# The Art of Protecting Electrical Systems, Part 14 - Single-Phase Short Circuit Calculations: A Step-By-Step Guide

## Part 14 of this continuing series provides detailed information on how to calculate four types of single-phase faults.

* 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,e Farrell, passed away in early 2006. Part 14 of this continuing series provides detailed information on how to calculate four types of single-phase faults.*

Previous articles in this series have presented several methods for calculating available short-circuit current in three-phase systems with three phases shorted. While the fundamental principles are the same for single-phase short circuits, whether the fault is line-to-line or line-to-neutral, a slightly different approach is required to compensate for the different voltages and the effect of single-phase impedance.

Table 14.1 is a continuation of the transformer tables presented in the previous article (C-SE, January 1992). Those tables listed the faults at the secondary of three-phase transformers with “low” and “standard” impedance.

Table 14.1 gives the fault current at the secondary terminals of single-phase, low- or standard-impedance transformers at both 120 and 240 volts. Motor contribution is not considered, as it is seldom significant on single-phase systems. Transformers from 25 through 500 kVA are included. The impedance diagram for these tables is shown in figure 14.5.

This article and the next consider four types of single-phase faults as illustrated in figures 14.1 through 14.4. In each case, the utility source is a three-phase system. In making single-phase faults calculations, the same base kVA may be used as in three-phase calculations.

The following conventions apply to all calculations:

The impedance of the ground return path for line-to-neutral and line-to-ground faults varies widely and is almost impossible to predict accurately. For example, the neutral conductor may or may not carry part of the current or there may be a grounding conductor in parallel with metallic raceways. Therefore, in making single-phase short-circuit calculations necessary to select interrupting ratings and withstand ratings of system components, impedance of the neutral, the ground conductor or both is assumed to be equal to the line conductor. This approximates the most severe line-to-neutral fault conditions.

Resistance, reactance and impedance values used in three-phase calculations are line-to-neutral values for sources, devices and conductors. In typical three-phase calculations, they are combined vectorially. In single-phase faulted circuits, current flows “out” in one path and “returns” in the other (impedances are in series), so resistance, reactance and impedance values must be doubled.

**Line-to-line, single-phase faults**

Consider the simple 4,160Y/2,400-volt system shown in Figures 14.1 and 14.2. Assume the utility short-circuit kVA (kVAsc) is 500,000 and X/R ratio is 15. Refer to Figure 14.1 for calculation of single-phase, line-to-line faults on a three-phase system:

Source (utility) per-unit impedance to a 1,000-kVA base (kVA _{b} ) is:

PUZ _{UT} = kVA _{b} /kVA _{SC} = 1,000/500,000 = 0.002 and

Per-unit resistance (PUR _{UT} ) is

PUR _{UT} = PUZ _{UT} /Ö(1 + (X/R)²) = 0.002/Ö (1 + 15²) = 0.000133

Per-unit reactance (PUX _{UT} ) is:

PUX _{UT} = (X/R) x PUR _{UT} = 15 x 0.000133 = 0.001995

Conductor resistance and reactance values for the 4,160/2,400 volts incoming service are assumed; they are of no particular size, configuration or length. Resistance of the cable (R _{C} ) is 0.00433 ohms. Cable reactance (X _{C} ) is 0.00319 ohms.

Cable PUR (PURc) = (Rc x kVA _{b} )/(1,000 x kV

Cable PUX (PUXc) = (Xc x kVA _{b} )/(1,000 x kV

The totals on a line-to-neutral basis are:

PUR _{T} = 0.000133 + 0.000250 = 0.000383

PUX _{T} = 0.001995 + 0.000184 = 0.002179

Total per-unit line-to-neutral impedance is:

PUZ _{T} =Ö(0.000383² + 0.002179²) = 0.002212

As stated previously, this impedance value must be doubled in calculating single-phase fault current. Short-circuit current at 4,160 volts for a line-to-line single-phase fault on a three-phase system is calculated as follows:

