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IIEE-AUH 1st Technical Seminar December 16, 2016

Testing and Commissioning of Main and Back-up

Protections of a Power Transformer

By: Engr. Mark Anthony M. Galo, PEE

+971 56 2591 528

[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

TRANSFORMER PROTECTION:

• The considerations for transformer protection vary with the application and importance of the transformer.

• To reduce the effects of thermal stress and electrodynamic forces, the overall protection should minimize the time that a fault is present within a transformer.

Types of Transformer Faults:

• Winding and Terminal Faults

• Core Faults

• Tank and Transformer Accessory Faults

• On-Load Tap Changer Faults

• Abnormal Operating Conditions

• Sustained External Faults

Transformer Application: Transformers converts energy from one voltage level to another voltage level. Step-up power transformers in generating stations are used to reduced the losses in EHV and HV transmission lines since current is indirectly proportional to the voltage. Step-down power transformers are used in distribution substation to reduce the voltage level from EHV/HV transmission lines to MV distribution network to match the ratings distribution equipment. Step-down distribution transformers are used to reduce the MV level to LV level for home consumption.

Protection Single Line Diagram of Power Transformer

TR1

20-MVA

33/11.5 kV

%Z=11.5

1,250/1 A

0.2FS10

To Busbar Protection Relay

Q1

Q0

33-kV Bus

600-400/1 A

CL X

1,200/1 A

Class X

1

600-400/1 A

5P20

3

600-400/1 A

0.2FS10

6

1,200/1 A

0.2FS5

7 1,200/1 A

5P20

4

1-Differential Protection

2-Restricted Earth-Fault Protection

3-Primary Overcurrent Protection

4-Secondary Overcurrent Protection

5-Stand-By Earth Fault Protection

6-Primary Metering Device

7-Seconday Metering Device

5

1,200/1 A

5P20

1,250/1 A

CL X

2

1,200/1 A

Class X

Q0

Q1

11-kV Bus

Note:

Metering devices are not part of power transformer protection

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

Generators

Reactors

Overhead Line

Transformer Differential Protection • Differential protection are the principal form of fault

protection for transformers rated at 10-MVA and above.[1]

• Transformer Differential Protection is generally recommended for transformers rated 7.5 MVA and above and provides fast protection against faults within the protected zone.[2]

• Chapter 11 of the IEEE Buff Book [S8] has recommended differential protection for transformers rated 5.0 MVA and above.[3]

• IEEE C37.91 [S16] has recommended this for transformers rated 10.0 MVA and above.[4]

• The concept of differential protection uses current circulation scheme or current balance scheme.

• The first task of the power transformer differential protection is to determine whether a fault is within the protected zone, or outside of the protected zone.

• The protected zone is limited by the position of current transformers and in principle can include more objects than just a transformer. If the fault is found to be internal, the faulty power transformer must be quickly disconnected from the system.

Q0

Q1

TR1

20-MVA

33/11 kV

%Z=11.5

Q1

Q0

33-kV Bus

1

[1] ABB Protection Book, Walter Elmore, Page 145 of 367 [2] Industrial Power System-Shoaib Khan Page 260 of 455 [3] IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems, ANSI/IEEE Std 242-1986, Page 430 of 588

Primary and Secondary Currents of a Power Transformers:

Even in a healthy power transformer, the currents are generally not equal when they flow through it. This is due to the following:

• Ratio of the number of turns of the windings.

• Connection group of the protected transformer.

• Phase shift across the transformer windings.

• Varying currents due to on-load tap changer (OLTC)

• The possibility of zero sequence current destabilizing the differential for an external earth fault.

Therefore the differential protection must first correlate all currents to each other before any calculation can be performed.

PHASE SHIFT ACROSS THE TRANSFORMER WINDINGS:

As an example, we have a Delta-Wye Transformer:

In delta connected system, 𝑽𝑳 = 𝑽 while 𝑰𝑳 = 𝟑𝑰.

In wye connected system, 𝑽𝑳 = 𝟑𝑽 while 𝑰𝑳 = 𝑰.

VR

VY

VB

Vr

Vy

Vb

IR–IY

IY–IB

IB-IR

By KCL at Blue Nodes: IR-IRY +IYB=0 IR=IRY –IYB

IY-IYB +IBR=0 IY=IYB –IBR

IB-IBR +IRY=0 IB=IBR –IRY

IR

IY

IB

VRN

VYN

VBN VNY

120o

120o

30o

30o

60o 60o

120o

VRY

VRY

VRN

VNY

By Applying Cosine Law: VRY

2 = VRN2 + VNY

2 – 2 (VRN)(VNY)(Cos 120o) Assuming a balance phase to neutral voltages, VRN = VNY

VRY2 = VRN

2 + VRN2 – 2 (VRN)(VRN)(Cos 120o)

VRY2 = 2VRN

2 – 2VRN2(Cos 120o)

VRY2 = 2VRN

2 – 2VRN2(-0.5)= 3VRN

2

𝑽𝑹𝒀 = 𝟑𝑽𝑹𝑵

Using parallelogram method:

k(Ir)

k(Iy-Ib)

k(Ib-Ir)

60o

1. Ratio of the number of turns of the windings, k 2. Connection group of the protected transformer. 3. Phase shift across the transformer windings.

