{"id":4306,"date":"2017-07-18T09:05:07","date_gmt":"2017-07-18T16:05:07","guid":{"rendered":"https:\/\/brainfiller.com\/?p=4306"},"modified":"2025-11-21T14:24:52","modified_gmt":"2025-11-21T21:24:52","slug":"utility-short-circuit-data-different-formats","status":"publish","type":"post","link":"https:\/\/brainfiller.com\/es\/electrical-safety-codes-and-standards\/utility-short-circuit-data-different-formats\/","title":{"rendered":"Datos de cortocircuito de servicios p\u00fablicos: diferentes formatos"},"content":{"rendered":"\t\t
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National Electrical Code 110.9 Interrupting\u00a0Ratings<\/strong>\u00a0states that:<\/p>

Equipment intended to interrupt current at fault levels shall have an interrupting rating at nominal circuit voltage at least equal to the current that is available at the line terminals of the equipment.<\/em><\/p>

\"Arc
Download FREE Arc Flash Calculations<\/figcaption><\/figure>

To comply with this requirement, \u00a0a short circuit studies is typically performed to determine the available fault current \u00a0for comparison to the protective devices interrupting rating. \u00a0\u00a0The results of a short circuit study are also a critical component \u00a0for other studies such as an arc flash study. Requesting the available short-circuit data from the electric utility company should be one of the first tasks in performing the study. This information is very important because it defines the magnitude of current that could flow from the utility and is used as a starting point for arc flash calculations.<\/p>

In addition to requesting this data for normal operating conditions, for an arc flash study the request should also include minimum short-circuit current conditions, if available. The minimum condition could be for a utility transformer or transmission line out of service or similar scenario. The minimum value can then be used to determine if the lower current could result in a protective device operating more slowly, which may increase the total incident energy during an arc flash.<\/p>

Having been in charge of the Short Circuit Studies group for a very large electric utility company in a past life, the accuracy of the results depends on the accuracy of the data and, the system configuration at the time (which can seem like a moving target).\u00a0 The available short circuit current can change over time as well.\u00a0 New substations, transmission and distribution lines and other system enhancements can serve to increase the short circuit current for a given area so it is important to review this data periodically.<\/p>

No Standard Format \u2013 Now What?<\/strong><\/h2>

There is no “standard” format for short-circuit data from an electric utility company. \u00a0Different utilities may provide this data in a variety of different formats.\u00a0 Several of the more common formats include:<\/p>

Short-circuit amperes (A)
Short-circuit megavolt-amperes (MVA)
Per-unit and symmetrical components<\/p>

Of course, with multiple formats, confusion could (and often does) result. I will compare the different formats using a three-phase short-circuit current of 6,000A at the 23-kilovolt (kV) level. Since arc flash calculations are based on a three-phase model, only the three-phase short-circuit calculations are used. Some of the values are slightly rounded.<\/p>

Short-circuit ampere format<\/strong><\/h2>

This is the simplest format because it defines the short-circuit current in terms of amperes at a specified location. As an example, the utility has provided the following information:<\/p>

Short-circuit amperes\u00a0three-phase<\/sub>\u00a0= 6,000A
Voltage = 23 kV\u00a0line-to-line<\/sub><\/p>

Since the data is already in terms of amperes, no additional calculations are necessary.<\/p>

\"How
BUY NOW:<\/strong> How To Perform a Short Circuit Study<\/figcaption><\/figure>

Short-circuit MVA format<\/strong><\/h2>

Many electric utility companies often provide short- circuit data in terms of short-circuit MVA. This format combines the short-circuit current with the voltage and the square root of 3 (for a three-phase representation) to provide the data in terms of short-circuit power. Below is an example of the MVA format.<\/p>

Three-phase short-circuit
MVA = 240 MVA
Voltage = 23 kV\u00a0line-to-line<\/sub><\/p>

To convert three-phase short-circuit MVA to short-circuit current in amperes, use the following equations:<\/p>

