protection 2014

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Introduction

Why do we need protection systems

How do we protect – concepts of protection system design

Protection Schemes

Overcurrent

Distance

Differential

Udaya Annakkage Power System Protection

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Abnormal conditions in power systems

Examples of abnormal conditions

Large currents

Large voltages, low voltages

Presence of negative and zero sequence currents

What is the problem with abnormal conditions ?

Damage to equipment and people

Interruption to power supply

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Objectives

Prevent or at least minimize damage when abnormalconditions arise

Isolate only the directly affected part of the system

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Design Criteria

Reliability

Selectivity (Discrimination)

Simlicity and Economy

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Protection Zones

Each protection scheme is mainly responsible for clearingfaults within its zone

Some schemes provide backup protection to neighbouringzones

Some schemes do not provide backup protection (unitprotection schemes)

Zones are overlapped to avoid unprotected areas.

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Functions of a protection system

Monitor the power system variables – Transducers

Detect abnormal conditions – Relay

Isolate the faulty section – Circuit Breakers

Trip Signal

Transducer

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Transducers

Current Transformers (CT)

Electromagnetic type (most common)

Magneto–Optical type

Voltage Transformers (VT)

Capacitor divider type

Electromagnetic type

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Burden on a CT

The load on a current transformer is called the burden

This can be expressed either as a VA load or as animpedance

.Example..

5 VA burden on a 1 A current transformer means:I = 1 AVI = 5 VATherefore, V=5 Vi.e. R = 5 ohm

What is the equivalent load in ohms of a 5 A burden on a 5VA CT?

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Standard CT ratios:50:5, 100:5,150:5, 200:5, 250:5, 300:5, 400:5, 450:5,500:5, 600:5, 800:5, 900:5, 1000:5, 1200:5

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Relays

Electromechanical: attracted armature, induction discunit, induction cup unit etc.

Static: logic units (AND,OR,NOT), time delay units,magnitude comparators, phase comparators, phase shiftunits, amplification units.

Numerical Relays: Sampling, A/D conversion, filtering,signal processing (computation of peak, average, rms,fundamental component, dc offset etc)

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Circuit Breakers

Duties:...1 Carry full load current...2 Withstand normal voltages...3 Open and close the circuit on no load...4 Make and break normal operating currents...5 Make short circuit currents...6 Break short circuit currents

1–3 Isolators1–5 Switches1–6 Circuit BreakersTypes: Oil, Air, Air blast, Vacuum, SF6

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Logical Design of Relays

...1 Magnitude Relays (example: overcurrent relays)

...2 Directional Relays (example: mho relays)

...3 Ratio Relays (example: distance relays)

...4 Differential Relays (example: generator or transformerinternal fault protection)

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Overcurrent Relays

Instantaneousrelays

Inverse time relays

..Relay Current (A)

.Operatingtime(s)

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Inverse time overcurrent relay

Relay operating time = time taken for relay operation. i.e.time delay between the occurrence of the abnormal conditionand sending a trip signal to the circuit breaker.

Electromechanical type relay

Operating time depends on:...1 Initial position of the moving contact...2 Speed of the disc (depends on mmf of the coil)

Two settings:...1 Pick-up current (plug setting, current setting)...2 Time setting (time dial setting, Time Multiplier Setting,TMS)

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Example 1

.CT:400/5. If

The operatng time of the relay is given by:

t =0.14

I 0.02r − 1× TMS

where, Ir =IfIsis the ratio between the fault current If and the

current setting Is of the relay. If the current setting and TMShave been chosen to be 6A and 1.0 respectively, determine theoperating time of the relay for a fault current of 800A.

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Example 2

The following information is given:

CT Ratio: 400/5

Current setting (pick up current) = 5.0 A

TMS = 0.2

t = 3×TMSlog(Ir )

Determine the operating time of the relay for a fault current of500A.

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Example 3

The following information is given:

CT Ratio: 200/5

Current setting (pick up current) = 6.0 A

Fault current If is 1200A

t = 3×TMSlog(Ir )

Determine the TMS to get an operating time of 1.4 s for thisfault.

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Example 4: Back up protection

.CT:400/5.

.CT:400/5

Current settings of R1 and R2 are at 6A and TMS of R2 is 0.1.R1 is required to provide back up protection to R2 for a faultcurrent of 1200A (a fault near bus A).Allow 0.3 seconds for backing up and determine the TMS ofR1.The fault current for a fault near bus B is 1800A. What is theoperating time of R1 for this fault.

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Example 5: Overcurrent relay coordination

Grounded Y-Y

.j12.8

.Fault Currents:..

Bus # 1 2 3 4 5Max. Fault Current (A) 3187 659 431 301 203Min. Falut Current (A) 1380 473 329 238 165Max. Load Current (A)

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Example 5: Settings for R4

Minimum fault current seen by Relay at B4 is 165A

Therefore choose a setting current of one third of 165A(165/3 = 55A)

Use a CT ratio of 50/5

relay current corresponding to 55A is:Ip = 55× 5/50 = 5.5A

Choose the nearest setting current, i.e. 5A.

Use a TMS of 0.5 (smallest available value is chosen tohave the fastest relay operation)

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Example 5: Settings for R3.

