Short-Circuit Calculation

Determining the available short-circuit Amperes or kVA values is one of the most important aspects of the electrical power distribution system design. Short-circuits must be expected, and since no one knows just where in the system they will occur, every part of the system must have adequate interrupting capacity (IC). Inadequate devices subjected to severe faults can fail to clear the fault and become hazardous to persons and equipment. On the other hand, arbitrarily oversized protective devices are needless extra cost. A realistic estimate of the short circuit current protection is necessary.

The short-circuit current that a low-voltage protective device (600-V or less) must interrupt on the low-voltage (secondary) side of a transformer originates from two sources:

1. The major portion comes through the transformer from the primary system. This current is limited by the impedance of the transformer and the short-circuit capacity of the primary system.
2. The second source of current flowing through the protective device into the short-circuit comes from any motors connected to the low-voltage bus through other feeders. These motors act as current generators for a short while after the short circuit occurs, taking energy from the spinning motor and converting it into electrical energy.

The protective device must interrupt the current flowing through it at the instant that its contacts part after the fault. This depends upon the speed of operation, as shown here.

This figure shows the short-circuit current to be made up of two components:

1. The symmetrical alternating current component.
   
2. The direct-current component which decays with time.

This total current as pictured is called "the asymmetrical current." If the contacts are prevented from separating until the DC component has dropped to a low value (as is sometimes done with power circuit breakers), the AC component is what is used to define the interrupting duty. This is called "the symmetrical short-circuit current." It is determined by the AC voltage and the impedance of the system as shown at figure on left.

The magnitude of the DC component depends upon the exact point in the voltage wave corresponding to the instant the short-circuit happened. Obviously, this cannot be predicted; but the worst case condition can be - and the system should be designed to withstand that. This worst case (maximum value) DC component is 1.41 times the peak value of the symmetrical sine wave. This gives a value for the peak asymmetric current of 2.3 times the RMS symmetric short-circuit current when the system X/R ratio is 6.6 to 1.0. The rate of decay of the DC component is slowed when the X/R ratio is high.

When a fault occurs, the system voltage is pulled down by the short circuit, and the induction and synchronous motors act as amperage generators because their magnetic fields persist for some time while they continue to rotate. These motors deliver current to the short-circuit in an amount about equal to the combined starting currents for all of the motors running at the time of the short-circuit. This only lasts for a short time, about three cycles for typical induction motors and a few seconds for synchronous motors. This motor contribution is significant if the circuit breaker contacts begin to separate before this current decays. Many circuit breakers begin to open in less than one cycle. Therefore, this motor contribution and the DC contribution both need be considered when determining the maximum possible short-circuit current.

Since the short-circuit can occur at any time, it is not possible to know how many motors will be running. For estimating purposes, some rules of thumb are:

1. For systems predominantly supplying motor power at 240-Volts or higher, assume the kVA of running motors to be equal to the transformer nameplate kVA.
2. For 240-Volt systems predominantly supplying lighting, assume the motor kVA to be 50 percent of the transformer capacity.
3. For 208-Volt systems, assume the motor load to be 50% of the transformer installed capacity.

These values should be modified to fit the particular conditions in the plant.

Motor contributions to short-circuit requirements generally assume an impedance of 25% This means that the motors contribute four kVA to the short-circuit current for each kVA of motor load as calculated by the rule of thumb given above.

Frequently, the symmetric short-circuit current that a protective device must clear is calculated with only a single transformer as the limiting impedance, and the power system plus connected running motors considered as the sources of current. This results in a pessimistic, and somewhat exaggerated estimate of the interrupting duty required, but errs on the side of safety and is certainly easy to calculate. If a circuit breaker or fuse has an IC adequate for this current, it could be used. If this calculation calls for use of a huskier and more costly device, the designer should complete the calculation which includes the impedance of circuit elements and conductors upstream of that point. This more precise analysis might show that another device with lower IC and lower cost could be safely used.

These more precise calculations require that an impedance diagram be drawn showing the resistances and reactances of all significant elements of the system upstream of the point in question. The process is outlined in the protection handbook and in the IEEE Buff Book (and other sources). The Buff Book* lists three tables for "Available Three-Phase Symmetrical RMS Fault Currents" for 208-Volts, 240-Volts, and 480-Volts, based on the size of the transformer and typical conductor layouts.

*IEEE Std. 242-1975: Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems.