Power System Protection Course- FAULTS AND FAULT LEVELS - LEKULE BLOG

## Thursday, 25 February 2016

FAULTS AND FAULT LEVELS
CONTENTS
FAULTS AND FAULT LEVELS.......................................................................
SYMMETRICAL AND ASYMMETRICAL FAULTS.........................................
FAULT CALCULATIONS....................................................................................
IMPEDANCE.........................................................................................................
FAULT LEVEL.......................................................................................................
GENERATORS....................................................................................................
CABLES................................................................................................................
SOURCE IMPEDANCE......................................................................................
MOTORS...............................................................................................................
PRACTICAL EXAMPLES.....................................................................................
EARTH FAULTS..................................................................................................

# FAULTS AND FAULT LEVELS

Before selecting a protection system we must consider the kind of fault which may occur.
The principal types are:
·         3 Phase (with or without earth)
·         Phase - to - phase
·         Phase - to -Earth
·         Double Phase - to - Earth
Sometimes there are open-circuits involved.  Transformers and motors are also subject to short circuits between turns of the same winding.
Only the 3 phase short circuit is a balanced condition.  The others are unbalanced and require a knowledge of the method known as ‘symmetrical components’ before they can be fully analysed.

## SYMMETRICAL AND ASYMMETRICAL FAULTS

When a short-circuit occurs, it may be between two of the three lines of a 3-phase system, or it may involve all three. The fault current may pass between phases as an arc which has some resistance and so limits the current, or there may be metal-to-metal contact, a so- called 'bolted' fault, where the impedance is zero. As an item of switchgear must be able to deal with the most severe possible case it is always assumed that the fault is a 3-phase bolted one, and that the whole circuit is mainly inductive with little resistance.
FIGURE 4.1  -  PARTIAL CURRENT ASYMMETRY AT ONSET OF A FAULT

It is known that with an inductive fault the current which immediately follows is in general partially asymmetrical. The asymmetry will be complete (100%) if the fault occurs at the instant of a voltage zero. If it occurs at a voltage peak, positive or negative, the asymmetry is zero (0%) - that is to say, the current wave is then wholly symmetrical.
Figure 4.1 shows the general case where the asymmetry is partial (between 0% and 100%). The point on the voltage wave at which a fault may occur is of course entirely random. So therefore is the degree of asymmetry which will occur in any particular case.
This asymmetrical current wave is regarded as resolved into two parts: a symmetrical a.c. wave plus a steady but decaying 'd.c. component' whose rate of decay is mainly the R/L ratio of the fault circuit; the d.c. component is also shown in Figure 4.1. The quicker the decay of the d.c. component, the quicker the fault current resumes symmetry.
With complete asymmetry the first peak of the asymmetrical current wave is almost double the amplitude of the a.c. component at that time - that is 2 x Ã–2 (= 2.82) times its rms value. However by the time the first current peak is reached there has already been some decay of the d.c. component, and it is usual to take the first current peak as approximately 2.55 times the rms value of the a.c. component. This figure however may differ slightly in special cases.
FIGURE 4.2  -  THREE PHASE CURRENT ASYMMETRY (GENERAL CASE)

