141849 Bai Giang an Toan Dien

7/26/2019 141849 Bai Giang an Toan Dien

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1

BI GING MN HCAN TON IN

2

Ni dung mn hc

Phn 1. Khi nim v phn tch an ton trong cc mng in Chng 1. Nhng khi nim c bn Chng 2. Phn tch an ton trong mng in n gin Chng 3. Phn tch an ton trong mng in 3 phaPhn 2. Cc bin php k thut an ton Chng 4. Bo v ni t Chng 5. Bo v ni dy trung tnh Chng 6. Bo v an ton bng thit b chng dng in r

Chng 7. Cc bin php an ton khc Chng 8. X l, cp cu ngi b in git

Chng 9. Phng chng in t trng


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Ti liu tham kho1. TS Nguyn nh Thng, TS Nguyn Minh Chc

K thut an ton in - NXB HBKHN

2. Nguyn Xun Ph (Ch bin)

K thut an ton trong s dng v cung cp in - NXBKHKT, 2003

3. Titres d'habilitation lectrique

4. RCD protection

4

Phn 1.Khi nim v phn tch an ton

trong cc mng in

Chng 1.CC KHI NIM C BN

1.1. CC TAI NN V IN1.1.1. Phn loi tai nn in

Cc tai nn in

in git

Phng inHo hon chy n do in


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1.1.2. Nguyn nhn dn n tai nn in

Chm vo cc phn tbnh thng c in p

Chm in gin tipChm in trc tip

Nguyn nhn dn n tai nn in

Chm vo cc phn t bnhthng khng c in p

Khc

HQ inXuthintrongKV intrngmnh

6

tip xc trc tip

Ph

N

t

Pha - Trung tnh Pha - t

Ing

. . . .


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Chm vo thanh ci

8

TIP XC GIN TIP

Ph

N

tIng

. .


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Ph

N

t

Ing

. .

TIP XC GIN TIP

10

1.1.3. S liu thng k tai nn in

S liu thng ktai nn in

a. Theo cpinp:U 1kV: 76,4%U > 1kV: 23,6%

b. Theo nghnghip:Thucngnh in: 42,2%Cc ngnh khc: 57,8%

c. Theo nguyn nhn tipxc in:Trctip: 55,9%Gin tip: 42,8%HQ in: 1,12%

Xuthintrong KV intrngmnh:0.08%

d. Theo nguyn latui:Di20: 14,5%21-30: 51,7%31-40: 21,3%

Trn 40: 12,5%


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1.2. TC DNG CA DNG IN

Khi ngi tip xc vi cc phn t c in p (k c tip xc trc tip hoc gintip), s c dng in chy qua c th, cc b phn ca c th phi chu tc ngnhit, in phn v tc dng sinh hc ca dng in lm ri lon, ph hu cc bphn ny, c th dn n t vong.

a) Tc ng v nhit:ca dng in i vi c th ngi th hin qua hin tng gybng, pht nng cc mch mu, dy thn kinh, tim, no v cc b phn khc trn cth dn n ph hu cc b phn ny hoc lm ri lon hot ng ca chng khi dngin chy qua.

b) Tc ng in phn:ca dng in th hin s phn hu cc cht lng trong cth, c bit l mu, dn n ph v cc thnh phn ca mu v cc m trong c th.

c) Tc ng sinh hc:ca dng in biu hin ch yu qua s ph hu cc qu trnhin - sinh, ph v cn bng sinh hc, dn n ph hu cc chc nng sng.

Mc nguy him ca dng in i vi c th ngi tu thuc vo tr s ca dngin, loi dng in (dng in mt chiu hoc dng in xoay chiu) v thi gianduy tr dng in chy qua c th (IEC 60479-1).

12

Standard IEC 60479-1Time/current zones defining the effects of AC current

(15 Hz to 100 Hz)

Vng 1:Khng cm nhn c

Vng 2:Cm thy kh chu

Vng 3:Co cc c, b (10 mA) let-gothreshold

Vng 4:Rung c tim, c1(30 mA)

b -c1: probability 0 %

c1 -c2: probability ~ 5 %

c2 -c3: probability ~ 50 %

>c3: probability > 50 %

a

ng a - Ngng cm nhn c dng in qua ngi

ng b - Ngng bung - nh


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Standard IEC 60479-1: Ngng dng in ti hn(Critical current thresholds)

Tim ngng p

Tim p mnh - Ngng RCT

T lit c quan h hp-Nght th

Bt u co c - Ngng bung nh

C cm gic nhi nh - Ngng cm nhn

AC

14

Ngng dng in ti hn

5

?

