2015 Third Quarter Technical Seminar, 8th Annual General Assembly and 2016 BOD

ElectionAl Rayhaan Hotel, Al Ghurair Centre

Dubai, U.A.E.September 18, 2015

Testing and Commissioning of High Impedance Busbar Differential Protection System for

HV/MV Primary Substations

By: Mark Anthony M. Galo+971 56 2591 528mark_galo_2603@yahoo.com.ph

Objectives of Electrical Power System

• The purpose of an electrical power system is

to generate and supply electrical energy to

consumers reliably, safely and economically.

• Keeping the power system operation

continuously without major breakdown to

ensure the maximum return on the large

investment in the equipment, which goes to

make up the power system and to provide

maximum customer satisfaction for reliable

service.

Power System Reliability

• The reliability of the substation depends on the reliability of associated equipment such as busbars, circuit breakers, transformers, isolators and controlling devices.

• The above substation equipment should be protected from electrical short circuits to keep the operation continuously without major breakdowns.

Power System Protection

• Protection cannot prevent faults but can minimize the consequences.

• Main Objective of Power System Protection

To safeguard the entire system to maintain the continuity of supply.

To minimize the damage of the associated substation equipment and repair cost.

To ensure safety of personnel.

Basic Requirements of Power System Protection

Selective-To detect and isolate the faulty item only.

Stable-To keep the healthy sections operational.

Fast-To initiate and operate promptly to prevent the major damage and breakdown of equipment and thus ensuring the safety to personnel.

Power System Major Equipment

Generators

Power Transformers

Reactors

Capacitors

Switchgears/Busbars

Cables

Overhead Line Conductors

Auxiliary Transformers

Motors

UPS (Rectifiers, Inverters and Battery Banks)

X

Switchgear

Switchgear is the combination of electrical disconnect switches, fuses or circuit breakers used to control, protect and isolate electrical equipment. Switchgear is used both to de-energize equipment to allow work to be done and to clear fuses downstream.

Metal-enclosed Switchgear

Metal-enclosed switchgear assemblies have metallic enclosures. These switchgears generally have three (3) high voltage compartments, namely circuit breaker compartment, busbar compartment, current transformer and cable compartment, which are separated by partitions.

Switchgears which have compartments with metal partitions that are intended to be earthed are called metal-clad switchgears.

Gas Insulated Switchgear

• GIS Equipment features vacuum as the interrupting medium and SF6 gas as the main insulation.

Air Insulated Switchgear

Unit CompartmentsA-Circuit Breaker compartmentB-Busbar CompartmentC-Cable CompartmentD-Low Voltage CompartmentE-Compact Gas Duct Channel

Current Transformers

Voltage Transformers

Courtesy of ABB

Air Insulated Switchgear

Courtesy of SIEMENS

Overview of Gas Insulated SwitchgearEquivalent One-Line

DiagramBusbar Compartment

Circuit Breaker Compartment

Courtesy of SIEMENS

SF6 Gas as an Insulating Medium

SF6 gas has excellent insulating qualities which is ideally suited for high voltage applications.

SF6 gas extinguishes arcs rapidly than other insulating medium.

SF6 gas is costly and is not discharged into the atmosphere.

Parts of the circuit breaker and busbar compartment of the GIS must be sealed to prevent leakage of SF6 gas.

During the interruption, ionized fluorine may combine with metal vapors and not recombine into SF6 gas. The by-products of arc interruption from SF6 gas are corrosive when exposed to the water which will damage the components of the circuit breaker.

GIS Busbar Compartment

Courtesy of SIEMENS

Arrangement of Gas Compartment

Courtesy of SIEMENS

GIS Current Transformer Mounting and Installation

Courtesy of SIEMENS

Double Busbar GIS

1. Three position disconnector2. Bay Control Protection Unit3. Gas density sensor and filling valve4. Vacuum circuit-breaker5. Cable socket6. Inner cone cable connector7. Plug-in voltage transformer-feeder8. Pressure relief-disk9. Current transformer10. Pressure relief duct11. Measuring sockets for capacitive voltage indicator system12. Busbars13. Plug-in voltage transformer-busbar

Courtesy of ABB

GIS Plug-in Components

Courtesy of ABB

Busbar Protection

• A bus is a critical element of a power system since it is the point of convergence of many circuits, transmission, distribution, generation or loads.

IncomingFeederOutgoing

Feeders

IncomingFeeder Outgoing

Feeders

Bus-Coupler(Tie-Breaker)

Busbars are frequently left without protection due to the following reasons:

The busbars and switchgear have a high degree of reliability, to the point of being regarded as basically safe.

