Distance Protection

1. BEF43303
POWER SYSTEM ANALYSIS AND PROTECTION
WEEK 10
DISTANCE PROTECTION

2. 10.0 CONTENTS
10.1 Introduction to system protection
10.2 System protection components
10.3 Instrument transformers

3. 10.1 INTRODUCTION TO SYSTEM
PROTECTION
• System protection is required to protect the system from abnormal
system operating condition.
• The typical causes of short circuit:
 Equipment insulation fails,
 Overvoltage caused by lightning or switching surges,
 Insulation contamination,
 Natural causes.
• Impossible to eliminate can be minimized through proper
maintenance.

4. 10.1 INTRODUCTION TO SYSTEM
PROTECTION
• The impact of short circuit:
 Insulation damage,
 Conductor melting,
 Fire and explosion,
 Mechanical damage to winding and busbar.
• EHV protection system is designed to clear fault within 3 cycles.
• LV protection system typically operates within 5-20 cycles.

5. 10.1 INTRODUCTION TO SYSTEM
PROTECTION
• The basic idea of protection system is:
 To define the undesirable conditions and look for differences
between the undesirable and permissible conditions that
relays or fuses can sense.
• It is important to remove only the faulted equipment from the
system.
• The unfaulted system needs to be retained as much as possible.
• For this course, we primarily focus on circuit breaker and relays
which are used to protect HV (115-230kV) and EHV (345-765kV).

6. 10.1 INTRODUCTION TO SYSTEM
PROTECTION
• The definition of relay according to IEEE:
• A device whose function is to detect defective lines or
apparatus or other power system conditions of an abnormal
or dangerous nature and to initiate appropriate control
action.
• In practice, a relay is a device that closes or opens a contact when
energized.

7. 10.1 INTRODUCTION TO SYSTEM
PROTECTION
• Problems with the protection equipment may occur within itself.
• A backup relays may be used to protect the primary relays.
• In HV and EHV systems, separate protective elements may be used
for backup relays.
• Various protective devices in the system must be properly
coordinated.
• The primary relays must operate first.
• If the primary relays fail, then backup relays should operate after a
specified time delay.

8. 10.2 SYSTEM PROTECTION
COMPONENTS
• Protection systems have three
basic components:
 Instrument transformers,
 Relays,
 Circuit breakers.
• Figure 10.1 shows a simple
overcurrent protection
schematic.
Figure 10.1 – Overcurrent protection schematic

9. 10.2 SYSTEM PROTECTION
COMPONENTS
Current transformer
 Function: to reproduce in its secondary winding a current I’ that is
proportional to the primary current I.
 Operation: Converts primary currents in the kA range to secondary
currents in the 0–5 ampere range.
 Advantages:
 Safety - Provides electrical isolation from the power system.
 Economy - Relays are smaller, simpler, and less expensive.
 Accuracy - Reproduce V and I over wide operating ranges.

10. 10.2 SYSTEM PROTECTION
COMPONENTS
Relay
• Function: To discriminate between normal operation and fault
conditions.
• Operation:
 When the variables being monitored exceeds a specified
‘‘pickup’’ value,
 The operating coil causes the NO contacts to close,
 The trip coil of the circuit breaker is energized,
 The circuit breaker to open.

11. 10.2 SYSTEM PROTECTION
COMPONENTS
Circuit breaker
• Function:
 A mechanical switch capable of interrupting fault currents and of
reclosing.
 Circuit breaker contacts are separated while carrying current.
• Operation: Based on information from instrument transformers, a
decision is made and ‘‘relayed’’ to the trip coil of the breaker, which
actually opens the power circuit.

12. 10.2 SYSTEM PROTECTION
COMPONENTS
The design criteria of system protection system:
• Reliability: Operate dependably when fault conditions occur, even
after remaining idle for months or years. Failure to do so may result
in costly damages.
• Selectivity: Avoid unnecessary, false trips.
• Speed: Operate rapidly to minimize fault duration and equipment
damage. Any intentional time delays should be precise.
• Economy: Provide maximum protection at minimum cost.
• Simplicity: Minimize protection equipment and circuitry.

