# The Art of Protecting Electrical Systems, Part 12: Approximating Short-Circuit Calculations for Conductors

## Short-cut methods provide a quick alternative to ohmic and per-unit methods for determining fault currents in many simple radial systems

09/21/2007

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, P.E., reprised the series in an updated version beginning in the Feb. 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. Mr. Valvoda passed away in Dec. 2001, and his long-time friend and editorial partner, George Farrell, passed away in early 2006.

By GEORGE W. FARRELL and FRANK R. VALVODA, P.E.

The ohmic and per-unit methods, covered in previous articles, are the most accurate and flexible means of fault calculation. However, they are also the most time-consuming.

Large or complex radial systems, closed-loop systems and networks require analysis by the per-unit method. However, for many straight-forward radial systems, short-cut methods provide a quick, safe approach to determining fault currents for equipment application. Often, these estimates are valid even for final calculations—depending on the project’s size and the fault current available from the utility. This article describes a short-cut method for calculating the decrease in fault current caused by circuit conductors.

If short-cut methods indicate an available fault for 75,000 amperes at the line side of a fused switch, and the switch and fuses are rated 200 kA, further calculations for the switch location generally are unnecessary. Similarly, if busway with a withstand rating of 22 kA will be used and calculations show an available fault current of 15,000 amperes, further calculations usually are unnecessary. However, if estimated available fault current is within 70% of a component’s short-circuit rating, a more exact study should be made.

Fault calculations of any type are truly accurate only when exact parameters are known. Up until the time a project is installed, many factors affecting fault current can only be estimated. For example, exact conductor lengths cannot be determined, motor horsepower is subject to change and actual transformer impedances are not known until the transformers are manufacturered.

At best, therefore, all fault calculations are intelligent estimates that tend to overstate the amount of fault current. However, because fault calculations are made primarily to determine the minimum interrupting capacity of protective devices and minimum short-circuit withstand of other system components, such overstatements err on the “safe” side.

Point-to-point
One common short-cut method is the “point-to-point” or “LICE” method (an acronym derived from quantities used in the calculations). The method is based on circuit parameters of conductor length (L), initial fault current (I), reciprocal of the conductor’s impedance (C) and operating voltage (E). The “C” value is the reciprocal of the impedance of one foot of conductor. Fault calculations are made point to point (circuit by circuit) using the known fault current available at the beginning of a circuit and calculating the decreased fault current at the end of the circuit due to conductor and other component impedances.

This method was developed many years ago and has been reviewed in numerous trade papers and articles. Early C values for wire and cable were developed from tables of cable resistance and reactance, originally developed for voltage-drop calculations. Conductor resistances in these tables were calculated at the conductors’ operating temperature (some as high as 90

Most references today, including the Institute of Electrical and Electronic Engineers (IEEE) “color” books, base cable resistances for short-circuit calculations on a 25

Probably 75% or more of the 600-volt cable used today is THHN or THWN. This thinner insulation results in cable runs with lower reactance. One factor determining reactance is the distance between conductors. For conductors in a given raceway, insulation thickness is the principle variable.

These two factors, lower reactance and resistance, combine to reduce the impedance of cable runs substantially, which is reflected in Table 12.1 showing 600-volt-cable C values used in this article.

Reactance and resistance values were derived from tables in IEEE Standard 242-1986 (The Buff Book). Buff Book resistance values are calculated at 25

Z = (R2+ X2)1/2

C Values are the reciprocal of Z.