The system depends on back-up protection such as overcurrent protection.

Expensive busbar protection system due to the required number of current transformers.

Maloperation of the busbar protection system.

However, busbar faults do occur.

Typical Single Line Diagram of 20-MVA, 33/11 kV Primary Substation (One Section)TRIP Logic of Busbar Protection

11-kV, 31.5 kA (In = 1,250 A)

33-kV, 31.5 kA (In = 1,250 A)

Q5

Q1

Q0

CT2600-300/1 ACL 5P20CL XCL X (1,250/1 A)

Q5

∆T101

20-MVA33/11 kV%Z=11.5

CT61,250/1 ACL 5P20

Q0

Q1

Q5

Q1

Q0

CT1 600-300/1 ACL 5P20CL XCL X (1,250/1) A

Q5

Q1

Q0

CT1 600-300/1 ACL 5P20CL XCL X (1,250/1) A

Q5

Q0

Q1

CT71,250/1 ACL 5P20

Q5

Q0

Q1

CT8600-400/1 ACL 5P20

Q5

Q0

Q1

CT91,250/1 ACL 5P20

Q5

Q1Q0

Q1

CT5100/1 ACL 5P20

To Other Section

Q0

Q11

CT31,250/1 ACL X

Q16A

Q12

CT4 1,250/1 ACL 5P20CL X

Q15

To Other Section

X X X

X

X

Differential Protection• Differential Protection applies the

Kirchoff’s Current Law (KCL) and compares the currents entering and leaving the protected equipment of the substation such as the following:

Power Transformers

Cables

Busbars

Capacitor Banks

Stability of High Impedance Busbar Differential Protection

Q5 Q1

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q1

Q5

Q1 Q1

Q5 Q1

Q1

Q5

Q1RST

M87B

33-kV, 31.5-kAH01 H02 H03 H04

In=1,250 A

Incomer Feeder

Outgoing Feeder

Outgoing Feeder

Outgoing Feeder

CT Ratio1,250/1 A

P1

P2

S1

S2

P1

P2

S1

S2

P1

P2

S1

S2

P1

P2

S1

S2

F

6,515 A

6,515 A 6,515 A

6,515 A

5.2 A5.2 A

X

Let:I1=H01 CT Secondary CurrentI2=H04 CT Secondary CurrentIr=Current flowing in the relay

Applying KCL at Node X:-I1+I2-Ir=0Ir=I2-I1Ir=5.2 A-5.2 AIr=0 A

Ir

M

Stability of High Impedance Busbar Differential ProtectionAir-Insulated Switchgear

6,515 A

M

P1

P2

S1

S2

M

P1

P2

S1

S2

M

P1

P2

S1

S2

87B

Ia = 5.21 A

Ib =5.21 A

Outgoing Feeder-1

11-kV, 25 KA In = 1,250 A

M

P1

P2

S1

S2

S2

S1

11-kV Incoming Feeder

x

F

Fault A

6,515 A

IrIb = 5.21 A

Outgoing Feeder-2

Outgoing Feeder-3

Outgoing Feeder-4

P2

P1

6,515 A

Ib =5.21 A

Sensitivity of High Impedance Busbar Differential Protection

Q5 Q1

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q1

Q5

Q1 Q1

Q5 Q1

Q1

Q5

Q1RST

M87B

33-kV, 31.5-kAH01 H02 H03 H04

In=1,250 A

Incomer Feeder

Outgoing Feeder

Outgoing Feeder

Outgoing Feeder

CT Ratio1,250/1 A

P1

P2

S1

S2

P1

P2

S1

S2

P1

P2

S1

S2

P1

P2

S1

S2

F

6,515 A

6,515 A

6,515 A

5.2 A

X

Let:I1=H01 CT Secondary CurrentIr=Current flowing in the relayApplying KCL at Node X:-I1-Ir=0Ir=I1Ir=5.2 A

Ir

M

Sensitivity of High Impedance Busbar Differential ProtectionAir-Insulated Switchgear

6,515 A

M

P1

P2

S1

S2

M

P1

P2

S1

S2

M

P1

P2

S1

S2

87B

Ia = 5.212 A

Ib =0 A

Outgoing Feeder-1

11-kV, 25 KA In = 1,250 A

M

P1

P2

S1

S2

S2

S1

11-kV Incoming Feeder

x

F

Fault A

6,515 A

IrIb = 0A

Outgoing Feeder-2

Outgoing Feeder-3

Outgoing Feeder-4

P2

P1

Busbar Protection CT

Busbar Protection CT

Substation B (Open-Point)

Q1 Q5

Q0

Q1

Q1Q5

Fail-Safe Logic of Busbar Protection CTs

F Three-Phase Fault

To prevent false tripping of the busbar protection due to remote energizing of locally earthed outgoing feeders, a fail-safe logic shall be provided for those circuits to short and isolate automatically the related main CT when the feeder is being earthed.

