Substation Testing and Commissioning
Sunday, September 30, 2012
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 ≥ I f * ( Rct + (2 * RL))
IF - Maximum fault current
Rct - CT secondary Resistance
RL - Lead resistance of cable
The maximum fault current (I F )
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The maximum fault current (I F ) 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
I F - 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 ≥ I F * ( 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
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= (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
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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
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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.
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SIVAKUMAR
THIS IS SIVA T&C ENGINEER IN NCC LTD - SAUDI.
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