• “A wire or cable fault is a partial or total local failure in the insulation or continuity of a conductor.”
• “A component fault is the physical condition that causes a device, a component, or an ele- ment to fail to perform in a required manner; for example, a short circuit, a broken wire, or an intermittent connection.”
All faults within these two definitions fall within one of two major categories: an open cir- cuit or a short circuit. An open circuit is any cir- cuit in which the normal continuity of the circuit is interrupted. The IEEE dictionary defines a short circuit as “an abnormal connection (includ- ing an arc) of relatively low impedance, whether made accidentally or intentionally, between two points of a different potential.” Within the same definition, there is a note that the term fault or
short-circuit fault is used to describe a short cir-
cuit.
Open circuits typically do not lead to damage to the electrical system. In addition, normally available protective sectionalizing devices used on electrical distribution systems do not typically detect open circuits. Frequently, the word fault is associated with its short-circuit definition only, and is used interchangeably for short circuit. Throughout the rest of this section, the word
fault will be used to mean short circuit. Al-
though protective relays that detect open circuits to some degree are available (and others are currently being developed), they are outside the scope of this section.
Description of Faults
Some of the phenomena associated with a fault are listed below.
• Very little current flows past a fault point, leading to loss of service to loads beyond the fault.
• Voltage at the fault and beyond decreases significantly. The voltage between the generation source and the fault decreases proportionally to the inverse of the line impedance.
• Faults typically lead to current levels that exceed the thermal rating of conductor and
other system components, causing damage within a fraction of a second.
• The abnormal low-impedance path can include nonutility property or human beings, causing damage, injury, and even fatalities.
Causes of Faults
Causes of common mechanical failures of under- ground cables are dig-ins, rodent damage, and improper handling and installation. This last cause includes sharp bending of cable, excessive pulling force during installation, driving vehicles over laid cable, walking on cable in a trench, placing or leaving rocks in a position to cause future cable damage, and allowing nails in reels to damage cable. Principal causes of electrical faults to underground systems include lightning, insulation treeing, and thermal insulation failure caused by overloading.
In addition, during single-phase faults on three-phase circuits, the phase-to-neutral voltage on the two unfaulted phases can sometimes in- crease to a level that can approach the normal phase-to-phase voltage. This increased voltage on the unfaulted phases stresses the insulation and can lead to failure. Failure of splices and el- bows is also either electrical or mechanical fail- ure, depending on the cause.
For a comparison of the sectionalizing of overhead and underground systems, it is useful to examine the many causes of faults on over- head distribution lines. Some of the more com- mon causes are:
• Lightning,
• Squirrels or large birds,
• Extreme weather conditions,
• Tree limbs or trees falling on the lines, and
• Vehicular damage.
Although the intent of this section is to focus on the protection of underground systems, overhead lines in many instances are connected either on the source side or, less frequently, on the load side of underground lines. In these cases, the protective devices often protect mixed line sections. Also, un- derground devices on systems served by over- head feeders must coordinate with those devices protecting the overhead portions of the system.
Symmetrical Versus Asymmetrical Faults
The terms symmetrical currents and asymmetri-
cal currents refer to the symmetry of the peaks
of the current waves about the zero current line. A symmetrical current is symmetrical about the zero current line, as shown in Figure 3.1. Such current symmetry would typically be found in a system under normal operating conditions. Dur- ing an asymmetrical current, the current wave is not symmetrical about the zero current line and can be completely above or below the zero line. Figure 3.2 shows a typical current curve immedi- ately before and after a fault initiation. As the curve shows, the current is symmetrical before the fault initiation. Immediately after the fault initiation, the current is asymmetrical for approx- imately the first three cycles before returning to a symmetrical waveform.
The degree of asymmetry in the current curve immediately after the initiation of a fault de- pends on two considerations. The first is the
FIGURE 3.1: Symmetrical Current.
time within a cycle that the short circuit occurs. If the fault is initiated during a voltage peak, then the resulting fault current will be totally symmetrical. If the fault is initiated near a volt- age zero, then the initial fault current will be highly asymmetrical. As the point on the voltage curve moves from the voltage zero point to the maximum voltage point, the degree of current asymmetry decreases accordingly.
The other consideration that affects the de- gree of asymmetry of a fault current is the reac- tance/resistance (X/R) ratio of the equivalent impedance circuit at the fault location. A high X/R ratio means the inductance of the circuit is greater than the resistance. The higher the X/R ratio is, the greater the asymmetry of the initial fault current is, all other conditions being con- stant. Using a standard symmetrical component notation, Equation 3.1 shows the X/R ratio for a three-phase fault. Equation 3.2 shows the X/R ratio for a single-phase fault. The positive se- quence impedance data (X1and R1) and zero se- quence impedance data (X0and R0) should be available from a system fault study.
Total Asymmetrical Current DC Component
AD Component
FIGURE 3.2: Asymmetrical Short-Circuit Current.