Q1 Q5

Q0

Substation AOutgoing Cable Feeder

Earth-switch will be closed prior to the execution of the cable works.

Busbar Protection will operate and trip all the respective circuit breakers of Substation A

Zone of Protection (One Section)

Q5

Q1

Q0

T1 600-300/1 ACL 5P20CL X

Q5

Q1

Q0

T1600-300/1 ACL 5P20CL X

Q1

Q0

T1600-300/1 ACL 5P20CL XCL X

Q5

∆TR1

20-MVA33/11 kV%Z=11.5

T11,250/1 ACL 5P20

Q0

Q1

Q5

Q1

Q0

T1 600-300/1 ACL 5P20CL X

33-kV, 31.5 kA In = 1,250 AFF1

F F2

F

F3F

F4

• For F1, F2, F3 and F4, the incomer feeder (H01) busbar protection CTs will sense the short circuit current which will not be sensed and measured by the outgoing feeders (H02, H03 and H04) busbar protection CTs since the fault is within the zone of protection. Accordingly, the Numerical High Impedance Relay will initiate and trip all the 33-kV circuit breakers to protect and isolate the faulted busbars.

FF5

H01 H02 H03 H04

FF6

• For F5 and F6, the incomer feeder (H01) and the outgoing transformer feeder (H02) busbar protection CTs will sense and measure the same magnitude of short circuit current but 180o out of phase due to the CT flow of current with respect to the CT polarities. The vector summation of the CT secondary currents of H01 and H02 is equal to zero. Hence, the Numerical High Impedance Relay will not initiate and operate since the fault is outside the zone of its protection.

P2

P1

P2

P1

P2

P1

P2

P1

Q5

Q11

T1

Q16

Q12

T1

Q15

Q0

Zone of Protection (Two Sections)

Q5

Q1

Q0

T1

Q5

Q1

Q0

T1

Q1

Q0

T1

Q5

∆TR2

20-MVA33/11 kV%Z=11.5

T11,250/1 ACL 5P20

Q0

Q1

Q5Q5

Q1

Q0

T1

Q1

Q0

T1

Q5

∆TR1

20-MVA33/11 kV%Z=11.5

T11,250/1 ACL 5P20

Q0

Q1

Q5 Q5

Q1

Q0

T1

87BB1 87BB2

H01 H02 H03 H04A H04B H05 H06 H07

UNPROTECTED

Q11

T1

Q16

Q12

T1

Q15

Q0

Zone of Protection (Two Sections)

Q5

Q1

Q0

T1

Q5

Q1

Q0

T1

Q1

Q0

T1

Q5

∆TR2

20-MVA33/11 kV%Z=11.5

T11,250/1 ACL 5P20

Q0

Q1

Q5Q5

Q1

Q0

T1

Q1

Q0

T1

Q5

∆TR1

20-MVA33/11 kV%Z=11.5

T11,250/1 ACL 5P20

Q0

Q1

Q5 Q5

Q1

Q0

T1

87BB2

H01 H02 H03 H04A H04B H05 H06 H07

87BB1

• Two sets of CT's connected in an over lapping arrangement and mounted on either side of the bus-coupler which shall be connected.

Isolator Q12

Isolator Q11

CT Zone-2

Circuit Breaker

Q0

CT Zone-1

184

Reserve 186

Main1

186

184

Main2

104

106

284

206

204

286

204

206 286

284

180

106

104

105

Reserve Bus

Main 1 Bus Main 2 Bus

Courtesy of SIEMENS

High-Impedance Double Busbar Differential Protection

IR

IY

1

2

3

4

5

67

89

1

2

3

4

5

6

Numerical High

Impedance Relay

-87B

Test Block

RST

M

Q1 Q5

Q0

Q1

Q1

Q5K1

S5

X1-3

33-kV, 31.5 kA

In = 1,250 A

CT Terminal Blocks

X1-1

X1-2

X1-4

The CT Secondary Circuit

Note: Terminal Block X1 is the paralleling point/node of all busbar protection CTs.

