Symmetrical Fault Current Calculations ECG 740

Symmetrical Fault Current Calculations ECG 740

Introduction A ault in a circuit is any ailure that intereres with the normal system operation. Lighting strokes cause most aults on highvoltage transmission lines producing a very high transient that greatly exceeds the rated voltage o the line. This high voltage usually causes lashover between the phases and/or the ground creating an arc. Since the impedance o this new path is usually low, an excessive current may low. Faults involving ionized current paths are also called transient aults. They usually clear i power is removed rom the line or a short time and then restored.

Introduction I one, or two, or all three phases break or i insulators break due to atigue or inclement weather, this ault is called a permanent ault. Approximately 75% o all aults in power systems are transient in nature. Knowing the magnitude o the ault current is important when selecting protection equipment (type, size, etc..)

3-Phase ault current transients in synchronous generators When a symmetrical 3-phase ault occurs at the terminals o a synchronous generator, the resulting current low in the phases o the generator appear as shown. The current can be represented as a transient DC component added on top o a symmetrical AC component. Beore the ault, only AC voltages and currents are present, but immediately ater the ault, both AC and DC currents are present.

Fault current transients in machines When the ault occurs, the AC component o current jumps to a very large value, but the total current cannot change instantly since the series inductance o the machine will prevent this rom happening. The transient DC component o current is just large enough such that the sum o the AC and DC components just ater the ault equals the AC current just beore the ault. Since the instantaneous values o current at the moment o the ault are dierent in each phase, the magnitude o DC components will be dierent in dierent phases. These DC components decay airly quickly, but they initially average about 50-60% o the AC current low the instant ater the ault occurs. The total initial current is thereore typically 1.5 or 1.6 times the AC component alone.

Symmetrical AC component o the ault current: There are three periods o time: Sub-transient period: irst couple o cycles ater the ault AC current is very large and alls rapidly; Transient period: current alls at a slower rate; Steady-state period: current reaches its steady value. It is possible to determine the time constants or the sub-transient and transient periods.

Fault current transients in machines The AC current lowing in the generator during the sub-transient period is called the sub-transient current and is denoted by I. The time constant o the sub-transient current is denoted by T and it can be determined rom the slope. This current can be as much as 10 times the steady-state ault current. The AC current lowing in the generator during the transient period is called the transient current and is denoted by I. The time constant o the transient current is denoted by T. This current is oten as much as 5 times the steady-state ault current. Ater the transient period, the ault current reaches a steady-state condition I ss. This current is obtained by dividing the induced voltage by the synchronous reactance: I ss E X A s

Fault current transients in machines The rms value o the AC ault current in a synchronous generator varies over time as I t I " I ' e I ' I e I t T " t T ' ss The sub-transient and transient reactances are deined as the ratio o the internal generated voltage to the sub-transient and transient current components: ss X " E A I " X ' E A I '

Fault current calculations Example 1: A 100 MA, 13.8 k, Y-connected, 3 phase 60 Hz synchronous generator is operating at the rated voltage and no load when a 3 phase ault occurs at its terminals. Its reactances per unit to the machine s own base are and the time constants are X 1.00 X ' 0.25 X " 0.12 s T ' 1.10 s T " 0.04 s The initial DC component averages 50% o the initial AC component. a) What is the AC component o current in this generator the instant ater the ault? b) What is the total current (AC + DC) in the generator right ater the ault occurs? c) What will the AC component o the current be ater 2 cycles? Ater 5 s?

Fault current calculations The base current o the generator can be computed as I L, base Sbase 100, 000, 000 4,184 A 3 3 13,800 L, base The subtransient, transient, and steady-state currents are (per-unit and Amps) EA 1.0 I " 8.333 pu34,900a X " 0.12 EA 1.0 I ' 4 pu,700a X ' 0.25 EA 1.0 Iss 1 pu 4,184 A X 1.0 s

Fault current calculations a) The initial AC component o current is I = 34,900 A. b) The total current (AC and DC) at the beginning o the ault is I 1.5 I " 52,350 A tot c) The AC component o current as a unction o time is t t T" T' 0.04 1.1 I t I " I ' e I ' I e I 18,200 e 12,516 e 4,184 A Ater 2 cycles t = 1/30 s and the total AC current is At 5 s, the current reduces to ss ss 1 I 7,910 12,142 4,184 24, 236 A 30 I 5 0 133 4,184 4,317 A t t

Fault current transients Example 2: Two generators are connected in parallel to the lowvoltage side o a transormer. Generators G 1 and G 2 are each rated at 50 MA, 13.8 k, with a subtransient resistance o 0.2 pu. Transormer T 1 is rated at 100 MA, 13.8/115 k with a series reactance o 0.08 pu and negligible resistance. Assume that initially the voltage on the high side o the transormer is 120 k, that the transormer is unloaded, and that there are no circulating currents between the generators. Calculate the subtransient ault current that will low i a 3 phase ault occurs at the high-voltage side o transormer.

Fault current calculations Let choose the per-unit base values or this power system to be 100 MA and 115 k at the high-voltage side and 13.8 k at the low-voltage side o the transormer. The subtransient reactance o the two generators to the system base is Thereore: Z new given new Zgiven new S given 2 S " " 13,800 100, 000 X1 X 2 0.2 0.4 pu 13,800 50, 000 2 The reactance o the transormer is already given on the system base, it will not change XT 0.08pu

Symmetrical ault current calculations The per-unit voltage on the high-voltage side o the transormer is pu Thevenin equivalent circuit: th = 1.044 pu Zth = j0.28 pu Short circuit current (pu) Isc = th/zth = 3.73 pu Base current on the high voltage side: Ibase = 502 A Short circuit current (A): Isc = 1,872 A actualvalue 120,000 1.044 pu basevalue 115, 000

Symmetrical ault current calculations To determine the ault current in a large power system: Create a per-phase per-unit equivalent circuit o the power system using either sub-transient reactances (i subtransient currents are needed) or transient reactances (i transient currents are needed). Find the Thevenin equivalent circuit looking rom the ault point, then divide the Thevenin voltage by the Thevenin impedance. You may use the Z-bus elements to determine the voltages and current lows elsewhere in the system (urther away rom the aulted bus) Today, sotware tools to do the calculations or us are readily available.

