141849 bai giang an toan dien
TRANSCRIPT
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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
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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
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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
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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:
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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:
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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. HIN TNG DNG IN TN VO TRONG T
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. HIN TNG DNG IN TN VO TRONG T
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
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1.5. HIN TNG DNG IN TN VO TRONG T
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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Cu hi v bi tp n tp chng 2
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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Cu hi v bi tp n tp chng 2
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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Cu hi v bi tp n tp chng 3
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?
Cu hi v bi tp n tp chng 3
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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
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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)
44
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
Cu hi v bi tp n tp chng 4
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
6.1.1. Cu to v nguyn l lm vic ca RCD
a. RCD 3 pha:
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6.1.1. Cu to v nguyn l lm vic ca RCD
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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TT system, exercise 1
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
TT system, exercise 2
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TT system, exercise 2
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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9/10/2014 67
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
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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