CURRENT VECTOR DIAGRAM OF DELTA-WYE TRANSFORMER

VR

VY

VB

Vr

Vy

Vb

IR–IY

IY–IB

IB-IR

k(Ir-Iy)

k(Iy-Ib)

k(Ib-Ir)

IR

IY

IB

IR

IB IY

120O

30O

120O

120O

IR

IB IY

-IY -IB

-IR

IR-IY

IY-IB

IB-IR

30O

30O

120O

By Applying Cosine Law: (IR-IY)

2 = IR2 + IY

2 – 2 (IR)(IY)(Cos 120o) Assuming a balance load, IR = IY

(IR-IY)2 = IR

2 + IR2 – 2 (IR)(IR)(Cos 120o)

(IR-IY)2 = 2IR

2 – 2IR2(Cos 120o)

(IR-IY)2 = IR

2 – IR2(-0.5)= 3 IR

2

𝑰𝑹 − 𝑰𝒀 = 𝟑𝑰𝑹

IR-IY

IB-IR IY-IB

PRIMARY AND SECONDARY CURRENT VECTOR DIAGRAM OF DELTA-WYE TRANSFORMER

Ir

Ib

Iy

30O

The vector group is the clock-face hour position of the LV A-phase voltage, with respect to the A-phase HV voltage at 12-o’clock (zero) reference. For a Dyn11 transformer, the hour position of LV phase-r current with respect to HV phase-R current is 11-o’clock. In terms of electrical degrees, it will be computed as: 30o x 11 = 330o

330O

Differential Protection CT Secondary Connection

In traditional transformer differential schemes, the phase and ratio correction were done by the following:

• Application of external interposing current transformers, as a secondary replica of the main transformer winding arrangements.

• The CT connection to the differential relay needs to be connected in Star for the transformer Delta winding and in Delta for the transformer Wye winding.

VR

VY

VB

Vr

Vy

Vb

IR–IY

IY–IB

IB-IR

Ir

Iy

Ib

IR

IY

IB

Ir’

Ir’

Ir’

Iy’

Iy’ Ib’

x

y z

Ir’-Iy’ Iy’-Ib’ Ib’-Ir’

IR’–IY’

IY’–IB’

IB’–IR’

Flow of Current in CTs

P1

S1

P2

S2

To Differential Relay

P2 P1

S2 S1


LoadFlow

Differential Protection CT Secondary Connection

• In Microprocessor-based differential relays or IEDs (Intelligent Electronic Devices), it is common to use star-connected line CTs on all windings of the transformer since the vector shift and ratios can be corrected internally through software and embedded algorithms.

• Furthermore, and compensate for the winding phase shift in software.

VR

VY

VB

Vr

Vy

Vb

IR–IY

IY–IB

IB-IR

Ir

Iy

Ib

IR

IY

IB

IR’–IY’

IY’–IB’

IB’–IR’

P1

S1

P2

S2


P1

S1

P2

S2


Ir’

Iy’

Ib’

Electro-mechanical Relays Solid State Relays

PTD SE PTI, Page 67 of 86Copyright © Siemens AG 2007.

MiCOM P631, P632, P633, P634-P63x/UK GS/A54, (GS) 3-25

Micro-processor Based Numerical Relays Intelligent Electronic Devices

Sample Transformer Differential Relay Configuration Micom P632

Sample Transformer Differential Relay Measurement Functions Micom P632

References: 1. SIEMENS 7UT612 Manual, C53000–G1176–C148–1 Page 259 of 346 2. Schneider Electric MiCOM P64x (P642, P643 and P645) Technical Manual, P64x/EN TD/B63, Page (TD) 2-23 3. ABB RET670 Buyer's Guide, Pre-configured, 1MRK 504 080-BEN, Revision: G, Page 29 4. THYTRONIC NT10 BIASED DIFFERENTIAL FOR TWO WINDING TRANSFORMERS- Manual - 01 – 2015, Page 29 of 322

SIEMENS 7UT612 MiCOM P64x (P642, P643 and P645)

ABB RET670 THYTRONIC NT10

OPERATING TIME OF TRANSFORMER DIFFERENTIAL RELAYS

References: 1. Schweitzer Engineering Laboratories, Inc., SEL-487E Transformer Differential Relay, SEL-487E Data Sheet, Page 27 of 32 2. GE Grid Solutions, 845 TRANSFORMER PROTECTION SYSTEM – INSTRUCTION MANUAL, Product version: 1.6x, GE publication code: 1601-0651-A3 (GEK-119651B)

