = 0,9. 1860,709 = 1665,638 (kVar)

=> Sttpx = = =2368,787KVAXc nh my b: ci thin h s cng sut ca mng in, cn c b t in lm ngun pht cng sut phn

khng. Cch gii quyt ny c gi l b cng sut phn khng.

Hnh gin m t nguyn l b cng sut Qc = P ( tg - tg ' )

cos= = =0,711 => tan =0,99

Chn cos sau khi b l 0.95 => tg =0.33=> Cng sut my b cn thit l:

Qb = Pttpx ( tg ttpx - tg) = 1684,281 . (0,99 0.33) = 1111.444 (kWar)Vi cng sut trn ta chn c: 23my b KC2-0.38-50-3Y3



31/96


=> Qsau b = 1665,638 23.50 = 515,638 (kVar)Chn MBA cho trng hp qu ti s c, t 2 MBA song song ngoi tri.

Ta c Sttpx =2368,787 (kVA)Sau khi b cng sut:

Sttpx = = = 1761,444KVAcossaubu =0.95

SmMBA = = =1258,174(kVA)

=> Chn SmMBA = 1500 kV

MBA c:P0 = 3300 (W) I0 = 1.2%

PN = 18 kW UN = 7%

RMBA= = =1,28.

ZMBA= = =7,35.

XMBA = = =7,35.

VI. TNH TON CHO T PHN PHI

Dng in tnh ton:

Ittpp =



32/96


H s s dng ca v h s cng sut t phn phi:

Ksd= = =0,625

Dng in nh nhn:Inpp = Inmaxnhm + Ittpp Ksdpp. Ittmaxnhm = 734 + 2676,237 0.625 .604.42 = 3032,475 (A)

VII.CHN DY CHO T PHN PHI V T NG LC

1. CHN DY T MBA N T PHN PHI CHNHTa c: Ilvmax = 2676,237 (A)Cch i dy: i dy trn mng cp. Chn cp ng 1pha cch in PVC do Lens ch to.V dng in ln nn ta chn 5 cp cho mi pha.

H s hiu chnh: K = K1 .K2 .K3 = 0.82K1 = 1 h s nh hng cch thc lp tK2 = 0.75 v xem nh c 5 cp 3 pha t trong hngK3 = 1 tng ng nhit mi trng l 300C

Icp

=> Chn dy c tit din F = 500mm2, Icp = 760 (A)2. CHN DY T T PHN PHI CHNH N T NG LC 1

Ilvmax = Ittnhm = 105,195 (A)iu kin chn dy:

Icp =126,131A

Cch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.



33/96


H s hiu chnh: K = K4 .K5.K6 .K7 = 0.8.1.1.05.1 = 0.84K4 = 0.8 cp lp trong ngK5 = 1 v ch c 1 cp 3 phaK6 = 1.05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy c tit din F = 25mm2, Icp = 144 (A)3. CHN DY T T PHN PHI CHNH N T NG LC 2

Ilvmax = Ittnhm = 604,42(A)

iu kin chn dy: Icp =899,43A

Cch i dy: chn trong t. Chn cp ng 1 pha cch in PVC do Lens ch to. Chn 2cp cho mi pha.H s hiu chnh: K = K4 . K5 .K6 .K7 = 0,8. 0,8.1,05.1 = 0,672

K4 = 0.8 cp lp trong ngK5 = 0.8 v xem nh c 2 cp 3 pha

K6 = 1.05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy c tit din F = 240mm2, Icp = 550 (A)4. CHN DY T T PHN PHI CHNH N T NG LC 3

Ilvmax = Ittnhm = 604.42(A)iu kin chn dy:

Icp =899,43A

Cch i dy: chn trong t. Chn cp ng 1 pha cch in PVC do Lens ch to. Chn 2 cp cho mi pha.H s hiu chnh: K = K4.K5. K6 .K7 = 0.8. 0.8.1.05.1 = 0.672

K4 = 0.8 cp lp trong ngK5 = 0.8 v xem nh c 2 cp 3 phaK6 = 1.05 tnh cht t mK7 = 1 tng ng nhit t l 200 C



34/96


=> Chn dy c tit din F = 240mm2, Icp = 550 (A)5. CHN DY T T PHN PHI CHNH N T NG LC 4

Ilvmax = Ittnhm = 514,453 (A)


Cch i dy: chn trong t. Chn cp ng 1 pha cch in PVC do Lens ch to. Chn 2cp cho mi pha.

H s hiu chnh: K = K4 .K5 . K6 . K7 = 0,8. 0,8. 1,05. 1 = 0,672K4 = 0,8 cp lp trong ngK5 = 0,8 v xem nh c 2 cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy c tit din F = 185mm2, Icp = 443 (A)6. CHN DY T T PHN PHI CHNH N T NG LC 5

Ilvmax = Ittnhm = 514,453 (A)iu kin chn dy: Icp =765,555A


H s hiu chnh: K = K4 .K5 . K6 . K7 = 0,8. 0,8. 1,05. 1 = 0,672K4 = 0,8 cp lp trong ngK5 = 0,8 v xem nh c 2 cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy c tit din F = 185mm2, Icp = 443 (A)7. CHN DY T T PHN PHI CHNH N T NG LC 6




35/96


iu kin chn dy: Icp Cch i dy: chn trong t. Chn cp ng 1 pha cch in PVC do Lens ch to.Chn 2cp cho mi pha.

H s hiu chnh: K = K4 .K5 . K6 . K7 = 0,8. 0,8. 1,05. 1 = 0,672K4 = 0.8 cp lp trong ng

K5 = 0.8 v xem nh c 2 cp 3 phaK6 = 1.05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy c tit din F = 185mm2, Icp = 443 (A)8. CHN DY T T PHN PHI CHNH N T NG LC 7


iu kin chn dy: Icp =787,969ACch i dy: chn trong t. Chn cp ng 1 pha cch in PVC do Lens ch to. Chn 2

cp cho mi pha.