100

130

Khng xc nh

Dng in xoay chiu:Icp= 10 mA

Dng in mt chiu:Icp= 50 mA

DC


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1.3. IN P TIP XC & TNG TR C TH NGI

inp tipxc v tngtrc th l hai i lng dng xc nh tr s dng inqua ngi.

1.3.1.inp tipxc Utx:Lin p gia hai im trn ngi ca dng in qua c th ngi(hay chnh l in p t ln cth ngi khi ngi tip xc in)thng l gia tay vi tay hoc

gia tay v chn.

ZT= Zng= Zp+ Zi1.3.2. Tng tr c th ngi:

16

Zng

inp tx

ngin

Dintch,psut

NhitTnhtrng

da

Thigianiqua


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1.3.3. in p tip xc cho php Utxcp

Nh xng

Ngpnc

Utx = Ung = Rng.Ing1200 * 10 mA = 12 V

mt 2500 * 10 mA = 25 V

Khro 5000 * 10 mA = 50 V

Utxcp12 V

24 V

48 V

Ngpnc 1200 * 10 mA = 12 V

mt 2500 * 10 mA = 25 V

Khro 5000 * 10 mA = 50 V

12 V

25 V

50 V

Tiu chun Php:

Tiu chun IEC:

18

ChcvctcchDngc

Nhngphngphp

Nnglc

Mitrng

Lutlaong Nhngquyphm

inp

CngtcAnton

1.4. CC YU T NH HNG N CNG TC AT


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Khi TBc dng chmv,ngdy intrixungt, ti ch chm t s c dng in tn vo trong t.Dng inny tnntn vo trong t?tr licu hiny lmt vn ht sc phc tp, nhng c th hnh dung mtcch ngin: Xt TH dng inny tnvo trong tthngqua mtbn cukim loic bn knh r0chn st mtt. Vigithit:Mi trngchn inccc intrsutl thunnht.

Dng in chmt I i t tm bn cu to vo trong ttheo ngbn knh.Trngcadng in I l dng trng tnh (tc l tphpcc ngscv ngngthcachng gingnhau).

1.5. HIN TNG DNG IN TN VO TRONG T

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1.5.1. S phn b th ti ch dng in chm t

2

d

x2

I

j

dxx2

.IJdxEdxdu

2

d

L m di dng vi phn: J = E hay E = J

x2

.Idx

x

1

2

.IduUUU d

x

2

d

x

xx


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1.5.2. in tr tnKhi dng in i vo trong t, b in tr ca in cc v t cn tr. intr ny gi l in tr tn R:

1.5.3. in p tip xc Utx

,r2

R

0

d

d

d

I

U

J

Ud

Utx

u (V)

l (m)l (m)

Utx

TBId

a)

0

l (m)

u (V)

Ud

20

b)

Utx= Ud

x,

x2

.I

r2

.IU-UU-UU d

0

d

xdchantaytx

22


1.5.4. in p tip xc Ub

T CT ta thy rng cng ngxa ch dng in chm t (incc ni t) in p bc cng ctr s nh. Khi ngi ng cchch chm t trn 20 m c th coiin p bc bng 0.

J

Ud

Ub

u (V)

l (m)l (m)

TBId

a

x

a)(x2

.a.I

a)x(2

.I

x2

.IU-UU ddd

axxb

x

V d: I = 1000A; = 102mv a = 0,8m th Ub= 30,6 V

Nh vy in p bc v in p tip xc thay i hon ton tri ngc nhaukhi khong cch n ch chm t thay i.


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Chng 2. PHN TCH AN TON TRONG MNG IN N

GIN

2.1. KHI NINCHUNG

- Khi nimvmnginngin

- Phn loimnginngin

+ Theo indung c: Mngindung nhvmngindung ln

+ Theo ch lm vic c: Mngnitvmngcch invit.

- Gc chmindnnmtan ton introng cc mngnginc thdo chmintrctiphocgn tip.

+ Chmvo hai dy: Rtnguy him

+ Chmvo 1 dy: Nguy himtuthucvotngloimnginv chmvo dy no.

