11- overcurrent protection

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Fundamental Principle of Overcurrent Relays Dr. Fouad Zaro Assistant Professor Electrical Power Engineering Palestine Polytechnic University

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Page 1: 11- Overcurrent Protection

Fundamental Principle of Overcurrent Relays

Dr. Fouad Zaro

Assistant Professor

Electrical Power Engineering

Palestine Polytechnic University

Page 2: 11- Overcurrent Protection

Time Multipier Setting

1. Primary protection should be fast.

2. Back up protection should act if and only if primary protection has failed. It is intentionally slow.

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Types of Overcurrent Relay

1. Instantaneous Relay • No intentional time delay. • The operating time a few milliseconds. • Has only the pick-up setting and it does not have any

TMS. • it's action is fast. • not suitable for backup protection. • common to set an instantaneous relay about

– 125–135% above the maximum value for which the relay should not operate.

– 90% of the minimum value for which the relay should operate.

• Solid-state or digital relays can be set closer, e.g. 110% above the maximum no-go value. Power System Protection 3

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2. Time delayed Definite Time Relay :

• can be adjusted to issue a trip output after a specified delay when the relay picks up (PSM>1).

• This delay is fixed and it is independent of PSM value.

• it has a adjustable time setting as well as a pick up adjustment.

• It is used for short length feeders where the fault current does not change significantly with the location of the fault across the feeder.

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Time delayed Definite Time Relay…. contd

• as we move along towards source, the relaying action slows down.

• There is an upper limit on any fault clearing time in the system and it equals approximately 1sec.

• This limit would be hit near the relay close to source. Power System Protection 5

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Time delayed Definite Time Relay…. contd

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Example:

• Consider a CTI of 0.3sec. Then what is the maximum length of a radial system of a feeder that can be protected by overcurrent relay. Assume, that primary protection uses DT relays and primary protection time should not be more than 1sec.

• Let 'n' be the maximum number of feeder sections that can be protected by overcurrent relays and let TOCmaxbe the upper limit on the speed of primary protection. Then . Thus overcurrent relays should be used over a limited length in the 3 feeder sections.

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3. Inverse definite minimum time (IDMT) Relay

• the most widely used characteristic.

• It is inverse in the initial part and tends to approach a definite minimum operating time characteristic as the current becomes very high.

• traditionally PSM of an overcurrent relay is set above 1.5.

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Back up protection by time discrimination

Relay setting and coordination steps: 1. Identify all possible Primary-Back-up relay pairs.

2. Decide the correct sequence for coordination of relays.

3. Decide the pickup value and hence PSM for relays.

4. Compute the TMS to meet the coordination.

5. Validation of the results.

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PSM setting for primary and back-up protection

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Step 1 : set relay R1 Choose for relay R1

TMS = 0.025. No intentional time delay is provided because R1 does not have backup responsibility. Relay 1 (R1) pickup current of R1 = 160A. For fault on section AB (Ifmax = 500 A): PSM = Fault Current / Actual Pick up = 500/160 = 3.125 TMS = 0.025

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Relay 2 (R2) Actual Pick up = 167 A. We co-ordinate R2 with R1 for close in fault for relay R1. PSM = Fault Current / Actual Pick up = 500/167 = 2.99 Expected operating time for relay 2 = Operating time of relay 1+ CTI = 0.15 + 0.3 = 0.45sec.

TMS = 0.07

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Now for maximum fault current on section BC (1200A) PSM = Fault Current / Actual Pick up = 1200/167 = 7.185 with TMS = 0.07 operating time of relay 2

Operating time of relay 2 = 0.24sec

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It is clear that slowest relay in the system is R4. To compute its worst case performance, we should evaluate its fault clearing time with minimum fault current at remote bus D for primary protection and bus C for backup protection. Time of operation for fault current of 1600A (bus D) = 0.82sec. Time of operation for fault current of 1100A (bus C) = 1.5sec. Since primary protection is always cleared within 1sec, we can consider the protection system to be satisfactory.

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