# The art of protecting electrical systems, part 15—calculating fault currents

## Part 15 of an ongoing series about protecting electrical systems offers tables for a quick method of determining fault current at the ends of feeders of various size and length.

01/17/2008

Part 15 of an ongoing series about protecting electrical systems offers tables for a quick method of determining fault current at the ends of feeders of various size and length.

Previous articles in this series have covered per-unit calculation of fault currents for both single-phase and three-phase systems. We have also presented several short-cut methods of estimating fault current using point-to-point calculations and transformer short-circuit tables.

This installmentsecondary voltages of 208, 240 and 480.

The second set has calculations for 120-V faults on 120/240-V, single-phase transformers from 25 kVA through 500 kV. An impedance diagram is shown in Figure 15.1.

Primary fault power for the tables is 500,000 kVA. Transformer impedances are the same as used in the transformer tables. Cable resistance and reactance values are based on THHN/THWN insulated copper conductors in plastic conduit at 77 F. These values, taken from the IEEE/ANSI Buff book, are the same as used in previous articles.

Conductor sizes begin at #14 wire and are shown for many typical values. In many cases, the largest conductors shown have ampacity significantly larger than the full-load current of the transformer. These were included in case conductors are being oversized for voltage drop or to provide for future installation of larger transformers.

No motor contribution is included in these calculations. When significant motor load is present in the system, it must be included. Motor load may be added at the end of the feeders, or more complete short-circuit calculations can be made, as previously described. A value for four times connected motor full-load current is a suitable approximation.

Example: Consider a 250 kVA, 120/240-V, single-phase 1.9% impedance transformer. The available root-mean square symmetrical fault current for a phase-to-neutral fault at the end of a 50-ft run of 1/0 copper conductor in non-magnetic conduit can be determined using the table for a 250-kVA, low-impedance transformer from the second set of tables.

Note that the fault current at the transformer terminals (zero feet) is 86,000 amps. Read down the left side of the table to find 1/0 conductor. Read horizontally to the right until you intersect the column headed “50 feet” and read available fault current as 65,700 amps. If there is less than 500 mVA available at the primary of transformer, or if the conductor is in magnetic conduit, available fault current will be somewhat less.

Both the transformer short-circuit tables included in Part 13 and Part 14 and the cable tables included here provide an excellent method of estimating the short-circuit current in a variety of circumstances quickly. However, they must be used with caution.

These tables are most useful in preliminary studies to help select types of system components and to analyze the effects of various circuit arrangements (making “what if” comparisons). However, they do not replace more complete studies in any but the simplest systems.

In particular, if tables values indicate an available fault current within 70% of interrupting or withstand ratings of components being considered, either careful point-to-point or per-unit calculations must be performed. This will be treated in depth in a future installment to this series discussing UL test standards for various components. Upcoming installments also will discuss the method of symmetrical components.

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