Chng 2.PTAT TRONG MNG IN N GIN

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2.2. PHN TCH AN TON TRONG MNG IN N GINC IN DUNG NH

cdng

R2R

U

ngI

2.2.1. Mng 2 dy cch in vi t

* Nh vy, mc nguy him ph thuc vo:

-in p ca mng U

-in tr c th ngi Rng

-in tr cch in ca mng Rc

2.2.2. Mng Mng ch c 1 dy:

* Ch : Khi 1 dy chm t m ngi chm vo dy cn li s rt nguy him.

cd2ocd20ng

cd2

R.RRRR

U.RI

ng

* Khi R0= 0 th:ng

ngRUI


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2.2.3. Mng 2 dy c 1 dy ni tTH chm vo dy khng ni t: Ung U

TH chm vo dy ni t: Ungmax= 5%U

* Ch :- Khi dy 1 chm vo dy 2 v tit din 2 dy nhnhau th U

ngmax

= 0,5U

- Khi dy ni t t pha u ngun th Ung U.

U

Rng

1

2

B

R0

Zt

CA

Ilv

Ilv

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2.3. PHN TCH AN TON TRONG MNG IN N GINC IN DUNG LN

2.3.1. S nguy him ca in tch tn d

a. Ngi chm vo 2 cc ca ng dy ct in:

12ngCR

t

ng

0ng .e

R

Ui

b. Ngi chm vo 1 cc ca ng dy ct in:)C(2CR

t

ng

0

ng

112ng

.e2R

U

i


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2.3.2. Chm vo 1 cc ca ng dy xoay chiu ang vn hnh

2

ng

22ng

RC41

CUI

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Cu hi v bi tp n tp chng 2

1.Phn tch an ton khi ngichm vo mtcc camnginngin c indung nh?

2.Phn tch an ton khi ngichm vo mtcc camnginngin c indung ln?

3.Hy xc nhdng inqua ngimngin2 dy cch inivitindung nhtrong cc trnghpngichmvo:

ngthi2 dy? Mtdy?

V cho bitngic nguy himkhng trong tngtrnghp,giithch?Bit: - Mnginc inp U = 220V;

-intrcch inRc= 30 k;4.Hy xc nh dng in qua ngi trong mng in 1 pha ca nc ta trong cc

trng hp ngi chm vo: ng thi 2 dy: dy pha v dy ni t (dy trung tnh)? Dy pha?

V cho bit ngi c nguy him khng trong tng trng hp, gii thch? Bit: - Mng in c in p U = 220V, f =50Hz;

-in tr ni t u ngun R0= 4 ;-in tr ngi Rng= 1000.


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5.*Hy xc nhdng inqua ngikhi ngichmvo dy trung tnh camngin1 pha ncta trong cc trnghp:

Chchmcch ngunin1 khongL1 = 30m?

Chchm ngay imuviphti?

Chchmcch ngunin1 khongL1 = 30m trong khi mngxy ra ngnmchtiphti?

Ch chm ngay im u vi ph ti khi dy trung tnh b t ti ungun?

Chmkhi dy pha bt?

*Cho bitngic nguy himkhng trong cc trnghptrn, giithch? * So snh

mcnguy himkhi chmintrong cc trnghptrn?Bitrng:

- Mng in c in p U = 220V, f =50Hz; dng dy ng mm M22,5 (r0 =8,06/km) di L = 50m cpincho phtic cng sut5,5 kW, cos= 0,85;

- GithitintrnitungunR= 0 ; intrngiRng= 1000.

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6.Hy xc nhdng inqua ngikhi ngichmngdy tiincao p tithiimvactra khingunc chiudi 1km ktngunnchchmintrongtrnghp:

Chmvo mtdy? Chmvo chai dy?

Bit: -inp gia2 dy tithiim t = 1s ngichminl 6kV;- Githitindung gia2 dy v 2 dy vitcng bng0,3F/km.-intrngiRng= 1,5k

7.Hy xc nhdng inqua ngikhi ngichmvo hai cccamttinngaysau khi ctra khiliin?

Bit: -inp gia2 cctithiim t = 0,5s ngichminl 3kV;- Githitindung catbng3F.

8.Hy xc nhdng inqua ngikhi ngichmvo mtdy camngin2 dycch invitcpincho 1 phtianglm viccch ngun500m?

Bit: -inp ngun6kV, f = 50Hz;- Chchmin: tiimuviphti.- Githitindung giacc dy vitbngnhau v bng0,3

F/km.

-intrngiRng= 1,5k.


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Chng 3. PHN TCH AN TON

TRONG MNG IN BA PHA3.1. KHI NIN CHUNG

Khi nim v mng in 3 pha

Mng c dng rng ri trong cng nghip

Phn loi mng in 3 pha

- Theo cp in p:

- Theo ch lm vic ca trung tnh:

Cc tnh hung chm in dn n tai nn in git:

- Chm trc tip: 1 pha; 2 pha; 3 pha

- Chm gin tip: Thng 1 pha b hng cch in nn cth coi trng hp ny nh trng hp chm trc tip vo 1pha.