Stabilizing Resistor, RST

When a CT becomes total saturated, its secondary winding can be considered as a resistance rather than a current source.

The value of this resistance is equal to the CT secondary resistance, RCT, and will be considerably larger than the resistance of the Relay analogue inputs. This means that most of the unbalanced currents from the other CTs will flow through the Relay and these may be of sufficient magnitude to operate the protection.

RCT2 Rr

RL

RL

RCT1 Vk’

IS1

RCTn

ISn

RCT > Rrelay

Rr

RL

RL

RCT1 Vk’

IS1

RCT2

IS2

RCTn

ISn

(b) Equivalent CT Secondary Circuit (Paralleled CTs)

(c) Equivalent CT Secondary Circuit of Paralleled CTs showing saturated current transformer, CT2.

S5

H01Q0

+DC

TC2TC1

-DC

87B

K86.1 H01

87B 87B

TC2TC1

-DC

K86.2 H02

H02Q0

TC2TC1

-DC

K86.n H…n

H…nQ0

Trip Circuit of Busbar Differential Protection

Test Block

Trip Links Trip Links Trip Links

Effect of Saturated CTs on BBP Performance

The following phase (Y) load currents were downloaded from the 13.2-kV feeder revenue meters of Substation A as follows:

Feeder No.

Primary Load

Current (A)

CT RatioCT

Secondary Current (A)

F1 115.7

1250

0.09256F2 128.1 0.10248F3 196.8 0.15744F4 86.5 0.0692

Incomer 527.1 0.42168

S1

S2

S1

S2

S1

S2

S1

S2

S1

S2

RELAY

Inco

mer

F1

F2

F3

F4

0.093 A 0.102 A 0.157 A 0.069 A0.4212 A Ir = 0 A

Equivalent CT Secondary Circuit of Busbar Differential Protection under normal condition.

S1

S2

S1

S2

S1

S2

S1

S2

S1

S2

RELAY

Inco

mer

F1

F2

F3

F4

0.093 A 0.102 A 0.157 A0.421 A Ir = 0.069 A

RCT>Rrelay

CT Secondary current distribution with one saturated CT.

Q0

Q1

T11,250/1 Class X

115.7 A

Q0

Q1

T11,250/1 Class X

128.1 A

Q0

Q1

T11,250/1 Class X

196.8 A

Q0

Q1

T11,250/1 Class X

86.5 A

Q0

Q1

T11,250/1 Class X

527.1 A

11-kV, 31.5 kA In=1,250 A

Relay as a High-Impedance Path

• The solution is to load the Relay circuit by

adding a series resistor such that most of the

unbalance current due to the CT becoming

saturated will instead flow through the

saturated CT secondary.

• Since this resistor will make the protection

stable for all through faults, it is termed the

Stabilizing Resistor, Rs.

• Similarly, it is this additional resistance which

makes the Relay a “High Impedance” path.

Stability Limit Voltage• The protection relay must remain stable under maximum

through fault condition, when a voltage is developed across the protection due to the fault current.

RST

Rr

RL

RL

RCT

VSIS

VS

By KVL:

VS = ISCMAX(RCT+2RL)

Where:

VS stability limit voltage

ISCMAX Maximum through fault current referred to secondary

RCT CT winding resistance

RL Wire lead resistances

Equivalent CT secondary circuit showing the CT winding resistance, lead wire resistances, stabilizing resistor and relay current coil resistance.

Quiz:

Calculate the stability limit voltage of the 33-kV high-impedance busbar protection using the following information:

CT Nameplate Data:

Class TPS (X)

CT Ratio=1,600/1 A

RCT = 4.2 Ω

Size of CT Cable: 4.0 mm2

Copper Resistance at 75oC = 0.00541 Ω/m

Distance from switchgear to BBP panel=50 m

BBP Relay Current Coil Resistance=0.2 Ω

Maximum through fault current=31.5 kA

Sample:

Calculate the stability limit voltage of the 33-kV high-impedance busbar protection using the following information:

CT Nameplate Data:

Class TPS (X)

CT Ratio=1,600/1 A

RCT = 4.2 Ω at 75oC

Size of CT Cable: 4.0 mm2

Copper Resistance at 75oC = 0.00541 Ω/m

Distance from switchgear to BBP panel=50 m

BBP Relay Current Coil Resistance=0.2 Ω

Maximum through fault current=31.5 kA

RST

Rr

RL

RL

RCT

VSIS

VS

2RL = 2[0.00541(50)]=0.541 Ω

ISCMAX=19.6875 A

By KVL:

VS = ISCMAX(RCT+2RL)

VS = 19.6875(4.2+0.541)

VS = 93.33 Volts

Calculation of Stability Limit Voltage

Required Current Transformers forHigh Impedance Busbar Differential Protection

• The knee point voltage of each CT should be at least 2 x Vs.

