# grounding and grounding fault analysis

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Grounding Grounding FaultAnalysisHow to calculate Grounding FaultHow to understand Grounding Fault

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• ( 3 )

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2013. 12

www.p7c.co.kr

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-ANSI/IEEE Std. 141-1986[page 172, 347]

-ANSI/IEEE Std. 141-1986[page 255~258]

-ANSI/IEEE Std. 141-1986[page 286]

-IEEE Std. 142-1991[page 20]

-ANSI/IEEE Std. 242-2001[page 238~239]

- IEC / , [2009]

- [2] / , [2003]-

• - 3 -

Profile

1. - 56 ( 4 1984)- (1984~1992) - (1992~1996) -( ) (1996~2003) - , (~2012) - (2004~ )

2. - (1998) - ( 14-06-003 )

3. - [1] (2003) - [2] (2003)- - [3] (2010) - ! (2013)

4. : www.p7c.co.kr

• - 4 -

1. 2. 3.

/

GPT

4. RN NGR(Neutral ground resistor) 3 GTR(Grounding transformer)

5. 1

6. 7. Next to : 4

• - 5 -

3

1.[ 1]

2

1 : 380mA ~ A : ~ A ( )

: 5~10A / : A(IR 3IC0 )

A ( 3 )

1

( E0) ( arc 6~8 )

( E0) ( 2.6 )

(1.4E0) [ ] 2.

- - -

(GPT, ZCT, SGR )

.

[ ] 1. (E0 ) (E0 ) . 2. 1000V IEC 60364 [ 4] .

3. arc . ( ) 4. -> ->

• - 6 -

3

2. ( )

[ 1] (ANSI/IEEE Std 141-1986[172] Fig 47)

[ 2] [ 3]

• - 7 -

3

3.

[ 4]

• - 8 -

3

3.

/ 1IC1 = IC2 = IC3 = 2fcE

, E = V3

: (3 ICg = IC1 + IC2 + IC3 = 0A)

C : 1 ( / )

a , a [ 4] 3

I

C2 = I

C3 = 2fc3 V

3= 2fcV

, jIC?jIC =I C2+I C3 =

3 2fcV =3 2fcE ( A )

GPT

IGF = (IN+ jIC )tan 1 ICIN (A ) (GPTIN 380 GPT .

70~90 , .)

• - 9 -

3

3.

[ 2] 6.6kV

-

-

PM1

PM1 9.12A 87.6 -PM1 5.55A 82.1 -

0.83 62.92 -PM1 MAIN 15.48A 82.9 -

PM3

PM3 10.3A 87.8 12.91A 88.3PM3 6.71A 86.7 9.32A 87.6

6.56 86.6 -PM3 MAIN 23.6A 86.3 31.44A 87.2

[ ] 1. , 2. .

• - 10 -

3

3.

, a ?

Vb =(a 2 1 )Z0 + (a 2 a )Z2

Z0 + Z1 + Z2Ea

VC =(a 1 )Z0 + (a a 2 )Z2

Z0 + Z1 + Z2Ea

, : 1 = 10 = 1 + j0 , a = 1120 = 12 + j3

2, a 2 = 1240 = 12 j

3

2, 1 + a + a 2 = 0

: : : :

[ 5] [ 6] , : [ 5] , , , , : [ 6] ,

• - 11 -

3

3.

[ 6] , Z0 = XCO = 1wC 0 ,

+ 3 .

Z 0 + Z 1 + Z 2 0 , (-2) .

[ 7] [ 7] , . , 1

6~8 , . .

• - 12 -

3

3.

, 6~8 . Surge .

GPT

/ .

6~8 . Capacitance Capacitance Inductance , .

If this ground fault is intermittent or allowed to continue, the ungrounded system could be subjected to possible severeover-voltages to ground, which can be as high as six or eight times phase voltage. Such overvoltages can puncture insulation andresult in additional ground faults. These over-voltages are caused by repetitive charging of the system capacitance or by resonancebetween the system capacitance and the inductances of equipment in the system. [ : ANSI/IEEE Std. 242-2001 8 page238~239]

• - 13 -

3

3.

GPT

[ 8] GPT [ 9]

2011.6.22 : 6.6kV 190kW , SGR -> GPT 3

• - 14 -

3

4.

[ 10]

• - 15 -

3

4.

RN

3 , RN . -Y , - .

, [ 11] IR = IN 3IC0 .

[ 11]

• - 16 -

3

4.

NGR(Neutral ground resistor)

: NGR , IN 3IC0 : IEEE 32 10 , 1 , 10 , 30 .

: 13

.

: 10 , 1 760 , 560

( )

R N =EIN

=6600V/

3

200A= 19,

, 6.6kV/3 , 200A, 19 , 30 , ( 200A )

[ ] , .

• - 17 -

3

4.

3 GTR(Grounding transformer)

Ground Transformer 3kVA = 3 (V /

3 )

Ig3

k3

, V : (kV), Ig : (A), k 3 : (%)

[ 3] k3 factor 3 Y- k3 factor

10 -1 0.1702 0.2403 0.2954 0.3405 0.380

[ ] IEC 1 95 / 3

: Q = 6, 6003

200 0 .240 = 183 [kVA ] 200250 [kVA ] , ( 200A )

: Pri. 6.6kV/3 (Wye connection with / neutral terminal ), Sec. 380V (Delta connection ) %ZPS =5.58% .

