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Short Circuit Current for the Fire Investigator

by Bernard Béland

Béland, Bernard. Short Circuit Current for the Fire Investigator.
Fire and Arson Investigator. Vol. 46, No. 4 (June 1996). p 19-21.

The computation of the short-circuit current is often required by the electrical designer for the proper selection of protection devices for the circuit and other considerations. Fire investigators sometimes require that information. The purpose of this article is to give the fire investigator some basic background in that field. While these computations are simple in many cases, there are other instances which could be very complex. The discussion will be limited to simple situations.

Theory

A short-circuit happens when two conductors at different potentials come into contact. The contact is such that the current takes a path that is not intended for the current to flow. The intensity of the current can be anywhere from a few times the normal current to hundreds of times that value. In some instances of ground faults or arcing conditions, that fault current could be only around the normal value.

Both too small or too high a short-circuit current could cause a dangerous situation. If the short-circuit current is only about the normal value for the circuit, then the fault current could flow indefinitely in an unintended path, since the protection would not open the circuit. Too high a short-circuit current implies that the protection devices (fuses or circuit breakers) may not be able to open the circuit under fault condition. A protective device has not only a nominal value for the current it can carry, but also a limited current-interrupting capacity. For example, a 15-A breaker is intended to carry 15-A on a permanent basis and opens the circuit under overload or short-circuit conditions. However, its maximum interrupting capacity is often limited to 10,000-A. This implies that if the short-circuit current is higher, the breaker would not perform its duty correctly. It will fail to open the circuit and the current will continue to flow between the open contacts for a short time to the complete destruction, arcing and, possibly, an ensuing fire.

To compute the short-circuit current, one just divides the source voltage by the total impedance of the circuit. The problem is to evaluate that impedance. This is simple in many instances. However, in complex circuits involving both resistances and inductances, the problem is more complex. In three-phase circuits, the problem is difficult to analyze, except for the simple case of a three-phase short-circuit, which seldom happens in practice. In this article, we will restrict ourselves to simple cases. In residences, three cases need to be considered:

1. Line-to-line fault
2. Line-to-neutral fault
3. Line-to-ground fault

The most common of these faults is the last one. The first one is very rare. Only the case of a solid fault will be considered. In the case of an arcing fault, the voltage drop in the arc limits the current to a smaller value than that of a short-circuit. The arcing current is usually two or more times smaller than that of a short-circuit. An arc is quite unpredictable and no accurate guideline can be given.

Impedance Computation

The total impedance to be considered is that of the transformer, the line from the transformer to the house, the service drop and any cable up to the fault point. The impedance of the power company transmission line is neglected. This is almost always justified, at least under residential and other low power systems. In this analysis, to be accurate, one should consider both the resistances and reactances. This will not be done so the results should be considered as approximative. However, the results are accurate enough in most cases.

Transformer

The impedance of a transformer is given by:

V2 (%Z)

Z= __________________ (1)

100,000 (KVA)

in which:

Z is the impedance in ohms referred to the secondary (low-voltage) side

V is the secondary (low-voltage) side in volts (for a 120/240-V transformer, the whole winding capacity, which is 240, should be used)

%Z is the percentage impedance of the transformer given on the name plate; it is usually between 2 and 4% for small distribution transformers. If that information is not available, 2.5% would be reasonably accurate in most instances. For large transformers (1,000 KVA), that impedance could be higher.

KVA is the power of the transformer in kilovolt-amperes. It is usually printed in large letters on the transformer.

Equation (1) gives the equivalent impedance of the transformer referred to on the secondary (low-voltage) side. For a 120/240-V transformer, 240 should be used for V. The computed impedance applies for a 240-V fault or a line-to-line fault. In the case of a 120-V fault (line-to-ground or line-to-neutral fault), the result of equation (1) should be multiplied by 3/8 or 0.375 to obtain the proper impedance.

For a 120/240-V transformer, equation (1) becomes:

Z=0.576(%Z)/KVA
for a 240-V fault

and

Z=0.216(%Z)/KVA
for a 120-V fault.

(2)

 

 

(3)

Conductor

Then, the impedance of all the conductors to the point of the short-circuit should be evaluated. This is quite simple if all the resistances of the conductors are known in ohm/foot. However, one must not forget to include the two conductors to the fault and back. Occasionally, the two conductors could be of different sizes. Table I gives the resistance of different sizes of copper conductors.

Examples

Let us consider a 2.5%Z, 50-KVA, transformer that feeds a house with 50 feet of No. 3 aluminum triplex cable. The service has a length of 20 feet of No. 0 copper conductors. A short-circuit is produced in a No. 18 copper flexible cord, 5 feet from an outlet. That outlet is feed by 30 feet of a No. 14 copper cable. Obviously, the fault is from line-to-neutral, at 120-V. The computation is as follows:

for the transformer (equation (3)),

Zx = 0.216 x 2.5/50 = 0.0108;

for the No. 3, aluminum triplex,

Z3=

0.202 x 1.6 x 50 x 2
________________

1,000

= 0.032;

for the No. 0 copper conductors,

Z0=

0.101 x 20 x 2
________________

1,000

= 0.004;

for the No. 14 copper cable,

Z14=

2.54 x 30 x 2
________________

1,000

= 0.15;

and for the No. 18 flexible cord,

Z18=

6.4 x 5 x 2
________________

1,000

= 0.064.

