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Industrial and Commercial Power Systems
Fault Calculation Methods
There are two major problems that can occur in electrical systems: these are open circuits and short circuits. Of the two, the latter is the most dangerous because it can lead to very high fault currents and these currents can have very substantial effects (thermal heating and electromechanical forces) on equipment that may require replacement of equipment and may even cause fires and other similar ensuing effects in the electrical system.
To prevent problems from short circuits, it is necessary to design electrical protection systems that will be able to detect abnormal fault currents that may occur and then take remedial action to isolate the faulty section of the system in as short a time as is consistent with the magnitude of the short circuit fault current level. This requires that the fault current be predicted for a fault in any particular location of the circuit system. We thus need to establish methods of fault calculation.
Fault calculation is not simple for a number of reasons:
▪ There are many different types of fault in three phase systems
▪ The impedance characteristics of all electrical items in the system must be known
▪ The fault impedance itself may be non-zero and difficult to estimate
▪ There may be substantial fault current contribution from rotating machines etc.
▪ The initial cycles of fault current may be asymmetric with substantial DC offset
▪ The earth impedance in earth faults can be difficult to estimate accurately
▪ DC system faults also include inductance effects in fault current growth
For example, the possible fault types that may occur in a three-phase system are:
▪ Three phase (symmetrical) faults (the most severe in terms of current)
▪ Phase to phase fault
▪ Single phase to earth fault
▪ Three phase to earth fault
Knowing the fault level, the impedance of the upstream circuit and devices (e.g. transformers, conductors) can be derived. The prospective fault current varies at different points in the supply:
▪ At the supply transformer terminals, it is limited by the impedance of the distribution transformer and feeder conductors. ▪ At the main switchboard, the fault current is further reduced because of the additional cable impedance of the consumer’s mains. ▪ At the distribution board, the fault current is further decreased by the cable impedance of the submains.
Supply Transformer MSB^ DB
Utility cables/lines Consumer cable^ Submain^ cables
Example:
An 11kV to 400/230V transformer has a prospective fault current of 32kA at the secondary terminals. The consumers mains circuit has a route length of 25m, using single-core 120mm^2 active conductors and 70mm^2 neutral. The submains circuit has a route length of 35m, using 16mm^2 multi-core cables.
We want to determine the prospective fault current at the main switchboard and the distribution board (for the purpose of selecting appropriately rated protection devices and switchgear).
Transformer impedance:
ZTX = 230(V) ÷ 32000(I) = 0.00718 Ω
Assume cable temperature is 45oC, from Table 34 of AS3008.1, impedance of consumers mains (1 phase):
ZCM = 0.170 ÷ 1000 × 25 = 0.00425 Ω
Prospective fault current at main switchboard:
ISC = 230 ÷ 0.00718 + 0.00425 = 20.2 kA
Assume cable temperature is 45oC, from Table 35 of AS3008.1, impedance of the submains (1 phase):
ZSM = 1.26 ÷ 1000 × 35 = 0.0441Ω
Prospective fault current at the distribution board:
ISC = 230 ÷ 0.00718 + 0.00425 + 0.0441 = 4.158 kA
Note that the above calculations are for a balanced three phase short-circuit fault across all three phases with the result that there is no neutral impedance to consider. A single phase short-circuit from a single phase conductor to neutral will produce a lower fault current. Here, we need to include the impedance of the neutral cable because the (unbalanced) fault current will flow in the neutral.
MVA is chosen), although a common method is to use the rating of a major element in the system such as a transformer or generator as the base SB.
For balanced symmetrical three phase faults the fault calculation is able to be done on a single phase basis using the per unit phase impedances in the one-line diagram of the fault circuit.
Some care must be taken to use the proper phase kVA and voltage levels in the single-phase circuit to calculate the appropriate base values of current and impedance.
IB = B
ZB = V²B ÷ SB
Where VB and SB are the line voltage and three phase kVA Values.
In the fault calculation the impedances in the fault circuit must include all significant components and all of these must have their impedance expressed in per unit terms using the appropriate base value. This requires changes in some per unit values if they are already expressed (for example on the name plate) using different base values. This may commonly occur with transformer impedances. To change per unit impedances from one base value to another we have to use the following equation as the basis for change:
Z pu = Z ohms ÷ Z B = Z ohms S B ÷ V²B
Thus: (i) For change of kVA base ( SB ), the new Z pu is given by: Z pu(new) = Z pu(old) × S B(new) ÷ S B(old)
(ii) For change of voltage base (VB) Z pu(new) = Z pu(old) × V²B (old) ÷ V² (new)
(iii) For change of both kVA and voltage bases at the same time: Z pu(new) = Z pu(old) S B(new) ÷ S B(old) × V²B (old) ÷ V² (new)
In most cases the impedances of items such as transformers, generators, motors etc, will be given on name plates in per unit or percentage terms based on the equipment’s rated voltage and power levels. These given values must be adjusted to the base values chosen for fault calculations if these are different from the nameplate values.
For cables, overhead lines, busbars, etc, the impedances will most likely be given or obtained in ohmic values. These must then be used with the appropriate base values to get their per unit values referred to the common bases. Thus the appropriate operating voltage and chosen SB must be used to get ZB and IB.
The base impedances and currents for a 1 MVA (1000 kVA) base and typical common voltage levels are shown below, using VB equal to rated voltage:
= Z ÷ ZB = j100 ÷ 174 = j 0.
Per-unit reactance of sending-end transformer: = Z pu(old) × S B(new) ÷ S B(old) = j 0.1 × 100 ÷ 50 = j 0.