• Vk=2VS

• The knee-point voltage, Vk is defined as the point at which a further increase of 10% of secondary voltage would require an increment of exciting current of 50%. This is also known as the CT saturation point.

• Class PX is the definition in IEC 60044-1 for the quasi-transient current transformers formerly covered by Class X of BS 3938, which is usually used for the main protection relays. RS

T

Rr

RL

RL

RCT

VSIS

VS

1.0 A 10.0 A 100.0 A

1.0 V

10.0 V

100.0 V

1000.0 V

Excitation curve data

L 1L 2L 3(Ikn,Vkn)/L1(Ikn,Vkn)/L2(Ikn,Vkn)/L3(Ikn2,Vkn2)/L1 VS = 93.33 Volts

Vk =2 VS

Vk =2 (93.33 Volts) = 186.66 Volts

Sample Nameplate of a Current Transformer

CT Ratio

Class

Knee-point Voltage

Exciting Current

Winding Resistance

Busbar Protection Relay Setting Calculation

• To achieve correct sensitivity to in-zone faults, the protection scheme must typically operate for a primary current of 10-30% of the minimum primary fault current.

• IP=N(IS+nIe+Im)

Where:

N=CT Ratio

IS=Relay Setting Current

n=Number of Current Transformers connected in parallel

Ie=Exciting Current of the CTs at VS

Im=Metrosil Current at VS

As per ADWEA standard specification, the BBP relay should not pick-up at 125% of the transformer full-load current. Furthermore, the pick-up current can be set based on the nominal current of the busbar being protected.

•

Stabilizing Resistor Setting Calculation

By KVL:VS = IS(RST+Rr)To calculate the value of the stabilizing resistor, RST:

• RST

Rr

RL

RL

RCT

Vk’IS1

Vk’

Metrosils• Metrosils are used to limit the peak voltage developed by the

current transformers under internal fault conditions, to a value below the insulation level of the current transformers, relay and interconnecting leads, which are normally able to withstand 3000V peak.

Calculation of Peak Voltage Across the Relay Circuit

• The following formulae should be used to estimate the peak transient voltage that could be produced for an internal fault.

• The peak voltage produced during an internal fault will be a function of the current transformer knee-point voltage and the prospective voltage that would be produced for an internal fault if current transformer saturation did not occur.

•

Current Flowing in the Metrosil

RST

Rr

RL

RL

RCT

Vk’IS1

Vk’ M

Courtesy of

Im

Ie

Iµ

Testing and Commissioning of High Impedance Busbar Differential Protection

Part of Switchgear Testing

• Contact Resistance Measurement

• Routine Test on Current Transformers

-Insulation Resistance Measurement

-Polarity

-Class Verification

-CT Saturation

-Winding Resistance

-Turns Ratio Test

• CT Secondary Circuit Burden Measurement

• Secondary current injection in CT secondary circuit up to the final scheme.

• Insulation Resistance Measurement on Main Busbars.

• HV Test on Main Busbars.

• Insulation Resistance Measurement on Main Busbars.

• Primary current injection test on each feeder of the switchgear.

Stability Test on High Impedance Busbar Differential Protection

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q1

Q5 Q1

Q1

Q5 Q1

Q1

Q5 Q1

RST

M87B

33-kV, 31.5-kAH01 H02 H03 H04

In=1,250 A

Q5 Q1P1

P2

P1

P2

S1

S2

S1

S2

Sensitivity Test on High Impedance Busbar Differential Protection

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q5

Q0

Q1 Q1

Q5 Q1

Q1

Q5 Q1

Q1

Q5 Q1

RST

M87B

33-kV, 31.5-kAH01 H02 H03 H04

In=1,250 A

Q5 Q1P1

P2

P1

P2

S1

S2

S1

S2

P1

P2

S1S2

S2

S1

Reversing the connection of CT secondary terminals.

Final Trip Test on High Impedance Busbar Differential Protection(Secondary Current Injection Method)

Final Trip Test on High Impedance Busbar Differential Protection(Secondary Voltage Injection Method)