Short-circuit amperes = [MVA x 1,000] \/ [kV\u00a0line-to-line<\/sub>\u00a0x the square root of 3]
where 1,000 is the conversion from MVA to kVA
Short-circuit amperes = [240 MVA x 1,000] \/ [23 kV\u00a0line-to-line<\/sub>\u00a0x 1.732]
Short-circuit amperes = 6,000A<\/p>

Per-unit and symmetrical components format<\/strong><\/h2>

The per-unit and symmetrical component format can appear to be the most complex of all. The term \u201cper-unit\u201d is simply the decimal equivalent of percent, i.e., 50 percent is equal to 0.5 per unit. In general, the per-unit method takes every electrical quantity and scales it by a reference value known as a base quantity. The utility derives the base values from two numbers: the<\/p>

MVA\u00a0base<\/sub>\u00a0and kV\u00a0base<\/sub>.<\/p>

Symmetrical components is a method used for solving complex unbalanced power system problems. Such terms as positive, zero and negative sequence are part of the vocabulary of this method, and although the actual theory can be quite complex, calculating the short-circuit current using this approach is not that difficult.<\/p>

The example below illustrates short-circuit data using the per-unit system and symmetrical components:<\/p>

MVA\u00a0base<\/sub>\u00a0= 100 MVA
kV\u00a0base<\/sub>\u00a0= 23 kV\u00a0line-to-line<\/sub>
Z1\u00a0<\/sub>= 0.418 p.u.<\/p>

Z1<\/sub>\u00a0is referred to as the positive sequence impedance and represents the equivalent impedance of the utility in this case. One hundred MVA and 23 kV are the base power and voltage used to determine the \u201cbase values\u201d necessary for the calculations.<\/p>

For the three-phase short-circuit current, only three steps are needed to convert the per-unit and symmetrical component values to short-circuit current in amperes:<\/p>

\"AnnouncementStep 1:<\/strong>\u00a0Calculate the base current (I\u00a0base<\/sub>) using the following equation:<\/p>

I\u00a0base<\/sub>= [MVA\u00a0base<\/sub>\u00a01,000] \/ [kV\u00a0base<\/sub>\u00a0x the square root of 3]
= [100 MVA x 1,000] \/ [ 23kV x the square root of 3]
= 2,510A<\/p>

Step 2:<\/strong>\u00a0Calculate the per-unit three-phase short-circuit current (I p.u.) with the following equation:<\/p>

I\u00a0p.u.<\/sub>\u00a0= V\u00a0p.u.<\/sub>\u00a0\/ Z1<\/sub><\/p>

V\u00a0p.u.<\/sub>\u00a0in the equation above is the per-unit voltage. In the absence of being provided the per-unit voltage, which is usually the case, it is common to assume it is 1.0 p.u. This means the actual voltage is 100 percent of the base voltage, so for this example:<\/p>

V\u00a0p.u.<\/sub>\u00a0= 1.0
I\u00a0p.u.<\/sub>\u00a0= 1.0 \/ 0.418 = 2.39\u00a0p.u.<\/sub><\/p>

Step 3:<\/strong>\u00a0Convert per-unit short-circuit current to amperes with the following equation:
I\u00a0amperes<\/sub>= I\u00a0p.u.<\/sub>\u00a0I\u00a0base<\/sub>
= 2.39\u00a0p.u.<\/sub>\u00a02,510A
= 6,000A<\/p>

Different methods = same results<\/strong>
Although the three methods seem quite different from each other and some are more complicated, they all produce the same result, which can be used as the starting point for arc flash calculations.<\/p>

For more about how to perform short circuit calculations, check out our 4 DVD set:<\/p>

How to Perform Short Circuit Calculations<\/a><\/strong><\/span><\/p>

You can also check out (and subscribe to) our YouTube channel for more video instruction about per unit and symmetrical components.<\/span><\/p>