Per-unit reactance of receiving-end transformer: = Z pu(old) × S B(new) ÷ S B(old) = j 0.12 × 100 ÷ 50 = j 0.
Load current (using formula P = √ 3 VL IL cosφ ):
Base current for the 33kV line:
Hence, per-unit load current is:
= I ÷ IB = 1203 ÷ 1750 = 0.687 pu
Per-unit voltage of load busbar:
= V ÷ VB = 30 ÷ 33 = 0.91 pu
The equivalent circuit is shown below:
Hence,
V S = 0.687 (0.8 - j 0.6) ( j 0.2 + j 0.575 + j 0.24) (0.91 + j 0.0)
V S = 1.328 + j 0.558 pu
V S = 1.44 pu or 1.44 × 11kV = 15.84kV
2. Fault Calculation Effects and Requirements
Fault levels in a power system are required to be determined at the design stage to allow determination of the following parameters: (i) overcurrent protection requirements (ii) peak electromagnetic forces
(iii) thermal heating effects
(iv) the maximum fault current (and the minimum fault current) (v) the (time) discrimination requirements of protection operation (vi) the touch voltages on earthed objects (personnel safety)
2.1 Sources of fault currents
In a complex electrical system, there are a number of potential sources of fault current when a short circuit occurs in the system. These are:
(i) the electrical utility supply grid system
impedances may need to be used: (i) sub-transient reactance ( Xd” ) (ii) transient reactance ( Xd’ ) (iii) synchronous reactance ( Xs )
We must use the sub-transient reactance for the fault current during the first few cycles, the transient reactance for the fault current up to a fraction of a second and the synchronous reactance for very long duration faults (usually synchronous reactance is not necessary as the protection should operate before it comes into effect).
For synchronous motors only the sub-transient and transient reactance are normally used before the exciting field dies away and the fault current contribution is then effectively reduced to zero. For induction motors, only the sub-transient reactance is used before the fault current contribution dies to zero. The combined overall fault current will thus decrease in the manner shown below. Examples of fault current waveforms
2.2 DC Offset (See figure below)
This must be included in fault calculations, particularly in low voltage systems as the offset can increase the initial current levels substantially and this is an important consideration for the circuit breaker (the momentary current rating): a very high initial current even for less than a cycle can open breaker contacts prematurely and result in contact welding. The magnitude and duration of the DC offset level is governed primarily by the X / R ratio of the faulted circuit. [The offset magnitude is also dependent on the angle on the voltage waveform at which the fault occurs. However the worst-case situation is always assumed in the fault calculation].
The following diagrams show some of the above effects of fault currents.
Fault type Magnitude
3-phase (most severe) Line-to-line Line-to-ground (usually least severe)
(E/Z) x multiplier About 0.87 x 3-phase fault Depends on system grounding
3. Fault Calculation Methods
For the simple fault calculations that we will cover here, we assume the following:
(i) The fault is balanced 3-phase symmetrical. (ii) The per unit impedances are pure reactances: any resistance is neglected, i.e. it is effectively a DC analysis. This is not a very good approximation for LV systems where the resistance can be significant (see item (vii) below). (iii) All significant component impedances are included (iv) The fault itself has zero impedance [that is, it is a “bolted” short circuit] (v) Earth circuit impedance is neglected because of the balanced 3-phase nature which eliminates the earth impedance. (vi) The appropriate rated voltage is used as the voltage base value. (vii) For LV systems where resistance is important, we
use the impedance determined by Z = and a DC analysis.
(viii) Record X R for all equipment, if necessary, to calculate the level of the DC offset multiplier after the symmetrical fault current has been calculated. It is necessary to know R and L separately.
4. Faults in DC Systems
DC systems are becoming increasingly common with the use of power electronics and the calculation of fault currents in such systems is also necessary to consider in modern commercial and industrial systems.
In DC systems the impedance elements which determine the steady state fault current level are only resistance elements. However in most cases the system inductance will also have a significant effect in that it will determine the rate of increase of the fault current level in DC system faults. The L / R time constants of such systems are usually long enough that the steady state fault current will not be reached before protection operates and the protection will thus be interrupting current when that current is still rising. Thus DC fault calculations are not necessarily simple to perform. The sources of DC fault currents are, typically, any of the following:
▪ DC generators
▪ Synchronous converters
▪ DC motors
▪ Rectifier systems
▪ Battery banks
▪ UPS systems
Another factor that must be considered in the design of the
protection system is that DC arc currents are more difficult to interrupt than AC arc currents. An AC circuit breaker has 100 current zeroes per second to interrupt the fault current, while a DC breaker has none. Thus the arc interruption is much more difficult for DC than for AC. In a DC breaker the arc voltage developed is an important factor in the protection design and in determining fault current levels. As a result of the difference between AC and DC faults, either specialised DC breakers or fuses must be used or, more commonly, if AC breakers are used they must be de-rated for use on a DC system.
The fault calculation procedure must involve the determination of the time constant and thus the initial exponential rate of rise of current as it is most likely that interruption will occur during this period.
A DC fault is modelled by a DC supply in series with a fixed circuit resistance, a fixed circuit inductance and a variable resistance in the form of the circuit breaker arc when its contact open (see figure over page).
The governing equation during the initial transient is:
V S = V R + Va + L × dI ÷ dt
or L × dI ÷ dt = ( V S − V R ) − Va
Initially, when V a is small or zero, ( VS − VR ) > Va and dI / dt
is positive and current increases, but later as the arc develops
and lengthens, ( VS − VR ) < Va and dI / dt is negative and
current decreases. The typical behaviour is shown below.