OPERATING TIME OF TRANSFORMER DIFFERENTIAL RELAYS

SEL-487E GE 845

Schematic Diagram of Transformer Differential Protection

Q0

Q1

TR1

20-MVA

33/11 kV

%Z=11.5

Q1

Q0

33-kV Bus

P1

P2

S1

S2

P2

P1

S2

S1

+110 VDC

Test Socket

Test Socket

IN/OUT Selector Switch

TRIP COIL 1 Q0 (33-kV)

-110 VDC

+110 VDC

Test Socket

Test Socket

IN/OUT Selector Switch

LOCK-OUT RELAY

-110 VDC

STABILITY OF TRANSFORMER DIFFERENTIAL PROTECTION FOR AN EXTERNAL FAULT (Fed from one side only with an electrical fault out of its zone of protection.)


F

ISC

Let: ISC=Fault Current I1=Secondary Current of CT1 I2=Secondary Current of CT2 Id=Differential Current in the relay

P1

P2

S1

S2

P2

P1

S2

S1

CT1

CT2

I1

I2 ≠ 0

Id I1

I1

I2 ≠ 0

I1

I2 ≠ 0

Q0

Q1

TR1

20-MVA

33/11 kV

%Z=11.5

Q1

Q0

33-kV Bus

Q0

Q1

IP

IS

M

By Applying KCL at Node M:

I1 – I2 – Id = 0

Id = I1 – I2 = 0

Therefore, the transformer differential protection relay will not send trip command to the associated circuit breakers.

STABILITY OF TRANSFORMER DIFFERENTIAL PROTECTION FOR AN EXTERNAL FAULT (Fed from one side only with an electrical fault out of its zone of protection.)


F

ISC=7.086 kA


P1

P2

S1

S2

P2

P1

S2

S1

400/1 A

1,200/1A

I1

I2=5.905 A

Id I1=6.173 A

I1=6.173 A

I2=5.905 A

I1=6.173 A

I2=5.905 A

Q0

Q1

TR1

20-MVA

33/11.5 kV

%Z=11.5

Q1

Q0

33-kV Bus

Q0

Q1

IP=2.469 kA

IS=7.086 kA

M

By Applying KCL at Node M:

I1 – I2 – Id = 0

Id = 6.173 A – 5.905 A = 0.268 A

Therefore, the transformer differential protection relay will not send trip command to the associated circuit breakers.

𝐼2 =7,086 𝐴

1,200= 5.905 𝐴

𝐼𝑃

𝐼𝑆=

𝑉𝑆

𝑉𝑃

𝐼𝑃 =𝑉𝑆

𝑉𝑃𝐼𝑆

𝐼𝑃 =11.5

337,086 𝐴

𝐼1 =2,469 𝐴

400= 6.173 𝐴

RESTRAINED CHARACTERISTIC OF TRANSFORMER DIFFERENTIAL RELAY (BIASETTING)

The intended application of restrained characteristic (bias setting) of differential relays is to ensure stability for external faults while allowing sensitive settings to pick up internal faults. The situation is slightly complicated if a tap changer is present.

Reference: 2015 Siemens Protection Devices Limited, 7SR242 Duobias Description Of Operation, Chapter 1 Page 24 of 60

Siemens 7SR242 Duobias y=mx+b Equation of a line. Taking the derivative with respect to x: 𝒅

𝒅𝒙𝒚 = 𝒎𝒙 + 𝒃

𝒅

𝒅𝒙

𝒅𝒚

𝒅𝒙= 𝒎 𝟏 + 𝟎

𝒅𝒚

𝒅𝒙= 𝒚′ = 𝒎

RESTRAINED CHARACTERISTIC OF TRANSFORMER DIFFERENTIAL RELAY (BIAS SETTING)

Reference: ABB RET670, 1MRK 504 113-UEN C, Technical reference manual, Page 120 of 1,097

ABB RET670

ABB RET670 Matrices for Differential Current Calculation

Reference: ABB RET670, 1MRK 504 113-UEN C, Technical reference manual, Page 114 of 1,097

RESTRAINED CHARACTERISTIC OF TRANSFORMER DIFFERENTIAL RELAY (BIAS SETTING)

References: 1. SIPROTEC, Differential Protection, 7UT612 Manual, C53000–G1176–C148–1, Page 39 of 346 2. P64x/EN AP/B63, Page (AP) 6-21

SIEMENS 7UT612 MiCOM P64x (P642, P643 & P645)

SENSITIVITY OF TRANSFORMER DIFFERENTIAL PROTECTION FOR AN INTERNAL FAULT (Fed from one side only)

Q0

Q1

TR1

20-MVA

33/11 kV

%Z=11.5

Q1

Q0

33-kV Bus


F

ISC


P1

P2

S1

S2

P2

P1

S2

S1

CT1

CT1

I1

I2 = 0

Id I1

Therefore, Id=I1 which is equal to the short

circuit current divided by the CT ratio.