Although Figure 4.2. shows the a.c. component as having constant amplitude, it does in fact gradually reduce in size as the current moves from its initial subtransient value towards the transient.
For reasons which it is not necessary to go into here, an asymmetrical current is less difficult to break than a symmetrical one.  Therefore, in order that the circuit-breaker is able to deal with the most difficult case, it is required, when testing, that the bolted fault shall continue long enough for the d.c. component to decay to a specified level.  This level depends on the opening time of the breaker itself (i.e. from instant of trip signal to separation of contacts) and may be of the order of 15%, after which the current is regarded as 'symmetrical'.  At Switchgear Testing Stations a deliberate delay is introduced between the onset of the fault and the trip signal to the breaker on test to ensure that this is so.
In 3-phase switching the asymmetry will in general be partial in all phases, as shown in Figure 4.2, and the percentages, taking account of sign, will always add up to zero.  (In Figure 4.2 they are -27 +97 -70 = 0.) If one phase happens to be symmetrical (0% asymmetry) the other two, being displaced 120°, must both be partially asymmetrical.
The breaking capacities of circuit-breakers are always rated in kA (or MVA) 'rms symmetrical'.  The peak asymmetrical current rating in kA may additionally be given.  All system fault calculations are made to determine the required rms symmetrical breaking current rating of the switchgear to be installed.
This analysis is necessary if the amount of fault current that will flow is to be correctly predicted but is beyond the scope of this course.
However, if a fault should develop somewhere in the system, that is to say a phase-to-phase short-circuit or a phase-to-earth breakdown, then all connected generators will at first feed extremely high currents into that fault, which will be limited only by the impedance of the complete circuit from generator to fault.  Fault currents can be ten or more times the normal full-load current.
Such currents will quickly cause intense overheating of conductors and windings, leading to almost certain breakdown unless they are quickly disconnected.  They will also give rise to severe mechanical forces between the current-carrying conductors or windings.  All such apparatus must be manufactured to withstand these forces.
A fault current of 50000A (rms) flowing in two busbars 3 inches apart will produce between them a peak mechanical force of nearly half a ton-force per foot run of bars.
The purpose of automatic protection is to remove the fault from the system and so break the fault current as quickly as possible.  Before this can be achieved, however, the fault current will have flowed for a finite, if small, time, and much heat energy will have been released.  Also the severe mechanical forces referred to above will already have occurred and will have subjected all conductors to intense mechanical stress.

## FAULT CALCULATIONS

In order to predict the performance of a protection scheme it is necessary to know what the fault conditions will be.  Although some relays will be required to deal with overloads, undervoltages, etc., the majority will be concerned with the detection of short-circuit conditions.  To determine the fault level when a short-circuit occurs requires a knowledge of the impedance of the various components of the power system and the ability to calculate the current in every part of the system.

## IMPEDANCE

Although an impedance consists of a resistance and a reactance it is usually sufficient to take only the reactance into consideration in fault calculations.  If a computer is used for the calculation it is just as simple to include resistance but if other means are used its inclusion is an unnecessary complication.  Inmost cases the exclusion of resistance is justified in that the resistance is only a small fraction of the impedance and even if it were as high as 20% it would only change the impedance by about 2%.
The exception is in cables where, if the cross-sectional area is small, the resistance is of the same order as the reactance.  However, as cables have a very low impedance compared to transformers and generators the overall effect of ignoring resistance is small.

## FAULT LEVEL

When evaluating relay performance it is usual to use the three phase fault level and, if earth-fault relays are involved, the earth fault level.  It is appreciated that a phase-phase fault is far more likely than a three-phase fault; however, the three-phase value is used on the basis that it is the most onerous condition.
Calculation of a three-phase fault is fairly straightforward as it is a balanced fault.  That is, the current in each of the three phases has the same magnitude and they are 120° apart.  Therefore all that is required is to calculate the current in one phase using the phase-neutral voltage and the impedance per phase.
For example, an 11kV generator has an impedance of 1.61W/phase:

 11,000 Ã–3

phase voltage =                    =      6350V,

 6350 1.61

fault current =                      =      3944A,
Although current is used in determining relay settings it is more usual to perform fault calculations in MVA as this avoids complications when there is a change in voltage, i.e.  when transformers are involved.

Therefore, fault level = 3 x 6350 x 3944 x 10-6 = 75MVA,
or Ã–3 x 11kV x 3.94kA = 75MVA.
A quicker way would be to perform the calculation in one operation,

 11,000 3 x 1.61

viz.  3 x 11,000 x                        x 10-6 = 75MVA
or in symbols

 3V x V x 10-6 3Z

 V2 Z

=             x 10-6
or if V is in kV

 V2 Z

fault MVA =
If the generator was rated as 15 MW, 0.8 power factor then the rating would be:

 15 0.8

= 18.75 MVA.
The rating as a fraction of fault level would be

 18.75 75

= ¼ or 25%
This ratio is known as the percentage impedance or Z%.  Generator and transformer impedances are generally expressed in this way

 MVA rating fault level

 MVA rating V2

Z% =                       x 100% =                     x Z x 100%.
Another example  Generator 12.5 MVA, 20%:

 12.5 20%

three-phase fault level =                      x 100% = 62.5 MVA.
Rather than calculate in percentages, and multiply by 100% every time, it is convenient to use per unit values.  For instance, 20% means every 20 in 100 and it could be written 0.2 p.u., i.e.  0.2 in every 4.
Therefore, for the above generator Z p.u.  = 0.2:

 12.5 0.2

three-phase fault level =                      = 62.5 MVA.
or the previous generator Z p.u.  = 0.25:

 18.75 0.25

three-phase fault level =                      = 75 MVA.

It is also more convenient to convert all per unit impedances to a common base, say 10 MVA, in the following manner:

 10 18.75

18.75-MVA generator Z p.u.  =                      x 0.25 = 0.133 p.u..,

 10 12.5

12.50-MVA generator Z p.u.  =                      x 0.2 = 0.16 p.u..
The reason for this is so that the relative values of impedance can be attributed to every component in the circuit and therefore allow easy calculations.
If a transformer rated at 4 MVA having an impedance of 6% is connected to the 18.75 MVA generator and both impedances are converted to a base of 10 MVA, then the generator impedance is 0.133 p.u. and the transformer is

x                     =0.15 p.u. .
The fault level on the secondary side of the transformer is

 10 0.283

 10 MVA 0.133 + 0.15

=                     =35.3MVA.
If there were two 4 MVA transformers in parallel each having an impedance of 6% then the total current impedance for a fault on the secondary side would be

 0.15 2

0.133   +                     = 0.208 p.u.
and the fault level would be

 10 MVA 0.208

= 48 MVA.
If one of the transformers instead of being 4MVA was 3MVA with an impedance of 6% then the system would be as Figure  4.3:

FIGURE 4.3  POWER SYSTEM DIAGRAM

Zgen = 0.133 p.u.
ZT1 =  0.15 p.u.

 10 3

 6 100

ZT2 =    x                      = 0.2 p.u.
Z = 0.086 p.u..
and the fault level

 10 0.133 + 0.086

= 45.7 MVA.
By the application of Ohm's law the fault current for any power system can be calculated by constructing an impedance network in which all the components are represented by a per unit impedance and the fault level is the "current" which is determined by dividing the MVA base  the "voltage"-by the per unit impedance.
In the example the fault level is 45.7MVA.  Across the two transformers in parallel the "voltage" is 45.7 x 0.086 = 3.93 and therefore the contribution to the fault through the 4MVA transformer is

 3.93 0.15

= 26.2MVA
and through the 3MVA transformer

 3.93 0.2

= 19.5MVA
Figure 4.3 shows the steps of calculation starting with the system diagram with reactances at (a), the impedance diagram at (b), the circuit reduction at (c) and (d) and the establishment of MVA flow at (e) and (f) culminating in the system diagram with current flow at (g) calculated from I = MVA/Ã–3V.  In an actual calculation some of these steps would be omitted but the object remains the same.  No matter how complicated the network is, the object is to reduce it to a single impedance from which the fault MVA and its flow in various parts of the circuit is determined so that the performance of the protection can be predicted.
The elements of a power system are specified as follows:
(a)        Generators and Transformers-per cent impedance on rating.
(b)        Feeders and Interconnectors-actual impedance/phase.
(c)        Reactors-voltage drop at rated current.

Typical impedance values can be attributed to all components of a power system in the absence of definite information.  Transformer impedances are usually easy to determine as the value is marked on the rating plate.  The impedance of generators is usually of secondary importance as most distribution systems generally have a much higher infeed and fault contribution from the public electricity supply system.  There is, however, a continuing increase in offshore installa­tions which there is no external supply.  In this case the performance of the generator is of prime importance.

## GENERATORS

The performance of a generator under fault conditions is more complicated than that of any other part of the distribution system.
The fault current is initially about 8 times full-load current decaying rapidly to 5 times full-load current and then decaying less rapidly to less than full-load current.  The three stages are known as sub-transient, transient and synchronous respectively.
The synchronous or steady-state reactance of a generator is high because of armature reaction and is in the range of 4.5 to 2.5 p.u. at the machine rating.  The value used is made up of two components the actual reactance of the machine which is small and a fictitious reactance.  When a fault occurs the current lags the voltage by 90° and the position of the field is such that it is demagnetised by the current flowing in the stator conductors so that the air-gap flux and therefore the generated e.m.f.  is low.  Rather than use this low e.m.f. and calculate the fault current by dividing it by the actual reactance it is more convenient to use the initial e.m.f., E, and divide it by a fictitious value to obtain the same result.  The rapid change of flux due to the demagnetising effect of the stator current results in an induced current in the field which opposes the change and tends to maintain the field flux.  Thus the initial flux, e.m.f. and fault current are somewhat higher than the steady-state value, and decay exponentially towards the steady-state value.  Once again a fictitious value of reactance coupled with the e.m.f. is used in calculation-the reactance being termed the transient value.  There is one other effect and that is the damper winding in the pole face will also produce a flux opposing demagnetisation and will result in a fault current slightly higher than that produced under transient conditions.  This fault current is of very short duration, it decays exponentially, and the fictitious value associated with it is known as the subtransient reactance.
The reactance values associated with a generator are typically
·         Subtransient reactance  Xd" value 0.12 p.u.
·         Transient reactance    Xd' value 0.16 p.u.
·         Synchronous reactance  Xd value 2 p.u.
The e.m.f.  at no load would be the same as the system voltage, V, which at the nominal value is 4.  At any other load the e.m.f.  would be greater:
E"  = [(V + X"d I sin f)2 + (X"d I cos f)2]½,
E'  = [(V + X'd I sin f)2 + (X'd I cos f)2]½,
E  = [(V + Xd I sin f)2 + (Xd I cos f)2]½,

At normal system voltage where I is the p.u. value of load = 1 at rated MVA and cosf  is the power factor of the load.
E"  = [(1 + X"d I sin f)2 + (X"d I cos f)2]½,
for the value given at full load 0.8 p.f.
E"  = [(1 + 0.12 x 1 x 0.6) 2 = (0.12 x 1 x 0.8)2] ½,
E"  = 4.076.

 1.076 0.12

Subtransient current (or MVA) =                    = 8.97 X FL current
or 8.97 x rated MVA.
Another example:
cosf= 0.9, f = 25.8°,
sinf= 0.436,
E' = [(1 + 0.16 x 0.7 x 0.436)2 + (0.16 x 0.7 X 0.9)2]½,
E' = 1.054.

 1.054 0.16

Transient current (or MVA) =                  = 6.58 x FL current
or 6.58 x rated MVA.
The above are the initial values of the current under short-circuit conditions.  The value would disappear in a fraction of second whilst it would take several seconds for the transient value to decay.
The decay is exponential and typical time constants are
T"d0 = 0.1s,
T'd0 = 5s.
These are open-circuit time constants, under short-circuit conditions the value is modified as follows:

 X"d Xd

T"d =                T"d0.

 X'd Xd

T'd =                 T'd0.

These would be the time constants for a terminal fault.  If the fault was on the secondary side of a transformer, reactance XT, connected to the generator then the short-circuit time constants would be

 X"d + XT Xd + Xt

T"d =                            T"d0.

 X'd + XT Xd + Xt

T'd =                             T'd0.
The subtransient current is of interest only to the switchgear designer to determine closing duty.  As far as protection is concerned it has disappeared before any relay operation.
Both the transient and synchronous values are used to determine the performance of the protection.  The transient value for high-speed and instantaneous schemes and the synchronous value for any scheme that has a time delay.
In practice as soon as the fault occurs there would be a reduction in voltage which would cause the automatic voltage regulator to increase the field thus increasing the synchronous value from no-load to the value for full load.  Figure 4.5 shows the values calculated above in a graphical form.

FIGURE 4.5  -  PERFORMANCE OF A 15MW GENERATOR UNDER FAULT CONDITIONS

## CABLES

The resistance of a cable is determined by the cross-sectional area of the conductors but the reactance depends on the distance between the conductors, i.e.  the insulation thickness which depends on the voltage.  The inductance can be calculated from

 d r

L = 0.46 log     mH/m
where d is the distance between conductor centres and r is the conductor geometric mean radius.  It should be remembered that the cable made from a number of strands and the radius which is calculated from area = pr2 is not the geometric mean radius which is approximately 78% of that value.  Also the actual radius which is used in conjunction with the insulation thickness to determine d is 15% larger than the calculated value.
Example  400 mm2 cable, 3-core screened, 6350/11000V
 400 p

re =   ½ = 11.28mm
r = 0.78 x 11.28 = 8.8
insulation thickness 5.6mm
d = (1.15 x 11.28)2 + 5.6 = 31.54 mm

 33.8 8.8

L = 0.46 log      = 0.255mH/m
and the reactance at 50Hz,  X = 2pfL =  2p x 50 x 0.255 = 80 mW/m
Where three single-core cables are used there is an increase in reactance because the distance between the conductors is increased.  A 400 mm2 single-core cable has all overall diameter of 39 mm and therefore if the three cables are  mounted in trefoil formation (Fig 4.6(a)) then

 39 8.8

L = 0.46 log      = 0.297mH/m
and the reactance at 50Hz     X = 93.4 mW/m
If the cables are laid flat as in Figure 4.6(b) then d is the geometric mean distance which is:
d =  (d1d2d3)1/3 = (39 x 39 x39 x 2)1/3 = 49.2 mm

 49.2 8.8

L = 0.46 log      = 0.343mH/m
and the reactance at 50Hz
X = 108 mW/m

There is no need to calculate the value in every instance, a close approximation can be obtained by using typical values.
FIGURE 4.6 CABLE FORMATION
Reactance in mW/m
 Three-core Trefoil Flat 11kV 80 95 110 415V 75 87 100
From the actual reactance the per unit reactance at the chosen base can be calculated from

 MVA base V2

Z p.u. =                             x Z  (where V is in kV)
for example
1 km, 11kV three-core cable
X = 1000 x 80 x 10-6 = 0.08W

 10 112

X p.u. =             x 0.08 = 0.0066pu
15m, 415 V, three-core cable
X = 15 x 75 x 10-6 = 0.001125W

 10 .4152

X p.u. =                x 0.001125 = 0.065pu
Comparison of the per unit reactance values shows that the 415V cable will have a much greater effect on the fault current than the much longer 11kV cable.

## SOURCE IMPEDANCE

This is merely a value which represents the impedance between the system under consideration and the source.  The value is determined by the fault level at the incoming busbar.  If the actual fault level is not known then a value based on the switchgear rupturing capacity is used.  For example, if the fault level or rupturing capacity is 250 MVA then the source impedance on a 10 MVA base is

 10 250

= 0.04pu
Figure 4.7(a) shows part of a typical distribution system and Figure 4.7(b) the impedance diagram.  As can be seen an impedance of 0.04 has been included to limit the fault level at the plant substation 11kV busbars to 250 MVA.

## MOTORS

There is also a contribution to a fault from any induction motors which are connected at the time of the fault.  The initial value will be roughly equal to the motor-starting current but will decay rapidly to zero.  It is mainly of interest to switchgear and power system designers as the affect on differential protection is small and the current will have disappeared by the time overcurrent relays operate.  It could affect the operation of fast-acting devices such as fuses or miniature circuit-breakers but the accuracy of these devices is not of high order and therefore a precise appraisal is unnecessary.
Synchronous motors behave in the same way as generators, the fault passing through the subtransient, transient to the synchronous stage.

### PRACTICAL EXAMPLES

Figure 4.7(a) shows part of a typical distribution system.  There is an incoming 11 kV supply to the plant substation.  Two 11kV interconnectors to the pump house substation where there are two 11/0.415 kV transformers
The first step is to construct an impedance diagram to a common base, say 10 MVA.  The fault level at the plant substation 11 kV busbars is 250 MVA and therefore the source impedance

 10 250

Xs =                 = 0.04 p.u.
The two interconnectors are each two 300 mm2 cables in parallel and therefore the reactance of each interconnector is
½ x 600 x 80 x 10-6 = 0.024W

 10 112

XI = 0.024 x     = 0.002pu

FIGURE 4.7  SYSTEM, IMPEDANCE, MVA AND CURRENT FLOW DIAGRAMS

The two 1.25 MVA transformers are each

 10 1.25

 6 100

XT =           x        = 0.48pu
The interconnecting cables to the switchgear are assumed to be flat in configuration:
½ x 25 x 100 x 10-6 = 0.00125W

 10 0.4152

XC = 0.00125 x                         = 0.0726pu
From this diagram the fault level at any particular part can be determined.  In more complicated arrangements it may be necessary to calculate the combined impedance of various parts of the system and redraw the impedance diagram to simplify it to the extent where the calculation is straightforward.  It may be that more than one redraw is necessary before the calculation can be made.
Returning to the impedance diagram of the system shown in Figure 4.7(b):
a fault at the plant substation 11 kV busbar is

 10 0.04

= 250 MVA, of course,
a fault at the pump house substation 11kV busbar is

 10 0.04 + ½(0.002)

 10 0.041

=            = 244 MVA,
a fault at the 415V busbar is

 10 0.041 + ½(0.48 + 0.0726)

 10 0.3173

=              = 31.5 MVA,
with two transformers, and

 10 0.041 + 0.48 + 0.0726

 10 0.5936

=              = 16.8 MVA,
with one transformer.
Note that if only one transformer is connected the current per transformer is greater than if two transformers were connected.  The system diagram and the flow through the various parts of the system is as shown in Figure  4.7(c).  Alternatively the actual current flow, marked in brackets, can be shown.

## EARTH FAULTS

The earth-fault level of a distribution system is determined by the method by which it is earthed.  Although earthing at each substation is by means of electrodes driven into the ground, very little of the earth-fault current flows via this route.
In distribution system at the higher voltages, i.e.  33kV, 11kV and 6.6kV, the main earth-fault current flow is via the cable sheath and armouring whereas at the utilisation voltages of 3.3kV and below the main earth fault is usually a direct bonded conductor from the equipment to the distribution transformer.
Because the cable sheath and armouring are used on the higher voltages the earth-fault path has a higher impedance than if it was directly bonded.  This means that for an earth fault at a location removed from the substation where the distribution transformer is installed a higher proportion of the voltage will be dropped in the return path of a value such that the voltage at the fault would be unacceptably high.  It is for this reason that all metalwork at each location must be earthed.  When this is done it means that the whole area is at high voltage and as such does not constitute a danger.  It does, however, stress the insulation of any connections between the fault area and the distribution point, e.g.  pilot wires or telecommunications circuits.  Because of the latter there is a requirement that the rise of earth voltage shall not exceed 430 V at any point.  To meet this requirement, in general, requires that 33 kV, 11 kV and 6.6 kV systems be earthed via a neutral earthing resistor.
The effect of a neutral earthing resistor is to limit the earth-fault current to a relatively low value, which means that during an earth fault most of the phase/neutral voltage is dropped across this resistor.
The reduction in fault current is also necessary so that the earth-fault current does not exceed the current-carrying capability of the sheath and armour.  Table 4.1 gives typical resistance values for cable sheath and armour.
 Conductor size (mm2) Lead sheath Steel armour Combined 50 1250 700 450 70 1050 650 400 95 950 600 370 120 870 570 340 150 760 540 310 185 700 500 290 240 570 460 250 300 500 340 200 400 430 310 180
TABLE 4.1.  -  RESISTANCE OF LEAD SHEATH AND STEEL WIRE ARMOUR FOR THREE-CORE PILCSWA, 6350/11,000V CABLE

The value of resistance chosen for the neutral earthing resistor is such that the earth-fault current is limited to around the full-load current of the transformer and so, for the purpose of assessing relay performance, it can be assumed that this will be the earth-fault level of the whole system.
Usually each transformer will have its own neutral earthing resistor and all transformers in a group must be earthed.  There will be, therefore, an earth-fault level throughout the system which, as far as the protection is concerned, is dependent only on the number of transformers connected.
The earth-fault level of a 415 V system is almost indeterminate.  The rupturing capacity of the switchgear is usually 31 MVA which is a maximum fault current of 43,000 A and yet if the fault path has an impedance of only 0.1W the fault current is reduced to almost a twentieth of that value.
FIGURE  4.8  IMPEDANCE DIAGRAM SHOWING EARTH RETURN IMPEDANCE
As shown earlier cable impedance does have a large effect on the fault level at 415V and a sufficiently accurate value of earth-fault level can be obtained if the reactance of the return path is assumed to be the same as the cable reactance from the transformer to the fault.  In other words, in the system shown in Figure  4.8 for a fault at B the three-phase fault level would be:

 1 0.002 + ½(0.06 + 0.004) + 0.01

 1 0.044

=                      = 22.7MVA (31.6kA)
and the earth-fault level approximately

 1 0.044 + ½(0.004) + 0.01

 1 0.056

=                      = 17.9 (24.9kA)