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Last updated on 2024年6月30日

What is the formula for calculating short circuit current in a power system?

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Short circuit current, also known as fault current, is the maximum current that flows in a power system when a fault occurs, such as a short circuit between two phases or a phase and ground. It is important to calculate short circuit current for various reasons, such as designing protective devices, assessing system stability, and evaluating equipment ratings. In this article, you will learn the basic formula for calculating short circuit current in a power system and how to apply it to different types of faults and configurations.

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Ali Sepehri

??????????????????????????????????.?????? platform & groups founder/Switchgear Senior Sales Engineer/ Switchgear…
Ahmadreza Eslami

PhD in Electrical Engineering, University of Tasmania

1 Assumptions and simplifications

To simplify the calculation of short circuit current, it is often assumed that the power system is balanced and symmetrical, with all three phases having equal voltage, current, and impedance. Additionally, it is assumed that the power sources are ideal with zero internal impedance, and that the transmission lines are lossless with zero resistance and only reactance. It is also assumed that the transformers are ideal with no leakage flux, no magnetizing current, and no losses, as well as that the loads are negligible and do not affect the short circuit current. Although these assumptions may not accurately reflect the real conditions of the power system, they enable a swift and straightforward estimation of the short circuit current.

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Ali Sepehri

??????????????????????????????????.?????? platform & groups founder/Switchgear Senior Sales Engineer/ Switchgear Technical & Marketing Consult
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The short circuit current (Isc) is the maximum current that can flow in a power system when a fault occurs. It depends on the voltage (V), the total impedance (Z), and the type of fault. Here is a creative poem to help you remember the formula for Isc: Short Circuit Current Poem When a fault occurs in a power system The current can rise to a very high level To calculate the short circuit current You need to know the voltage and the impedance The formula is simple and easy to recall Just divide the voltage by the impedance, that’s all But remember, the impedance is not a constant value It changes with the fault type and the circuit components too.

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2 Basic formula

The basic formula for calculating short circuit current in a power system is Isc = V / Z, where Isc is the short circuit current in amperes (A), V is the pre-fault voltage in volts (V), and Z is the total impedance in ohms (Ω) between the fault point and the power sources. To calculate Z, all components in series between the fault point and the power sources must be taken into account, such as transmission lines, transformers, and generators. The impedance of each component can be expressed as a complex number with a real part (resistance) and an imaginary part (reactance). Additionally, it can be expressed as a magnitude (impedance) and an angle (power factor angle).

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Ahmadreza Eslami

PhD in Electrical Engineering, University of Tasmania
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I think the calculation is wrong. As transformer is modeled as an impedance in pu, on the High V side, the impedance is 0.1and the SC level is Isc1=1/0.1=10pu and on the Low V side: the impedance is 0.1+0.2 the SC level is Isc2=1/.3=3.333pu which is Isc1_HV=5.26 KA in HV and Isc2_LV=48.3. If the SC level in LV transformed to HV, it equals Isc2_HV=1.756 kA. The SC level has reduced because of the impedance introduced by the transformer.

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Shadi Abdelkareem aleryani

+26Million Impressions | +45kFollower |Professional Electrical Engineering at OMV
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In the event that manufacturer data is provided, there is an additional formula for approximate calculations. Equipment that rotate are sources of short circuit current. Isc might just be a normal or rated current/sub-transient reactance, if the values of the sub-transient reactances are known. Every single rotating equipment needs to have this done. The total is represented by the final Isc at the fault point.

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3 Types of faults

Different types of faults can occur in a power system, such as a single line-to-ground (SLG) fault, where one phase is shorted to ground, a line-to-line (LL) fault, where two phases are shorted to each other, a double line-to-ground (DLG) fault, where two phases are shorted to each other and to ground, and a three-phase (3P) fault, where all three phases are shorted to each other and to ground. The type of fault significantly affects the calculation of short circuit current due to the different equivalent circuits and impedances. For instance, a 3P fault has the lowest impedance and the highest short circuit current, whereas a SLG fault has the highest impedance and the lowest short circuit current.