Equation 3.1
where: X1 = Positive sequence reactance R1 = Positive sequence resistance
Ratio = X1÷ R1 X
R
Three-Phase Fault
Equation 3.2
where: X1 = Positive sequence reactance R1 = Positive sequence resistance X0 = Zero sequence reactance R0 = Zero sequence resistance
Ratio = [(2 × X1) + X0] ÷ [(2 × R1) + R0] X
R
symmetrical current interrupting rating and a corresponding maximum X/R ratio for the circuit in question. Likewise, switches and sectionaliz- ers will have a close-and-latch rating expressed as amperes symmetrical with a maximum X/R ra- tio. The asymmetrical rating is based on the rms (root mean square) value of the maximum asym- metrical fault during the first half cycle of fault current. The X/R rating shows that the device is able to successfully interrupt or close into the maximum asymmetrical fault current expected for a system with the following:
• A maximum available fault current less than or equal to the symmetrical current rating of the device, and
• An X/R ratio less than or equal to the rating of the device.
Where an X/R ratio is used to show the maxi- mum asymmetrical interrupting rating of a de- vice, this value is usually fairly conservative. In other words, most distribution system X/R ratios would be expected to be less than the rating of this device and fall within its capabilities. Table 3.1 should be useful where devices are rated in asymmetrical currents or where devices are rated in maximum X/R ratios and the actual X/R ratio exceeds the rated value.
X/R Ratio “Maximum RMS” Factor
for 1/2 Cycle, Mrms* 1.0 1.002 1.5 1.015 2.0 1.042 2.5 1.078 3.0 1.116 4.0 1.189 5.0 1.253 6.0 1.305 8.0 1.383 10.0 1.438 15.0 1.522 20.0 1.569 40.0 1.646 100.0 1.697
* Multiply per-phase symmetrical rms short-circuit current by Mrmsto obtain momentary per-phase asymmetrical rms fault current.
TABLE 3.1: Multiplying Factors to Determine Asymmetrical Fault Currents Where
Symmetrical Fault Currents Are Known.
EXAMPLE 3.1: Device Rated in Maximum Asymmetrical Current Capacity.
The calculated maximum symmetrical fault on a sys- tem is 8,000 amperes. The X/R ratio at this location is 10 and the fuse being considered for this location has a symmetrical interrupting rating of 8,600 amperes and an asymmetrical interrupting rating of 12,000 am- peres. The multiplying factor Mrmsis 1.438 for an X/R ratio of 10.0. The maximum asymmetrical fault for this location is 1.438 × 8,000 amperes, or 11,504 am- peres. The maximum symmetrical fault of 8,000 in this location is less than the interrupting rating of 8,600 amperes, and the maximum asymmetrical fault of 11,504 amperes is less than the asymmetrical inter- rupting rating of 12,000 amperes; therefore, the de- vice is acceptable.
The rate at which a fault current decays from its asymmetrical waveform to an essentially sym- metrical waveform also depends on the X/R ratio. A circuit that has a low X/R ratio (one that is mostly resistive) will decay very quickly. A cir- cuit with a high X/R ratio (one that is highly in- ductive) will take much longer to decay.
Typical protective devices such as fuses, break- ers, and reclosers are rated in maximum sym- metrical fault-interrupting capability, although some fuses may be rated for maximum asym- metrical fault-interrupting capability. In addition, they will have either a maximum asymmetrical current interrupting capability or a maximum
Maximum Available Fault
The maximum available fault current is used to determine if the interrupting capacity of a device is adequate. The maximum
fault current is also the current magnitude at which the coor- dination of devices is checked for adequate time clearance. Maximum faults should be cal- culated for both three-phase faults and single-phase-to- ground faults. Maximum faults are calculated using those con-
ditions that will lead to the maximum available faults. Typical conditions are as follows:
• The maximum fault is available from the power supplier. In this case, the power supplier is operating its system with maximum generation and with its transmission system intercon- nected to result in a maximum available fault.
• Substation transformers and buses are inter- connected to produce the maximum available fault. A common example is two substation
transformers operating in parallel if such an arrangement is possible and usual.
• A bolted fault (both three-phase and phase-to- ground) is applied at each location to be evaluated. A bolted fault has zero fault resis- tance (or reactance).
The system engineer should take some pre- cautions when calculating maximum faults:
• Do not calculate maximum faults for system configurations that cannot actually exist because of operating restrictions.
• When determining the interrupting capability of devices, use the maximum expected fault, even if it would occur only under unusual or emergency conditions.
• When considering the coordination of devices, calculate the maximum fault under normal conditions. In other words, devices should be coordinated under normal system configuration. It may not be possible to coor- dinate devices under emergency conditions (such as when a circuit is backfed from a nearby substation).
