GENERATOR FOR BATTERY CHARGER

1. Generator for Battery charger commissioning should be more then 220KVA

2. Normally generators are Thyristor control and voltage will regulate according with load .

3. Battery chargers contain RLC circuit and it will rise the generator voltage abnormal value.

4. Connect some external lighting loads rather then battery charger to avoid this abnormal condition.

RESISTORS & DIODES & CAPACITORS

POWER TRANSFORMER ERECTION

115kv 67MVA Transformer

1.Erection takes total 8 days with 15 man power
2.Vaccuming of main tank take 3 days upto  -0.5tar
3.Oil circulation 3 days
4.Oil Filling and setteling 1 days

=8+3+3+1=15 days / Transformer  

SAS GIS EQUIPMENT OPERATION

As per SEC requirement the GIS operations ( Open /Close) should be like below.

GIS Local Cubicle Control panel contain one multi position switch
There are four position available

1.Emergency

All operation should be in LCC panel Open/close switches

2.Local

All operation should be through Bay Control relay mounted in LCC panel

Its checks all hoard wire interlock and BCU software interlock

3. Remote

All operation from HMI and SCADA

4. OFF

There is no operation from any where.
 

BREAKER FAILURE INITIATION

Transformer Mechanical Protection BF Initiation

The Breaker failure protection is used for protect the breaker when fail to open for a Protection trip.

If any protection Lock out operate it will give a trip to CB and initiate the BF relay.

After minimum time BF relay will check the CB status and fault current flow then relay will give one more trip to the CB. Now CB is still fail to open then BF relays initiate BF second stage lock out and give trip to adjacent CBs and save faulty CB.

But transformer Mechanical protection ( Main tank Bucholz relay Operate and winding temperature high and oil temperature high, PRD ) Operation independent of fault current.

Therefore BF relay should not consider fault current for transformer Mechanical protection initiation.

This initiation separate given as digital input in Numerical relays REB 500 (ABB),7VK61 ( Siemens)

But this initiation will initiate the first and second stage timers in normal non numerical relays ( RAICA - ABB).
 
 

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.

DIRECTIONAL OVERCURRENT RELAY FUNTION

How to calculate the Operating region for Directional over current relay ?

Relay characterstic Angle (RCA)

The Relay characterstic value is used to define the region to operate the relay in the
specified direction.

Example :-

Let RCA = 45 ͦ
In normal condition angles between Current & Voltages

IR IY IB
0 ͦ 120 ͦ 240 ͦ
VR VY VB
0 ͦ 120 ͦ 240 ͦ
VRY VYB VBR
120 ͦ 360 ͦ 240 ͦ

If any fualt occure in any phase the voltage will be drop and heavy current will flow.

So to see the direction of the current flow relay will take other helth phase voltages.

Operating region = ( Phase displaced + RCA) to ( Phase displaced + RCA +180 )

Let's take R phase grounded and relay will take VYB voltage value is referance
VYB displaced 0 or 360 ͦ from IR
Operating region = ( 0 + 45 ͦ to 45 ͦ + 180 ͦ)
= 45 ͦ to 225 ͦ

Let's take Y phase grounded and relay will take VBR voltage value is referance
VYB displaced 240 ͦ from IR

= (240 ͦ + 45 ͦ ) = 285 ͦ and
= 285 ͦ + 180 ͦ = 465
= 465 - 360 = 105 ͦ
Operating regien = 105 ͦ to 285 ͦ

Let's take B phase grounded and relay will take VBR voltage value is referance
VRY displaced 120 ͦ from IB

= (120 ͦ + 45 ͦ ) = 165 ͦ and
= 165 ͦ + 180 ͦ = 345 ͦ
Operating regien = 165 ͦ to 345 ͦ

CT TEST IN LIVE STATION

How to test the CT's in live station ?

