In-depth explanation of the short-circuit behaviors of five important types of electrical equipment

I described in a previous article the benefits of learning how to do short-circuit analysis by hand. There are five primary elements that you need to know how to model when doing so. I go into the details here . . . .

UTILITY

The off-site utility is the primary source of fault current during a short circuit. In a hand calculation, it can be an actual off-site utility or an approximation of an upstream bus in the same plant that’s just modeled as a utility. Its function is to essentially supply an “infinite bus” from which an infinite amount of fault current can be supplied to downstream faults. In terms of a Thevenin equivalent circuit, assuming an “infinite bus” simply means assuming an ideal voltage source that has zero source resistance. The effect of this is that the source voltage does not decline no matter what value of current the source supplies. The utility characteristics are provided by the power company, and sometimes an MVA rating is all that is available.

TRANSFORMERS

Transformers “let through” fault current and are usually supplied on their primary side by a very large utility bus, or at least an assumed infinite bus. Along with their primary and secondary-side voltage ratings and overall rated power, there is typically a per-unit impedance associated with a power transformer. It’s denoted as “Z” and is given as a percentage of the rated primary voltage, such as Z = 5.25%. It acts as a multiplier on the full-load current during a short circuit. The short-circuit current available on the transformer’s secondary side is calculated as ISC = IFLA ÷ Z, as in 100 full-load amps ÷ 0.0525 = 1905 short-circuit amps. As a multiplier, it becomes 1 ÷ 0.0525 = 19.05, as in 100 amps x 19.05 = 1905 amps. The impedance, Z, is usually listed on a plant’s single-line diagrams near the transformer’s symbol.

“Z” is a measured value. A transformer’s nameplate is stamped with its measured Z value before it leaves the factory (though preliminary values that are close to the actual values are given on datasheets). It’s measured by attaching a voltage transformer with a voltmeter to the transformer’s input (primary) and an ammeter in series with its short-circuited output (secondary). The voltage on the primary is increased until the ammeter on the secondary reads the rated full-load current. To give an example of how this process works, if we had a 4160-to-480-volt transformer with a Z of 5.25%, that means an input voltage of 218.4 volts produced the secondary side’s full-load current. In other words, 218.4 ÷ 4160 = 0.0525, or 5.25%.

A three-phase transformer’s full-load current is calculated using the following formula:

I_{FL} = \frac{kVA}{kV\cdot \sqrt{3} }

Where kVA is the transformer’s rated power and kV is the line-to-line secondary voltage. The equation for a single-phase transformer is similar and given by:

I_{FL} = \frac{kVA}{kV }

Where kVA is the transformer’s rated power and kV is the secondary voltage.

MOTORS

Motors act as a source of short-circuit current at the beginning of a fault. Though they consume power during normal operation, because of Ampere’s Law and Faraday’s Law they tend to become momentary generators when they resist the voltage disruptions created during a fault by generating opposing voltages that produce currents until they dissipate their latent momentum. They contribute a peak amount of short-circuit current at the beginning of the fault that decays to zero shortly afterward. The amount of fault current they contribute is usually denoted as a multiple of their full-load current (typically around six times), and this multiple varies depending on the type of motor.

CABLES

Cables reduce the amount of fault current delivered to a branch circuit because they increase the circuit’s impedance. In accordance with Ohm’s Law, current is controlled by a circuit’s impedance, and current and impedance are inversely proportional to one another. The longer a cable, the higher its impedance, and the lower the fault current.

GENERATORS

Generators act as a source of fault current. Even though a “utility” is ultimately powered by a generator (i.e. a power plant, whether it is coal, hydro, or nuclear uses its energy production to spin a generator), in the context of calculating short-circuit currents it usually means a local generator whose characteristics are known, like an emergency diesel generator.

The initial magnitude of the generator’s fault current depends on its sub-transient reactance, X’’d. This number, again given as a per-unit value, is often stated on one-line diagrams near the generator symbol. The generator’s short-circuit (or instantaneous) fault current is usually calculated as ISC = IFLA ÷ X’’d .

Calculating a generator’s instantaneous short-circuit current using only its reactance to the neglect of any resistive component is an approximation, but it’s realistic. Physically, it’s really the impedance, Z, that controls the current value. But since impedance is made up of a resistive component and a reactive component (Z = R + jX), and since in the transient cases considered here (a fault) the reactive component becomes dominant compared to the resistive component, only the reactances are used.

For a more detailed explanation of the source of a generator’s subtransient reactance, click here.