HIGH IMPEDENCE BUSBAR PROTECTION

High Impedance Bus bar Protection

Two type of relays

1. Voltage Operated
2. Current Operated


Basic circuit of High Impedance Bus bar Protection


CURRENT OPERATED

 


 VOLTAGE OPERATED


 


Stabilizing voltage

The voltage developed across the CT secondary during maximum fault condition.

Vs ≥ If * ( Rct + (2 * RL))

IF                       -  Maximum fault current

Rct           - CT secondary Resistance

RL                  -  Lead  resistance of cable

The maximum fault current (IF )

The maximum fault current (IF ) or Maximum short circuit current is the base value for design the substation system. The Equipment specifications and relay settings are calculated based on the Maximum fault current value of the system.
The Maximum fault current is specified in the standard for each system voltage

For Example : In SEC standard TES - P-119.02 CLAUSE 8.2.4

SYSTEM VOLTAGE                        SHORT CIRCUIT CURRENT

380 kV                                                              63kA
230 kV                                                              63kA 
115 kV                                                              40 kA
34.5 kV                                                             25 kA
13.8 kV                                                             21 kA

Consider 380kV system

IF - 60kA

If - Maximum fault current in CT secondary = 21 A for CT 3000/1 Tap selection


Rct - CT secondary Resistance

Rct can get from the CT specification

Consider 3000/1 Tap used for Bus bar protection and Rct = 4.65 ohms 

RL - Lead resistance of cable

Example:

6 Sqmm cable   RL = 0.00308 ohm/meter at 20 dec C

RL at 75 dec C = RL ( 1+ ∞ 20 (T2-T1))

Consider length = 500m

RL at 75 dec C =1.88 ohms

 

Stabilizing voltage

Vs ≥  IF * ( Rct + (2 * RL))

Vs = 21 X ( 4.65 + ( 2 X 1.88 ))

     = 177 volts

Consider 20% Margin
    
 = 177 x 1.2 = 211.932 volts

Hence select Vs = 275 volts

Therefore in maximum fault current the voltage developed across the CT secondary at the relay end is 275 V.
                                                                                                                                                            

VOLTAGE OPERATE RELAY  (MFAC - AREVA)

Calculate voltage actually need to operate the relay or total voltage drop in the secondary circuit during the Maximum fault condition

Is = Total secondary current drop

Is = CT Magnetizing current +   CT supervision relay coil current ( MVTP) + Protection relay coil 
       current for set value (MFCA SETTING RANGE 25-325 V) + Metrosil leakage current

   =  Im + I mvtp +Ir + Imetrosil

Consider 8 CTs are connected parallel

   =  (0.005 x 8) + 0.003 + 0.03 + 0.018

 Is  = 0.091 A

This is the secondary current drop during the fault condition. In other words this is the current enough to operate the protection relay.

Convert this current to Primary value in 3000/1 ratio CT

Ip = Is x 3000 = 273 Amps

Voltage for 63000A is 275V
Voltage for 273A is ?

63000 / 273 =275/ Vsec

Vsec = 1.2 volts This voltage is enough to operate the relay.

This voltage and current  is too small compared to Maximum fault current and corresponding voltage.

This secondary operating current can achieve even leakage or through fault.


So we have to decide  the fault current at which the relay should operate

Shunt Resistor

Say The primary fault current 742A at which the relay should operate.

This is called bias value and up to this value relay should not operate.

To achieve this condition add some resistance and increase the Is secondary leakage current value.

This External resistor should perllaly  connect with Relay. 

Ip = (Is + (Vs/Rsh)) x 3000 

742 = (Is + 275/Rsh) x 3000

Rsh = 1759.7 ohms

So select the 2700 ohms Variable resister and adjust the value.

Adjust the resistance value according to our primary operating fault current requirement .


Example Manifa Final relay setting for MFAC relay

Vs = 175 V

Ir = 0.20A

Rsr = 850 Omhs

Therefore Ip = (Is +Vs/Rsh) x 3000

Ip = (((0.005 x 8) + 0.003 + 0.03 + 0.018) + (175/850) ) x 3000)

    =888 Amps should be develop in primary and 0.266 Amps should be flow through the Secondary circuit to operate the relay with this setting.

CURRENT OPERATE RELAY (MCAG - AREVA)

All other calculations are same except Stabilising Resistor

Relay setting Voltage Vs = 275 Volts

Relay setting current    Ir = 0.2 Amps

Relay burden at 275V setting is consider as 1 VA

Stabilising resistor required Rsr = ( Vs/Ir) - ( Relay burden - Ir^2)

Rsr = 1350 Omhs

Select the resistor range 0-1500 Ohms


CT SUPERVISION RELAY

This relay parallely connected with Main protection relay

 

Consider  'R' phase wire open in CT -2.

The CT-1 load current will start to flow through both relays.