• Calculate both maximum three-phase and phase-to-ground faults. This must be done be- cause phase-to-ground faults typically exceed three-phase faults in and near delta-to-wye-
connected substation trans- formers, whereas three-phase faults typically exceed phase- to-ground faults further out on the circuit. Furthermore, some devices have different operat- ing characteristics for phase- to-ground faults than for three- phase faults. Another reason for calculating both types of faults is that most systems have single-phase taps for which only phase-to-ground faults should be used when devices are coordinated. When coordinating devices on vee-phase lines, calculate phase-to-phase-to-ground faults.
Minimum Available Fault
The term minimum available fault current does not accurately describe the desired value. The actual minimum fault current on any circuit ap- proaches zero. For example, if a broken conductor
EXAMPLE 3.2. Device Rated for Maximum Circuit X/R Ratio.
In this application, the location being considered has a maximum available sym- metrical fault current of 2,500 amperes with an X/R ratio of 20. The device being considered is a recloser with a maximum interrupting rating of 3,000 amperes symmetrical and a maximum circuit X/R ratio of 12. The Mrmsfactor for the cir- cuit X/R ratio of 20 is 1.569. The Mrmsfactor of 1.569 times the maximum sym- metrical fault current of 2,500 amperes yields a maximum asymmetrical fault current for the circuit of 3,922. Although Table 3.1 does not list an X/R ratio of 12, interpolation can be used to calculate an Mrmsfactor, which, although not exact, will be within acceptable limits.
The Mrmsvalue of 1.4716 × 3,000 amperes symmetrical equals an asymmetri- cal interrupting rating of 4,415 amperes. The maximum fault conditions of 2,500 amperes symmetrical and 3,922 amperes asymmetrical are less than the device ratings. Therefore, the recloser is acceptable. If the circuit’s X/R ratio had been 12 or less, there would have been no need to calculate the respective asym- metrical fault current.
Equation 3.3 × (1.522 – 1.438) + 1.438 = 1.4716 (12 – 10) (15 – 10) Mrmsfor X/R of 12 =
Maximum available
fault current should
be used to check
interrupting ratings.
falls on dry sand or a dead, bone-dry tree, the effective fault resistance approaches infinity, causing a fault that approaches zero amperes. However, the concept of a minimum fault cur- rent actually involves calculating the minimum fault current that can be expected during most of the faults on a system. The variables that typi- cally affect the calculated minimum fault are the following:
• Available fault current from the source utility or transmission system, which is mainly controlled by the amount of genera- tion online and the transmission system and bus configuration;
• The configuration of the distribution system and substation buses; and
• The fault resistance, which is the resistance between the faulted conductor and the return path that must be added to the known imped- ances of the source, transformers, circuit, and other system components.
Although the effects of the first two variables should not be discounted, they frequently either do not vary significantly from the maximum fault configuration or are not available in the minimum fault configuration. The third variable (fault resistance) usually has the greatest influ- ence on the difference between the maximum and minimum faults.
Many field measurements made on utility sys- tems in the 1930s were used to develop a plot of apparent fault resistances versus a percentage of faults at that resistance level. The results showed that the median level of fault
resistance was 25 ohms and the average level was 35 ohms. A commonly used value of fault resistance for overhead circuits is 40 ohms. For substa- tions of greater than 5,000-kVA base capacity operated in the 15-kV distribution class, a val- ue of 30 ohms is often used. These values are for faults that
occur on the overhead portion of the system. For faults on underground systems with concentric neutrals or metallic shields, some parties recom- mend a value of zero to 10 ohms to calculate min-
imum faults, with 10 ohms giving more conserva- tive results. Where circuits are composed of inter- connected sections of underground and overhead, it may be necessary to make two sets of fault cal- culations using the underground fault resistance in one run and the overhead fault resistance in the other run. It is also important to note that site conditions vary widely between utilities and within each distribution system. This variability should always be considered when determining the system standard protection parameters.
DESIRABLE LOCATIONS FOR SECTIONALIZING DEVICES Beginning of UD Cable
It is normally desirable to place sectionalizing devices at the beginning of underground cables, that is, any location where a transition from over- head to underground cable takes place or in a substation or step-down transformer where the underground circuit originates (see Figure 3.3). Doing so will minimize restoration time and help distinguish between overhead and under- ground faults.
Faults on overhead lines are usually temporary and are best protected by reclosing devices such as breakers or reclosers. Since faults on under- ground lines are usually permanent; they are best protected by nonreclosing devices such as fuses. Of course, there are exceptions to this recom- mendation, such as where a circuit is mostly overhead with a short section of underground (for instance, under a river, highway, transmission line, or airport glide path). Coordinating a fuse with in-line reclosers on the source side and the
load side of the fuse might be impossible. In this case, re- duced protection of the under- ground line section is more desirable than frequent opera- tion of the fuse caused by tem- porary faults on the load-side overhead line.
To compensate for the re- duced protection of the under- ground line section, the engi- neer could design the system with a spare cable (or cables), install the primary cable in conduit, or both. This reduces the time needed to restore service in case of a failed cable.