H s hiu chnh: K = K4 .K5 . K6 . K7 = 0,8. 0,8. 1,05. 1 = 0,672K4 = 0.8 cp lp trong ngK5 = 0.8 v xem nh c 2 cp 3 phaK6 = 1.05 tnh cht t m.K7 = 1 tng ng nhit t l 200 C=> Chn dy c tit din F = 185mm2, Icp = 434 (A)9. CHN DY T T PHN PHI CHNH N T NG LC 8

Ilvmax = Ittnhm = 520,228 (A)



H s hiu chnh: K = K4 . K5 . K6 . K7 = 0,8. 0,8. 1,05. 1 = 0,672K4 = 0.8 cp lp trong ng



36/96


K5 = 0.8 v xem nh c 2 cp 3 phaK6 = 1.05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy c tit din F = 185mm2, Icp = 443 (A)10. CHN DY T T NG LC N THIT B

10.1. Chn dy t t ng lc 1 n cc thit b nhm 1Chn dy n thit b 1Ilvmax = Im = 37,984(A)

iu kin chn dy: Icp =45,219ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4 .K5 . K6 . K7 = 0,8. 1. 1,05. 1 = 0,84

K4 = 0,8 cp lp trong ng

K5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy 3G 4 c tit din F = 4mm 2, Icp = 53 (A)

Chn dy n thit b 2Ilvmax = Im = 37,984 (A)



H s hiu chnh: K = K4 . K5 . K6 . K7 = 0,8. 1. 1,05. 1 = 0,84K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C



37/96


=> Chn dy 3G 2,5c tit din F = 2,5mm2 ,Icp = 41 (A)Chn dy n thit b 3Ilvmax = Im = 25,322 (A)


Cch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4 . K5 . K6 . K7 = 0,8.1. 1,05. 1 = 0,84K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy 3G 1.5 c tit din F = 1.5mm2, Icp = 31 (A)

Chn dy n thit b 4lvmax = Im = 21,705 (A)

iu kin chn dy: Icp =25,839ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4 . K5 . K6 . K7 = 0,8.1.1,05. 1 = 0,84

K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m. 200CK7 = 1 tng ng nhit t l=> Chn dy 3G 2.5 c tit din F = 1,5 mm2, Icp = 31(A)

Chn dy n thit b 5

Ilvmax = Im = 30,386 (A)iu kin chn dy: Icp =36,174A




38/96


H s hiu chnh: K = K4 . K5 .K6 .K7 = 0,8.1. 1,05. 1 = 0,84K4 = 0.8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1.05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy 3G 2,5 c tit din F = 2,5mm2, Icp = 41 (A)

Chn dy n thit b 6Ilvmax = Im = 25,322(A)

iu kin chn dy: Icp =30,145ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4 .K5 . K6 . K7 = 0,8. 1. 1,05. 1 = 0,84

K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 pha

K6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy 3G 1,5 c tit din F = 1,5mm2, Icp = 31 (A)10.2. Chn dy t t ng lc 2 n cc thit b nhm 2

Ilvmax = Im = 29,12 (A)

Icp =34,667ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4 . K5 . K6 . K7 = 0,8. 1. 1,05 . 1 = 0,84

K4 = 0.8 cp lp trong ng

K5 = 1 v ch c 1cp 3 phaK6 = 1.05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy 3G 2.5 c tit din F = 2,5 mm2, Icp = 41 (A)10.3. Chn dy t t ng lc 3 n cc thit b nhm 3



39/96


Ilvmax = Im = 29.12 (A)

Icp =34,667ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4 . K5 . K6 . K7 = 0,8. 1. 1,05 . 1 = 0,84

K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy 3G 2.5 c tit din F = 2.5 mm2, Icp = 41 (A)10.4. Chn dy t t ng lc 4 n cc thit b nhm 4

Ilvmax = Im = 19,534 (A)

iu kin chn dy Icp =23,255ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.

H s hiu chnh: K = K4 . K5 . K6 .K7 = 0,8.1. 1,05 . 1 = 0,84K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy 3G 1.5 c tit din F = 1,5mm2, Icp = 31 (A)10.5. Chn dy t t ng lc 5 n cc thit b nhm 5

Ilvmax = Im = 19,534 (A)

iu kin chn dy Icp =23,255ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4 . K5 . K6 . K7 = 0,8.1. 1,05 . 1 = 0,84

K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 pha



40/96


K6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C=> Chn dy 3G 1.5 c tit din F = 1,5mm2, Icp = 31 (A)10.6. Chn dy t t ng lc 6 n cc thit b nhm 6

Ilvmax = Im = 22,790 (A)

iu kin chn dy: Icp =27,131ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4. K5 . K6 . K7 = 0,8. 1. 1,05 . 1 = 0,84

K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200 C=> Chn dy 3G 1.5 c tit din F = 1.5mm2, Icp = 31 (A)

10.7. Chn dy t t ng lc 7 n cc thit b nhm 7Ilvmax = Im = 24,961 (A)

iu kin chn dy: Icp =29,71ACch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4. K5 . K6 . K7 = 0,8. 1. 1,05 . 1 = 0,84

K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C

=> Chn dy 3G 1,5c tit din F = 1,5 mm2, Icp = 31 (A)10.8. Chn dy t t ng lc 8 n cc thit b nhm 8

Ilvmax = Im = 29,12 (A)




41/96


Cch i dy: chn trong t. Chn cp ng 3 li cch in PVC do Lens ch to.H s hiu chnh: K = K4 . K5 . K6 . K7 = 0,8.1. 1,05 . 1 = 0.84

K4 = 0,8 cp lp trong ngK5 = 1 v ch c 1cp 3 phaK6 = 1,05 tnh cht t m.K7 = 1 tng ng nhit t l 200C Chn dy 3G 2.5 c tit din F = 2,5 mm2, Icp = 41 (A)

Phn on S dy x M dy Bn knh (m) Ichophp (A)

MBA TPPC 15xCVV - 1x500 0.0366 760TPPC TL1 1xCVV 1x25 0.727 144TPPC TL2 6xCVV 1x240 0.0754 501TPPC TL3 6xCVV 1x240 0.0754 501TPPC TL4 6xCVV 1x185 0.0991 443

TPPC TL5 6xCVV 1x185 0.0991 443TPPC TL6 6xCVV 1x185 0.0991 443TPPC TL7 6xCVV 1x185 0.0991 443TPPC TL8 6xCVV 1x185 0.0991 443TL1 THIT B 1 1xCVV 3x4 4,61 53TL1 THIT B 2 1xCVV 3x2,5 7,41 41TL1 THIT B 3 1xCVV 3x1.5 12.1 31TL1 THIT B 4 1xCVV 3x1.5 12,1 31TL1 THIT B 5 1xCVV 3x2,5 7,41 41TL1 THIT B 6 1xCVV 3x1.5 12,1 31

TL2 THIT B 1xCVV 3x2.5 7.41 41TL3 THIT B 1xCVV 3x2.5 7.41 41TL4 THIT B 1xCVV 3x1.5 12.1 31TL5 THIT B 1xCVV 3x1.5 12.1 31TL6 THIT B 1xCVV 3x1.5 12.1 31



42/96


TL7 THIT B 1xCVV 3x1.5 12,1 31TL8 THIT B 1xCVV 3x2.5 7.41 41

VIII.TON ST P V NGN MCHA. TNH ST P

Kim tra st p nhm kim tra cht lng mng in c thit k. st p cn tha:Ch lm vic bnh thng: UUcp= 5% Um

Ucp =5%380 =19(V)

Khi khi ng: U Ucp = 10% UmUcp=10%380= 38 (V)U = (2767,237 7,32 10-5 0,95+2767,2371,610 -40,31)=0,57V



43/96


B.TNH ST P T MBA N TPPC

Khong cch t MBA n TPPC l 10m

Ch lm vic bnh thng:Upp == (2676,2377,32.10-50,95+2627,4751,610 -40,31=0,57V

Khi khi ng:Upp = .(Idnpp.Rd.cos+Idnpp.Xd.sin)= (3032,4757,32.10-50,95+3032,4751,610 -4.0,31)=0,625VC.TNH ST P T TPPC N T NG LC V N THIT B1. St p t TPP chnh n t ng lc 1 v n thit b 1 ca nhm 1

Khong cch t TPPC n TL1 l 25m, TL1 n thit b 1 l 12m

U= .