The total resistance is:

ZT = 0.0108 + 0.032 + 0.004 + 0.15 + 0.064 = 0.2608.

The fault current is then:

ISC=

V
_____

Z

 120
_____

0.2608
= 460 - A

In the above equations, the subscripts X and T stand for the transformer and total impedances, respectively. The number indices stand for the size of the conductors. ISC stands for the short-circuit current.

Remarks

The above example is quite typical of actual situations. It will be noted that the transformer and the service do not contribute much to the total impedance. Much of the total impedance is due to the triplex and the cables inside the house. A 120-V fault at the transformer would have resulted 120/0.0108 = 13,000-A. A short-circuit at the main switch or the distribution box would have produced a current of:

120/(0.0108 + 0.032 + 0.004) = 2564-A

At the wall outlet, the fault current would have been:

120/ (0.0108 + 0.032 + 0.004 + 0.15) = 610-A

Clearly the fault current decreases quickly as a function of the distance from the power source.

Household circuit breakers usually have an interrupting capacity of 10,000-A. It is sometimes claimed that short-circuit current of that level is not exceptional. This is unfounded. Under industrial and commercial conditions, this could be the case in some places close to large transformers. Under household conditions, this is possible only if a house is close to a very large transformer and fed by a short, large size cable. Even then, the fault would have to be in the distribution box. Let us consider an infinite power source that feeds a distribution box. The short-circuit current will be infinite. If three feet of No. 14 copper cable is added, the resistance of which is 3 x 0.0025 x 2 = 0.015 ohm, then, the short-circuit current is reduced to 8,000-A. That length of conductor (6 feet) is often used inside the box. Obviously, the short-circuit current decreases quickly with the distance. With 10 feet of cable, the short-circuit current is already reduced to 2,400-A or less. With that last current, the breaker will have to open in about one cycle if damages to the insulation is to be avoided.

TABLE I
Resistances of copper conductors in / 1,000 ft. at 25°C
Size (AWG) 18 16 14 12 10 8 6 4 3 2 1 0
R(1,000 ft.) 6.51 4.1 2.57 1.62 1.02 .645 .403 .254 .202 .16 .127 .101

Notes:

  • for aluminum conductors, multiply these values by 1.6.
  • For resistance in / ft. divide these values by 1,000.
  • For intermediate sizes, take the average value of the one above and below.
  • Add 4% to these values for each 10°C above 25°C or subtract 4% for each 10°C below 25°C.

From the above considerations, it should be clear that, under household conditions, the fault current is normally a few hundred amperes. In a recent study by UL, on 15-A circuits, out of 943 measurements at different outlets, in numerous households, it was found that the short-circuit current: 1

1. Was less than 100-A in 2% of the cases
2. Exceeded 1,000-A in 1% of the cases
3. The highest current was 1,650-A.
4. The lowest current was 77-A
5. The current was between 100 and 500-A in 87% of the cases.
6. The average short-circuit current was 300-A.

Clearly, most short-circuit currents at outlets are at a few hundred-ampere level. Short-circuit current above 1,000-A are rare and 2,000-A is extremely unlikely to happen in practice. In the distribution panel or main switch of a house, the short-circuit current is usually between a few hundred to a few thousand amperes. A fault current of 10,000-A is extremely unlikely. It is possible only in extreme circumstances such as with a 200-KVA transformer with 20 feet of No. 0 aluminum triplex.

Obviously, under arcing conditions, the fault current will be even smaller than that under a solid short-circuit. In large buildings or under industrial conditions, the fault current could be much higher in the 100,000-A. Even under these circumstances, these large currents are possible only close to large power sources. Most of the lighting and small power distribution systems, even under industrial conditions, have a fault current comparable to that of many hours. For example, the school at which this author taught has about 15,000 students. The highest short-circuit current in the power laboratory, at 120/240-V, is just under 2,000-A. Any test that requires a higher power has to be conducted elsewhere.

Arching at 120-V with 1,000-A short-circuit current produces a voltage of about 70-V with 500-A for a power of about 30,000 watts. That is still an impressive power that can do a lot of damage if it is sustained for some time. However, in most circumstances, the power will be cut off quickly because of the blown fuse. However a 200-A fuse will take many seconds before it opens. For comparison purposes, most arc welding is done at about 50-V with a current of around 100 to 200-A. The destructive power of arc welding (useful in this case) is due to the long duration of the arc. Arc welding can melt a lot of steel; however, if the arc lasts for a second or less, the amount of melting is very minimal and will not pierce a hole in steel plates.

Conclusion

In this article, it was seen how to evaluate short-circuit current in simple situations that cover most residential cases and many others. It was seen that, in most cases, the available short-circuit current is usually a few hundred amperes and is seldom above a few thousand amperes.


Reference

1. "Fact-Finding Report on an Evaluation of Branch Circuit, Circuit-Breaker Instantaneous Trip Levels," Underwriters Laboratory, File E87837, Project 92ME51901, October 1993.

Reprinted with permission.

 
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