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3.2. PHN TCH AN TON KHI NGI CHM VO 1 PHATRONG MNG IN 3 PHA TT CCH IN VI T

3.2.1. Trng hp chung:

2 2

B C C B B C C Bngng 2 22

A B C ng A B C

3 g g 3 C C 3 g g 3 C C U.gI

2g g g g C C C


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3.2.2. Mng h p U 1kV:

3.2. PHN TCH AN TON KHI NGI CHM VO 1 PHA

TRONG MNG IN 3 PHA TT CCH IN VI T

cdng

P

ngRR

UI

3

3

3.2.3. Mng cao p U >1kV: ng2 2 2

ng

3 CUI

1 9 C R

Ch :TH ngi chm 1 pha trong khi 1 trong hai pha cn li

chm t Rt nguy him.

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3.3.1.ivimngcao p:

Vicnittrung tnh chyubil do kinh t. V mnginTT nich

chncch intheo inp pha, trong khi mngintrung tnh cchinchntheo inp dy.

3.3.2.ivimnghp:

Vicnittrung tnh chyuvil do an ton cho ngiv thitb.

C thso snh tnghpdigc an ton giamngTT nitvimngTT cch inbngsau:

3.3. PHN TCH AN TON KHI NGI CHM VO 1 PHATRONG MNG IN 3 PHA TT NI T


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Mng trung tnh cch in i t

V c thnh phnindung v indngiacc pha vitnn dng inqua nginh,c thkhng nguy himntnh mng.

- Cc pha cn li,inp pha tngln inp dy. Dnginchmtnhcc thitbbov(cuch, ptmt...)khng tc ngdnns chmt duy tr v ba phamtixngqugiihncho php. V th:+Phtimtpha nidy trung tnh vipha khng chmtc thbph hng.+Ngichmvo pha khng chmtsnguy himhnnhiuso vimngtrung tnh nitcng cpinp.

Phtimtpha nidy trung tnh vidy pha bngngcpinKhng mbotnh cung cpinlin tc.

Trung tnh sphichuin p pha bn trung p (hocchu sng inp khi b st nh) rt nguy him chongiv thitb.

Mng trung tnh ni t

Dng in qua ngi lnhnnhiumng trung tnh cchin (v ngignnhphi chu ton bin p pha tvo), nguy himntnh mng.

- Cc pha cn li,inp cgignnhkhng thay i.Dng inchmtln,thitbbovddng tc ngctphntbchmtra khimnginm khng nhhngnthitbkhc. V th:+ San ton cho ngiv thitbkhi c chmt.+ Ph timt pha ni dy trung tnh vi pha khng chmtvnlm viccbnh thng.+ Ngi chm vo pha khng chm t th mc nguyhimgnnhlc chac mtpha chmt.

Ph timt pha ni dy trung tnh vi dy pha khng bngngcpin(v cn c nitlpli)mbo tnhcung cpinlin tc.

V trung tnh cnitviintrnhnn inp trungtnh nhAn ton hncho ngiv thitb.

Khi dy trung tnh b t (pha u ngun)

Khi ngi chm vo mt pha trong ch lm vic bnhthng

Khi c s xm nhp t in p cao sang in p thp (cch in trung p v h p ca MBA bhng hoc khi mng b st nh)

Khi c mt pha chm t

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1.Phn tch an ton trong cc mngin3 pha?2.So snh mngin3 pha trung tnh cch invitv mngtrung tnh nitdigc

an ton in?3. *Hy xc nhdng inqua ngikhi ngichmvo 1 pha camngin3 pha trung

tnh cch invittrong cc trnghp:

Ngichmintrong chmnginlm vicbnh thng? Ngichmintrong chmnginangxyra chmtpha khc?

*C nhnxt g sau khi tnh ton 2 trnghptrn?Bit: - Mngc ininp 380/220 V, f = 50Hz;

-intrcch inRc= 40k; indung khng ngk;-intrngiRng= 1k.

* Hy xc nhRctithiungic Rng = 1000chmvo 1 pha vnan ton?4. *Hy xc nhdng inqua ngikhi ngichmvo 1 pha camngin3 pha trung

tnh cch invittrong chmnginlm vicbnh thng.Bit: - Mngc ininp 10 kV, f = 50Hz; c chiudi L = 10km;

-indncch ingc0; indung nvC0= 0,3F/km;-intrngiRng = 2k.