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4 Fault analysis methods

Analyzing and calculating short circuit current in a power system can be done by using different methods, such as the per-unit method, the symmetrical components method, and the direct method. The per-unit method normalizes voltages, currents, and impedances to a common base value, such as the rated power of the system, making it easier to calculate and compare different components and scenarios. The symmetrical components method decomposes an unbalanced three-phase system into three balanced systems, known as positive, negative, and zero sequences. This allows for the calculation of short circuit current for any type of fault using the superposition principle. Lastly, the direct method calculates short circuit current directly from a basic formula using actual values of voltages and impedances. However, this method may not be accurate or applicable for all types of faults and configurations.

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UMUT DONER

ELECTRICAL ENGINEER
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Manual or hand calculation can be performed through IEC 60909 standard. Simulation software such as short circuit tool in ETAP is convenient way to analyze short circuit in power systems. You can select related industrial standard in the tool. You have to purchase or obtain IEC Standard which gives calculation methods, some examples with single lines, electrical parameters of sample values ( particularly impedance of power system components).

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Edgar Soares

Electrical Engineer, EIT
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One more important method to list along the per-unit method (specially if you are interested in the PE exam) is the apparent power method (aka "MVA Method"). In this more-expedient short-circuit calculation method, we can calculate the total available power contribution based on each power sources and line impedances on the oneline diagram, and add them in series or in parallel, depending on the topology of the oneline as well as the location of the fault. . For transformers, S(sc) = S(rated) / %Z . For rotating machines (generators and motors), S(sc) = S(rated) / %X" . For line contributions, S(sc) = V(line)^2 / Z(line)

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5 Example calculation

To illustrate the calculation of short circuit current in a power system, consider the following example: a power system consists of a 10 MVA, 11 kV generator, a 10 MVA, 11/0.4 kV transformer, and a 0.4 kV bus. A 3P fault occurs at the bus with the generator having a subtransient reactance of 0.2 pu, the transformer having a leakage reactance of 0.1 pu, and the pre-fault voltage at the bus being 0.4 kV. To use the per-unit method, we must first calculate the base values - base power: 10 MVA; base voltage: 11 kV for the high side and 0.4 kV for the low side; base current: 10 MVA / 11 kV = 0.526 kA for the high side and 10 MVA / 0.4 kV = 14.5 kA for the low side; base impedance: 11 kV^2 / 10 MVA = 1.21 Ω for the high side and 0.4 kV^2 / 10 MVA = 0.016 Ω for the low side - then we can calculate the per-unit impedances - generator: 0.2 pu; transformer: 0.1 pu * (11 kV / 0.4 kV)^2 = 7.59 pu; total: 0.2 pu + 7.59 pu = 7.79 pu - and finally, we can calculate the per-unit short circuit current at the bus as 1 pu / 7.79 pu = 0.128 pu with an actual short circuit current at the bus of 0.128 pu * 14.5 kA = 1.86 kA.

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Edgar Soares

Electrical Engineer, EIT
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To solve the example with the MVA method: ? Both the generator and the transformer are in series with the bus (= series addition); ? 0.2pu and 0.1pu in the example are 20% X" and 10% %Z, respectively; Individual contributions: ? S(gen) = 10 MVA / 20% = 50 MVA ? S(xfmr) = 10 MVA / 10% = 100 MVA Power sources add in the same way capacitors do, so to add the above contributions in series: ? S(total) = 1 / ((1 / S(gen) + (1 / S(xfmr)) = 33.3 MVA Therefore, 33.3 MVA is the total power available at the bus, where the fault occurred. Finally, the short-circuit current can be calculated in the following way: ? I(sc) = S(total) / (sqrt(3) * V(bus)) = 48.1 kA

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6 Here’s what else to consider

This is a space to share examples, stories, or insights that don’t fit into any of the previous sections. What else would you like to add?

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What is the formula for calculating short circuit current in a power system?

1

2

3

4

5

6

1 Assumptions and simplifications

2 Basic formula

3 Types of faults

4 Fault analysis methods

5 Example calculation

6 Here’s what else to consider

Electrical Engineering

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What is the formula for calculating short circuit current in a power system?

1

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3

4

5

6

1 Assumptions and simplifications

2 Basic formula

3 Types of faults

4 Fault analysis methods

5 Example calculation

6 Here’s what else to consider

Electrical Engineering

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