Ch lm vic bnh thng:Ul-1= (105,1950,7270,0250,621+37,9844,610,0120,6)=4,24VU=Ul-1+Upp=4,24+0,57=4,81VKhi khi ng: chn cos =0.35U1-1= (276,4650,7270,0250,35+537,9844,610,0120,35)=9,4VU=Ul-1+Upp=9,4+0,625=10,025 V



44/96


=> Dy dn chn t yu cu.2. St p t TPP chnh n t ng lc 1 v n thit b 2 ca nhm 1

Khong cch t TPPC n TL1 l 25m, TL1 n thit b 2 l 15mUpp =Ch lm vic bnh thng:Ul-2= (105,1950,7270,0250,621+37,9844,610,0150,7)=6,44VU=Ul-1+Upp=6,44+0,57=7,01V

Khi khi ng: chn cos =0.35U1-2= (276,4650,7270,0250,35+537,9844,610,0150,35)=14V

U=Ul-1+Upp=14+0,625=14,625 V=> Dy dn chn t yu cu.3. St p t TPP chnh n t ng lc 1 v n thit b 3 ca nhm 1

Khong cch t TPPC n TL1 l 25m, TL1 n thit b 3 l 20mUpp =Ch lm vic bnh thng:Ul-3= (105,1950,7270,0250,621+25,32212,10,020,6)=8,42V

U=Ul-1+Upp=8,42+0,57=8,99V

Khi khi ng: chn cos =0.35U1-3= (276,4650,7270,0250,35+525,32212,10,020,35)



45/96


=21,62 VU=Ul-3+Upp=21,62 +0,625=22,245 V

=> Dy dn chn t yu cu.

4. St p t TPP chnh n t ng lc 1 v n thit b 4 ca nhm 1

Khong cch t TPPC n TL1 l 25m, TL1 n thit b 4 l 12mUpp =Ch lm vic bnh thng:Ul-4= (105,1950,7270,0250,621+21,70512,10,0120,7)=5,88VU=Ul-4+Upp=5,88+0,57=6,45V

Khi khi ng: chn cos =0.35U1-4= (276,4650,7270,0250,35+521,70512,10,0120,35)=12,6VU=Ul-4+Upp=12,6+0,625=13,225 V=> Dy dn chn t yu cu.5. St p t TPP chnh n t ng lc 1 v n thit b 5 ca nhm 1

Khong cch t TPPC n TL1 l 25m, TL1 n thit b 5 l 15m

Upp =Ch lm vic bnh thng:Ul-5= (105,1950,7270,0250,621+30,3867,410,0150,5)=4,98 V



46/96


U=Ul-5+Upp=4,98 +0,57=5,55 V

Khi khi ng: chn cos =0.35U1-5= (276,4650,7270,0250,35+530,3867,14.0,0150,35)

=13,28 VU=Ul-5+Upp=13,28 +0,625=13,905 V=> Dy dn chn t yu cu.6. St p t TPP chnh n t ng lc 1 v n thit b 6 ca nhm 1

Khong cch t TPPC n TL1 l 25m, TL1 n thit b 6 l 20mUpp =Ch lm vic bnh thng:Ul-2= (105,1950,7270,0250,621+25,32212,10,020,621)

=8,42VU=Ul-6+Upp=8,42+0,57=8,99V

Khi khi ng: chn cos =0.35U1-2= (276,4650,7270,0250,35+525,3227,140,020,35)=21,62VU=Ul-6+Upp=21,62+0,625=22,245V=> Dy dn chn t yu cu.7. St p t TPP chnh n t ng lc 2 v n thit b 11 xa nht ca nhm 2

Khong cch t ng lc chnh n ng lc 2 l 40mKhong cch t ng lc 2 n thit b 11 xa nht l 35mUpp =



47/96


Ch lm vic bnh thng:U2= (604,42.0,07540,50,040,6+604,420,080,040,8+29,127,410,0350,6)=10,13VU=U2+Upp=10,13+0,57=10,7

Khi khi ng: chn cos =0.35U2= (7340,07540,50,040,35+7340,080,50,040,936 +529,127,410,0350,35)=25,46 VU=U2+Upp=25,46+0,625=26,085 V=> Dy dn chn t yu cu.

8. St p t TPP chnh n t ng lc 3 v n thit b 12 xa nht ca nhm 3

Khong cch t ng lc chnh n ng lc 3 l 60mKhong cch t ng lc 3 n thit b 12 xa nht l 35mUpp =Ch lm vic bnh thng:U3= (604,420,07540,50,060,6+604,420,080,50,040,8+ 29,127,410,035.0,6)=11,27VU=U3+Upp=11.27+0,57= 11,84 V

Khi khi ng: chn cos =0.35

U3= (7340,07540,060,50,35+734 0,080,50,060,936 +529,127,410,0350,35)=26,75 VU=U3+Upp=26,75+0,625=27,375 V=> Dy dn chn t yu cu.



48/96


9. St p t TPP chnh n t ng lc 4 v n thit b 10 xa nht ca nhm 4

Khong cch t ng lc chnh n ng lc 4 l 80mKhong cch t ng lc 4 n thit b 10 xa nht l 35mUpp =Ch lm vic bnh thng:U4= (514,4530,09910,080,50,7+514,5130,080,50,080,71 +19,53412,10,0350,7)=14,54VU=U4+Upp=14,54+0,57=15,11 V

Khi khi ng: chn cos =0.35U4= (598,4790,09910,50,080,35+598,4790,080,50,080,936 +519,53412,10,0350,35)=29,62 V

U=U4+Upp=29,62+0,625=30,245 V=> Dy dn chn t yu cu.10. St p t TPP chnh n t ng lc 5 v n thit b 7 xa nht ca nhm 5

Khong cch t ng lc chnh n ng lc 5 l 70mKhong cch t ng lc 5 n thit b 7 xa nht l 35mUpp =Ch lm vic bnh thng:U5= (514,4350,09910,50,070,7+514,4350,080,50,070,71 +19,53412,10,035.0,7)