I _{SC} = (E _{L-L} x kVA _{b} ) / (2 x PUZ _{T} x kV

Another approach is:

kVAsc = (kVA _{b} /2) _{T} = 1,000/0.004424= 226,040

Symmetrical short-circuit current =

(1,000 x kVA _{SC} ) / E _{L-L} = (1,000 x 226,040) / 4,160 = 54,337 amps

**Line-to-neutral, single-phase faults**

Figure 14.2 illustrates short-circuit current for single-phase, line-to-neutral faults on three-phase systems. If the neutral conductor is the same size as the line conductor, the following formula applies:

Symmetrical short-circuit current L-N = I _{SCLL} /Ö3

In some systems, the neutral conductors are not the same size as the line conductors. Their size often is reduced when they carry only unbalanced current from a single-phase load.

However, to accommodate harmonic current being generated by solid-state devices and other nonlinear sources, neutral conductors are being specified larger than line conductors in an increasing number of systems. (Increased conductor ratings to handle harmonic currents will be treated in depth in a future article.)

When neutral impedance is different from line impedance, fault calculations must be modified. Instead of doubling line impedance, line and neutral impedances are handled as follows:

Consider the same system used in previous example, except in this example to illustrate the point, the neutral conductor PUR and PUX values are half the line conductor values: PUR _{UT} = 0.000133 and PUX _{UT} = 0.001995.

Cable values are: Line conductor PUR _{C} = 0.000250 and PUX _{C} = 0.000184. Neutral conductor PUR _{N} = 0.000125 and PUX _{N} = 0.000092.

Thus, total PUR and PUX are:

PUR _{T} = 2 x PUR _{UT} + PUR _{C} +PUR _{N} = 2 x 0.000133 + 0.000250 + 0.000125 = 0.000641

PUX _{T} = 2 x PUX _{UT} + PUX _{C} + PUX _{N} = 2 x 0.001995 + 0.000184 + 0.000092 = 0.004266

Total per-unit impedance for a line-to-neutral fault is:

PUZ _{T} =Ö(0.000641² + 0.004266²) = 0.004314

Short circuit current for a line-to-neutral single-phase fault on a three-phase system (4,160Y/2,400 volts):

I _{SC} = (E _{L-N} x kVA _{base} ) / _{T} x kV

**Single-phase transformers**

Figure 14.3 illustrates the calculation of fault current at the terminals of single-phase transformer served by a three-phase system. The single-phase, 4,160-to-120/240-volt transformer is rated 100 kVA and has a 1.3% impedance and an X/R of 2. It is located at the end of a 75-foot single-phase tap from a three-phase line.

System PUR and PUZ to the point where the tap is made from the three-phase line are the same as used in the previous examples:

PUR _{SYS} = 0.000383 and PUX _{SYS} = 0.002179

The single-phase tap conductors are 75 feet of #6 copper spaced one foot apart on overhead poles. Conductor resistance (R _{C} ) is 0.033975 ohms. Conductor reactance (X _{C} ) is 0.009075 ohms. Thus:

PUR _{C} = (R _{C} x kVA _{b} ) / (1,000 x kV

PUX _{C} = (X _{C} x kVA _{b} ) / (1,000 x kV

Transformer per-unit values are:

PUZ _{TR} = (kVA _{b} x Z%) / (kVA _{TR} x 100) = (1,000 x 1.3) / (100 x 100) = 0.130000

PUR _{TR} = PUZ _{TR} /Ö (1 + (X/R)²) = 0.130/Ö (1 + 2²) = 0.058138

PUX _{TR} = PUZ _{TR} x X/R = 0.116276

Total PUR and PUX values are:

PUR _{T} = PUR _{SYS} + PUR _{C} + PUR _{TR} =

0.000383 + 0.001963 + 0.058138 = 0.060484

PUX _{T} = PUX _{SYS} + PUX _{C} + PUX _{TR} =

0.002179 + 0.000524 + 0.116276 = 0.118979

These values must be doubled:

PUR _{TF} = 0.120968

PUX _{TF} = 0.237958

PUZ _{TF} =Ö(PUR _{TF} _{TF}

Ö(0.120968² + 0.237958²) = 0.266941

kVA _{SC} = kVA _{b} /PUZ _{TF} = 1,000 / 0.266941 = 3,746.15 and

I _{SCLL} = 1,000 x kVA _{SC} / E _{L-L} = 1,000 x 3,746.15 / 240 = 15,610 amps

However, faults on the secondary of the single-phase transformers are most severe for line-to-neutral faults, as there is only one-half of secondary winding impedance.

