Fault Level Role in Selection of Protective Equipment in Power System

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Fault Level Role in Selection of Protective Equipment in Power System

Almost after a decade, US economy on rising trend, a very good news for all of us. India is also an emerging economy, where power sector has to play crucial role. After years of silence, Indian power sector has risen to embrace multi million rupees i.e. Rs 23000, crore deals in the last few months under the new regime at the center. In addition to this there is significant rise in monthly power generation too, since May 2014 as compared to year 2013 in India, according to CEA.

Over the years, tsunami of troubles hit the power sector like shortage of quality coal and gas supply, problems in land acquisition etc. But in spite of all these problems India was adding roughly 20 GW annually. Transmission and distribution losses 23.65% have been on reducing trend in last few years but still long way to go to match the level of developed countries like USA 5.04%, Canada 4.01%, Australia 4.53%, China 6.45%,South Africa 11.44%, world average 9.8% according to 2010 report of CEA. In India, in fact, some sectors it was found around 40-45% according to expert sources. Fault level is an important criterion in designing the basic parameters of an electrical setup. Fault Level at any given point of the electric power system is the maximum current that would flow in case of short circuit condition.


The aim of this article is to discuss the short circuit requirements of HV switchgear i.e. generator, distribution circuit breakers, both indoors and outdoors in brief for basic understanding of a crucial parameter. It identifies the various short circuit withstand capabilities, their significance and importance in relation to short circuit and other transient conditions that exist in a power system under fault conditions. Many electrical engineers often confuse the real meanings of these requirements and generally depend on manufacturers in selecting the right switchgear for the intended application.

Usual Causes and Nature of Faults

It may be of following types of faults –

  • L–E (Line to Earth)
  • L-L (Line to Line)
  • L-L-L (3 phase fault).

Short circuits are generally caused by insulation failure, flashovers, short circuits, broken conductors, physical damage or human error. Short circuits involving all three phases simultaneously are of symmetrical nature, whilst those involving only one or two phases are asymmetrical faults. The balanced three phase faults are normally analyzed using equivalent single phase circuits. Use of symmetrical components helps to resolve the asymmetrical system faults.

Short circuits do occur even in well-designed power systems, which result in disruptive electro-dynamic and thermal stresses that are potentially damaging. Fire risks and explosions are often inherent.

Sources of Short Circuit Current

There are four basic sources of short circuit current such as-

  • Generators
  • Synchronous Motors
  • Induction Motors
  • Electrical Utility systems.

What to do if it occurs?

It is essential to isolate the power source to the faulty section, and away from the healthy section as quickly as possible in order to protect both equipment and personnel. It is essential that the short circuit withstand ratings of the protected equipment such as transformers, reactors, cables and conductors are not exceeded and thus the consequential damages are either eliminated or limited. Fast isolation/interruption reduces the transient instabilities and the power system will remain in synchronism. The fault current interrupting device is the ‘switchgear’ or the ‘circuit breaker’ & it should withstand the dynamic effects of short circuits.

Purpose of Fault Level Calculations in Power System

  • For selecting short circuit protective device of adequate safety
  • For selecting Circuit Breaker and other switch gear of sufficient Short Circuit making and breaking capacity
  • For designing Bus bar and its supporting structure, cable and other switchgear to withstand thermal and mechanical stresses in case of fault
  • To do current based discrimination of protective switchgear.

Usual Methods for Finding Fault Level

  • Direct Method
  • Per Unit Method.

Short Circuit Analysis

There is plenty of software available in the market to calculate them today but as a design engineer, we should have alternative methods to cross check them too, because it is very crucial parameter in designing of entire electrical setup and its costing will be effected drastically with slight variation in calculated values.

Short Circuit analysis is used to determine the magnitude of short circuit current, the system is capable of producing and compares that magnitude with the interrupting rating of the over current protective devices. Since the interrupting ratings are based by the standards, the methods used in conducting a short circuit analysis must conform to the procedures which the standard making organizations specify for this purpose. Short circuit currents impose the most serious general hazard to power distribution system components and are the prime concerns in developing and applying protection systems. Fortunately, short circuit currents are relatively easy to calculate. The application of three or four fundamental concepts of circuit analysis will derive the basic nature of short circuit currents.

The three phases bolted short circuit currents are the basic reference quantities in a system study. In all cases, knowledge of the three phase bolted fault value is required and needs to be singled out for independent treatment. This will set the pattern to be used in other cases.

Device that interrupts short circuit current, is a device connected into an electric circuit to provide protection against excessive damage when a short circuit occurs. It provides this protection by automatically interrupting the large value of current flow, so the device should be rated to interrupt and stop the flow of fault current without damage to the over- current protection device. Here are reference values that will be needed in the calculation of fault current.

Impedance Values for Three phase transformers

  • HV Rating 2.4KV – 13.8KV 300 – 500KVA Not less than 4.5%
  • HV Rating 2.4KV – 13.8KV 750 – 2500KVA 5.75%
  • General Purpose less then 600V 15 – 1000KVA 3% to 5.75.