=13,97 VU=U5+Upp=13,97+0,57=14,54 V




49/96


U5= (598,4790,09910,50,070,35+ 598,4790,080,50,070,936+ 519,53412,10,0350,35)=29,05 VU=U5+Upp=29,05+0,625=29,675 V=> Dy dn chn t yu cu.11. St p t TPP chnh n t ng lc 6 v n thit b 8 xa nht ca nhm 6

Khong cch t ng lc chnh n ng lc 5 l 70mKhong cch t ng lc 6 n thit b 7 xa nht l 35m

Upp =Ch lm vic bnh thng:

U6= (531,9230,09910,50,070,675)+ 531,9230,080,50,070,738 +22,7912,10,0350,675)=16,5U=U6+Upp=16,5+0,57=17,07V

Khi khi ng: chn cos =0.35U6= (629,920,09910,50,070,35+ 629,920,080,50,070,936 +522,7912,10,0350,35)=34,63U=U6+Upp=34,63+0,625=35,255 V=> Dy dn chn t yu cu.12. St p t TPP chnh n t ng lc 7 v n thit b 13 xa nht ca nhm 7

Khong cch t ng lc chnh n ng lc 7 l 100m

Khong cch t ng lc 7 n thit b 13 xa nht l 35mUpp =

Ch lm vic bnh thng:U7= (529,5140,09910,50,10,7+529,5140,080,50,10,71 +24,96112,10,0350,7)



50/96


=18,6VU=U7+Upp=18,6+0,57=19,17V


U7

= (529,5140,09910,50,10,35+ 529,5140,080,50,10,934+524,96112,10,0350,35)=38,07 VU=U7+Upp=38,07+0,625=38,695 V=> Dy dn chn cha t yu cu.

Chn li dy c F=240mm2 c r0=0,0754/KmCh lm vic bnh thng:

U7= (529,5140,07540,50,10,7+5 29,5140,080,50,10,71 +24,96112,10,0350,7)=17V

U=U7+Upp=17+0,57=17,57V

Khi khi ng: chn cos =0.35U7= (529,5140,07540,50,10,35+ 529,5140,080,50,10,936 +524,96112,10,0350,35)=37 VU=U7+Upp=37+0,625=37,625 V=> Dy dn chn cha t yu cu.13. St p t TPP chnh n t ng lc 8 v n thit b 14 xa nht ca nhm 8

Khong cch t ng lc chnh n ng lc 8 l 110mKhong cch t ng lc 8 n thit b 13 xa nht l 25mUpp =

Ch lm vic bnh thng:



51/96


U8= (520,2280,09910,50,110,625+520,2280,080,50,110,78+29,127,410,0250,625)=14,34 VU=U8+Upp=14,34+0,57=14,91 V

Khi khi ng: chn cos =0.35U8= (520,2280,09910,50,110,35+ 520,2280,080,50,110,936 +529,127,410,0250,35)=28U=U8+Upp=28+0,625=28,625V=> Dy dn chn t yu cu.

BNG TNG KT CHN DY SAU KHI XT ST P

Phn on Ulvbt (V) Ucp lvbt

(V) Uk (V) U

cp k

(V)MBA TPPC TL1 THIT B 1 4,24 19 10,025 38MBA TPPC TL1 THIT B 2 7,01 14,625MBA TPPC TL1 THIT B 3 8,99 22,245MBA TPPC TL1 THIT B 4 6,45 13,225MBA TPPC TL1 THIT B 5 5,55 13,905MBA TPPC TL1 THIT B 6 8,99 22,245MBA TPPC TL2 THIT B 16,91 26,085MBA TPPC TL3 THIT B 11,84 27,375MBA TPPC TL4 THIT B 15,11 30,245MBA TPPC TL5 THIT B 14,54 29,675MBA TPPC TL6 THIT B 17,07 35,255



52/96


MBA TPPC TL7 THIT B 17,57 37,625MBA TPPC TL8 THIT B 14,91 28,625

IX.TNH NGN MCHA.Tnh ngn mch ti t phn phi chnhTa chn MBA c: SmMBA = 1500 kVA

P0 = 3300 (W) I0 = 1.2%

PN = 18 kW UN = 7%

RMBA= = =1,28

ZMBA= = =7,467

XMBA = =

=7,35

V ta t 2 MBA song song nn:

=> ZMBA= = =3,73

Chn chiu di dy t MBA n TPPC khong 10m



53/96


X0 = 0,08 ( / km ) v tit din dy F 50 mm2

Xd = 0,08 .0,01 = 8.10-4

Rd = 0,0366 .0,01 =3,6610-4 ( )

Zd = = =8,810-4

Zd0 = = =1,76.10-4

= = =45,84KA

iu kin chn CB: ImCB IlvmaxIlvmax Ir IcpddIcu

Imm Im

Ta c : Ilvmax =2676,237(A)=45,84(kA)Icpdd = k.Icpdd = 0,82. 760 .5 = 623,2 (A)

Ilvmax Ir Icpdd Ilvmax Kr .I mCB Icpdd



54/96


Kr.IdmCB

0.83 Kr 0.97

=> Chn Kr= 0.9 Ir = 0,9 .3200 = 2880 (A)Im = Km . Ir= 4. 2880 = 11520 (A)

=> Tra bng 8.27 chn c CB cho TPPC M32 c:UmCB = 690 kV; ImCB = 3200 (A); Icu = 75 kA

Tnh ngn mch ti t ng lc 1Chn chiu di dy t TPPC n t ng lc 1 khong 25m

X0 = 0 ( / km ) v tit din dy F 50 mm2

Xd = 0 ( )

Rd = 0,727.0,025=18,17510-3 ( )

=> Zd1 ==

18,17510-3 ( )

= =

=9,93KA


T C i L TP HCM N CUNG CP IN


55/96


Chn CB:

+ maxdmCB lvI I =

Ta c : maxlvI =105,195(A)

dmCBI =105,195 (A)

Chn CB: NC125H dmCBI =125(A)Trip Unit STR22SE CUI = 10(kA)

0r r dmCBI k k I=

,

cpdd cpdd I k I=

=0.76174=132.24(A)

*iu kin: ,maxlv r cpdd I I I

,

max cpddlv r

dmCB dmCB dmCB

II II I I

,

max

0

cpddlvr

dmCB dmCB

IIk k

I I

Chn :max

0

lv

dmCB

Ik

I =

K0 >=

=0,84

0k =0,85




56/96


,

max

0 0

cpddlvr

dmCB dmCB

IIk

I k I k

0.90 rk 1.24

rk =0.98

rI =0,850.98125= 104,125(A)