This is not a fault current but it causes  unnecessary trip. To avoid this trip CT supervision relay is used.

The voltage setting of this relay is very low with time delay.

So this relay will operated for leakage current and its contact short the CT.

Main relay setting is Grater then this value and instantaneous. So for heavy internal busbar fault Main protection will operate immediately.

Final relay setting Vs=14 v Time = 3 sec

Open 'R' phase CT will save by metrosil mounted near the CT marshaling box.


METROSIL

This is non liner resistor

During Heavy fault condition Protection relay will immediately clear the fault . If fault not clear and secondary fault voltage will damage the relay.

This kind of situation Metrosil will short the CT secondary circuit and save the relay coil.

Some times  circuit is open due to relay coil damage. In this situation if fault occur heavy voltage will develop and CT get damage. To avoid this metosil will short the circuit and save the CT.

Metrosil should withstand the for voltage develop due to maximum shout circuit stability voltage Vs.


Busbar Stability Test

1. Current operated Relay

1. Calculate the Primary operating current for relay setting value current

Ex: Relay setting 20%

CT ratio = 3000/1

20 % of secondary current 0.2Amps

Primary Current 600 Amps

2. Short the Stabilising resistor with small shorting wire

3. Drown out the CT supervision Relay to reduce the load for Primary injection kit.

4. Drown out the Main protection relay and circuit will short in side and get closed path.

5. Open the metrosil one side wire ( Don't short the Metrosil)

5. Slowly inject the primary current and reach 0.2 Ampls secondary

6. Now inset the Protection relay let it operate and measure the current.

7. Primary current value will if Metrosil , Resistor , CT supervision relay are include in the circuit.


2. Voltage operated Relay


1. Calculate the Primary operating current for relay setting value voltage

Ex: Relay setting 175v Rsh = 1500 Ohms

CT ratio = 3000/1

Ip = (Is +Vs/Rsh) x 3000 = (0.091+ 0.1166) x 3000 = 623 Amps


 Secondary current 0.2Amps

3. Drown out the CT supervision Relay to reduce the load for Primary injection kit.

4. Drown out the Main protection relay and circuit will short in side and get closed path.

5. Open the metrosil one side wire ( Don't short the Metrosil)

5. Slowly inject the primary current and reach 0.2 Ampls secondary

6. Now inset the Protection relay let it operate and measure the Voltage.



Don't inject the current continuously to the relay. Because relay will get damage.






 

TRANSFORMER TERTIARY WINDING

This is Power Transformer tertiary winding Bushing. The connection of tertiary winding is open delta in YNynd1. If tertiary winding voltage not used in site and it should be solidly ground as like pig. Should not ground only one bushing of open delta. Because at closed delta point the induced EMF in tertiary winding three phases are zero and the leakage current developed due to unbalance voltage will flow to ground. In open delta point the potential difference is very high and it will make heavy current flow ( Direct short) to the ground and protection will operate.

 

VT SECONDARY INJECTION

During VT secondary injection we faced simple and small problem.

A 10 X 2.5 Sq.mm cable is used between GIS "R" phase VT junction box to Local Control Cubical .

The VT is two winding VT and Secondary voltages as below

1a1- n = 64 v
1a2 - n = 64 v

2a1- n = 64 v
2a2 - n = 64 v

So in 10 X 2.5 Sq.mm cable 5 cores are used and all others are spare.

Now 64 V AC injected between terminal 1a1-n at VT junction box and measured at LCC and value is ok.

For cross checks we measured 2a2-n also and found same 60V voltage.

Witness says may be cable is damaged and cable is checked thoroughly and meggerd  and every think is normal.

Some times if secondary cables connected with VT the applied voltage in secondary first winding will reflect in primary and it will induced voltage in second winding (Back Energization).

But we removed all secondary wires from VT.

After long discussion the problem is found it is due to Mutual induction.

Instant of AC voltage DC voltage is applied and there is no voltage in 2a1-n.

CT TESTING IN OUTDOOR SUBSTATION

We are did CT testing in live outdoor substation with CT analyzer ( OMICRON)

CT was enclosed  with Oil Circuit Breaker.

The dis-connectors of CB both sides are opened and a personal Ground placed on the line side as shown in the fig.

The CB is closed and CT analyzer primary leads are connected as shown and secondary leads are connected with CT secondary terminals.

When equipment on it says " Secondary leads are open pls check".

Then we are checked completely and every think is ok and normal and CT secondary resistance are measured with same lead and its ok and perfect.

After long search we found 60 V in CT secondary leads without on the test kit.

Then we are inspect the substation and found one live line cross the bus as shown.
Due to this some induction voltage developed and leakage current flow through the personal ground ( As red line). This leakage primary current developed the voltage in secondary.

To rectify the problem we shifted the personal grounding  to before the CT -1 as shown in fig. and induced current grounded before the CT.

Now CT secondary voltage is reduced and test is competed successfully.