This installment of our ongoing series introduces short-circuit calculations, starting with the fundamentals and proceeding on through available computer programs.

In previous articles, we discussed the importance of overcurrent protection, presented fundamental theorems used in short-circuit calculations and examined component short-circuit ratings. This article begins the study of calculation methods.

The fundamental electrical quantities used in short-circuit calculations are voltage, current and impedance, i.e.,

Current = K1 x K2

where K1 is a constant, depending on whether the circuit is single-phase or three-phase, and K2 is a constant by which the various levels of voltage and kVA in a distribution system system are related to each other.

Typical short-circuit calculations are carried out using line-to-neutral values of impedance, with K1 = 1. Tabulated values of impedance for conductors and other circuit components are always given in line-to-neutral ohms. Both resistance and reactance are expressed this way.

K2 is selected usually by the per-unit method (discussed later) in which impedances will be adjusted for the various voltages and equipment kVA levels. In this way, they might be manipulated during calculations, reducing a complex system of impedances to the Thevenin equivalent of the faulted circuit.

Sometimes the resistance is ignored in short-circuit studies, with only the reactance considered. Calculations based only on reactance indicate larger short-circuit currents than may actually occur. In most systems, this is not significant, as the errors introduced are on the safe side.

Before the days of the digital computer, the great amount of calculations in dealing with the complex quantities of impedance (=(resistance2+ reactance2)) usually made the reactance-only calculations the method of choice. Now, however, by using computer programs, more accurate results may be quickly obtained with both resistance and reactance being studied.

The IEEE recommended practice for low-voltage short-circuit studies mandates that both resistance and reactance calculations be made. Also, calculations for comparison to electrical equipment ratings for alternating current (AC) high-voltage circuit breakers rated on both asymmetrical current and total current basis involve both resistance and reactance (per ANSI/IEEE standards).

Another important reason to use resistance in calculations is that an X/R ratio may be determined thereby providing a way to calculate asymmetrical current. To simplify this discussion of the per-unit method, however, only the reactance of systems will be illustrated; resistance calculations use the same formulas and methodology.

Using the per-unit method

In the per-unit method, all impedances are assigned a value to a chosen kVA base. Generally, the kVA base is related to the magnitude of values found in the system, typically, 1,000, 10,000 and 100,000 kVA. Once all impedances have been converted to the chosen base they may be combined without further consideration of voltage or kVA rating levels.

To better understand the per-unit method, consider an example from another field. In this case, the Widget Sales Company tracks its sales every month by total sales and by sales per sales person. Figure 8.1 shows the sales figures for the first six month of the year. Note that although Smith’s sales for May were only $7,500, his per-unit sales for the month on a $1 million base were $75,000—by far his best month. Thus, the per-unit method has provided a method by which an appraisal of Smith’s performance may be readily achieved.

Similarly, in short-circuit calculations, the impedance of all elements of the system must be considered (sales for all months and a total sales base). This includes the equivalent utility reactance, reactances of all static items (cable, busway, current transformers, switches) and reactances of all dynamic items (transformers, motors, generators). Formulas for application of the per-unit method to electrical systems for short circuit calculations are indicated below.

For cable, bus and busway, overhead lines, current transformers, switches, etc., where the reactance is usually given in ohms:

FORMULA 1

PUX = per-unit reactance = percent reactance

FORMULA 2

PUX (on chosen kVA base) = ohms reactance x kVA base 2

where ohms is the line-to-neutral value for a single conductor, kVA is the chosen three-phase base kVA and kV is the line-to-line kilovolts.

For example, the reactance of single-conductor, non-shielded 500-kcmil, 600-volt cable in magnetic raceway is 0.0466 ohms per thousand feet. The per-unit reactance of 500 feet of cable operating at 480 volts on a kVA base of 1,000 is:

PUX = 0.0466/2 x 1,000 = 0.1011 pu (per unit)

Working with generators, motors, transformers, etc., where the reactances are usually given in percent to their own kVA rating, the reactance values are first changed to a per-unit value on their own kVA base by the use of Formula 1, and then changed to a per-unit value on the chosen kVA base by use of Formula 3:

FORMULA 3

PUX (on chosen kVA base) = PUX on kVA rating x kVA base

For example, given a 5.75%-reactance, 1,500-kVA transformer what is the per-unit reactance to a 10,000-kVA base? From Formula 1:

PUX (on kVA rating) = 7.75% = 0.0575 pu

and from Formula 3:

PUX (on base kVA) = 0.0575 x 10,000 = 0.3833 pu

It is necessary to obtain the available fault current from the utility company when making a short-circuit study. The data are usually in the form of symmetrical amps at a given X/R ratio at the supply voltage, although it may be furnished as short-circuit kVA.

If the available fault current is given in asymmetrical amps, the symmetrical fault current may be calculated using the following:

FORMULA 4

Symmetrical RMS amps = assymetrical amps

where the X/R ratio multiplier may be obtained from the tabulation given in the third part of this series. Conversion to PUX may be made using the following:

FORMULA 5

PUX (on base kVA) = 3 x kV

where kV = line to line utility company voltage

Example:kVA base = 10,000

Fault current = 65,000 amps @ 208 volts

PUX = = 0.4270 pu3 x 0.208

If the available fault energy is given in short-circuit kVA, convert to PUX by the use of the following formula:

FORMULA 6

PUX =

Example: base kVA = 10,000. Utility short-circuit = 350 mVA

PUX = 10,000 = 0.02857 pu

Sometimes the available fault current is given as a percent reactance to some base kVA. In this case, convert to a per-unit on a chosen kVA base by Formula 3.

As shown in the discussion of the Thevenin Theorem, it is necessary to combine reactances to obtain the total (or net) reactance (see the Formula Box for Formula 7 through Formula 11 ).

When all the reactances in the system have been reduced to one reactance, PUX, the available fault current in symmetrical ampsis obtained by the following:

FORMULA 12

RMS symmetrical amps = 3 x kV

Example:base kVA = 10,000kV = 4.16PUX = 0.0694

RMS symmetrical amps = 3 x 4.16

And finally, if desired, FORMULA 13:

Symmetrical short-circuit kVA= base kVA

Example:

Symmetrical short-circuit kVA= 10,000

Figure 2 presents these formulas in a convenient way. Other per-unit method formulas are included in the figure for reference. Future articles in this series will show how all these formulas are applied.

The next article in this series will demonstrate manual calculations for a typical faulted electrical system.

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Related Stories:

The Art of Protecting Electrical Systems, Part 1: Introduction and Scope

The Art of Protecting Electrical Systems, Part 2: System Analysis

The Art of Protecting Electrical Systems, Part 3: System Analysis

The Art of Protecting Electrical Systems, Part 4: System Analysis

The Art of Protecting Electrical Systems, Part 5

The Art of Protecting Electrical Systems, Part 6

The Art of Protecting Electrical Systems, Part 7: Equipment Short Circuit Ratings

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