## 1253 System Faults

For this discussion, a fault is a short circuit imposed on the power system. The short circuit can occur at any point in the circuit from the substation transformer to the light switch on the wall. The response of loads and DG units will differ as a function of fault location in the system. An understanding of Ohm's law is required to understand system response to a fault. Simply,

where,

E = Voltage I = Current Z = Impedance

Simply stated, the voltage at any point in a circuit is equal to the product of current flow in the circuit and the impedance to that point in the circuit. For example, if there were a source of 100 volts having an impedance of 1 ohm and a current flow of 10 amps, the total impedance of the circuit would be 10 ohms. The voltage at the terminals of the source would be 90 volts because the voltage drop across the source impedance is 10 A x 1 ohm = 10 volts.

If a fault occurred at a point in the circuit such that the total impedance reduced to 5 ohms, current in the circuit would increase to 20 A and source terminal voltage would drop to 80 volts. If the fault occurred at the terminals of the source, terminal voltage would reduce to zero volts and current would

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FIGURE 12.4

Simplified schematic of a radial distribution feeder.

increase to 100 A. Putting this into context and looking at Figure 12.4, a fault occurring at the substation would produce higher energy into the fault than one occurring at the far end of the feeder. Protective devices would clear this fault. Should the fault occur in the distribution feeder, the breaker (represented by the square at the beginning of the feeder) would open.

Typically, distribution feeder circuits are protected by high-speed reclosers. When a fault occurs, the recloser disconnects the feeder and then immediately recloses it. This operation is used because faults on a distribution circuit can have many causes. If the cause is temporary, when the recloser closes, the fault will have disappeared. If the fault were the result of an automobile accident that knocked a pole down, causing wires to come into contact with each other, upon reclosure, the fault would still be present and the recloser would trip again. Typically, a recloser will reenergize a feeder circuit one or two times before it locks the circuit out and requires resetting. This operation is used to ensure maximum continuity of power to the feeder.

The operation of the recloser on initial opening occurs within six cycles of fault initiation. Reclosing occurs in about another six cycles. Accordingly, the total outage would last 12 cycles or more. When the lights blink, it is typically because a fault has occurred in the power system and a fault protective device has operated to clear the fault. When the lights go out and stay out, it is likely the result of the fault remaining on the feeder after the reclosure.

The occurrence of a fault in any user's system can produce a flicker in the lights of his neighbors for as long as it takes his overcurrent device to operate. The impedance of the circuit to the point of fault in the user's system would typically be too high to allow sufficient current flow to cause the recloser to operate. Figure 12.5 is an example of a typical larger user facility. Such a system would be found in facilities with maximum demand of 3 MW. Facilities with larger demands would have several systems like that shown.

Evaluating the system for response to faults, four locations have been selected. A fault at the secondary side of T2, location 1, would produce the highest fault current available on the load side of this transformer. The value

Utility Utility

Service A Service B

Utility Utility

Service A Service B

FIGURE 12.5

Typical single line power diagram of a commercial or institutional user.

FIGURE 12.5

Typical single line power diagram of a commercial or institutional user.

of this fault would be equivalent to the secondary voltage of the transformer divided by the impedance of the transformer where the distribution feeder can be considered an infinite source. The available MVA at this point is reduced when the impedance of the feeder and its source cannot be ignored. For the sake of discussion, assume that the transformer is rated 2.5 MVA and its impedance is the normal value, 5.75%; where the impedance of the feeder source can be ignored, the available fault current at location 1 would be 3000/.0575 = 52 kA, 17.4 times the full load rating. At location 2, the fault current is reduced by the additional impedance in the circuit path to the point of fault.

It is necessary to make a similar evaluation of the on-site generation fault current capability. To understand this, the various generation schemes must be evaluated. For DG, the power will be generated either by a static or dynamic source. The static source will be an inverter because the prime movers of these sources will be generating DC power. These sources will be a fuel cell, microturbine, or photovoltaic system. In the case of the photovoltaic, the prime mover is the sun. As the sun's incident radiation on photo cells increases, the output current of the cell increases to some maximum value. This current is passed through an inverter that converts the DC to AC.