*Cho bitdng inny c nguy himivingikhng?* Theo bngimdng in qua ngi khi tip xc 1 pha trong mngny c cc bin

php no?


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5.Hy xc nhdng inqua ngikhi ngichmvo 1 pha camngin 3 pha trung tnh ni t trong ch mng in lm vic bnhthngv khi c chmt1 pha khc?Bit: - Mngc ininp 380 V, f = 50Hz;

-intrngiRng= 1k, intrnittrung tnh R0= 46. *Hy xc nhdng inqua ngikhi ngi(Rng= 1k) chmvo 1 pha

camngin3 pha trung tnh nit380/220 V chmnginlm vicbnh thngtrong trnghpngichm:

Tipxc trctipvit(ichn t)? igiyc intrRg= 10k?

igiyc intrRg= 10knhnglichmvo phnnh khitcamtktcukim loichn trctiptrong tgn?

Githit: intrnittrung tnh R0 = 4v intrcaktcukimloiR = 20.

*C nhnxt g trong cc trnghpktrn?


38

Phn 2.Cc bin php k thut an ton

Bo v chng in git

Chng tip xc in trc tip Chng tip xc in gin tip

KhongCch

an ton

S dngTn hiu,bin bov kha

lin ng

Cn tr,V ngn

cchbo v

Ngunin p

thp

Ni tbo v

Ni dyTT

bo v

T ngct mch

bo v

S dngCchin

S dngdng c,ph tinan ton


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4.1. KHI QUT CHUNG Trong HTtnti3 loinit:

- Nitlm vicR0:Thchinnicc imcamngin(thngl trung tnh mngin)vih thngnitnhmmbocc chlm viccamngin.

- Nitan ton (BV) R: Thc hin ni cc phn t bnh thngkhng mang inp (thngl vmy, khung my, chn s,) vihthngnitnhmmboan ton cho ngitipxc vicc phntny khi v l do no (thngl cch inbhng)chng c in.

- Nitchngst Rxk: Thc hin ni cc thit b chng st vi h

thngnitnhmmboan ton cho ngiv cc thitb, cngtrnh khi c st nh.Trong nidung mn hcny chyuchcpnitan ton. Tuynhin cc cng thc,trsintrnit,cch thctnh ton, thitkv lpttrnh by c thcp dngcho c3 loinitktrn.

Chng 4. BO V NI T

40

4.1. KHI QUT CHUNG

R0 R

BA

TB


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4.1. KHI QUT CHUNG Mththngnitc thl:

- Tnhin:Tndngcc bphnkim loic sntrong lng tlm hthngnit.

- Nhn to:Chnhdng cc incckim loi(bngngl ttnht)chn su trong tlm hthngnit.

- Hnhp:Kthp2 loinitny.

in ca mt h thngni t gm 2 thnh phn: in tr ca bnthn incckim loiv intrcakhittham gia qu trnh tndngin vo trong tcgi l in tr tn. in tr ny ph thuc vo

kch thc,chn su v intrsutcavng t.in trsutcatc nhhng lnnht ti trscain trtn. Do intrsutphthucvo rtnhiuyutnh: loit,thitit,cht,(trong cbit lu nyu t thi tit)nn khi tnh tonintrtn,intrsutcnchiuchnhtheo hsma km.

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4.1. KHI QUT CHUNG

B tr in cc chn su t, m km Ghi ch

Thanh ngang 0,5 4,5-6,5 Khi o m ly tr s lnv ngc li kh ly trs nh

0,8 1,6-3

Cc thng ng 0,8 1,4-2


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4.2. MC CH- NGHA CA BO V NI T

Mcch:

Nhmgimdng inqua ngintrsan ton;

Tngdng inscpha-vcc thitbbovqu dngtruynthng(CC, ATM, BVRL) ctphntny ra khimngin,an ton cho ngiv thitb.

ngha: Khi cch in gia pha v phn t bnh thngkhng mang inbhng,nits duy tr 1 inp gia

cc phn t ny vi t nh s an ton cho ngi chmphi.

(Xt v dchngminh)

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4.4. TNH TON, TK V LP T HT NI T

B1. Thu thpsliu

Loimngincung cp

Xc nhvtr v intrsutcavng tsthchinnitbov

B2. Xc nhintrnityu cuRyc(davo quy phmphlc1)

B3. Dkimcc loiinccdng trong hthngnitsau p dngcng thctnh ton intrnit(theo phlc1).

B4. So snh trsintrtntnh ton cB3 viRyc. Nu:

RHTRycChuynsang B5.