The differential relay will operate and will send

trip command to its respective Circuit Breaker.

Restricted Earth-Fault Protection of Power Transformer

TR1

20-MVA

33/11.5 kV

%Z=11.5

Q1

Q0

33-kV Bus








1,250/1 A

CL X

2

1,200/1 A

Class X

Q0

Q1

11-kV Bus

Restricted Earth-Fault Protection (REF) • This is a unit protection scheme for one winding of the transformer. REF

protects the power transformer winding against the faults involving earth.

• [1] REF protection is the fastest and the most sensitive protection a power transformer winding can have and will detect faults such as:

Earth faults in the transformer winding when the network is earthed through an impedance.

Earth faults in the transformer winding in solidly earthed network when the point of the fault is close to the winding neutral(star) point.

Inter-turn faults

VR

VY

VB

Vr

Vy

Vb

IR–IY

IY–IB

IB-IR

Ir

Iy

Ib

IR

IY

IB

Load Flow

IN

Reference: [1] ABB RET670 Application Manual, 1MRK504089-UEN rev. B, Section 4, IED application, Page 135 of 684

VR

VY

VB

Vr

Vy

Vb

IR–IY

IY–IB

IB-IR

Ir

Iy

Ib

IR

IY

IB

Ir’

Iy’

Ib’

Iy’+Ib’

Ir’

Load Flow

IN

Id’

In’ In’

Restricted Earth-Fault Protection (REF)

• Three (3) protection CTs are required for phase R, Y and B.

• The neutral of the star-connected winding of the power transformer is equipped with bushing CT with a minimum of two (2) cores. One for REF protection and the other one is for Stand-By Earth-Fault protection.

• The secondary of all CTs are connected in parallel observing the correct polarities.

• Single earth-point shall be implemented for all protection CTs.

The vector sum of the line currents IR, IY and IB is equal to zero considering that the loads are balanced. In a wye connected system, the neutral current, IN is equal to the vector sum of IR, IY and IB.

INN= IRR+IYY+IBB

INN= 200o+20240o+20120o=0 INN= 20-30o+20210o+2090o=0 INN= 10-30o+20165o+2090o=22.82117.85o

M

At Node M: -Id’-In’+ Ir’+ Iy’+ Ib’=0

REF Protection Relay

P2

S2

P1 S1

P1

S1

P2

S2

RST

M

M-Metrosil RST-Stabilizing Resistor

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.

(c) Equivalent CT Secondary Circuit of Paralleled CTs showing

saturated current transformer, CT2.

Rr

RL

RL

RCTn

Vk’

ISn

RCTb

ISb

RCTy

ISy

(b) Equivalent CT Secondary Circuit (Paralleled CTs)

ISr

RCTr

RCTb

Rr

RL

RL

RCTn

Vk’

ISn

RCTy

ISy

RCT > Rrelay

ISr

RCTr

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.

Non-Linear Resistor, Metrosils (Non-linear Resistors)

• Occurrence of over voltage across the relay terminals which is the

product of the CT secondary current and the equivalent circuit resistance

of the series connection of the stabilizing resistor and current coil of the

relays.

• The relay can only withstand a maximum of 3kV peak under fault

conditions.

• For safety reasons, a check is required to see if this voltage is exceeded

– if it is, a non-linear resistor, known as a Metrosil, must be connected

across the relay and stabilizing resistor.

Non-linear Resistor Metrosils

Stabilizing Resistors

𝑉𝑃 = 2 2𝑉𝐾(𝑉𝑓 − 𝑉𝐾)

Where: VP=Peak Voltage VK=Knee-point Voltage Vf=Stability Voltage

𝑉𝑓 =𝐼𝑓

𝐶𝑇𝑅𝑅𝐶𝑇 + 2𝑅𝑙 + 𝑅𝑆𝑇

Stabilizing Resistor Setting Calculation

By KVL:

VS = IS(RST+Rr)

To calculate the value of the stabilizing resistor, RST:

RST =VS

IS− Rr

RST =VS

IS− Rr

RST

Rr

RL

RL

RCT

Vk’

IS1

Vk’

VR

VY

VB

Vr

Vy

Vb

IR–IY

IY–IB

IB-IR

Ir

Iy

Ib

IR

IY

IB

5.9 A

0

0

0+0

5.9 A

Load Flow

IN

Id’

5.9 A 5.9 A

Stability of Restricted Earth-Fault Protection (REF)

• For a Single-Line-to-Ground fault in any point outside the zone of protection, let us check if the REF protection relay will operate.

M

At Node M: -Id’-In’+ Ir’+ Iy’+ Ib’=0 Id’=-5.9 A +5.9 A+ 0+ 0=0 Id’=-5.9+5.9) A -Id’= 0 A Therefore, the REF relay will not operate.