The newer type of “sandwich” busway construction has substantially lower reactance than earlier designs, and this also is reflected in the tables. Manufacturers’ tables of busway resistance are most often developed during heat-rise tests conducted for UL listing. These tests are based on a temperature rise of 55

C VALUES FOR LOW VOLTAGE BUSWAY

 Ampere Feeder Plug-In Rating Copper Aluminum Copper Aluminum 100 -- -- 8,014 4,480 225 -- -- 12,617 14,264 400 -- -- 34,384 24,669 600 -- 36,433 34,384 42,826 800 62,406 51,134 41,482 35,469 1,000 75,061 59,997 41,482 53,568 1,350 99,995 87,131 62,635 76,570 1,600 113,894 102,485 89,799 85,358 2,000 151,013 142,422 100,256 106,697 2,500 201,292 187,845 124,912 153,141 3,000 216,930 199,363 180,305 160,046 4,000 312,348 286,652 -- -- 5,000 413,096 -- -- --

Values for medium-voltage cables were taken from IEEE Standard 241 (The Grey Book). Grey Book resistance values for copper were calculated at 75

Copper Cable

R T2 = R T1 (234.5 + T 2 ) divided by

Aluminum Cable

R T2 = R T1 (228.1 + T 2 ) divided by 318.1

where:

R T1 = Resistance calculated at stated temperature (75

R T2 = Resistance at new temperature

T 2 = Temperature

234.5 = Base temperature of copper

309.5 = Base temperature of copper plus 75

228.1 = Base temperature of aluminum

318.1 = Base temperature of aluminum plus 90

The following example illustrates the principle: Given a 500-kcmil copper conductor with a resistance of 0.029 ohms per 1000 feet at 75

R T2 = T1 (234.5 + T 2 ) divided by

= 0.029 x (234.5 + 25)

These formulas may be used to convert any resistance data.

Calculating procedure
The following data are needed to start the calculating procedure:

1. Available rms symmetrical amperes at the start of the circuit to be calculated, which could be at the service entrance.
This information may be obtained from the utility company, from other system studies, or by calculating the maximum fault current available at transformer secondary terminals by the following method:

A) Determine the transformer full load current (I FL ) from the transformer nameplate or by using one of the following formulas:

For three-phase transformers:

I FL = 1,000 kVA

LL

For single-phase transformers:

I FL = 1,000 kVA

LL or E LN

where:

E LL = Line-to-line voltage
E LN = Line-to-neutral voltage

For single-phase, dual-voltage secondary transformers (such as 120/240), use E LL for full-winding calculations and E LN for line-to-neutral calculations.

B) Determine the transformer multiplier:

multiplier =

Use the percent impedance value provided by the manufacturer, if available. Otherwise, use 4.95% (90% of 5.5% nominal impedance) for 500 kVA and larger three-phase transformers and 1.8% (90% of 2% nominal impedance) for smaller three-phase transformers and all single-phase transformers. The 90% value of nominal impedance is used as tolerance is + 10%.

C) Determine maximum transformer short-circuit current at transformer secondary terminals from the following formula:

I SC = Transformer I FL x Multiplier

D) Note that an unlimited primary source is implied if only transformer impedance is used in the calculation. The next article in this series will provide tables of available fault current from various types of transformers with varying available transformer primary fault energy.

2. Determine the total connected motor load, in amperes, for the system or portion of a system to be studied.
If the actual load is unknown, 277/480-volt and 120/208-volt systems can be assumed to be 50% of motor load, and 240- and 480-volt, three-phase, single-voltage systems can be assumed to be 100% motor load. Single-phase systems are assumed to have negligible motor load.

A value of four times motor full-load amperes is added to the fault current determined in Step 1. This may overstate motor contribution slightly, especially if motors are located remotely. As experience is gained in making these calculations, motor contribution can be modified to allow for this. It may be added at points where it makes a significant contribution.

3. Determine the length, size and number of conductors per phase for each run to be calculated, including busway runs.

4. Determine the type of raceway for each run, i.e., steel, aluminum or nonmetallic (PVC).

The “f” factor
Beginning with the first run, determine a factor “f” using one of the following formulas:

Line-to-neutral
faults on single-phase, center tapped transformers:

f = 3 x L x I

LN

Note: Single-phase, line-to-neutral fault currents generally are higher than equivalent line-to-line faults. The above formula approximates this difference for faults at transformer terminals. When determining fault current at the end of single-phase circuits, use actual circuit voltage (either line-to-neutral or line-to-line) in the LICE calculations.