1 . Open the Isolator 1 & 2

2 . Close the Isolator 3

3 . Close the CB

4 . Close Line earth switch

5 . Close Maintenance earth switch 1 & 2

6 . Now the Induction voltage is zero in dead portion

7. Remove the earth link in the Maintenance earth switch

8. Connect the primary leads on that.

9. Start the testing

10. There is no leakage current in earth because primary current enter in to Main earth switch -1 and return at Main earth switch -2.

SIMPLE Q & A

1. Why DC supply used as control supply in all protection and control panels?

DC supply is storable and if any problem in HV side AC voltage then Alarm and protection is easy by using DC supply.

2. Why CT & VT wires are bigger in size comparatively other control cables?

The I²R loss is more when small Sq.mm cables used.
Resistance is inversely proportional to cross section area of the wire or cable.
When area increase Resistance decrees.
R ∞ 1/a
Example :
VT secondary voltage = 115v
Burden of the VT secondary circuit = 300 VA
2.5 Sq.mm cable RL = 2 Ω
Current flow = 300 / 115 = 2.6 Amps
Loss per meter = I² RL = 35 watts
Voltage drop = I X RL = 2.6 * 2= 5.2 volts

3. What is Inrush current in Transformer or motor?

During commissioning time the residual magnetic flux in the
Transformer and motor coil is almost zero. There is no back emf for opposing
the main emf in the coil. So the magnetic current very high during starting
stages.
During charging time the inrush current value is less when
HV primary voltage frequency cross the zero value and comparatively high during
frequency cross its peck value.
Inrush current contain 3rd and 5th harmonics and it is detected by protection relays and block the tripping.

4. Why most of the Power Transformer primary side connection arrangements Star connection?

Per Phase voltage value of star connection = V/ √3
But per phase voltage in Delta connection is same as Phase
to Phase voltage.
The insulation requirement of transformer is depending open the voltage value. If voltage reduces then insulation requirement reduced and size & cost of transformer reduced.

TRANSFORMER PROTECTION OVER CURRENT

CONSIDER YNyn0 TRANSFORMER

HV side Over current protection

1.Directional over current HV side (67HV)
2.Directional over current HV side (Neutral current) (67NHV)
3.Non directional Time delayed over current HV Neutral side (51GHV)

LV side Over current protection

(Incase Transformer to LV side Bus tie normally close ( NC ) )

1. Directional over current LVside (67LV)
2. Directional over current LVside (Neutral current) (67NLV)
3. Non directional Time delayed over current LV Neutral side (51GLV)

LV side Over current protection

(Incase Transformer to LV side Bus tie normally open ( NO ) )

1. Non directional over current LV side (50/51 LV)
2. Non directional over current LV side (Neutral current) (50/51 NLV)
3. Non directional Time delayed over current LV Neutral side (51GLV)


TV side Over current protection

1. Non directional Time delayed over current LV side (51TW HS &LS)

Why 67HV instant of 50/51HV on HV side?

The Transformer HV side protection should be directional over current because it should operate only for fault in the transformer HV side. If fault on HV bus then the Bas bar protection should operate.

For the above fault Bus bar protection will operate because 67HV relay is directional over current with curve setting but bus bar protection setting is instantaneous.

FAULT IN TRANSFORMER HV SIDE

Now 67HV transformer protection will operate and BB protection will not operate because BB protection CT will cover this fault.

In this cause first HV side relay will trip and open the HV side CB and full fault current will flow through the LV side and LV side relay will trip and open the LV side breaker.



FAULT IN TRANSFORMER LV SIDE

Consider a fault in transformer LV side and the current flow through LV side CT is more than current flow through HV side CT. Therfore first LV side relay will operate and open the LV side CB. Then all fault current flow through the HV side CT and 67 HV will operate and open the the HV side CB. In case LV side Tie open condition then there is no source to feed the fault from LV side. So for this condition 67LV will not operate and 67HV operate.

LV SIDE BUS TIE NORMALLY OPEN CONDITION

If LV side tie breaker normally open then 67LV is no need . Then non directional relay can use for protect the power cable between transformer LV side to LV SWG. And CT connection can take from the LV CT bushing CT.