Ta c: m m rI k I= ( i vi t ng lc mk =4)

mI = 4104,125 =416,5(A)Tnh ngn mch ti t ng lc 2Chn chiu di dy t TPPC n t ng lc 2 khong 40m

X0 = 0.08 ( / km ) v tit din dy F > 50 mm2

Xd = 0.080.04 = 3,210-3 ( )

Rd = 0.07540.04 = 3.01610-3 ( )

=> Zd1 ==

=4,410-3( )

=> Zd2 = Zd/2 = 2,210-3 ( )

= =

=35,93KA




57/96



dmCBI =604,42 (A)

Chn CB: M08 dmCBI =800(A)

CUI =40 (kA)

Ta c: r r dmCBI k I=

,cpdd cpdd I k I=

=0.765012=761,52(A)


,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max cpddlvr

dmCB dmCB

II kI I

0.75 rk 0.95

rk =0.9 rI =0.9800

= 720(A)Ta c: m m rI k I= ( i vi t ng lc mk =4)

mI = 4720 =2880Tnh ngn mch ti t ng lc 3Chn chiu di dy t TPPC n t ng lc 2 khong 60m



Trng C in Lc TP HCM N CUNG CP IN


58/96


Xd = 0.080.06 = 4,810-3 ( )Rd = 0.07540.06 = 4,52410-3 ( )

=>Zd1 =

=

=6,610-3( )

=> Zd3 = Zd/2 = =3,310-3 ( )

= =

=30.445

Ta c : maxlvI =604,42(A) dmCBI =604,42 (A)

Chn CB: M08 dmCBI =800(A)

CUI =40 (kA)


,cpdd cpdd I k I=

=0.765012

=761,52(A)*iu kin: ,maxlv r cpdd I I I

,

max cpddlv r

dmCB dmCB dmCB

II I

I I I




59/96


,

max cpddlvr

dmCB dmCB

IIk

I I

0.75 rk 0.95

rk =0.9 rI =0.9800


mI = 4720 =2880Tnh ngn mch ti t ng lc 4Chn chiu di dy t TPPC n t ng lc 4 khong 80m


Xd = 0.080.08 = 6,410-3 ( )Rd = 0.09910.08 = 7,9810-3 ( )

=> Zd1 ==

=10,18 10-3( )

=> Zd4 = Zd/2 = 5,0910-3 ( )

= =

=24,387KA




60/96


Chn CB:

+ maxdmCB lvI I =


dmCBI =514,453 (A) Chn CB: C801 dmCBI =800 (A)

CUI =50 (kA)


,cpdd cpdd I k I=

=0.76443 2=673(A)

*iu kin:,

maxlv r cpdd I I I

,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max cpddlvr

dmCB dmCB

IIk

I I

0.64 rk 0.95 rk =0.8 rI =0.8800




61/96



mI = 4640 =2560(A)Tnh ngn mch ti t ng lc 5

Chn chiu di dy t TPPC n t ng lc 4 khong 70mX0 = 0.08 ( / km ) v tit din dy F > 50 mm2

Xd = 0.080.07 = 5,610-3 ( )Rd = 0.09910.7 = 6,93710-3 ( )

=> Zd1 ==

= 8,91510-3( )

=> Zd5 = Zd/2 = 4,4510-3 ( )

= =

= 24(kA)

Chn CB:

+ maxdmCB lvI I =

Ta c :maxlvI =514,453 (A)

dmCBI =514,453 (A)

Chn CB: C801 dmCBI =800 (A)

CUI =25 (kA)




62/96



,cpdd cpdd I k I=

=0.76443 2=673(A)

*iu kin:,

maxlv r cpdd I I I

,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max cpddlvr

dmCB dmCB

IIk

I I

0.57rk 0.842

rk =0.8 rI =0.8800


mI = 4640 =2560(A)Tnh ngn mch ti t ng lc 6Chn chiu di dy t TPPC n t ng lc 6 khong 90m


Xd = 0.080.09 = 7,210-3 ( )Rd = 0.09910.09 = 8,91910-3 ( )




63/96


=> Zd1 ==

= 11,4610-3( )

=> Zd6 = Zd/2 = 5,7310-3 ( )

= =

=22,768KAChn CB:

+ maxdmCB lvI I =


dmCBI =531,923(A)


CUI =50 (kA)Ta c: r r dmCBI k I=

,cpdd cpdd I k I=

=0.76443 2=673(A)


,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max cpddlv r

dmCB dmCB dmCB

II I

I I I




64/96


,

max cpddlvr

dmCB dmCB

IIk

I I

0.64 rk 0.84

rk =0.8

rI =0.8800= 640(A)

Ta c: m m rI k I= ( i vi t ng lc mk =4)



Xd = 0.080.1 = 810-3 ( )

Rd = 0.07540.1 = 7,5410-3 ( )

Zd1 = ==10,99 10-3 ( )

=> Zd7 = Zd/2 = 5,49510-3 ( )

= == 23,337 (kA)

Chn CB:+ maxdmCB lvI I =





65/96

g

dmCBI =529,514 (A)


CUI =50 (kA)


,cpdd cpdd I k I=

=0.76443 2=673(A)


,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max cpddlvr

dmCB dmCB

IIk

I I

0.66 rk 0.84

rk =0.8 rI =0.8800







66/96

g

Xd = 0.080.11 = 0.0088( )Rd = 0.09910.11 = 10.90110-3 ( )

=>Zd1 ==

= 14.0110-3( )=> Zd8 = Zd/2 = 7,00510-3 ( )

= =

=20,107KA

Chn CB:

+ maxdmCB lvI I =Ta c : maxlvI =520(A)

dmCBI =520(A)

Chn CB: NS630N dmCBI =630(A)

Trip Unit STR23SE CUI = 45(kA)

0r r dmCBI k k I= ,

cpdd cpdd I k I=

=0.76443 2=673(A)





67/96

g

,

max cpddlv r

dmCB dmCB dmCB

II I

I I I

,

max

0

cpddlvr

dmCB dmCB

IIk k

I I

Chn :max

0

lv

dmCB

IkI

=

K0 >=

=0,84 0k =0,85

,

max

0 0

cpddlvr

dmCB dmCB

IIk

I k I k

0.97 rk 1,256

rk =0.98 rI =0.980.85630

= 524.79(A)Ta c: m m rI k I= ( i vi t ng lc mk =4)

mI = 4524.79=2099,16(A)B.TNH NGN MCH TI CC THIT B

Tnh ngn mch ti thit b 1 ca nhm 1Chn chiu di dy t t ng lc 1 n thit b 1 ca nhm 1 khong 12m

X0 = 0 ( / km ) v tit din dy F < 50 mm2

Xd = 0 ( )




68/96

Rd = 4,610.012= 0.0055 ( )

