ground fault current

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gGE Multilin

Technical Notes

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T60 Restricted GroundFault ProtectionCalculationsGE Publication No. GET-8427B

Copyright 2004 GE Multilin

DESCRIPTION The T60 Transformer Management Relay is equipped with number of RGF (RestrictedGround Fault) protection elements.

The RGF element uses a novel algorithm where the restraint current is a maximum of pre-calculated values based on positive-, negative-, and zero-sequence symmetricalcomponents derived from the measured winding phase currents. The actual ratio of theground differential and restraint currents is then compared to the quantity formed by thepickup and a single slope RGF characteristic.

SETTINGS CALCULATIONEXAMPLE

Consider a transformer with the following nameplate data:

Transformer type: Delta/Wye 30 Rated MVA: 10 MVA Nominal phase-phase voltage: 33 kV (Winding 1) / 11 kV (Winding 2) Phase and Neutral CT ratios: 600:1 Impedance: 10% Winding neutral grounded through resistance of R g = 6.3

The rated load current of the Wye connected winding is calculated as:

(EQ 1)

The maximum phase-to-ground primary fault current is calculated as:

(EQ 2)

I rated Rated MVA

3 Winding 2 Nominal Voltage-------------------------------------------------------------------------------------- 10 MVA

3 11 kV----------------------------- 525 A= = =

I gf (max)Winding 2 Nominal Voltage

3 R g ------------------------------------------------------------------------ 11 kV

3 6.3 ----------------------------- 1000 A= = =


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GET-8427B: T60 Restricted Ground Fault Protection Calculations

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For a winding fault point at 5% distance from the transformer neutral, the phase-to-ground primary fault current is calculated as:

(EQ 3)

Increases in phase current ( I ph ) due to the fault are negligible, meaning that .

The ground differential pickup setting can be calculated as:

(EQ 4)

Note that phase CT primary is used as a unit for calculating the RGF settings. Hence,magnitude scaling is applied to the measured ground current.

The new restraint calculation algorithm provides very secure behavior of the element onexternal faults and CT saturation, and high sensitivity on internal faults. The setting of theslope should be selected based on two criteria:

1. Reliable detection of the internal fault currents corresponding to the point of theselected distance from the grounded neutral.

2. Security on external faults with or without CT saturation.

Let us consider the case of an internal ground fault that occurs on in the winding at 5%distance from the transformer neutral. According to the data from the example, the pri-mary fault current of 50 A is:

(EQ 5)

Based on 100% transformer load currents of 525 A on the Wye winding, the phase unitcurrents are:

(EQ 6)

and the symmetrical components are:

(EQ 7)

The ground differential current is calculated according to the following equation.

(EQ 8)

The restraint current used by the relay algorithm is defined as the maximum from theIR 1, IR 2, IR 0 quantities, based on the following symmetrical components calculations:

(EQ 9)

The value of IR 1 is calculated using the positive-sequence current:

(EQ 10)

However, if , then .

I fault 0.05 I gf (max) 50 A= =

3 I 0 0

I GD 3 I 0 I g 0I fault

Phase CT Primary------------------------------------------------- 50 A

600------------ 0.08 pu= = = =

I G 0.08 pu=

I A

I B

I C

0.875 pu= = =

I 1 0.875 pu; I 2 I 0 0 pu= = =

I GD I G I N + 0.08 pu 0 pu+ 0.08 pu= = =

I GR max IR 1 IR 2 IR 0, ,( )=

If I _1 1.5 pu of phase CT, and I _1 I _0> , then IR 1> 3 I _1 I _0( )=

else IR 1 0=

I _1 1.5 pu< IR 1 I _18

-----------=


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The value of IR 2 is calculated using the negative-sequence current.

In this case, for the first two cycles, following complete de-energization of the winding (all three phase currents below 5% of nominal for at least 5 cycles). Other-

wise, .

The value of IR 0 is calculated using the zero-sequence current:

(EQ 11)

Applying the formulas from above to our example, gives us the following values:

(EQ 12)

(EQ 13)

(EQ 14)

Therefore, the ground differential current is:

(EQ 15)

and the ground restraint current is:

(EQ 16)

The RGF element would therefore calculate a ground differential/restraint ratio of:

(EQ 17)

Thus, the slope should be set at some level below 73.1%.

Ideally, if no CT saturation is involved, the RGF protection detects no differential currentduring ground faults that are external to the zone. However, the RGF slope settingshould be also selected to maintain the security on external faults with CT saturation.