R3 must provide primary protection to line 3–4 and backup protection for R4

Minimum fault current seen by R3 is 165A

Choose CT ratio of 50/5 and a setting current of 5A

The highest fault current seen by R4 is 301A.Corresponding relay current is 301× 5/50 = 30.1A

If /Ip = 30.1/5 = 6.

From the given relay characteristics (If /Ip = 6 and TMS= 0.5) operating time of R4 is 0.135s

For the same fault current the operating time of R3 mustbe 0.135 + 0.3 = 0.435 s

From the given relay characteristics, the required TMS is2.0

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Example 5: Settings for R2.

R2 must provide primary protection to line 2–3 and backup protection for R3

Minimum fault current seen by R2 is 238A

Choose CT ratio of 100/5 (238/3 = 79.3) and a settingcurrent of 4A (79.3× 5/100)

The highest fault current seen by R3 is 431A.Corresponding relay current is 431× 5/50 = 43.1A

From the given relay characteristics (If /Ip = 8.6 and TMS= 2) operating time of R3 is 0.31s

For the same fault current the operating time of R2 mustbe 0.31 + 0.3 = 0.61 s

From the given relay characteristics, the required TMS is2.6 (If /Ip = 431× 5/100× 1/4 = 5.4)

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TCCC – Time Current Characteristics Curves

.Fault Current (A)

Operatingtime(s)

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Directional Relay

−I12

θmin > θ > θmax TRIP

θmin < θ < θmax BLOCK

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Ratio Relay: Impedance Characteristics

.. R.T

|Z | < |Zr | TRIP

|Z | > |Zr | BLOCK

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Ratio Relay: Mho Characteristics

|Z − Zr | < |Zr | TRIP

|Z − Zr | > |Zr | BLOCK

|x + iy − xr − iyr | < |Zr |(x − xr )

2 + (y − yr )2 < |Zr |2

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Ratio Relay: Offset Mho Characteristics

|Z − Zr1| < |Zr2| TRIP

|Z − Zr1| > |Zr2| BLOCK

|x + iy − xr1 − iyr1| < |Zr2|(x − xr1)

2 + (y − yr1)2 < |Zr2|2

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Distance Protection

R1:Zone 1

R1:Zone 2

R1:Zone 3

Impedance

Operatingtime(s)

R1:Zone 1

R1:Zone 2

R1:Zone 3

R2:Zone 1

R2:Zone 2

R3:Zone 1

R3:Zone 2

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Example 6: Distance Protection (Impedance

Relays)

The impedance of a transmission line is 5 + j45 Ω. The CTratio is 400/5 and the VT ratio is 1000/1. A three phase faultoccurs at 60 % of the line. What is the impedance seen by therelay..

I= (5 + j45) ∗ 0.6 = 3 + j54

Vmeasured =V

Imeasured =5

Zmeasured =Vmeasured

Imeasured=

1000× 400

5= 0.24 + j4.32Ω

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Example 7

Determine the Zone 1 and Zone 2 settings for the relay inExample 6..Zone 1 setting..

Line Impedance seen by the relay = 0.4 + j 7.2 ΩZone 1 setting is 80 % of the line impedance =0.8×

√0.42 + 7.22 = 0.8× 7.21 = 5.77Ω

.Zone 2 setting..

Zone 2 setting is 120 % of the line impedance =1.2× 7.21 = 8.65Ω

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Example 8

Determine the Zone 3 setting for the relay in Examples 6 and7 to provide back up protection to the next line having animpedance of 10 + j100 Ω..Zone 3 setting..

Impedance of the second line as seen by the relay =(10 + j100)× 400

1000×5= 0.8 + j8Ω

Zone 3 setting is 7.21 + 1.2×√0.82 + 82 = 9.65Ω

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Differential Protection

The principle of discrimination is by comparison ofcurrents entering and leaving the protected zone.

Differential protection (also known as unit protection)schemes have absolute selectivity, fast operation and highsensitivity.

They provide protection only to a specied unit and do notprovide back up protection to other units of the system.

This is in contrast to all other schemes that operate onthe principle of time discrimination.

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Differential Protection

External Fault

Internal Fault

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Differential Protection: Spill Current

Non-ideal situation

CTs are not identical

Leads connecting the CTs to the relay are not identical

Therefore, the magnetizing currents are different.

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Differential Protection: High Impedance Scheme

Stabilizing Resistance Rs is chosen to prevent relay operationfor external faults under worst case scenario where only oneCT is saturated.

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Differential Protection: Low Impedance – Bias

|IR | > k |I1 + I2| TRIP

|IR | < k |I1 + I2| BLOCK

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State whether the following statements are True or

False and explain your answer

...1 Inverse time overurrent relays can be used for back upprotection

...2 Differential protection can provide primary protection to atransformer and back up protection to a nearby bus bar.

...3 Overlapping zones are used for providing back upprotection.

...4 Impedance relay characteristics are not suitable fordistance protection.

...5 An overcurrent relay is expected to operate undermaximum loading conditions.

...6 There will be no spill current in a differential protectionscheme if the CTs are identical.

protection 2014

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