CHANGE THE ANY ONE TAP POSITION OF TRANSFORMER AND MAKE THE CIRCULATING CURRENT TO TEST THE DIRECTIONAL OVER CURRENT FUNCTION


What is the meaning of 67NHV?

This is Negative Sequence Overcurrent Protection (NPS).

This scheme worked based on residual current produced during earth fault and phase to phase faults. In normal the vectorial sum of all three phase current at neutral point should be zero.
But any earth or phase to phase fault accure a residual current generate due to unbalance.So a leakage current flow through the CT neutral path.This current sence by the NPS element and initiat the trip or alarm.

Negative Phase sequence overcurrent elements give more sensitive to resistive P-T-P faults, where phase over current elements may not be operate.

In certain application residual current may not be detected by an earth fault relay due to the system configuration. Example :- an earth fault relay applied on the Delta Side of a Delta - Star transformer is unable to detect earth faults on on the star side. However negative sequence current will be present on the both sides of the transformer for any fault condition , irrrespective of the transformer configuration. Therefore, a negative phase sequence overcurrent element may be employed to provide time-delayed protection for any any uncleared asymmentrical faults.

% IMPEDANCE OF A TRANSFORMER

The value mention on name plate for % Z is the percentage of voltage dropped by the
transformer impedance.
Z=R + j(XL+XC)
Z - Impedance
R - Transformer coil resistance ( Fixed Value)
XL - 2∏fL
XC - 1/2∏fC
The impedance of the Transformer is depending upon voltage applied.

Example:

33 / 11 KV
87.5 /262.5 A 6.81 %Z

The value of voltage drop in nominal voltage = 33000X6.81/100 = 2247 V

The value of impedance on full load condition = 2247.3/87.5 = 25.82 Ω

Conform this % impedance by site testing

1.Short the secondary side terminals

2. Apply the 380 V 3 phase in the HV side.

3. Before that calculate the primary current for applied voltage

If applied voltage 2247.3 v to HV side then full loade current 87.5 A will flow on primary side ( Secondary shorted).

Calculate the drop voltage if we are apply 380v

X /380 = 87.5 / primary current

This current can measure in primary side = 14.8A.

X /380 = 87.5 /14.8

X = 2246.62 Volts ( This is the total voltge drop for transformer total impedance)

Then we have to calculate this value how mach % in the Rated primary voltage

33000 * x / 1000=2246.62

x=6.8

This value is % Z of the transformer the error not more than
± 0.1 %

AUTO TRANSFORMER 86LOR TRIP ARRANGEMENT

The Transformer protection tripping can grouped as follows

1. SET-1 Main Protection

The main protection can be divided in to two groups (CB TC-1 Tripping)

a. LOR 86-1 ( Neumarical relays protection)
1.Differential protection (86T)
2.REF protection (87 REF)
3.HV side Over Flux protection (24 - HV/T)
b. LOR 86-3 ( Mechanical protection)
1.Buchhloz Relay trip - Main tank (63 T)
2.WindingTemp High Trip - Serious Winding Main Tank (49 W-TS)
3.Winding Temp High Trip - Common Winding Main Tank ( 49W-TC)
4.Winding Temp High Trip - Tertiary Winding ( 49 W-TT)
5.Buchhloz Relay trip - OLTC (62-OLTC)
6.Buchhloz Relay Trip - Grounding Transformer (63- GT)

2. SET-2 Back up Protection

The back up protection can be divided in to two groups (CB TC-2 Tripping)

a. LOR 86-2 ( Numerical relays protection)

1.HV & MV side backup 3 phase & ground fault inverse time & inst O/C
relay unit and function (67HV & 67NHV and 67MV & 67NMV)
2.HV side Over Flux protection (24 - HV/T)
3. Neutral backup O/C relay ( 51 G-T)
4.Tertiary winding O/C relay ( 51TW/HS & 51 TW/LS)

b. LOR 86-4 ( Mechanical protection)