P2

S2

P1 S1

P1

S1

P2

S2

RST

M

F

ISC =7.086 kA 1,200/1 A

1,200/1 A

VR

VY

VB

Vr

Vy

Vb

IR–IY

IY–IB

IB-IR

Ir

Iy

Ib

IR

IY

IB

0 A

0

0

0+0

0 A

Load Flow

IN

Id’

5.9 A 5.9 A

Selectivity of Restricted Earth-Fault Protection (REF)

• For a Single-Line-to-Ground fault in the winding of the power transformer, let us check if the REF protection relay will operate.

M

At Node M: -Id’-In’+ Ir’+Iy’+ Ib’=0 -Id’-5.9 A+0+0+0=0 -Id’=-5.9 A Therefore, the REF relay will operate and trip the associated circuit breaker.


P2

S2

P1 S1

P1

S1

P2

S2

RST

M

1,200/1 A

1,200/1 A

F

ISC =7.086 kA

Overcurrent Protection of Power Transformer








5

1,200/1 A

5P20

TR1

20-MVA

33/11.5 kV

%Z=11.5

Q1

Q0

33-kV Bus

600-400/1 A

5P20

3

1,200/1 A

5P20

4

Q0

Q1

11-kV Bus

Overcurrent relays are also used on larger transformers provided with standard circuit breaker control.

The time delay characteristic should be chosen to discriminate with circuit protection on the secondary side.

11-kV Bus

P2

P1

DS

CB

ES

S2

S1

400 / 1 A

1

2

OCR

13

14

33 34

+110 VDC

TC1

200 A -110 VDC +110 VDC

TS

21

27

13 14

22

28

CT

CC Close

Open

+11

0 V

DC

-11

0 V

DC

-110 VDC

0.5 A

0.5 A

TC2

OVERCURRENT PROTECTION

• The CT Circuit

IP-R

IP-Y

IP-B

S2

S2

P2

P1 S1

P2

P1

S2

S1

P2

P1 S1

***

***

***

***

CT Terminal

Block

1

2

3

4

5

6

7

8

9

22

26

24

28

21

25

23

27

Test Socket

22

24

26

28

1

2

3

4

5

6

7

8

Overcurrent Protection

Relay

Phase and Earth-Fault Overcurrent Protection A multi-processor based numerical overcurrent relay consist of the following as a minimum protection elements:

1. Phase Overload Protection

2. Inverse Definite Minimum Time (IDMT) Protection for Phase and Earth

3. Instantaneous Phase and Earth-Fault Overcurrent Protection

Phase Overload Protection-protects the equipment from electrical overload not exceeding the continuous rating of the protected equipment.

IDMT Overcurrent Protection-the selectivity of overcurrent protection relays can be achieved by applying time or current grading in the Inverse Definite Minimum Time (IDMT) protection function of the relay.

Instantaneous Overcurrent Protection-protects the feeder and the transformer against short circuits, including faults at the bushing terminals.

The relay settings are first determined to give the shortest operating times at the maximum fault levels thus verifying the operation and response at the minimum fault current as calculated. The operating time of each relay will be based on the mathematical equation of the curve that will be selected for the relay. The inverse time characteristic curves in accordance with ANSI/IEEE and IEC are given below:

A discrimination time or time interval of 0.4 second will be applied between the downstream and upstream relays to ensure selectivity. In numerical relays there is no overrun, hence the discrimination time of 0.20 second can be applied.

t = TMS x K

IIP

∞

− 1

+ L seconds

t Tripping Time seconds

K Coefficient (see Table) -

I Value of Fault Current Amperes

IP Value of Programmed Threshold (Pick-up Value) Amperes

Coefficient (see Table) -

L ANSI/IEEE (Coefficient) Zero for IEC Curves -

Where:

Type of Curve Standard K Factor Factor L Factor

Short Time Inverse IEC 0.05 0.04 0

Standard Inverse IEC 0.14 0.02 0

Very Inverse IEC 13.5 1 0

Extremely Inverse IEC 80 2 0

Long Time Inverse IEC 120 1 0

Short Time Inverse ANSI/IEEE 0.2663 1.2969 0.03393

Moderately Inverse ANSI/IEEE 0.0103 0.02 0.0228

Long Time Inverse ANSI/IEEE 5.6143 1 2.18592

Very Inverse ANSI/IEEE 3.922 2 0.0982

Extremely Inverse ANSI/IEEE 5.64 2 0.02434

Protection Coordination Study

1,200/1 A

5P20

TR1

20-MVA

33/11.5 kV

%Z=11.5

Q1

Q0

33-kV Bus

400/1 A

5P20

1,200/1 A

5P20

Q0

Q1

11-kV Bus

The TRANSMISSION COMPANY required that the overcurrent protection of the 33-kV incoming feeder of a certain primary substation should not exceed 0.75 second. Calculate the overcurrent protection relay settings for the 20-MVA, 33/11.5 kV, Dyn11 power transformer having an impedance of 11.5%.