Fault current calculations using the impedance matrix Beore the ault, the voltage on bus 2 was. I we introduce a voltage source o value between bus 2 and the neutral, nothing will change in the system. Since the system operates normally beore the ault, there will be no current I through that source.

Fault current calculations using the impedance matrix Assume that we create a short circuit on bus 2, which orces the voltage on bus 2 to 0. This is equivalent to inserting an additional voltage source o value - in series with the existing voltage source. The later will make the total voltage at bus 2 become 0. With this additional voltage source, there will be a ault current I, which is entirely due to the insertion o the new voltage source to the system. Thereore, we can use superposition to analyze the eects o the new voltage source on the system. The resulting current I will be the current or the entire power system, since the other sources in the system produced a net zero current.

Fault current calculations using the impedance matrix I all voltage sources except are set to zero and the impedances are converted to admittances, the power system appears as shown. For this system, we can construct the bus admittance matrix as discussed previously: Y bus j16.212 j5.0 0 j6.667 j5.0 j12.5 j5.0 j2.5 0 j5.0 j13.333 j5.0 j6.667 j2.5 j5.0 j14.167 The nodal equation describing this power system is Ybus = I

Fault current calculations using the impedance matrix With all other voltage sources set to zero, the voltage at bus 2 is, and the current entering the bus 2 is I. Thereore, the nodal equation becomes Y Y Y Y 0 11 12 13 14 1 " Y21 Y22 Y23 Y 24 I Y31 Y32 Y33 Y 34 3 0 Y Y Y Y 0 41 42 43 44 4 where 1, 3, and 4 are the changes in the voltages at those busses due to the current I injected at bus 2 by the ault. The solution is ound as = Y I = Z I -1 bus bus

Fault current calculations using the impedance matrix Which, in the case considered, is Z Z Z Z 0 1 11 12 13 14 " Z21 Z22 Z23 Z 24 I 3 Z31 Z32 Z33 Z34 0 4 Z41 Z42 Z43 Z44 0 where Z bus = Y -1 bus. Since only bus 2 has current injected at it, the system (12.38.1) reduces to 1 12 22 3 32 4 Z I Z Z I 42 " I Z I " " "

Fault current calculations using the impedance matrix Thereore, the ault current at bus 2 is just the preault voltage at bus 2 divided by Z 22, the driving point impedance at bus 2. I The voltage dierences at each o the nodes due to the ault current can be calculated by substitution: " 1 12 2 3 32 4 42 Z 22 Z Z I " 12 Z22 Z Z I " 32 Z22 Z Z I " 42 Z22

Fault current calculations using the impedance matrix Assuming that the power system was running at no load conditions beore the ault, it is easy to calculate the voltages at every bus during the ault. At no load, the voltage will be the same on every bus in the power system, so the voltage on every bus in the system is. The change in voltage on every bus caused by the ault current I - so the total voltage during the ault is 12 12 Z 22 Z 22 1 1 0 2 2 Z 32 Z32 1 3 3 Z 22 Z 22 4 4 Z4 2 Z 42 1 Z 22 Z 22 Z Z 1 The current through a line between bus i and bus j is ound by I Y ij ij i j

Fault current calculations using the impedance matrix Example: The power system depicted earlier is working at no load when a symmetrical 3 phase ault is developed on bus 2. 1. Calculate the per-unit subtransient ault current I at bus 2. 2. Calculate the per-unit voltage at every bus in the system during the subtransient period. 3. Calculate the per-unit current I 1 lowing in line 1 during the subtransient period o the ault. Solution: 1. The per-phase per-unit equivalent circuit and the bus admittance matrix was previously calculated as Y bus j16.212 j5.0 0 j6.667 j5.0 j12.5 j5.0 j2.5 0 j5.0 j13.333 j5.0 j6.667 j2.5 j5.0 j14.167

Fault current calculations using the impedance matrix The bus impedance matrix calculated using Matlab as the inverse o Y bus is Z bus 0 j0.1515 0 j0.1232 0 j0.0934 0 j0.1260 0 j0.1232 0 j0.2104 0 j0.1321 0 j0.1417 0 j0.0934 0 j0.1321 0 j0.1726 0 j0.1282 0 j0.1260 0 j0.1417 0 j0.1282 0 j0.2001 For the given power system, the no-load voltage at every bus is equal to the preault voltage at the bus that is 1.000 pu The current at the aulted bus is computed as I ",2 1.00 0 4.753 90 pu Z j0.2104 22

Fault current calculations using the impedance matrix The voltage at bus j during a symmetrical 3 phase ault at bus I can be ound as Z 12 j0.1232 1 1 1 1.00 0.4140pu Z22 j0.2104 0.00pu 2 Z 32 j0.1321 3 1 1 1.00 0.3720pu Z22 j0.2104 Z 42 j0.1417 4 1 1 1.00 0.3270pu Z22 j0.2104 The current through the transmission line 1 is computed as I12 Y12 1 2 0.4140 0.00 j5.0 2.0790 pu That s it!