1.Pressure Relief - trip-main tank (63PR-T)
2.Oil Tem very high trip - main tank ( 26-T)
3.Oil Tem very high trip OLTC ( If any) (26- OLTC)
4.Pressure relief OLTC tank - (63-OLTC)
5.Oil pressure relay Main tank (63-T)
6.Pressure relief Trip grounding transformer ( 63 - PR-GT)

TEST FORMATS

1. AUX RELAY

2.MAIN RELAY

3.EQUIPMENTS

TRANSFORMER OIL LEVEL vs TEMPERATURE CURVE

This comman for all Power Transformers

RELAY SOFTWARES USED IN SUB-3

1. RED 670 - LINE PRO - ABB
PCM 600 - 2.3
2. RET 670 - TR PRO - ABB
PCM 600 - 2.3
3. REB 500 - BB PRO - ABB
REB WIN - 7.5
4 . 7SS52 - BB PRO- SIEMENS
DIGSI - 4.83
5.MICOM - AREVA
MICOM S1 STUDIO - 3.4
6. SEL METER - ENERGY METER - SEL
SEL SOFTWARE

AUTO RECLOSE

Consider a station (1 1/2 schemes ) feeds with two lines
1. Line -1 dead for maintenance work
2.In Line -2 there is a Zone -1 fault
Now The Auto reclose should block because already one Line -1 is out and may be fault exist in Line -2 now the condition is station totally dead. So the station should not energized by Line-2 Auto reclose with fault line.
If the Power Transformer connected with same cross bay then it should be energized with Bus voltage. If Bus side Transformer CB in open condition then the A/R operation should block for the line.Otherwise the Transformer also energized with fault line.

WARNING !!!!

ITS CAME TO MY MAIL
" MUST READ and Share" " MUST READ and Share" " MUST READ and Share"Can ELECTRICITY pass through Flash light of the Digital camera to your body??? Yes it is 100% true..!This is a true incidence reported of a boy aged 19, who was studying in 1st year of engineering, who died in Keshvani Hospital, Mumbai. He was admitted in the Hospital as a burned patient. Reason ??????... This boy had gone to Amravati (a place located in State of Maharashtra ) on a study tour, on their return they were waiting at the railway station to catch the train. Many of them started taking pictures of their friends using "Mobile Phones" and / or "Digital Camera". One of them complained that, he was unable to capture the full group of friends in one frame in the Digicam.This boy moved away to a distance to get the whole group.He failed to notice that at an angle above his head, 40,000 volts electrical line was passing through.As soon as he clicked the digital camera? 40,000 volt current passed through the camera flash light to his camera and then from his camera to his fingers & to his body. All this happened within a fraction of a second. His body was half burned.They arranged for an ambulance & his burned body was brought to Keshavani Hospital, Mumbai.For one & half days or so he was conscious & talking. Doctors did not have much hopes as there was a lot of complex issues in his body. He passed away later.Now how many of us are aware about these technical threats & dangers? Even if we are, how many of us are adhering??Now should we call ourselves as educated and knowledgeable people?▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼* Please avoid mobile phones on petrol outlets.* Please avoid talking on mobile phones while driving.* Change that "Chalta Hai Yaar Attitude".* Please avoid talking on mobile phones while kept in charging mode without disconnecting from wall socket.* Please do not keep mobile phones on your bed while charging and / on wooden furniture.* Avoid using mobile phones / Digital cameras near high voltage electrical lines like in railway stations and avoid using flash.

TRANSFORMER VECTOR GROUP

The vector group test is importent test in the transformer testing

The Test procedure

1. Connect the HV side "R" phase with LV side " R" phase. ( No need for Auto transformer )

2. Apply the 3 Phase voltage (380v) on HV side.

3. Measure the voltages in various combination on both side of the transformer ( HV- LV ).

4. For each vector group need diffrent type of voltage combination to conform the vector group ( refer drawings ).

For more information

http://electricalnotes.wordpress.com/2012/05/23/vector-group-of-transformer/

Jubail -2 Sub-3 Energization

This station successfully energization without any small problem 25 & 26 on June 2012.
Really i got good confident on the work and well experience.
We got good team during commissioning and its lead us to success.
Note:-
The total bay portion of Line and transformer feeders are protected by there own relays
The spare feeders are protected by only busbar protection and it will cover up to CB line side CT.
During first time charging we need to protect the spare feeders up to Air bushing.
Generally two methods are following to protect the spare feeders.
1. If short the CB line side Busbar CT then busbar protection will cover upto Air bushing.
2. Change the other end (Existing end ) protection relay Zone 2,3 time settings to zero and it will cover upto New station spare feeder air bushing.