Solution:

1. Compute the pick-up current of the overcurrent relays at 33-kV and 11.5-kV side of the power transformer. The pick-up current will be base on the full-load current of the transformer.

IP = 349.91 A

IP = 1,004.087 A

2. Calculate the intended operating time of each overcurrent relay.

tI/C = 0.75 s

Discrimination Time = 0.2 s

t33kV = (0.75 – 0.2) s = 0.55 s

t11.5kV = (0.55 – 0.2) s = 0.35 s

3. Select from the IDMT curves. Say, IEC Standard Inverse.

F

ISC =7.086 kA

F

ISC =8.744 kA

Protection Coordination Study

1,200/1 A

5P20

TR1

20-MVA

33/11.5 kV

%Z=11.5

Q1

Q0

33-kV Bus

400/1 A

5P20

1,200/1 A

5P20

Q0

Q1

11-kV Bus

4. Solve for the time multiplier setting for the overcurrent relays.

t = TMS x K

IIP

∞

− 1

+ L seconds

For the 33-kV Overcurrent Relay:

0.55 = TMS x 0.14

7,08633

11.5

I

349.91

0.02

− 1

+ 0 seconds

TMS33-kV = 0.156 s

For the 11.5-kV Overcurrent Relay:

0.35 = TMS x 0.14

7,0861,004.087

0.02

− 1

+ 0 seconds

TMS33-kV = 0.1 s

F

ISC =7.086 kA

F

ISC =8.744 kA

Operating Time of Overcurrent Relays for a fault at the 11.5-kV Bus:

Operating Time of Differential Relay and Overcurrent Relays for a fault in the Secondary Terminals of the Transformers:

Operating Time of Differential Relay and Overcurrent Relays for a fault in the 11.5-kV Bus:

Stability Test on Transformer Differential Protection

Q0

Q1

TR1

10-MVA

33/11.5 kV

%Z=8.43

Q1

Q0

0.415-kV Bus

1

1. A three-phase, 415-Volt supply will be injected in the primary winding of the power transformer. A three-phase fault will be simulated in the 11-kV busbar by shorting phases R, Y and B ensuring that the faulted section is outside the zone of protection of the differential relay.

2. Calculate the expected secondary voltage of the transformer using the transformation ratio. 𝑉1

𝑉2=

𝑁1

𝑁2

0.415 𝑘𝑉

𝑉2=

33

11.5

V2 = 0.14462 kV 3. Compute the short-circuit current which is the secondary current that will flow on the secondary windings of the power transformer.

𝑝. 𝑢. 𝐼𝑆𝐶 =𝑝. 𝑢. 𝑉𝑜𝑙𝑡𝑎𝑔𝑒

𝑝. 𝑢. 𝑍

𝑝. 𝑢. 𝐼𝑆𝐶 =

𝐴𝑐𝑡𝑢𝑎𝑙 𝑉𝑜𝑙𝑡𝑎𝑔𝑒𝐵𝑎𝑠𝑒 𝑘𝑉2

𝑝. 𝑢. 𝑍

Select Base kV1=33 and Base kV2=11.5 which is as per the

nameplate voltage of the transformer.

𝑝. 𝑢. 𝐼𝑆𝐶 =

0.1446211.5

0.0843

𝑝. 𝑢. 𝐼𝑆𝐶 = 0.149177

F

SHORT PHASE R, Y and B

Stability Test on Transformer Differential Protection

Q0

Q1

TR1

10-MVA

33/11.5 kV

%Z=8.43

Q1

Q0

0.415-kV Bus

1

4. Convert the per unit short circuit current to actual value.

𝐵𝑎𝑠𝑒 𝐼 = 𝐵𝑎𝑠𝑒 𝑀𝑉𝐴𝑥1,000

3𝑥(𝐵𝑎𝑠𝑒 𝑘𝑉2)

𝐵𝑎𝑠𝑒 𝐼 = 10𝑥1,000

3𝑥(11.5)

Base I = 502.0437 A

0.149177 = 𝐴𝑐𝑡𝑢𝑎𝑙 𝐼𝑆𝐶

502.0437 A

Actual ISC = 74.89 A 5. Solve for the primary current using the transformation ratio. 𝐼1

𝐼2=

𝑉2

𝑉1

I1 = 26.099 A 6. The differential relay should not initiate and operate since the simulated fault is outside the zone of its protection. 7. Measure and record the secondary currents flowing in the associated terminals. Furthermore, record the displayed current measurements in all protection relays, ammeters and transducers as applicable.