RHT> RycCntngslnginccv tnh liB3 sao tRyc.


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4.4. TNH TON, TK V LP T HT NI T

B5. V mt bng, mt ct ca h thng ni t v hnh v hng dn thicng, lp t

B6. Lp t

B7. Kim tra

46

4.5. PHM VI P DNG V 1 S IU CN LU KHI THC HIN BO V NI T

4.5.1. Phm vi ng dng

Phm vi p dng

Mng cao p(U>1000V)

Mng h p(U1000V)

Mi loi mng inu phi p dng

Mng TT ni t Mng TT cch in

Dng BVNDTT Khi in p 150V

Khi in p < 150V

+ N.Xngnguy himvAT

+ N.Xngnguy cchy ncao+ Cc thitbtngoi tri


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4.5.2. Mt s iu cn lu khi thc hin BVN

ivinhngvng tc intrsutln,c thsdngcc giiphpsau:

Lidngthm cc inccnittnhin

Thay tgcc intrsutcao bngtmic intrsutthphn

Sdngmuin

Sdngha chtgimintrsutcuc b:

- Vt liu tngcng tiptGEM (Earth Enhancing Metarial) gmcc iondninv httmn.

- Hp cht tng cng tip t EEC (Earth Enhancing Compound) gm 2thnh phn dng bt: Sumfat ng (15%) v Sudium Ferro Cyanide (85%)ha vo ncv rivo hchn inccnit.- Ha chtSan-Earth (Nhtbn)

Cng nghnittngsu

Nitngth

Vickimtra nhkHTN: 6 thng/ln

48


1. Cc loinittrong hthngin?

2. Mcch, ngha,trnh ttnh ton v phmvip dngcabovnit?

3. Tnh ton ni t lm vic v ni t an tonca mt trm binp 630kVA/35/0,4kV? Bit:o= 100m ma kh.


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Chng 5.BO V NI DY TRUNG TNH5.1. KHI QUT CHUNG

Khi nim: BVNDTT l thc hinni cc phn t bnh thng khngmang in p (thng l v, khungmy) vi dy trung tnh camng hp 3 pha 4 dy c trung tnh nit.

Mcch: Nhm bin s c chmv thnh s cngnmch1 pha, cc thitbbov(CC, ATM) ddng

ct cc thit b b s c chm v rakhimnginsan ton cho ngitipxc.

R0

50

5.1. KHI QUT CHUNG

Phmv i p dng:

Sdngtrong mngin3 pha hp c trung tnh nit;

Mngin1 pha hp c 1 dy nit(ncta);

KthpBVNv BVNDTT5.2. MTSIUCNCH KHI THCHINBVNDTT

Cnc nitlplitrn cc ondy trung tnh 280-300m (Nitviintrnhhn1 cpso viungun-xem PL3); ring nudng cp3 pha 4 dy th khng cnnitlpli.

Khng ct thitbngct,bov trn dy trung tnh. Munctphictngthicdy pha.

LptBVNDTT cngnhBVN,cnnicc v,khung my trong cng1 nh xngvinhau v nividy trung tnh (xem hnh v).


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5.2. MT S IU CN CH KHI THC HIN BVNDTT

52

Cu hi n tp chng 5

1. Khi nim,mch chv phmvi p dngcaBVNDTT?

2. Tisao ivimnginhp 3 pha trung tnh nit,ngita lidng BVNDTT thay BVN?

3. Nhngiucnch khi thchinBVNDTT?


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Chng 6. BO V CHNG IN GIT BNG RCD

6.1. KHI QUT CHUNG

6.1.1. Cu to v nguyn l lm vic ca RCD

a. RCD 3 pha:

54


a. RCD 3 pha:


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b. RCD 1 pha:

I1I2

I

In

56

Ci t gi tr tc ng:

Ngoi chc nng chng in git RCD cn c th chng c chy n


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6.1.2. Cu trc mng in (theo IEC 364-3)

TT system IT system TN system

58

6.2. THC HIN RCD TRONG CC S

6.2.1. S TT

u

L

R

UnI

RCD


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6.2.1. S TT

60

TT system, exercise 1

1) Tnh dng in s c If?

2) Tnh in p tip xc Ung?

3)tIncaRCD l bao nhiu?-nhy(sensitivity)?- (Thigian tr)time delay?

4) Nudng RCD c In= 30mA th intrnitvthitbbao nhiu khi c dngchmvRCD stc ng.