BUSBAR & BF PROTECTION TESTING

The testings in Busbar protection low impedance relays DOWNLOAD

The testings in Beaker Failure protection relays DOWNLOAD

BREAKERS RATINGS

230kV GIS BREAKER

1. Nominal system voltage = 230kV

2. Maximum system voltage = 242 kV

3. Maximum system voltage = 253 kV for 30 mim

4. Rated continuous current = 3150 A for Bus couppler

5.Rated continuous current = 2500 A for I/C & O/G feeders

6. Rated system short circuit interrupting current of the CB = 63kA

34.5kV BREAKER

1. Nominal system voltage = 34.5 kV

2. Maximum system voltage = 36.2 kV

3. Maximum system voltage = 38 kV for 30 mim

4. Rated continuous current = 2500 A for Bus Tie & Incomer

5.Rated continuous current = 1250 A for O/G feeders

6. Rated system short circuit interrupting current of the CB = 31.5 kA

SITE TESTING OF LINE DIFFERENTIAL RELAY

1.Analog input Measurements 

Inject the rated secondary CT & VT values and check the primary and secondary values and power values  in the relay.

2.Input and Out put card checks

Give the digital input commands check the response in the relay
Check the all output contacts of the relay

3.  LED status

Check the all LED status as per configuration

4. Secondary injection tests 

4.1 . Check all settings are applied for function
4.2.  Zone reach measurements
4.3. Zone timings
4.4. Phase selection function
4.5. Power swing
4.7. Switch On to faults
4.8. Scheme communication Logic
a. Permissive under reach
b. Permissive over reach
c. Blocking scheme
d. Direct inter trip scheme
e. DIST WEI ECHO
g. Current reversal Logic

5. Over current & Earth fault function 

6.DEF COMMUNICATION SCHEME

6.1. Permissive over reach scheme
6.2. Blocking scheme
6.3. DEF WEI ECHO
6.4. DIST WEI TRIP
6.5. Current reversal logic

7. Fault locator function

8. Differential Function

8.1. Communication failure checks
8.2. Pickup/ Drop off Id min
8.3. Time measurement for Id min &  Id min High
8.4. Alarm checks
8.5. Bias Characteristic Testing
8.6. Harmonics Restraint Checks

9. Auto Re-close Checks  

HV TEST FOR 230 KV GIS 

High Voltage level of injection is below 

Highest voltage level 380 kv is calculated from 80% of factory Hi Voltage injection Level 

304 kV is 80% of 380kv 

170 kv is 1.2X Un/ rot 3

132 kv is  Un/ rot 3 

PD measurement curve at 132kv & 170 voltage level 

THE STORY OF TRANSFORMER STABILITY

The transformer stability test used to conform the CT connections for Differential and REF protection. This test also used as a primary injection test for HV & LV side BCT's.

Example   Transformer Data

Voltage Ratio :             230/ 34.5/13.8 kV

MVA Rating :              100/133 MVA                                                                                                                  

Vector Group:             YNyn0(d5)

% Impedence :             12.24 @ 100 MVA BASE.