F SHORT PHASE R, Y and B

Sample Report-Stability Test on Transformer Differential Protection

Sensitivity Test on Transformer Differential Protection

1. After recording all the primary and secondary current values, switch off the three-phase, 415-Volt supply. The shorted phases R, Y and B at the 11-kV switchgear shall be retained.

2. Reverse the polarity of the 11.5-kV differential protection CT then switch on once again the 415-Volt power supply.

3. The current that will flow in the relay is expected to be doubled due to the reversal of CT secondary terminals S1-S2 or S1-S3 as applicable in the LV compartment of the 11-kV switchgear.

4. Sensitivity test can be done also by opening the links of the CT terminal blocks thus shorting terminals S1-S2 or S1-S3 as applicable. In ANSI CTs, the secondary terminals are markd as X1, X2, X3 and Xn depending upon the number of cores.

5. Depending on the type of the switchgear, for Air-Insulated Switchgear, there is a possibility to have an access to simulate a three-phase fault by shorting phases R, Y and B before the primary side of the differential protection CTs.

6. Three-phase fault simulation can be done also by shorting phases R, Y and B at the secondary terminals of the power transformers.

7. Measure and record the secondary currents flowing in the associated terminals. Furthermore, record the displayed current measurements in all protection relays, ammeters and transducers as applicable.

F 1. SHORT PHASE R, Y and B

F

5. SHORT PHASE R, Y and B

P2

P1

Q0

Q1

TR1

10-MVA

33/11.5 kV

%Z=8.43

Q1

Q0

0.415-kV Bus

1

P1

P2

S2

S1

S1

S2

F

6. SHORT PHASE R, Y and B

On-Line Stability Test on Transformer Differential Protection

Transformer Mechanical Protection

1. Buchholz Relay (Gas-detector relay)

A Buchholz relay is connected in the oil feed pipe connecting the conservator to the main tank.

Intended Application of Buchholz Relay:

1.1 To detect free gas being slowly produced in the main tank, possibly as a result of partial discharging.

1.2 To detect sudden surge movement of oil due to an internal transformer fault.

1.3 To provide a chamber for collection and later analysis of evolved gas.

2. Sudden-Pressure Relay

For sealed-tank design transformers, generally when rated 7.5 MVA and larger.

The relay is calibrated for mounting either in the oil or gas space above the oil, and is equipped with a micro switch for alarm and trip.

3. Winding-Temperature Indicator,

Responsive to the combination of top oil temperature and winding current.

The device is calibrated to follow the hottest spot temperature of the winding.

The temperature indicator, for forced-cooled transformers, shall be equipped with adjustable contacts for starting the cooling fans and pumps, while all transformers should be equipped with at least two contacts for alarm and trip.

These devices are fitted to the larger designs of transformer and some cases, are applied to reflect, separately, the primary and secondary winding temperatures.

4. Oil-Temperature Indicator,

The temperature sensor is located in the path of the hottest oil and and mounted adjacent to the winding temperature indicator. The temperature indicator is equipped with adjustable contacts that are normally used for alarm only, except for smaller transformers, which are not provided with winding-temperature devices, where the top oil temperature is used for tripping.

Questions?

MERRY CHRISTMAS TO ALL OF YOU

• God bless you all and your whole family.

Setting Guidelines for Transformer Biased Differential Protection

• From Micom P64x/EN AP/B63, (AP) 6 Application Notes, Application of Individual Protection Functions, Page (AP) 6-38

• The differential setting, Configuration/Diff Protection, should be set to Enable. The basic pick up level of the low set differential element, Is1, is variable between 0.1 pu and 2.5 pu in 0.01 pu steps. The setting chosen is dependent on the item of plant being protected and by the amount of differential current that might be seen during normal operating conditions. When the P64x is used to protect a transformer, a setting of 0.2 In is generally recommended.

• The second slope, K1, is user settable. K1 ensures sensitivity to internal faults up to full load current. It allows for the 15% mismatch which can occur at the limit of the transformer’s tap changer range and an additional 5% for any CT ratio errors. The K1 slope should be set above the errors due to CT mismatch, load tap changers and steady state magnetizing current. The errors slope, which is the combined tap changer (T/C) and current transformer (CT) error, should always be below the K1 slope to avoid mal operations. It is recommended to set K1 to 30%, as long as the errors slope is below the K1 slope by a suitable margin. The second slope, K2, is also user settable, and it is used for bias currents above the rated current. To ensure stability under heavy through fault conditions, which could lead to increased differential current due to asymmetric saturation of CTs, K2 is recommended to be set to 80%.


From Siemens Energy Sector, Power Engineering Guide, 5th Edition, Page 261 of 418

• Siemens Power Engineering Guide ·

Transmission and Distribution · 4th Edition, Page 6 of 63

• Differential relays (87) • Transformer differential relays are normally set

to pick-up values between 20 and 30% rated current. The higher value has to be chosen when the transformer is fitted with a tap changer.