=> 2 21 1d d dZ R X = + = 0.0055( )

=

==3,213KA



Xd = 0 ( )Rd = 7,410.015 = 0.111( )

=> 2 21 2d d dZ R X = + = 0.111( )

= =

=1,64KATnh ngn mch ti thit b 3 ca nhm 1Chn chiu di dy t t ng lc 1 n thit b 3 ca nhm 1 khong 20m


Xd = 0 ( )

Rd = 12.1

0.02 = 0.242 (

)=> 2 21 3d d dZ R X = + = 0.242( )




69/96

= =

=830,78ATnh ngn mch ti thit b 4 ca nhm 1Chn chiu di dy t t ng lc 1 n thit b 4 ca nhm 1 khong 12m

X0 = 0 ( / km ) v tit din dy F < 50 mm2Xd = 0 ( )Rd = 12,10.012 = 0.1452 ( )

=> 2 21 4d d dZ R X = + = 0.1452( )

=

=

=1,312KATnh ngn mch ti thit b 5 ca nhm 1Chn chiu di dy t t ng lc 1 n thit b 5 ca nhm 1 khong 15m


Xd = 0 ( )Rd = 7,410.015 = 0.111 ( )

=> 2 21 5d d dZ R X = + = 0.111( )

= ==1,648kA




70/96



71/96

Xd = 0 ( )Rd = 7.41 0.009 = 0.0667 ( )

=> 2 22 11d d dZ R X = + = 0.0667 ( )

= =

= 3,013 (kA)Chn CB:

maxlvI =29,12(A)

iu kin: maxdmCB lvI I =

dmCBI = 30.39(A)

chn CB: C60N dmCBI =32(A)

CUI = 10(kA)Ta c: r dmCBI I= =32(A)

,cpdd cpdd I k I=

=0.7641=21,16(A)

tha iu kin: ,maxlv r cpdd I I I

- Ta c: m m rI k I=

5mm dmtbI I= =5 29,12 mmI = 145.6(A)

iu kin: minmm m nmI I I




72/96

mm

m

r

Ik

I =

mk =

=4,56 mk = 5

mI = 5 32=160(A)

Tnh ngn mch ti thit b 12 ca nhm 3Chn chiu di dy t t ng lc 3 n thit b 12 gn nht ca nhm 3 khong 10m


Xd = 0 ( )

Rd = 7.41 0.009 = 0.0667 ( )=> 2 22 11d d dZ R X = + = 0.0667 ( )

= =

= 3,013 (kA)Chn CB:

maxlvI =29,12(A)


dmCBI = 30.39(A)

chn CB: C60N dmCBI =32(A)

CUI = 10(kA)




73/96

Ta c: r dmCBI I= =32(A)

,cpdd cpdd I k I=

=0.7641=21,16(A)

tha iu kin:,

maxlv r cpdd I I I

- Ta c: m m rI k I=

5mm dmtbI I= =5 29,12

mmI = 145.6(A)


mm

m

r

Ik

I =

mk =

=4,56

mk = 5

mI = 5 32=160(A)Tnh ngn mch ti thit b 10 ca nhm 4Chn chiu di dy t t ng lc 4 n thit b 10 gn nht ca nhm 4 khong 10m


Xd = 0 ( )Rd = 12.10.009 = 0.101 ( )

=> 2 24 10d d dZ R X = + = 0.101( )




74/96

= =

=1,99KAChn CB:

maxlvI =19,534(A)iu kin: maxdmCB lvI I =

dmCBI = 19,534 (A)

chn CB: C60H dmCBI =20 (A)

CUI =6(kA)

Ta c: r dmCBI I= = 20(A)

,cpdd cpdd I k I=

=0.76

41=31.16(A) tha iu kin: ,maxlv r cpdd I I I

- Ta c: m m rI k I=

maxdmtb lvI I= = 19,534(A)

5mm dmtbI I= =19,534 5=97,67A


mm

mr

I

k I = = mk = 4.88

mk = 5



75/96



76/96

- Ta c: m m rI k I=

5mm dmtbI I=

maxdmtb lvI I= =19,53(A)

mmI = 97,67 (A)


mm

m

r

Ik

I =

mk = 4.88

mk = 5

mI = 20 5=100(A)



Xd = 0 ( )Rd = 12.10.009 = 0.101 ( )

=> 2 24 10d d dZ R X = + = 0.101( )

==

= 1,98(kA)Chn CB:

maxlvI =22,79(A)



77/96



78/96

Xd = 0 ( )Rd = 12.10.009 = 0.101 ( )

=

==1,9787kA

Chn CB:

maxlvI =24,961(A)


dmCBI = 24,961(A)

chn CB: C60N dmCBI =25 (A)

CUI =6(kA)

Ta c: r dmCBI I= =32(A)

,cpdd cpdd I k I=

=0.7641=31.16(A)

tha iu kin: ,maxlv r cpdd I I I

- Ta c: m m rI k I=

5

mm dmtb

I I=

maxdmtb lvI I= = 24,961 (A)

mmI = 124,805(A)





79/96

mm

m

r

Ik

I = =

mk = 4.99

mk = 5

mmI = 5 25=125 (A)



Xd = 0 ( )Rd = 7.41 0.009 = 0.0667 ( )

=> 2 22 11d d dZ R X = + = 0.0667 ( )

==

=2,827KAChn CB cho cc thit b

iu kin : dmCBI = maxlvI

GVHD: Nguyn Hu Khoa

KH

MB

maxlvI M

CB

dmCBI ,

cpddI = k cpddI rI mk mI

13 24,96 C60N 25 22,58 25 5 12514 29,12 C60N 32 26,34 32 5 160

79



80/96

X.TNH TON AN TON

CC BIN PHP BO VBo v chng dhm in trc tipS dng cc phn dn in c in tr cch in ng theo cp in p yu cuLp t cc phn mang in trn cao, ngoi tm vi hoc che chn trnh ngi s dng c th s ti.




81/96

Lp t cc phn mang in khng c bc cch in trong t kn c kha v cha c gi bi ngi c chc nng.Thit k v lp t mng in c U Uchophp [ 40(V) hoc 24(V), 12(V), 6(V) ] ly ngun qua mng cch ly.Thit k v lp t mng ly ngun sau bin p cch ly c bit v m bo mc cch in ca mng in theo ng

yu cu an ton, thc hin ni ng th v cc thit b.Ngoi ra IEC v mt s nc c qui nh bt buc s dng bin php bo v ph nh thit b chng r RCD cc ni

c nguy c chm in i vi cc thit b in di ng.

