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The importance of overcurrent protection

Transmission and distribution systems are exposed to overcurrent flow into their elements. In an electric power system, overcurrent or excess current is a situation where a larger than intended electric current exists through a conductor, leading to excessive generation of heat, and the risk of fire or damage to equipment.

The fundamentals of protection relay co-ordination and time/current grading principles
The fundamentals of protection relay co-ordination and time/current grading principles (photo credit: JAL International)

Possible causes for overcurrent include short circuits, excessive load, transformer inrush current, motor starting, incorrect design, or a ground fault.

Therefore, for normal system conditions, some tools such as demand – side management, load shedding, and soft motor starting can be applied to avoid overloads.

In addition, distribution systems are equipped with protective relays that initiate action to enable switching equipment to respond only to abnormal system conditions. The relay is connected to the circuit to be protected via CTs and VTs according to the required protection function.

In order for the relay to operate, it needs to be energized. This energy can be provided by battery sets (mostly) or by the monitored circuit itself.

This article deals with co-ordination between protection relays in general and  principles of Time/Current grading used to achieve correct relay co-ordination.

  1. Co-ordination procedure
  2. Principles of Time/Current grading
    1. Discrimination by Time
    2. Discrimination by Current
    3. Discrimination by both Time and Current

1. Co-ordination procedure

Correct overcurrent relay application requires knowledge of the fault current that can flow in each part of the network. Since large-scale tests are normally impracticable, system analysis must be used.

The data required for a relay setting study are:

  1. Single-line diagram of the power system involved, showing the type and rating of the protection devices and their associated current transformers.
  2. The impedances in ohms, per cent or per unit, of all power transformers, rotating machine and feeder circuits.
  3. The maximum and minimum values of short circuit currents that are expected to flow through each protection device.
  4. The maximum load current through protection devices.
  5. The starting current requirements of motors and the starting and locked rotor/stalling times of induction motors.
  6. The transformer inrush, thermal withstand and damage characteristics.
  7. Decrement curves showing the rate of decay of the fault current supplied by the generators.
  8. Performance curves of the current transformers.

The relay settings are first determined to give the shortest operating times at maximum fault levels and then checked to see if operation will also be satisfactory at the minimum fault current expected.

It is always advisable to plot the curves of relays and other protection devices, such as fuses, that are to operate in series, on a common scale. It is usually more convenient to use a scale corresponding to the current expected at the lowest voltage base, or to use the predominant voltage base.

The alternatives are a common MVA base or a separate current scale for each system voltage.

The basic rules for correct relay co-ordination can generally be stated as follows:

RULE #1

Whenever possible, use relays with the same operating characteristic in series with each other.

RULE #2

Make sure that the relay farthest from the source has current settings equal to or less than the relays behind it, that is, that the primary current required to operate the relay in front is always equal to or less than the primary current required to operate the relay behind it.

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2. Principles of Time/Current grading

Among the various possible methods used to achieve correct relay co-ordination are those using either time or overcurrent, or a combination of both. The common aim of all three methods is to give correct discrimination.

That is to say, each one must isolate only the faulty section of the power system network, leaving the rest of the system undisturbed.

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2.1 Discrimination by Time

In this method, an appropriate time setting is given to each of the relays controlling the circuit breakers in a power system to ensure that the breaker nearest to the fault opens first.

A simple radial distribution system is shown in Figure 1, to illustrate the principle.

Radial system with time discrimination
Figure 1 – Radial system with time discrimination

Overcurrent protection is provided at B, C, D and E, that is, at the infeed end of each section of the power system.

Each protection unit comprises a definite-time delay overcurrent relay in which the operation of the current sensitive element simply initiates the time delay element. Provided the setting of the current element is below the fault current value, this element plays no part in the achievement of discrimination.

For this reason, the relay is sometimes described as an ‘independent definite-time delay relay’, since its operating time is for practical purposes independent of the level of overcurrent.

It is the time delay element, therefore, which provides the means of discrimination. The relay at B is set at the shortest time delay possible to allow the fuse to blow for a fault at A on the secondary side of the transformer. After the time delay has expired, the relay output contact closes to trip the circuit breaker. The relay at C has a time delay setting equal to t1 seconds, and similarly for the relays at D and E.

If a fault occurs at F, the relay at B will operate in t seconds and the subsequent operation of the circuit breaker at B will clear the fault before the relays at C, D and E have time to operate.

The time interval t1 between each relay time setting must be long enough to ensure that the upstream relays do not operate before the circuit breaker at the fault location has tripped and cleared the fault.

The main disadvantage of this method of discrimination is that the longest fault clearance time occurs for faults in the section closest to the power source, where the fault level (MVA) is highest.

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2.2 Discrimination by Current

Discrimination by current relies on the fact that the fault current varies with the position of the fault because of the difference in impedance values between the source and the fault.

Hence, typically, the relays controlling the various circuit breakers are set to operate at suitably tapered values of current such that only the relay nearest to the fault trips its breaker.

Figure 2 illustrates the method.