Reactance Values for Induction & Synchronous Machine X

  • Sub transient Salient pole Gen 12 pole 0.16, 14 pole 0.21
  • Synchronous motor 6 pole 0.15, 8-14 pole 0.20
  • Induction motor above 600V 0.17
  • Induction motor below 600V 0.25.

Transformer Fault Current

Calculating Short Circuit Current, When Transformer is in the circuit. Every transformer has ‘%’ impedance value stamped on nameplate.
So what does this mean for a 1000KVA 13.8KV – 480Y/277V.
First you need to know the transformer Full Load Amps
Full Load Ampere = KVA / 1.73 x L-L KV
FLA = 1000 / 1.732 x 0.48
FLA = 1,202.85
The 1000KVA 480V secondary full load ampere is 1,202A.

When the secondary ampere meter reads 1,202A and the primary Voltage Meter reads793.5V, the percent of impedance value is 793.5 /13800 = 0.0575. Therefore; % Z = 0.0575 x 100 = 5.75%.

This shows that if there was a 3-Phase Bolted fault on the secondary of the transformer then the maximum fault current that could flow through the transformer would be the ratio of 100 / 5.75 times the FLA of the transformer, or 17.39 x the FLA = 20,903A. Based on the infinite source method at the primary of the transformer, a quick calculation for the Maximum Fault Current at the transformer secondary terminals is –
FC = FLA / %PU Z FC = 1202 / 0.0575 = 20,904A

This calculation can help you determine the fault current on the secondary of a transformer for the purpose of selecting the correct over current protective devices that can interrupt the available fault current. The main breaker that is to be installed in the circuit on the secondary of the transformer has to have a KA Interrupting Rating greater then 21,000A. Be aware that feeder breakers should include the estimated motor contribution too. If the actual connected motors are not known, then assume the contribution to be 4 x FLA of the transformer. Therefore, in this case the feeders would be sized at 20.904 + (4 x 1202 = 25,712 Amps).

Generator Fault Current

Generator fault current differs from a Transformer. Below, we will walk through a 1000KVA example.

800KW 0.8% PF 1000KVA 480V 1,202FLA
KVA = 800 / .8
KVA = 1000
FLA = KVA / 1.732 x L-L Volts
FLA = 1000 / 1.732 x 0.48
FLA = 1,202
(As listed in the table for generator sub transient X” values is 0.16)
FC = FLA / X”
FC = 1202 / 0.16
FC = 7,513A

So, the fault current of a 1000KVA Generator is a lot less then a 1000KVA transformer. The reason is the impedance value at the transformer and Generator reactance values are very different. Transformer 5.75% vs. a Generator 16%.

System Fault Current

Below is a quick way to get a MVA calculated value. The MVA method is fast and simple as compared to the per unit or ohmic methods. There is no need to convert to an MVA base or worry about voltage levels. This is a useful method to obtain an estimated value of fault current. The elements have to be converted to an MVA value and then the circuit is converted to admittance values.

Utility MVA at the Primary of the Transformer

MVAsc = 500MVA

Transformer Data

13.8KV – 480Y/277V
1000KVA Transformer Z = 5.75%
MVA Value

1000KVA / 1000 = 1 MVA
MVA Value = 1MVA / Zpu = 1MVA / .0575 = 17.39 MVA
Use the admittance method to calculate Fault Current
1 / Utility MVA + 1 / Trans MVA = 1 / MVAsc
1 / 500 + 1 / 17.39 = 1 / MVAsc
0.002 + 0.06 = 1/ MVAsc
MVAsc = 1 / (0.002 + 0.06)
MVAsc = 16.129
FC at 480V = MVAsc / (1.73 x 0.48)
FC = 16.129 / 0.8304
FC = 19.423KA
FC = 19, 423 A

The 480V Fault Current Value at the secondary of the 1000KVA transformer based on an Infinite Utility Source at the Primary of the transformer as calculated in the Transformer Fault Current section in this article is 20,904A.The 480V Fault Current Value at the secondary of the 1000KVA transformer based on a 500MVA Utility Source at the Primary of the transformer as calculated in the System Fault Current section in this article is 19,432A.The 480V Fault Current Value at the secondary of the 1000KVA transformer based on a 250MVA Utility Source at the Primary of the transformer the calculated value is 18,790A When the cable and its length is added to the circuit the fault current in a 480V system will decrease to a smaller value. To add cable into your calculation, use the formula. Cable MVA Value MVAsc = KV2 / Z cable. Use the cable X & R values to calculate the Z value then add to the Admittance calculation as shown in this article. The conclusion is that you need to know the fault current value in a system to select and install the correct Over current Protective Devices. The available FC will be reduced as shown in the calculations when the fault current value at the primary of the transformer is reduced. If the infinite method is applied when calculating fault current and 4 x FLA is added for motor contributions, then the fault current value that is obtained will be very conservative. This means the calculated value in reality will never be reached, so you reduce any potential over current protection device failures due to fault current. In view of sizing an electrical setup and the related equipment (HV & LV), as well as determining the means required for the protection of life and property, short-circuit currents must be calculated for every point in the network.

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