• - 18 -

3

5.

[ 12]

, , 1 - , -

• - 19 -

3

5.

[ 4]

TT TN-C TN-S IT

IT

(N) 1.45Uo - 1.45Uo 1.45Uo -

(PE) 3Uo - 1.45Uo 3Uo 3Uo

PEN - 1.45Uo - - -

[ 1] Uo ( )

[ ] 2 (SPD)

• - 20 -

3

5.

1

[ 5] 1

TT TN-C TN-S IT Ungrounded

1. (In) TN A A 10A GPT 1380

2. (IC) 1A 1A 1A 1A 1A

[ ] 1. IC 1A , .ANSI/IEEE Std 141-1986[172, 347]

2. IT 10A . IEEE Std 142-1991[20] 3. Z0 < Z1 or Z2 , 1 3 125% .ANSI/IEEE Std 141-1986[286] 4. (Zero Sequence Current Transformer) ( : 480V 1000/5A ) ANSI/IEEE Std 141-1986[255~258]

[ ] (Equipotential bonding) , , 1 3 , (7. )

Ig = 3I0 =3E

Z 0 + Z 1 + Z2 + 3Rf Ig = 3 100%Z0 + %Z1 + %Z2 + 3Rf

IN ISC = 100%Z1IN IN : , 3Rf : 0

• - 21 -

3

5.

[ 13] 440V

VCB

ACB

MCCB

. . . . . . . . . . . . . .

Inverter

M150kW

160kW*

M112kW

DCM

112kW

DC

22.9kV

440V

M110kW

160kW*

M200kW

220kW*

440V

440V

* .

Inverter Inverter

Converter 170kVA

MCCB MCCB MCCB

Converter 170kVA

MCCB

MCCB MCCB

MCCB MCCB MCCB MCCB

. . . . . . . . . . . . . .

• - 22 -

3

5.

[ 6]

22.9kV/440-254V3 1800kVA

%Z=4.64,

SV2200IP5A-4OL440V, 220kW

2012.04.30~07.23

3

2012.07.24

2012.07.24~08.11 2012.08.11

2012.08.27~08.31 2012.08.31 TR ACB

2012.09.14~09.14 2012.09.14 1600rpm

SV1600IP5A-4OL440V, 160kW

2012.07.24~08.24

22012.08.31 2012.08.31 TR ACB

2012.09.14 2012.09.14

• - 23 -

3

: 3 Y , [ 14] 3 . .

[ 14] 3 94.22V ( ) [ 15] 3 0.19V ( )

• - 24 -

3

: 3 Y , N~M point 94.22V , over lap . -> 3 4 ,.

[ 16] 3 [ 17] 3 N~M point

[ ] 3 6 1, 3, 5 4, 6, 2 120 , 1~4, 3~6, 5~2 . 1, 3 (Over lap) AC . AC .

• - 25 -

3

6.

[ 18]

1

• - 26 -

3

6.

[ 19] ( )

[ 20] (Ring Core Balance)

[ 21] 3

• - 27 -

3

6.

[ 22] GPT [ 23]

R = n2 r9

= 302 509

= 5, 000 ( ) R : 1 , n : GPT , r :

GPT 3 (0.38A/3 ) 30 = 3.8A GPT 3.8A 190V = 1902/50 = 722VA (CLR) (3.8A )2 50 = 722W, . [ ] 440V .

ZCT 200/1.5mA : n0-, n0>100, n0>2000 ( 20A n0>100 )

• - 28 -

3

7.

[ 24]

25SGR CLR

200/1 .5m A SGR SGR

M

CLR

6.6kV BUS

6.6kV BUS

200/5A OCGR

M

6.6kV BUS

6.6kV BU S

NG R19.05200A

OCGR OCGR

G PT 500VA x 3

200/5A 200/5A200/1 .5m A

200/1.5m A

200/5A

380m A

380m A

( )

25

G PT 500VA x 3

G TR250kVA

OCGR

• - 29 -

3

2 100/5A

[ 25] 2

• - 30 -

3

7.

. 100/5A 1,000/5A , 2 . 1 : Ig = 3I0 = 3 100%Z0 + %Z 1 + %Z2 + 3Rf IN ISC =

100%Z1

IN =1006

2000 kVA3 0.38

50kA

: 1.0VA ( ) : 5P 10P (KSC IEC 60044-1 )

. (50,000A/100A)=500 (50,000A/1,000A)=50 , 1.0VA ? 1.0VA ?

1.0VA500=500[VA] . 1.0VA50=50[VA] . = = 20 ? 10 ?

500/20=25[VA] . 50/10=5[VA] .

. 2100/5A, 25VA, 10P20, 63kA 1,000/5A, 5VA, 5P10, 63kA

MCCB Trip Tap Range 0.5~2.0A Tap Range 0.5~2.0A

2.0A 1005

= 40 (A ) pick up 2.0A 1,0005

= 400 (A ) pick up

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