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1) inp pha l UP= 230V. Do dng inscl:

If= 230/(4+6) = 23 A

2)inp tipxc Ung= 23x6 = 138V.inpny nguy himcho ngiv Ung>> Utxcp.

3) -nhy: In< Utxcp/Ru

Utxcp= 50 V In< 8A;

Utxcp= 25 V In< 4 A.

- Thigian tr: tcthi4) Tra theo bnghoctnh theo:

u

L

R

Un

I

62



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64

6.2.2. S IT

Khi ch c chm t mt im

Ngi tip xc khng nguy him

Dng chm t nh, khng ckh nng gy ha hon

Ch cn lp b phn cnh bo cchm t


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6.2.2. S IT

Khi c 2 im chm t

Tr thnh NM 2 pha, cc thit bqu dng (CC, ATM) s tc ng.

66

6.2.2. S TN

a. S TN-C:y l mng 3 pha 4 dy

PEN = PE+N

Khng dng c RCD

RCD


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6.2.2. S TN

b. S TN-S

RCD

c. S TN-C-S

RCD

68

c. S TN-C-S (Ch )

TN-C TN-S


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Cu hi n tp chng 6

1. Cu to v nguyn l lm vic ca RCD?

2. Cc loi mng in theo tiu chun quc t v vic thc hinRCD trong cc s ny?

70

Chng 7.CC BIN PHP AN TON KHC(Phng nga ri ro)

7.1. CC BIN PHP K THUT

Bin php phng nga ri ro

m bo tt cch in cady dn, thit b

Cn m bo khong cch, bao che,ro chn cc b phn mang in

Theo quy nh

KHong cch, tr ngi

cch in

S dng tn hiu, binbo, kha lin ng

Theo quy nhS dng phng tin,dng c an ton

H thp in p, cch lyS dng mba


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Chng 7. CC BIN PHP AN TON KHC

(Phng nga ri ro)1) m bo tt cch in ca dy dn, thit b:

Dy dn:Bc cch in bn ngoi:

Thit b in:S dng cch in kp:

Ph

N

K hiu

72

Chng 7. CC BIN PHP AN TON KHC(Phng nga ri ro)

2) m bo khong cch, bao che, ro chn cc b phn mang in:

m bo khong cch: trnh va chm vi b phn mang in, quynh:


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2) m bo khong cch, bao che, ro chn cc b phn mang in

Bao che, ro chn: gy tr ngi, cch ly vi b phn mang in

Cao p:

Tn chn kn Tn chn h Lng chn

Chn li DCL

74

Bao che, ro chn: gy tr ngi, cch ly vi bphn mang in (tip)

Tmchn230V

H p:


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Ch s IP theo IEC 60364-5-51

Chng li vt rn t ngoi vo

Chng li dn gn tcc phn t nguy him

Mu bn tay

Dng c

Bi khng th xm nhp vo

Chng c git nc ri thng ng

Chng c phun nc

Chng c ngm nc chc lt

Chng c ngm nc duy tr

Chng c phun nc p lc

Chng c phun nc thnh tia

Chng c phun nc mi pha

S c trngth nht

S c trngth 2

Note: the IP code applies toelectrical equipment for voltagesup to and including 72.5 kV.

V d:ngcc IP44

76

CH SO BAO VE (O KN) IP#X

# KHANANG NGAN VAT RAN X KHANANG NGAN NC

0 Khong bao ve 0 Khong bao ve1 ngan chan vat > 50 mm 1 giot nc ri thang ng2 ngan chan vat > 12 mm 2 giot nc ri xien 15 o3 ngan chan vat > 2.5 mm 3 giot nc ma ri 60 o4 ngan chan vat > 1 mm 4 nc tat vao5 ngan chan bui bam 5 Tia nc phun vao6 ngan chan toan bo 6 song nc dap vao

7 b ngap trong nc

8 B ngam trong nc


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Bao che, ro chn: gy tr ngi, cch ly vi b

phn mang in (tip)

MG

NS80

12,5 mm

IP 0

IP 2

78

3) S dng bin bo, kha lin ng

ZONE de TRAVAIL

N 1 2 3 PE


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4) S dng phng tin, dng c an ton

80

4) S dng phng tin, dng c an ton

V d


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5) S dng my bin p cch ly

Ph

N

230V 230V

Ph

N

230V 12/25/50V

Ph

N

230V 5/12/15V

PE

82

Ch

Ph

N

230v 230v

Mybinpbitlphaycchin

Minguyhimkhitipxctrctip

. .


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7.2. CC BIN PHP T CHCLun phic thcchphnh nghim chnhniquy, quy nhcacquan; quy trnh, quy phm;tiu chunra.

