# The art of protecting electrical systems, part 13: calculating short-circuit current at secondary of transformers

## In this 13th installment of an ongoing series, the authors present calculations for available fault currents at the secondaries of three-phase transformers having secondary voltages of 120/208 to 600 V.

10/16/2007

Editor’s Note: From 1965 through 1970, Consulting-Specifying Engineer ’s predecessor, Actual Specifying Engineer, ran a series of articles on overcurrent protection. Due to the immense popularity of the 31 installments in the series, the authors, George Farrell and Frank Valvoda, PE, reprised the series in an updated version beginning in the Februart 1989 issue of CSE. Over the years since the last installment ran in the late ’90s, we have received many requests to re-run this series. Valvoda passed away in December. 2001, and his long-time friend and editorial partner, George Farrell, passed away in early 2006.

In the last part (Part 12) of this series, we presented a point-to-point method of calculating short-circuit currents. Known as the LICE method, this shortcut is based on circuit parameters of conductor length, initial fault current, reciprocal of the conductor’s impedance (C value), and operating voltage.

Part 12 presented the C-values for wire and cable of various sizes and constructions. This article presents available fault currents at the secondaries of three-phase transformers having secondary voltages of 120/208 to 600 V. Short-circuit contribution to the transformers’ primary terminals are included, allowing rapid determination of available root-mean-square (rms) symmetrical fault current for systems with available primary fault energy from 50,000 kVA through unlimited kVA.

In addition, motor contribution is indicated. This may be added to transformer fault current when applicable. Grounded neutral systems are assumed to have 50% connected motor load (208- and 480-V, 3-phase, 4-wire systems), while other systems are assumed to have 100% motor load (240-, 480- and 600-V, 3-phase, 3-wire systems). As covered in earlier articles, the contribution from motors may be added at the main switchboard or at the point where the motors are connected to the system, depending on the accuracy desired.

Values can be calculated directly using the following formulas. This is particularly useful in cases where transformer impedances are other than those assumed in the IEEE “Buff” book tables. A detailed discussion of the formulas appears in Part 8 in this series.

Data required:

Short circuit kVA at primary of transformer (kVA SC )

Source X/R (X/R SOURCE ) assumed at 15 for these calculations.

Transformer rated kVA (kVA T )

Transformer percent impedance (%Z T )

Transformer X/R (X/R T )

Transformer secondary volts line-to-line (E L-L )

A kVA base (kVA base ) of 1,000 is used in these calculations.

Source per unit impedance =

PUZ SOURCE = kVA base / kVA SC

Source per unit resistance =

SOURCE = ((PUR SOURCE )2)/(1 + (X/R SOURCE )2)0.5

Source per unit reactance =

SOURCE = X/R SOURCE X PUR SOURCE

Transformer per unit impedance =

T = (kVA base x %Z T )/(kVA T x 100)

Transformer per unit resistance =

PUR T = ((PUZ T )2)/(1 + (X/R T )2)0.5

Transformer per unit reactance = PUX T = X/R T x PUR T

PUR TOTAL = PUR SOURCE + PUR T

PUX TOTAL = PUX SOURCE + PUX T

PUZ TOTAL = ((PUR T )2 T )2)0.5

Short-circuit current =

I SC = (kVA base /(30.5x (E L-L /1,000)))/PUZ TOTAL

X/R at transformer secondary = PUX TOTAL /PUR TOTAL

Sample calculation

Assume a system with the following parameters: 1,500-kVA transformer, 480 V, 3-phase, 3-wire, %Z = 5.75%, transformer X/R = 7, utility fault available = 150,000 kVA = 150 mVA, utility X/R = 15, kVA base (assumed) = 1,000.

PUZ SOURCE = 1,000/150,000 = 0.006667

PUR SOURCE = (0.0066672/(1 +152))0.5= 0.000443

PUX SOURCE = 0.000443 x 15 = 0.006652

PUZ T = (1,000 x 5.75)/(1,500 x 100) = 0.038333

PUR T = (0.0383332/(1 + 72))0.5= 0.005421

PUX T = 0.005421

PUR TOTAL = 0.000443 + 0.005421 = 0.005864

PUX TOTAL = 0.006652 + 0.037948 = 0.044600

PUZ TOTAL = (0.0058642 + 0.0446002)0.5= 0.044984

I SC = 1,000 / (=3 x 480/1,000) x 1/0.044984 = 26,739 amperes

X/R TOTAL = 0.044600 / 0.005864 = 7.61

I FL =

Motor contribution at 100% = 4 x 1,804.2 = 7,217 amperes

I SC TOTAL =

Low-impedance transformers

Recent years have seen an increasing use of low-impedance transformers by both electric utilities and users. In particular, underground distribution has resulted in the universal application of pad-mounted transformers, which may have impedances as low as 1.6% to 2% through 500 kVA.

The result has been a marked fault-current increase for even moderately sized installations. For example, a popular fast food chain uses an 800-ampere, 120/208-V service for its restaurant locations. In many areas of the country, utility companies have been indicating fault currents in excess of 40,000 amperes. This amount of fault current at a small, free-standing restaurant would have been rare 20 years ago.

The next article in this series will contain tables for fault at the secondary terminals of single-phase transformers, and will begin the presentation of tables that provide fault currents at the end of radial feeders of varying length and size originating at transformers.

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