Bo v chng chm in gin tipBo v chng chm in gin tip bao gm cc bin php an ton c thc hin nhm ct thit b chm v hoc bo

tn hiu khi c dng r xung t ln. Tiu chun Vit Nam, IEC v mt s nc qui nh cc hnh thc bo v thng quavic ni v kim loi thit b in theo cc s ni t v vic s dng thit b bo v thch hp.THIT K BO V AN TONChn s ni t

Chn s ni t cho xng theo s TN-C-SCc u im: s n gin, kinh t.Cc khuyt im: khi dng chm v ln nguy c chy n cao, nguy c nhiu in t ln, iu ny c th lm thit b

hot ng sai.Tuy nhin khi ta t cc thit b bo v, c ni t lp li an ton ta s khc phc tnh trng UN-t 0 khi c hin

tng chm v, nng tnh an ton cao


A

B

C

N

PE

RnHT



82/96

Chn thit b bo v an tonThit b bo v l CB/ cu ch. Cc thit b ny c chn trong phn thit k mng cung cp in.Chn dy bo v theo tiu chun IECKhi: Spha 16 mm2 (Cu) v 25mm2 (Al) => SPE = SphaKhi: 16 mm2 < Spha 35 mm2 (Cu) => SPE = 16mm2

Hoc: 25 mm2 < Spha 50 mm2 (Al) => SPE = 25mm2

Cc trng hp cn li: => SPE = 0.5 Spha

BNG CHN DY CHO CC PHN ON

Phn on Dy pha R pha ( /m) Dy pha R PE( /m)

MBA TPPC

15xCVV

1x500 0.0366

5xCVV

1x300

0.0601

TPPC TL1 1xCVV 1x25 0.72701xCVV 1x16

1.15

TPPC TL26xCVV 1x240

0.07542xCVV 1x120

0.153

TPPC TL36xCVV 1x240

0.07542xCVV 1x120

0.153

TPPC TL46xCVV

1x185

0.09912xCVV

1x120

0.153

TPPC TL56xCVV 1x185

0.09912xCVV 1x95

0.193

TPPC TL6 6xCVV 0.0991 2xCVV 0.193




83/96

1x185 1x95

TPPC TL76xCVV 1x185

0.09912xCVV 1x95

0.193

TPPC TL86xCVV 1x185

0.09912xCVV 1x95

0.193

TL1 THITB 1

1xCVV 3x4 4.61 1xCVV 1x4 4.61

TL1 THITB 2

1xCVV 3x2,5 7.411xCVV 1x1,5

12,1

TL1 THITB 3

1xCVV 3x1.5 12.11xCVV 1x1.5

12,1

TL1 THITB 4

1xCVV 3x1,5 12.11xCVV 1x1.5

12,1

TL1 THITB 5

1xCVV 3x4 4.61 1xCVV 1x4 4.61

TL1 THITB 6

1xCVV 3x2.5 7.411xCVV 1x2.5

7.41

TL2 THITB

1xCVV 3x2.5 7.411xCVV 1x2.5

7.41

TL3 THITB

1xCVV 3x2.5 7.411xCVV 1x2.5

7.41

TL4 THITB 1xCVV 3x1.5 12.1

1xCVV 1x1.5 12.1

TL5 THITB

1xCVV 3x1.5 12.11xCVV 1x1.5

12.1



84/96



85/96

3 -3 3 2 3 3 2(3.73 10 0.11 10 0.301 10 ) (3.675 10 0.24 10 )td

Z = + + + +

= 5.69910-3 ( )

3

0.95 220

7.599 10chamvo

I

=

= 36.67 (kA)

=> Kim tra kh nng bo v ca CB:m

I = 11.52(kA) chamvoI CB chn tha iu kin .2) Tnh dng chm v t ng lc1:

pp dlR r l

= =0.524 0.025 = 13.110-3 ( )

pp dlX = 0 ( )

2PENR r l= =1.150.025 = 0.02875 ( )

2 21 2( ) ( )td MBA M BA pp pp dl PEN PEN MBA MBA pp pp dl Z R R R R R X X X = + + + + + + +

2 20.046 0.00392td

Z = + = 0.0462( )

=>0.95 220

0.0462chamvo

I

= = 4.528 (kA)

Kim tra kh nng bo v ca CB:

mI = 0.392(kA) chamvoI

CB chn tha iu kin .3) Tnh dng chm v t ng lc 2:

Xd = 0.08

0.05 = 4

10

-3

( )Rd = 0.07540.05 = 3.7710-3 ( )

pp dlR r l

= =0.0754 0.05 = 0.00054( ) ( r =0.0283 \km)

0.08pp dlX l = =0.080.019=0.0015( )




86/96

PER r l= = 0.0190.0576 =0.012( )2 2

1 2( ) ( )td MBA M BA pp pp dl PEN PEN MBA MBA pp pp dl Z R R R R R X X X = + + + + + + +

2 2(0.001443 0.00054 0.012) (0.0087 0.0015)td

Z = + + + +

2 20.014 0.0102td

Z = + = 0.017( )

209

0.017chamvo

I = =12.07(kA)



1.Tnh dng chm v t ng lc 3:l = 30(m)

pp dlR l r =

=0.03

0.47=0.0141(

)( F=400

2mm

,r =0.47

\km) 0.08pp dlX l = =0.080.03=0.0024( )

2PENR l r= = 0.0300.094=0.0028( )2 2


2 2(0.001443 0.0141 0.0028) (0.0087 0.0024)td

Z = + + + +

2 20.018 0.011td

Z = + = 0.021( )

209

0.021chamvo

I = =9.95(kA)



1) Tnh dng chm v t ng lc 4: pp dlR l r = =0.024 0.0366=0.0141( )




87/96

0.08pp dlX l = =0.080.03=0.0024( )

2PENR l r= = 0.0300.094=0.0028( )2 2

1 2( ) ( )

td MBA M BA pp pp dl PEN PEN MBA MBA pp pp dl Z R R R R R X X X

= + + + + + + +

2 2(0.001443 0.0141 0.0028) (0.0087 0.0024)td

Z = + + + +

2 20.018 0.011td

Z = + = 0.021( )209

0.021chamvo

I = =9.95(kA)



2) Tnh dng chm v t ng lc 5: pp dlR l r = =0.036 0.47=0.017( )( F=400 2mm ,r =0.47 \km)

0.08pp dlX l = =0.080.036=0.0029( )

2PENR l r= = 0.0360.094=0.0034( )2 2

1 2( ) ( )


= + + + + + + +

2 2(0.001443 0.017 0.0034) (0.0087 0.0029)td

Z = + + + +

2 20.022 0.012td

Z = + = 0.025 ( )

209

0.025chamvo

I = =8.36(kA)

Kim tra kh nng bo v ca CB:m

I = 2.22(kA) chamvoI

3) Tnh dng chm v t ng lc 6:




88/96

pp dlR l r = =0.034 0.47=0.016( )( F=400 2mm ,r =0.47 \km)

