APPLICATION
Important issues to consider when rating
medium-voltage switchgear for fault duties
by Dr. Ian Boake, DRPA Consulting
The correct rating of switchgear for fault-currents is especially relevant now with increasing X/R ratios of supply networks and the real possibility
of large generators (which are electrically close) being connected to such networks.
The rating of switchgear for fault currents is
important for two major reasons:
Firstly, under-rating of circuit-breakers poses
a serious risk of failure, for example excessive
electromagnetic force effects greater than
those for which the circuit breaker has been
designed, type-tested and those guaranteed
by the manufacturers. The higher DC-offsets
which a breaker may experience in the field as
a consequence of time constants exceeding
45 ms will result in greater arcing times in
the breaker with a very real risk of vacuuminterrupter explosions (these time constants will
be explained in greater detail later). The higher
short-circuit making and breaking values,(see
Fig. 6 to see the relative relationships of these
fault parameters), from this higher dc-offset
and ac fault current contributions due to
machines may lead to the asymmetrical
current peak factor, (APF) being as large as
3,02. The safety aspects of such a possible
event are of paramount importance and the
financial implications of the loss of supply from
such an event are potentially very large.
Secondly, if the switchgear is excessively
over-rated this may result in excessive project
capital costs.
The so called “rules of thumb” employed
until now should be carefully evaluated in
light of: increasing use of local generation,
the tendency toward low loss transformers,
the increasing use of fault limiting reactors,
t r a n s f o r m e r s w i t h h i g h s h o r t- c i r c u i t
reactances,(on transmission networks) and
higher conductor bundle arrangements
on overhead lines (e.g. quad and twin
conductors).
make the asymmetrical breaking current
higher if this DC-offset has a very high value.
Take for example the breaking current at
60 ms in Fig.1, here the current still has a
component at this time.
Further to this, the latest IEC specification
(IEC62271-100) to which switchgear should
conform puts a greater onus on the
purchaser’s applications engineer than ever
before. This is particularly true in respect to
asymmetrical breaking current and ‘making
onto a fault’ current ratings.
The duration of this DC-offset is a function of
what is called the “system time constant”. The
slower this current “eases-off”’, the higher the
current will be when it is to be interrupted by
the circuit-breaker. A measure of how slowly
the current eases-off is called the system
time-constant. The higher the time-constant
the slower the current eases-off. The slower
the current eases-off then the higher the
portion of transient current that will be present
at the point of current breaking of the circuit
breaker.
What is meant by “DC-offset”, “system
time-constant” and “AC-decays”
The DC-offset (also called the DC-decay) is
simply seen in the waveform of Fig. 1.
The blue curve is the DC-offset (also called
the “transient part”) which is purely a result
of magnetic fields in the supply network
not being able to collapse immediately
under fault conditions and are thus subject
to a ‘gradually easing off” effect. The red
curve is the steady-state current added to
theDC-offset. The fault thus consists of two
portions added together. This current is not
symmetrical about the time axis until the DCoffset component is zero and is thus called
an asymmetrical fault current.
It is immediately obvious from this curve that
the DC-offset can make the peak value
during the first half-cycle higher. It could also
In Fig. 2 it can be seen that a breaker
applied in a network where the system time
constant is 120 ms has a much harder job
to break the fault current than a breaker
applied in a network with a time constant
of 45 ms.
How is this system time constant calculated?
Well for the circuit breakers doing service at
your main intake substation, it is simply found
by substitution of the X/R ratio (obtained from
Eskom at your consumer substation) into the
following expression:
(1)
For brevity the mathematical derivation of
this system time constant is included at the
end of this article. For any other point in your
own network add all the reactances together
(including Eskom’s) and divide it by the sum
of all the resistances (including Eskom’s). This
value divided by 314,159 (for 50Hz systems)
gives the time constant in seconds. That
takes care of the DC-offset and the system
time constant.
Fig. 1: Fault current with DC offset.
energize - Jan/Feb 2008 - Page 72
Now for the AC-decay. This is purely as a result
of more current being contributed to the
fault by generators and motors. In Fig. 3 this
additional current is added to that in Fig. 1 to
give the waveform shown in Fig. 3.
APPLICATION
The asymmetrical peak factor (APF)
τ4 = 120 ms
τ3 = 75 ms
τ2 = 80 ms
This value is simply defined as the ratio of
the peak value of the current divided by
the root-mean-square (rms) value as given
in Equation 2
τ1 = 45 ms
Fig. 2: "Fig. 9 in IEC62271."
Fig. 3: Fault current with DC and AC decay.
It isn’t obvious without closely comparing
Fig. 1 and Fig. 3 that the current in Fig. 3 is
in fact higher. A dead give away is the peak
value in the first half cycle which is much
higher in Fig. 3. This is as a result of more ac
current being present from the machines
close to the fault. This additional AC current
component is symmetrical about the x-axis
and the overall AC current is the sum of the
AC current from the utility (Eskom) and these
machine fault currents. The machine currents
do change with time and eventually die
away and they are no longer present in the
steady-state AC waveform. This is the reason
it is given the name AC-decay.
It will also be noticed that the DC-offset is
still present. This is true for all faults. The only
difference between Fig. 1 and Fig. 3 is that in
Fig. 3, local machine contributions to the fault
are present (theAC-decay effect). The term:
“machines close to the fault” needs further
clarification. Only the machines that are
connected close to the fault will contribute this
additional ac-current. So any machine which
has low impedance between it and the fault
will contribute AC fault current. The larger the
machine, the larger the current contribution
(the AC-decay of induction machines smaller
than 37 kW dies away before 1 s). See Fig. 4 in
order to calculate the current contributions for
various machine sizes. By “low impedance”
would be meant:
• Supplied from the same busbars where
the cable has a length of only a few
hundred meters or less
• Supplied from a transformer connected
to the same bus.