Line-to-line
faults on single-phase, center tapped transformers:

f = x L x I

LL

Three-phase
faults:

f = 1.73 x L x I

LL

A multiplier “M” may then be determined by the formula:

M 1

M also

Initial short circuit current, I, used in the LICE formula, is then multiplied by M. The result is the available fault current at the end of the conductor.

Using these simple techniques, available short-circuit current may be determined readily at any point in the system, from the service entrance to the smallest branch circuit.

Motor contribution
Motor contribution may be added to the bus service motors, or it may be added at the service entrance. Adding it at the service entrance results in a small error, which is insignificant, especially when calculations are limited to simple radial systems with available fault not exceeding 70% of a system’s component short-circuit ratings. The following examples compare both ways of treating motor contribution for the example shown in the figure.

The first approach is the more exact, albeit more time-consuming, method where motor contribution is added to Bus #2. Full-load current for a 600-hp motor load (at 1 kVA per hp) is 600/(1.732 x 0.48) = 722 amperes. Thus, motor fault contribution at four time full-load current is 2,888 amperes.

Solve for fault at Bus #1:
Available fault current

Thus:

f = 1.73 x L x I 1.73 x 45 x 68,000

LL 3 x 26,706 x 480

Then determine M from the formula:

M = 1 = =

Therefore, fault current at Bus #1 from the utility is 68,000 x 0.8790 = 59,772 amps.

Solve for motor contribution to Bus #1:
Motor fault current at Bus #2 is 2,888 amps. The value of C for 500 Kcmil copper conductors in nonmetallic duct is 26,706, which is multiplied by two for two parallel conductors.

f = 1.73 x 25 x 2,888

Then determine M from the formula:

M

Thus, fault current contribution at Bus #1 from Bus #2 is 2,888 amps times 0.9952, or 2,874 amps. The total fault on Bus #1 is 59,772 amps plus 2,874 amps, or 62, 646 amps.

Solve for fault at Bus #2:
(contribution from Bus #1 plus motor contribution from load directly connected to Bus #2):

f = 1.73 x 25 x 59,772

Then determine M from the formula:

M

Thus, fault current at Bus #2 from Bus #1 is 59,772 amps times 0.9084, or 54,297 amps. The motor contribution is 2,888 amps. Therefore, total fault current at Bus #2 is 54,297 amps plus 2,888 amps, or 57,185 amps.

Solve for fault at Bus #3:
From Table 12.1, the value of C for #1 AWG copper conductor in steel duct is 7,293.

f = 1.73 x 20 x 62,646

Then determine M from the formula:

M

Thus, fault current at Bus #3 is 62,646 amps times 0.6176, or 38,690 amps.

Faster method
A less exact, but faster, method is to add a contribution (in the example, 2,888 amps) to the available current from the utility:

68,000 amps + 2,888 amps = 70,888 amps

Solve for fault at Bus #1:

f = 1.73 x 45 x 70,888

Then determine M from the formula:

M

Thus, fault current at Bus #1 from the utility is 70,888 amps times 0.8745, or 61,992 amps.

Solve for fault at Bus #2:

f = 1.73 x 25 x 61,992

Then determine M from the formula:

M

Thus, fault current at Bus #2 is 61,992 amps times 0.9053, or 56,123 amps.

Solve for fault at Bus #3:
From Table 12.1, the value of C for #1 AWG copper conductor in steel duct is 7,293.

f = 1.73 x 20 x 61,992

Then determine M from the formula:

M

Thus, fault current at Bus #3 is 61,992 amps times 0.6201, or 38,439 amps.

All calculations in these examples were made with a hand calculator carrying decimals to eight places.

The next article in the series will present newly calculated tables indicating fault current available at the secondaries of various types and ratings of transformers.

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