EXTERNAL SOLID PHASE ATO GROUND FAULT WITHNO CT SATURATION

Consider an external solid phase a to ground fault with no CT saturation. Given the fol-lowing currents and corresponding secondary values:

IR 2 I _2=

IR 2 3 I _2=

IR 0 I G I N I G I A I B I C ( )= =

IR 1 I _18

----------- 0.875 pu8

----------------------- 0.194 pu (since I _1 1.5 pu< )= = =

IR 2 3 I _2 3 0 pu 0= = =

IR 0 I G I N 0.08 pu 0 0.08 pu= = =

I GD I G I N + 0.08 pu 0+ 0.08 pu= = =

I GR max I R 1 I R 2 I R 0, ,( ) max 0.194 pu 0 0.08 pu, ,( ) 0.194 pu= = =

I GDI GR --------- 100% 0.08 pu

0.194 pu----------------------- 100% 73.1%= =

Primary Currents Secondary Values

I A = 1000 A I A = 1000/600 = 1.67 pu 0

I B = 0 A I B = 0 puI C = 0 A I C = 0 pu

I N = 1000 A I N = 1000/600 = 1.67 pu 0

I G = 1000 A I G = 1000/600 = 1.67 pu 180


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The symmetrical components derived from the phase unit currents are:

(EQ 18)

The ground differential current will be:

(EQ 19)

The IR 1, IR 2, and IR 0 currents are:

(EQ 20)

(EQ 21)

(EQ 22)

The restraint current is the maximum of the three currents IR 1, IR 2, and IR 0:

(EQ 23)

Therefore,

(EQ 24)

and the protection element is effectively in non operating stage.

EXTERNAL PHASE A TOGROUND FAULT WITHSATURATION ON THE

GROUND CT

Suppose an external phase A to ground fault where the ground CT saturates as a result,producing only 5% of the total fault current on its secondary winding. As per the previousexample, the phase fault currents (primary and secondary) would be:

The symmetrical components derived from the secondary phase currents are:

(EQ 25)

The differential ground current is calculated as:

(EQ 26)

The IR 1, IR 2, and IR 0 currents are:

(EQ 27)

I 1 I 2 I 0 0.557 pu= = =

I GD I G I N + 1.67 pu 0 1.67 pu 180 + 0 pu= = =

IR 1 I _18

----------- 0.557 pu8

----------------------- 0.07 pu= = =

IR 2 3 I _2 3 0.557 pu 1.67 pu= = =

IR 0 I G I N 1.67 pu 180 1.67 pu 0 3.34 pu= = =

I GR max IR 1 IR 2 IR 0, ,( ) 3.34 pu= =

I GDI GR --------- 100% 0 pu

3.34 pu-------------------- 100% 0%= =

Primary Currents Secondary Values

I A = 1000 A I A = 1000/600 = 1.67 pu 0

I B = 0 A I B = 0 pu

I C = 0 A I C = 0 pu

I N = 1000 A I N = 1000/600 = 1.67 pu 0

I G = 50 A I G = 0.05 1000/600 = 0.0835 pu 180

I 1 I 2 I 0 0.557 pu= = =

I GD I G I N + 1.67 pu 0 0.0835 pu 180 + 1.587 pu= = =

IR 1 I _18

----------- 0.557 pu8

----------------------- 0.07 pu= = =


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(EQ 28)

(EQ 29)

The restraint current is the maximum of the three currents IR 1, IR 2, and IR 0:

(EQ 30)

Therefore,

(EQ 31)

This is well above the restraint characteristic in the operating region if no special treat-ment to the restraint is provided.

For cases of CT saturation and switch off conditions, the ground restraint I GR is set todecay slowly. Since the CTs do not immediately saturate at the same instant the faultoccurs, the restraint current will be equal to the initial fault current value of 3.34 pu (as inthe no saturation case). The restraint will drop to 50% of its original value after 15.5cycles, so that if the worst case of saturation occurs after a cycle, the restraint current willbe approximately 3.23 pu, giving the following ratio:

(EQ 32)

SUMMARY These examples can help the user identify the boundaries from the security/sensitivityrange for setting the RGF slope. In our case, the slope would have to be above 50%(assuming very severe saturation leading to only 5% current from the saturated groundCT), and below 70% (assuming detection of faults on the winding at 5% distance fromthe neutral point).

IR 2 3 I _2 3 0.557 pu 1.67 pu= = =

IR 0 I G I N 0.0835 pu 180 1.67 pu 0 1.753 pu= = =

I GR max IR 1 IR 2 IR 0, ,( ) 1.753 pu= =

I GDI GR --------- 100% 1.587 pu

1.753 pu----------------------- 100% 90.5%= =

I GDI GR --------- 100% 1.587 pu

3.23 pu----------------------- 100% 49%=

ground fault current

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