13. 10.3 INSTRUMENT TRANSFORMERS
• There are two basic types of
instrument transformers:
 Voltage transformers (VTs),
 Current transformers (CTs).
• Figure 10.2 shows a schematic
representation for the VT and
CT.
Figure 10.2 - VT and CT schematic

14. 10.3 INSTRUMENT TRANSFORMERS
• For system-protection purposes, VT is usually modeled as an ideal
transformer as 𝑉𝑉′ = ⁄
1 𝑛𝑛 × 𝑉𝑉.
• 𝑉𝑉𝑉 is a scaled-down representation of 𝑉𝑉 and is in phase with 𝑉𝑉.
• A standard VT secondary voltage rating is 115 V (line-to-line).
• Standard VT ratios are given in Table 10.1.
Table 10.1 - Standard VT ratios

15. 10.3 INSTRUMENT TRANSFORMERS
• The primary winding of a CT consists of a single turn, obtained by
running the power system’s primary conductor through the CT core.
• The normal current rating of CT secondary is standardized at 5A in
the United States, and 1A in Europe.
• Standard CT ratios are given in Table 10.2.
Table 10.2 - Standard CT ratios

16. 10.3 INSTRUMENT TRANSFORMERS
• An approximate equivalent
circuit of a CT is shown in
Figure 10.3, where:
 𝑍𝑍𝑍 - CT secondary leakage
impedance,
 𝑋𝑋𝑒𝑒 - CT excitation reactance,
 𝑍𝑍𝐵𝐵 - Impedance of
terminating device.
Figure 10.3 – CT equivalent circuit

17. 10.3 INSTRUMENT TRANSFORMERS
• 𝑍𝑍𝐵𝐵 is also called the burden.
• It is typically less than an ohm.
• It may also be expressed as VA
at a specified current.
• The relationship between the
CT secondary voltage 𝐸𝐸𝐸 and
excitation current 𝐼𝐼𝑒𝑒 is given by
an excitation curve shown in
Figure 10.4.
Figure 10.3 – CT equivalent circuit

18. 10.3 INSTRUMENT TRANSFORMERS
Figure 10.4 - Excitation curves for a multi-ratio bushing CT with a C100 ANSI accuracy classification

19. 10.3 INSTRUMENT TRANSFORMERS
• 𝑍𝑍𝐵𝐵 is also called the burden.
• It is typically less than an ohm.
• It may also be expressed as VA
at a specified current.
• The relationship between the
CT secondary voltage 𝐸𝐸𝐸 and
excitation current 𝐼𝐼𝑒𝑒 is given by
an excitation curve shown in
Figure 10.4.
Figure 10.3 – CT equivalent circuit

20. 10.3 INSTRUMENT TRANSFORMERS
Using the CT equivalent circuit and excitation curves, the following
procedure can be used to determine CT performance.
1) Assume a CT secondary output current 𝐼𝐼′.
2) Compute 𝐸𝐸′
= 𝑍𝑍′
+ 𝑍𝑍𝐵𝐵 𝐼𝐼𝐼.
3) Using E′, find 𝐼𝐼𝑒𝑒 from the excitation curve.
4) Compute 𝐼𝐼 = 𝑛𝑛 𝐼𝐼′ + 𝐼𝐼𝑒𝑒 .
5) Repeat Steps 1–4 for different values of I’, then plot I’ versus I.

21. 10.3 INSTRUMENT TRANSFORMERS
• For simplicity, calculations are made with magnitudes rather than
with phasors.
• The CT error is the percentage difference between 𝐼𝐼′ + 𝐼𝐼𝑒𝑒 and 𝐼𝐼𝐼,
given by:
CT error =
𝐼𝐼𝑒𝑒
𝐼𝐼′ + 𝐼𝐼𝑒𝑒
× 100%

22. 10.3 INSTRUMENT TRANSFORMERS
Example 10.3.1
Evaluate the performance of the multi-ratio CT in Figure 10.4 with
a 100:5 CT ratio, for the following secondary output currents and
burdens:
a) 𝐼𝐼𝐼 = 5𝐴𝐴 and 𝑍𝑍𝐵𝐵 = 0.5Ω (Ans.: 2.91𝑉𝑉, 0.25𝐴𝐴, 105𝐴𝐴, 4.8%);
b) 𝐼𝐼𝐼 = 8𝐴𝐴 and 𝑍𝑍𝐵𝐵 = 0.8Ω (Ans.: 7.06V, 0.4A, 168A, 4.8%);
c) 𝐼𝐼𝐼 = 15𝐴𝐴 and 𝑍𝑍𝐵𝐵 = 1.5Ω (Ans.: 23.73𝑉𝑉, 20𝐴𝐴, 700𝐴𝐴, 57.1%).
Also, compute the CT error for each output current.

23. 10.3 INSTRUMENT TRANSFORMERS
Example 10.3.2
An overcurrent relay set to operate at 8𝐴𝐴 is connected to the multi-
ratio CT in Figure 10.4 with a 100:5 CT ratio. Will the relay detect a
200A primary fault current if the burden 𝑍𝑍𝐵𝐵 is
(a) 0.8Ω (Ans.: relay will operate),
(b)3.0 Ω (Ans.: relay will not operate).
Note: For additional exercises, students may refer to Power System
Analysis and Design by Glover, Problems 10.1-10.6.