Radial system with current discrimination
Figure 2 – Radial system with current discrimination

For a fault at F1, the system short-circuit current is given by:

I = 6350 / (ZS + ZL1) A

where:

  • ZS = source impedance = 112 / 250 = 0.485 Ω
  • ZL1 = cable impedance between C and B = 0.24 Ω

Hence,

I = 6350 / 0.725 = 8800 A

So, a relay controlling the circuit breaker at C and set to operate at a fault current of 8800A would in theory protect the whole of the cable section between C and B.

Points affecting this method

However, there are two important practical points that affect this method of co-ordination:

Point #1 – It is not practical to distinguish between a fault at F1 and a fault at F2, since the distance between these points may be only a few metres, corresponding to a change in fault current of approximately 0.1%.

Point #2 – In practice, there would be variations in the source fault level, typically from 250MVA to 130MVA.

At this lower fault level the fault current would not exceed 6800A, even for a cable fault close to C. A relay set at 8800A would not protect any part of the cable section concerned.

Discrimination by current is therefore not a practical proposition for correct grading between the circuit breakers at C and B. However, the problem changes appreciably when there is significant impedance between the two circuit breakers concerned.

Consider the grading required between the circuit breakers at C and A in Figure 2. Assuming a fault at F4, the short-circuit current is given by:

I = 6350 / (ZS + ZL1 + ZL2 + ZT)

where:

  • ZS = source impedance = 112 / 250 = 0.485 Ω
  • ZL1 = cable impedance between C and B = 0.24 Ω
  • ZL2 = cable impedance between B and 4MVA transformer = 0.04 Ω
  • ZT = transformer impedance = 0.07 × (112/4) = 2.12 Ω

Hence,

I = 6350 / 2.885 = 2200 A

For this reason, a relay controlling the circuit breaker at B and set to operate at a current of 2200A plus a safety margin would not operate for a fault at F4 and would thus discriminate with the relay at A.

Assuming a safety margin of 20% to allow for relay errors and a further 10% for variations in the system impedance values, it is reasonable to choose a relay setting of 1.3 x 2200A, that is, 2860A, for the relay at B.

Now, assuming a fault at F3, at the end of the 11kV cable feeding the 4MVA transformer, the short-circuit current is given by:

I = 6350 / (ZS + ZL1 + ZL2)

Thus, assuming a 250MVA source fault level:
I = 6350 / (0.485 + 0.24 + 0.04) = 8300A

Alternatively, assuming a source fault level of 130MVA:
I = 6350 / (0.93 + 0.214 + 0.04) = 5250A

For either value of source level, the relay at B would operate correctly for faults anywhere on the 11kV cable feeding the transformer.

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2.3 Discrimination by both Time and Current

Each of the two methods described above has a fundamental disadvantage. In the case of discrimination by time alone, the disadvantage is due to the fact that the more severe faults are cleared in the longest operating time.

On the other hand, discrimination by current can be applied only where there is appreciable impedance between the two circuit breakers concerned.

It is because of the limitations imposed by the independent use of either time or current co-ordination that the inverse time overcurrent relay characteristic has evolved.

With this characteristic, the time of operation is inversely proportional to the fault current level and the actual characteristic is a function of both ‘time’ and ‘current’ settings.

Figure 3 shows the characteristics of two relays given different current/time settings.

Relay characteristics for different settings
Figure 3 – Relay characteristics for different settings

For a large variation in fault current between the two ends of the feeder, faster operating times can be achieved by the relays nearest to the source, where the fault level is the highest.

So, by using both functions the disadvantages of grading by time or current alone are overcome!

Variations of current/time tripping characteristics of IDMT relays will be discussed in some of the coming technical articles.

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References //

  • Network Protection & Automation Guide by Alstom Grid
  • The Basics Of Overcurrent Protection – Seminar Paper by Genc Baruti

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author-pic

Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry fascilities. Professional in AutoCAD programming. Present on

8 Comments


  1. R H Magem
    Aug 26, 2018

    Your portal is very educating


  2. İRFAN SELVİ
    Aug 16, 2018

    Hello Edward
    Which brand(s) and model do you recommend


  3. senthil kumar.V
    Aug 14, 2018

    hi .,

    6350 is 11KV / root 3., I = V/Z


  4. Jürgen Verpaele
    Aug 14, 2018

    Hello,

    You talk about:
    “The time interval t1 between each relay time setting must be long enough to ensure that the upstream relays do not operate before the circuit breaker at the fault location has tripped and cleared the fault.”

    Are there some typical time values for this? Or how can this be selected?

    Thanks

    • Edvard
      Edvard
      Aug 17, 2018

      It depends on many things in your network. The best would be to perform a protection study using some testing device and its protection relay software (OMICRON for example) and see the results.


  5. Madhav
    Aug 14, 2018

    Hello Edvard,
    No where in this article you have mentioned as to how the figure 6350 is arrived in the calculation of fault current… I wonder how and why?


    • Heidar Ghamsarian
      Aug 16, 2018

      LL=11000 so that LN=11000/square root3=6350 volt


  6. wagner jirschik
    Aug 13, 2018

    Hi, thank you for the explantion.

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