Vd:Lm victheo phiucng tc

PHIU THAO TCS........

Ngy................ Thi gian bt u.................. ........Thi gian kt thc.........................

Nhim v: Ct in v ni t ng dy s 2-110kV.Trnh t thao tc:

1. Ct my s :................2. Kim tra trng thi ct ca my ct in.

3. Kim tra cch in ca dao cch ly ng dy.4. Ct dao cch ly ng dy.5. ng dao ni t ca ng dy.6. Ct dao cch ly thanh gp ca h thng thanh gp....

Ngi thao tc Ngi duytK K

84

Chng 8. X L, CP NGI B IN GIT

Khi thy ngi b tai nn in git, bt c ai cng phi ctrch nhim tm mi bin php cu ngi b nn.

Vic x l, cp cu cng tin hnh nhanh th t l nn nhn

c cu sng cng cao. Theo thng k, trong 1 pht nu nn nhn c tch ra khi

ngun in v c cp cu kp thi th t l cu sng 98%,nhng nu n 6 pht t l ny ch l 10%.

Vic s l, cp cu ngi b in git ng cch cn thchin theo 2 bc c bn:

Tch nn nhn ra khi ngun in, v Cp cu nn nhn ngay sau khi tch ra khi ngun in.


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8.1. PHNG PHP TCH NN NHN RA KHI NGUN IN

Cn phi phn bit ngi b in git mng in cao p hay h p

86

8.2. PHNG PHP CP CU NGI B IN GIT


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Chng 9. PHNG CHNG IN T TRNG

(Jack ie LaMuth - Ohio Universi ty) in t trng (Electromagnetic Fields; EMFs) l g?Dng in l nguyn

nhn sinh ra in t trng. in t trng c phn lm 5 loi theo tn s can:

Loi ELF (tn s cc thp; extremely low frequencies) - cc thit b in gia dng,ng dy in.

Loi HF v LF (tn s cao [high frequencies] v tn s thp [low frequencies]) -sng radio AM

Loi VLF (tn s rt thp; very low frequencies) - tivi v video Loi VHF (tn s rt cao; very high frequencies) sng tivi v radio FM Loi SHF (siu tn s; super high frequencies) tn s ca microwave Con ngi tip xc vi EMFs nh th no?

Con ngi tip xc vi nhiu ngun EMFs khc nhau, trong c ngun tnhin v ngun nhn to, do rt kh xc nh mi lin h liu lng - hu quca mt ngun EMFs duy nht no .

Trong t nhin cc EMFs c to ra bi qu trnh sm chp v t trngca Tri t.

Cc ngun EMFs nhn to sinh ra r qu trnh sn xut, truyn ti v sdng in.

88

Chng 9.PHNG CHNG IN T TRNG

Tc ng ca in t trng i vi c th ngi:

Nu ngi tip xc vi nhiu ngun in t trng khc nhau v cng ln hn gii hn cho php ko di s dn n n. S thay i mt s

chc nng ca c th, trc h l h thn kinh trung ng (ch yu lmri loi h thn kinh thc vt v h thng tim mch). S thay i ny cth lm:

Nhc u, d mt mi, kh ng hoc bun ng nhiu, suy yu ton thn.

Lm chm mch, gim p lc mu, au tim, kh th, lm bin i gan vl lch.

Ngoi ra, nng lng in t trng tn s cao (trn 50-60Hz) gi l bcx ion, n c nng lng tch electron ra khi nguyn t. Tia X c nng lng ph hy cc phn t cha gene. Nu con ngi tipxc nhiu vi bc x ion c th b ung th, bin i mu, gim s thnh

mi, bin i nhn mt.


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Chng 9. PHNG CHNG IN T TRNG

Bin php phng chng in t trng:

Tun th nghim tc cc quy tc v tiu chun ca ngnh v nh nc.

Khng ng qu gn cc ngun pht sinh in t trng, s c thgim c phn ln cc nh hng.

Khng nn ng gn cc thit b in, t bit l cc thit b c motor.

Gi khong cch vi u my video t nht l 18 inches (18*2,54cm),hy tt u my khi khng s dng. Khng ngi gn pha sau hoc bncnh mn hnh vi tnh (thm ch khi cch mt vch phng).

Nu c th hy tt thit b si ging, chn in, trc khi i ng.

Gi khong cch vi feet (1 feet = 12 inches) i vi ti vi (k c michiu)

Hn ch s dng chn in v my sy tc

Ht

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141849 Bai Giang an Toan Dien