0.08pp dlX l = =0.080.034=0.0027( )

2PENR l r= = 0.0340.094=0.0032( )2 2


2 2(0.001443 0.016 0.0032) (0.0087 0.0027)td

Z = + + + +

2 20.021 0.0114td

Z = + = 0.024( )

209

0.024chamvo

I = =8.7(kA)



CB chn tha iu kin ngn mch

4) Tnh dng chm v t ng lc 7: pp dlR l r = =0.042 0.0366=0.00154( )

0.08pp dlX l = =0.080.042=0.0034( )

2PENR l r= = 0.0420.094=0.0028( )2 2

1 2( ) ( )


= + + + + + + +

2 2(0.001443 0.00154 0.0028) (0.0087 0.0034)td

Z = + + + +

2 20.0058 0.0904tdZ = + = 0.091( )209

0.091chamvo

I = =2.29(kA)



89/96



90/96

Nh cch chn dy PE trn ta c : dl dc PE R R R = =2 2

1 2( ) ( )td MBA M BA pp pp dl dl dc PEN PEN PE MBA MBA pp pp dl dl dcZ R R R R R R R X X X X = + + + + + + + + + +

2 23(0.0076 ) (0.0094 )td dl dc PEN dl dcZ R R X = + + + +

209chamvo

tdI Z=

(1. Tnh ngn mch choan t phn phi- t

ng lc1

1) Tnh dngchm v ti

cc thit bnhm 2:Nhm 2 gm cc

thit b ging nhau nn v mt chn dy v CB l nh nhau

GVHD: Nguyn Hu Khoa

KHMB

l(m)

M CB mI

(A)

R

X

chamvoI

kAm c ha mv o

I I

1 6 C60L 315 0.005 0.0005

10.3 Tha

2 8 C60H 315 0.007 0.0006

8.78 Tha

3 10 C60A 100 0.053 0.0008

1.83 Tha

4 6 C60A 80 0.057 0.0005

1.71 Tha

5 8 C60A 80 0.075 0.0006

1.32 Tha

6 10 C60A 65 0.181 0.0008

0.57 Tha

90



91/96

nn khi tnh dng chm v thit b ta ch xt thit b xa nht c chamvoI b nht

l = 13(m) dl dcR l r = =0.0132.31=0.03( ) (r =2.31 \km)

0.08dl dcX l = =0.080.013=0.001( )

PER l r= Nh cch chn dy PE trn ta c : dl dc PE R R R = =

2 2


2 2(0.014 0.03 0.03) (0.0102 0.001)Z = + + + +

= 0.075( )

209

0.075chamvo

I =

= 2.79(kA)


mI = 160(A) chamvoI

CB chn tha iu kin

2) Tnh dng chm v ti cc thit b nhm 3:

Nhm 3 gm cc thit b ging nhau nn v mt chn dy v CB l nh nhaunn khi tnh dng chm v thit b ta ch xt thit b xa nht c chamvoI b nht

l = 13(m)




92/96

dl dcR l r = =0.0133.4=0.044( ) (r =3.4 \km)

0.08dl dcX l = =0.080.013=0.001( )

PER l r=

Nh cch chn dy PE trn ta c : dl dc PE R R R = =

2 21 2( ) ( )td MBA M BA pp pp dl dl dc PEN PEN PE MBA MBA pp pp dl dl dcZ R R R R R R R X X X X = + + + + + + + + + +

2 2(0.018 0.044 0.044) (0.011 0.001)Z = + + + +

= 0.11( )

209

0.11chamvoI =

= 1.9(kA)Kim tra kh nng bo v ca CB:

mI = 125(A) chamvoI

CB chn tha iu kin

1) Tnh dng chm v ti cc thit b nhm 4:Nhm 4 gm cc thit b ging nhau nn v mt chn dy v CB l nh nhau


l = 13(m) dl dcR l r = =0.0132.31=0.03( ) (r =2.31 \km)

0.08dl dcX l = =0.080.013=0.001( )




93/96

PER l r=



2 2(0.018 0.03 0.03) (0.011 0.001)Z = + + + +

= 0.079( )

209

0.079chamvoI =

= 2.6(kA)


mI = 160(A) chamvoI

CB chn tha iu kin2) Tnh dng chm v ti cc thit b nhm 5:

Nhm 4 gm cc thit b ging nhau nn v mt chn dy v CB l nh nhau


l = 13(m)

dl dcR l r

= =0.0133.4=0.044( ) (r =3.4 \km)

0.08dl dcX l = =0.080.013=0.001( )

PER l r= Nh cch chn dy PE trn ta c : dl dc PE R R R = =

2 2





94/96

2 2(0.022 0.044 0.044) (0.012 0.001)Z = + + + +

= 0.11( )209

0.11chamvoI =


mI = 125(A) chamvoI

CB chn tha iu kin

1) Tnh dng chm v ti cc thit b nhm 6:Nhm 6 gm cc thit b ging nhau nn v mt chn dy v CB l nh nhau


l = 13(m) dl dcR l r = =0.0133.4=0.044( ) (r =3.4 \km)

0.08dl dcX l = =0.080.013=0.001( )

PER l r=



2 2(0.021 0.044 0.044) (0.0114 0.001)Z = + + + +

= 0.11( )




95/96

209

0.11chamvoI =

= 1.9(kA)


mI = 125(A) chamvoI CB chn tha iu kin

2) Tnh dng chm v ti cc ng c nhm 7:Nhm 7 gm cc thit b ging nhau nn v mt chn dy v CB l nh nhau


l = 13(m) dl dcR l r = =0.0132.31=0.03( ) (r =2.31 \km)

0.08dl dcX l = =0.080.013=0.001( )

PER l r=



2 2(0.0058 0.03 0.03) (0.0904 0.001)Z = + + + +

= 0.11( )

209

0.11chamvoI =




96/96


mI = 160(A) chamvoI

CB chn tha iu kin

1) Tnh dng chm v cho ng c:Nhm 8 gm c 2 lai ng c,ta xt 2 ng c xa t ng lc nht.

dl dcR l r =

dl dcX = 0.08 l

PER l r=


1 2( ) ( )

td MBA M BA pp pp dl dl dc PEN PEN PE MBA MBA pp pp dl dl dcZ R R R R R R R X X X X

= + + + + + + + + + +

2 2(0.032 ) (0.013 )td dl dc PE dl dc

Z R R X = + + + +

=>209

chamvo

td

IZ

=

KHMB

l(m)

M CB mI (A) R( ) X( ) chamvoI (kA) m c ha mv oI I

14 9 C60N 160 0.021 0.0007 2.78 Tha

13 6 C60N 125 0.02 0.0005 2.85 Tha

Xac Dinh Tam Phu Tai

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