Similarly, a fuel cell produces DC output, except its output current is derived from the processing of a fuel to release electrons that form the current. The microturbine is an extremely high-speed generator producing AC current. Because of its high speed, it cannot be directly connected to the grid.

Its output is rectified to produce DC and then inverted to produce AC at the grid frequency. In some cases, storage batteries will be included in these inverter power systems. Where there are no storage batteries, the transient output capability of these systems is quite limited. When batteries are included, the transient output can be higher than the full load rating. In most cases, the magnitude of transient output current of an inverter is limited to protect the solid state switching devices that produce the AC output. Typically, operating on the inverter alone, these power sources cannot be used to start motors of equivalent size. One can assume that transient output current will be limited to less than six times the full load current rating.

The dynamic sources will have rotating generators producing AC outputs. The prime movers of these sources will be combustion, steam, wind or hydraulic turbines, or reciprocating engines. These machines will drive either induction or synchronous generators. The induction generator will depend upon the grid for excitation. The synchronous generator will have its own excitation system. For the case of the induction generator, the prime mover is started, and, as it approaches synchronous speed, it is connected to the grid. The terminal voltage of the induction generator is determined by the grid voltage. When the speed of this generator exceeds synchronous speed, it begins to generate power. The higher the speed, the higher the output power will be. To avoid overloading the induction generator, its speed is closely controlled by metering the power stream (fuel, steam, wind, or water). The synchronous generator produces terminal voltage when it is running.

A characteristic of the dynamic generator is that it stores electric energy in its air gap. This energy can help the generator respond to fluctuations in load. This stored energy also contributes current flow into a fault. Recall that when conducting a short circuit study of an AC power system, the designer must add the contribution from all running motors. In practice, this can be about four times the full load current of the motor. A motor having a 100 A full load current can contribute up to 400 A to a fault on its power system. The actual value will be a function of the motor impedance and circuit impedance between the motor and the fault. Similarly, the synchronous generator will also produce a significant contribution to a fault on its power system. In each case, the magnitude and duration of this contribution will be a function of machine and control characteristics.

Because the induction generator excitation is derived from the grid, when the voltage reduces, the energy is drained from the machine. This will happen in a very short period of time for a close in fault. For example, if the fault causes the grid voltage to go to zero at the location of the induction generator, one can expect the fault current contribution to decay from its initial value of 4 times full load current to 1.5 times full load current in about 20 to 30 milliseconds, 1.25 to 1.9 cycles when the generator was operating at full load at the time of the fault. Were the generator operating at less than full load, the contribution to the fault current would be less.

In the case of the synchronous generator, because it has an excitation system, the duration of the fault current contribution would be somewhat longer. Excitation systems for synchronous generators are typically designed to produce up to 300% full load current for up to 10 seconds. To accomplish this, generator excitation is supported. However, initial fault current flow will be limited by the load on the generator at the time of the fault and the generator's subtransient reactance. Generators rated up to 12.5 MVA* can have a subtransient reactance of 10% or more. This translates to a maximum contribution to fault current flow of ten times full-load rating if the generator is operating at full load when the fault occurs. The subtransient time constant is the most significant determining factor in initial fault current magnitude. Generators of the sizes discussed herein will have subtransient time constants of 20 to 30 milliseconds. At the end of the second cycle after initiation of the fault current flow, the generator contribution will typically be less than half its initial value.

Given the preceding conditions, going back to Figure 12.3 and determining the fault current contribution from a single generator at 2 MW and 10% sub-transient reactance, the initial maximum fault current contribution from a single generator would be 30 kA. With zero impedance to the paralleling bus, the available fault at point 3 would be 90 kA. At point 4, this fault current would be reduced by the impedance of the path from the paralleling bus to the location of the fault in the feeder.

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