If actual transformer values are available, they should be used; national standards do not exist for single-phase transformer resistance and reactance values. If actual values are not available, multiplying PUR _{TR} by 1.5 and PUX _{TR} by 1.2 approximates actual impedance. These approximations are accurate enough for fault calculations.

Calculate line-to-neutral faults at the secondary terminals of single-phase transformers in the following manner (all PUR and PUX values are the same as determined for Figure 14.3) :

1.5 x PUR _{TR} = 1.5 x 0.058138 = 0.087207

1.2 x PUX _{TR} = 1.2 x 0.116276 = 0.139535

PUR _{T} = 0.000383 + 0.001963 + 0.087207 = 0.089553

PUX _{T} = 0.002179 + 0.000524 + 0.139535 = 0.142238

Again these values must be doubled:

PUR _{TF} = 0.179107

PUX _{TF} = 0.284476

PUZ _{TF} =Ö(PUR _{TF} _{TF}

Thus:

kVA _{SC} = kVA _{b} /PUZ _{TF} = 1,000/0.336163 = 2,974.75 and

I _{SCLN} = (1,000 x kVA _{SC} ) / E _{L-N} = (1,000 x 2,974.75) / 120 = 24,790 amps

**Faults at the ends of feeders**

Single-phase faults at the ends of feeders originating at single-phase transformers, whether line-to-line or line-to-neutral are determined by adding feeder PUR and PUX to the line-to-line or line-to-neutral PUR and PUX at the transformer secondary terminals. These values are then doubled, and calculations proceed as in the previous examples. Consider the fault illustrated in Figure 14.4.

To calculate for a line-to-neutral fault at the end of a 50-foot 4/0 THHN copper feeder in plastic conduit with values at transformer terminals the same as in previous example (PUR = 0.089553 and PUX = 0.142238):

Resistance in ohms per 1,000 feet for this cable is 0.0511 and resistance is 0.0314. For 50 feet, R _{C} = 0.002555 and X _{C} = 0.001570. Thus:

PUR _{C} = (R _{C} x kVA _{b} ) / (1,000 x kV

PUX _{C} = (X _{C} x kVA _{b} ) / (1,000 x kV

Note that the secondary kV rating of 0.120 is used in these caluculations.

The system and the cable values are added:

PUR _{T} = 0.089553 + 0.177431 = 0.266984

PUX _{T} = 0.142238 + 0.109028 = 0.251266

Again, these values must be doubled:

PUR _{TF} = 0.533968 PUX _{TF} = 0.502532

PUZ _{TF} =Ö(PUR _{TF} _{TF}

Thus:

kVA _{SC} = kVA _{b} /PUZ _{TF} = 1,000 / 0.733253 = 1,363.79 and

I _{SCLN} = (1,000 x kVA _{SC} ) / E _{L-N} = (1,000 x 1,363.79) / 120 = 11,365 amps

This concludes the study of per-unit short-circuit calculations. The next article will introduce tables that state the fault current at the ends of various feeders fed from both three-phase and single-phase transformers. The subsequent article will provide introduction to symmetrical components.

**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 **

** The Art of Protecting Electrical Systems, Part 8: Short-Circuit Calculations **

** The Art of Protecting Electrical Systems, Part 9: Assigning Impedance Values **

** The Art of Protecting Electrical Systems, Part 10: Assigning Impedance Values **

** The Art of Protecting Electrical Systems, Part 11: Impedance in Systems with Rotating Machinery **

The art of protecting electrical systems, part 13: