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 [email protected]
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
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.
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
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.
•
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)