CALCULATION TO FIND THE PRIMARY AND SECONDARY CURRENT


PRIMARY SIDE : -

HV side full load current  =  Power /  √3 x Voltage = 100×10^3            =  100000000 = 251 Amps

                                                                                √3 × 230× 1000        399050

HV side Impedence    =   HV side Voltage              =   230× 1000  = 916 Ohms
                                          HV side full load current       251

HV side Impedence/ Phase =  HV side Impedence X % Impedence = 916 X 12.24  =  112 Ohms
                                                                                                                          100

HV side current / Phase    =  Applied Voltage / HV side Impedence/ Phase

                                        = 380 / 112 =  3.38 Amps


 LV side SIDE : -

Transformer Ratio = 230/ 34.5 = 6.66

LV side current = 6.6666* 3.38 = 22.53 Amps
                                                                                                             

TEST -1 Three Phase Balanced Injection

Test connection: Three phase short circuit outside protected zone on 34.5kV side. Three phase voltage injected on HV side = 380V

The balanced current can measured in the relay as follows 

                                                            

Relay - Primary Current (A)
HV	LV
3.326<0°	22.201<181.141°
3.205<241.5°	22.935<59.28°
3.288<121.5°	23.573<299.34°

Note : -


1. LV current 180 dec phase sifted from HV current.

2. No return current in Neutral 

TEST -2 Three Phase Unbalanced condition

To make this condition open any one of CT FT fingers and short by insert the FT jack. 

The Unbalanced current can measured in the relay as follows 

Relay- Primary Current (A)
HV	LV
0.00	21.985
0.00	22.226
0.00	22.265

TEST -3 Three Phase REF balanced condition

Test connection: Three phase short circuit outside protected zone on LV side. Single Phase Injection was done one by one  on the primary to check the stability and sensitivity.

1. No return current in Neutral 

2. During single phase injection voltage will developed  in the one phase Tertiary winding. If winding is connected with ground  a circulating current will flow through ground during stability.

TEST -4 Three Phase REF Unbalanced condition

 Polarity of the neutral BCT was reversed at the protection panel to simulate the unstable condition.

TEST -5 BCT primary injection

PRIMARY INJECTION HV & LV  WINDING :

Test connection: Three phase short circuit outside protected zone on LV side. Three phase voltage injected on HV side = 380V 

PRIMARY INJECTION TERTIARY  WINDING :

Test connection : The shorting and grounding bar was removed from the Delta winding and current was injected with Severker Test Kit and secondary currents were measured in the Transformer LCP panel.

                                     

TIMINGS & TIME SETTINGS

ABTS LOGIC

ABTS logic truth table for two transformer arrangement
Closing Logic











Tripping Logic
















Note:-
ABTS Block
1. Transformer Timed Over current Trip ( This trip non directional some times happend for low side fault also. So its block the ABTS)
2. Lower side BB protection Operated in Unhealthy Bus
3. Over current operate in Lower side Bus Tie
4. FFR operated in Healthy Bus
ABTS Initiation
1. Low side Incomer Manual opens
2. High side instantaneous transformer protection trip
3. Transformer Mechanical protection trip
4. High side BB protection trip
General Conditions
1. All low side breakers should be in service conditions
2. One Bus should be unhealthy condition
3. One Bus should be healthy condition
4. ABTS in service position

INTER LOCK FOR TRANSFORMER HI & LOW SIDE INCOMER

LOW SIDE CB RACK IN BLOCK
1.Low Side Inter locks
1.1 CB Aux Contact - CB should be open Condition
1.2 Bus Earth & feeder line side switch should be Open condition
2 Hi Side Inter Locks
2.1 Transformer side ES should be open condition

LOW SIDE CB CLOSE BLOCK COIL
1. Transformer Protection all lockouts should be normal position
2. 34.5kv Bus earth switch should be open

LOW SIDE CB CLOSEING COIL
1.Hi side breaker should be in close condition
2.Transformer side Disconnector switch should be in closed condition.

LOW SIDE CB TRIPPING

1.All protection LOR

2.High side CB service condition and open

LOW SIDE INCOMER EARTH SWITCH
1.Hi side breaker should be in open condition
2.Transformer side Disconnector switch should be in open condition.

HI SIDE TRANSFORMER FEEDER DS INTERLOCK
1.Low side CB open
2.Low side ES open
HI SIDE TRANSFORMER FEEDER ES INTERLOCK
1. Low side CB open
2. Low side CB in test position