From Technical Manual Chapters of 7SR242 Duobias Transformer Protection Relay, ©2015 Siemens Protection Devices Limited, Chapter 7 Page 12 of 64

87BD Initial Setting (0.1 to 2.0 x In)

This setting is selected to ensure stability in the presence of CT and relay errors when low levels of bias current are present i.e. low load levels.

This is the minimum level of differential current at which the relay will operate. Typically this setting is chosen to

match the on load tap-change range. For example if the tap change range is +10% to –20%, a setting of 0.3In is selected.

87BD 1st Bias Slope Setting (0.0 to 0.7)

Steady state unbalance current will appear in the differential (operate) circuit of the relay due to the transformer tap position, relay tolerance and to CT measurement errors. The differential current will increase with increasing load or through fault current in the transformer so, to ensure stability, the differential current required for operation increases with increasing bias current. The bias slope expresses the current to operate the relay relative to the biasing (restraint) current.

The Bias slope setting chosen must be greater than the maximum unbalance, it is selected to ensure stability when through fault or heavy load current flows in the transformer and the tap changer is in its extreme position.


The recommended setting is 1 x the tap change range. As the protection is optimized around the center tap position then using the total tap change range includes for a 100% safety margin, this provides contingency for CT and relay tolerances. For example if the tap change range is +10 to –20%, the overall range is 30% so a 0.3x setting is chosen.

87BD 1st Bias Slope Limit Setting (1 to 20 x In)

• Above this setting the ratio of differential current to bias current required for operation is increased. When a through fault occurs, saturation of one or more CTs may cause a transient differential current to be detected by the relay. The bias slope limit is chosen to ensure the biased differential function is stable for high through fault currents coincident with CT saturation. This setting defines the upper limit of the bias slope and is expressed in multiples of nominal rated current i.e. the lower the setting the more stable the protection. The three phase through fault current can be estimated from the transformer impedance. For a typical grid transformer having a 15% impedance, the maximum through fault will be 1/0.15 = 6.66 x rating. A setting value is chosen that introduces the extra bias at half of the three phase through fault current level of the transformer, so 6.66/2 = 3.33 and a setting of 3 would be selected as the nearest lower setting available.


87BD 2nd Bias Slope Type (Line, Curve)

87BD 2nd Bias Slope Setting (1.0 to 2.0 – applied to ‘Line’ only)

• These settings are chosen to ensure the biased differential function is stable for high through fault currents coincident with CT saturation.

87HS Setting (1 to 30 x In)

• The 87HS element is set as low as possible but not less than the maximum three phase through fault current and not less than half the peak magnetizing inrush current.

• For almost all applications a setting of 7 or 8 x In has shown to be sufficiently sensitive for internal faults as well as providing stability during external faults and transient system conditions.

• A Differential High set Setting of 7 x In will be stable for a peak magnetizing inrush levels of 14 x rated current.

• Smaller transformers generally will have lower impedance and therefore greater three phase through fault levels and magnetizing inrush currents.

• A setting of 8 x can be used as CT saturation is reduced as system X/R is usually very low and the peak level of magnetising current does not usually ever exceed 16 x rating.


From Application Manual, 1MRK504089-UEN rev. B, Section 4, IED application

Restrained and unrestrained differential protection

The usual practice for transformer protection is to set the bias characteristic to a value of at least twice the value of the expected spill current under through faults conditions.

These criteria can vary considerably from application to application and are often a matter of judgment.

The second slope is increased to ensure stability under heavy through fault conditions which could lead to increased differential current due to saturation of current transformers. Default settings for the operating characteristic with IdMin = 0.3pu of the power transformer rated current can be recommended as a default setting in normal applications. If the conditions are known more in detail, higher or lower sensitivity can be chosen. The selection of suitable characteristic should in such cases be based on the knowledge of the class of the current transformers, availability of information on the on load tap changer (OLTC) position, short circuit power of the systems, etc.


From Application Manual, 1MRK504089-UEN rev. B, Section 4, IED application

Inrush Restraint Methods

With a combination of the second harmonic restraint and the waveform restraint methods it is possibly to get a protection with high security and stability against inrush effects and at the same time maintain high performance in case of heavy internal faults even if the current transformers are saturated. Both these restraint methods are used by RET 670.

The second harmonic restraint function has a settable level. If the ratio of the second harmonic to fundamental harmonic in the differential current is above the settable limit, the operation of the differential protection is restrained.

It is recommended to set parameter I2/I1Ratio = 15% as default value in case no special reasons exist to choose other value.

Required Current Transformers for High Impedance REF 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.

RST

Rr

RL

RL

RCT

VS IS

VS

1.0 V

10.0 V

100.0 V

1000.0 V

0.0001 A 0.001 A 0.01 A 0.1 A 1.0 A 10.0 A 100.0 A

Excitation curve data

L 1

L 2

L 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

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