To take account of the generators the
following is a conservative methodology:
This constitutes the explanations for: DC-offset,
system time constant and AC-decay.
Generator AC-decay component in the first
half-cycle:
10 times the rated current of the generator is the ACdecay component of current from this generator.
Generator AC-decay component at the
instant of opening of a medium voltage
breaker:
6,7 times the rated current of the generator is the ACdecay component of current from this generator.
Table 1: Estimating fault current contributions from nearby generators.
energize - Jan/Feb 2008 - Page 74
(2)
A little word of caution is warranted here, as
this definition only holds true in the first cycle of
fault inception. This means that if there are AC
-decays (machine contributions) present these
must be brought into account. To ensure that
this distinction is clearly understood, the term Ik’’
is used from the standard for calculating shortcircuits (IEC60909) which is called the initial
symmetrical root-mean-square fault current.
This is a root-mean-square value of the AC
current component during the first cycle, which
by definition includes the ac-decays. So in all
cases Ik’’ is equal to irms. The AC component
value (also an root-mean-square value) when
all the AC-decays have died away is given the
symbol Ik in IEC60909. In general Ik is not equal
to irms because of the AC-decays. An example
may help to emphasize this distinction. In
Fig. 5, from an actual fault recording, Ik
(at 600 ms) = 700A rms (no AC-decay).
Whilst Ik’’ (first cycle) = 1800 A rms (because
of significant AC-decay current). So it is clear
that Ik is not equal to Ik’’. The irms is always equal
to Ik’’, i.e. 1800 A rms.
The peak value in the first cycle is 4600 A
peak, thus applying the definition for
APF = 4600/1800 = 2,55.
The reality however in this particular case
is that this peak value is 6,57 times greater
than the Ik value. So if an engineer wanted
to predict the peak value if he had only
the Ik value, at his disposal he wouldn’t be
able to use the commonly known factor
of 2,55 as this will only give a value of
1785 A peak when the actual value is
4600 A peak, a massive error indeed!!
Also if he used 6,57 times each time he
wanted to predict the peak value he may
be significantly over-rating the switchgear
when the AC-decays are lower than in this
fault recording example. Thus using the APF
to find the peak value should be undertaken
with great care.
Rating the switchgear
With all the definitions and explanations
behind us now, a methodology for rating
switchgear for fault duties is required. We are
primarily only interested in the making current
IMA and Ib the breaking current. The making
current is the peak value of the first half cycle
and the breaking current is the root-meansquare current at the time instant given by
the vertical line EE’ in Fig. 6.
APPLICATION
This methodology is conser vative and
represents the worst case conditions:
• Calculate the fault current from the
utility:
(3)
where ∑ Z represents all the impedances
that limit the fault current.
The factor 1,1 is to account for capacitor
banks and tap-changers which may lift
the voltage above its nominal value.
Ask the utility (Eskom, etc.) for their
portions of the R and X values, as they
simulate this fault current at your main
intake busbar and these values are
readily available from them. In the past,
only the three-phase fault current and
single-phase fault current were asked
for but the X/R ratio is needed for the
dc-offset. Asking for their R and their X
values for particular faults gives you all
the necessary information. If you don’t
get the X/R ratio, be conservative and
assume it to be 15.
AC Fault current contributions from induction machines
Induction Machine sizes
½ Cycle
At instant of breaker opening
(instantaneous setting on MV breaker)
>750 kW , <= 1500 rpm
Locked-rotor current
66% of Locked-rotor current
>185 kW, at 3000 rpm
Locked-rotor current
66% of Locked-rotor current
All others, >= 37 kW
83% of Locked-rotor current
33% of Locked-rotor current
< 37 kW
59% of Locked-rotor current
n/a
Fig. 4: Taking account of induction motor AC fault current contributions.
Fig. 5: Actual current recording showing that Ip is greater than a factor 2,55 times the Ik value.
• If you don’t want to have to make use of
fault limiting reactors in the future when
Eskom’s fault levels rise, multiply the
current Iutility above by a factor (greater
than one) to account for this continually
increasing fault current from Eskom. For
example use a factor of 1,4.
• Estimate all the AC-decay components in
the first half-cycle (from Fig. 4 above and
from Table 1). Take the highest values from
all the induction machines and generators
nearby and add all of these together.
• Taking the current from point 2 and from
point 3 above, add these together to give
the initial symmetrical root-mean-square
value Ik’’.
• Calculate all the X/R ratios of all the
machines and cables to the fault
location and the X/R ratio provided by
the utility. Take the highest of all these
values which is most likely the utilities X/R
ratio.
• Multiply Ik’’ by the APF for the X/R ratio
calculated above. The APF is given in
Table 1. This then gives the IMA making
current for the breaker (this is a peak
value).
• Check with the manufacturer of the
switchgear you intend to use whether
it can withstand your calculated peak
making current IMA as the breaker has
to withstand the electromechanicalforces of the fault associated with your
calculated I MA when closing onto a
fault.
• Taking this worst case X/R ratio above,
calculate the “system time constant”
from Eqn. 1, or read if off Table 2.
Fig. 6: Fig. 8 in IEC62271.
• Being conservative, assume that: the
breakers contacts will separate within
55 ms and that the relay needs 10 ms
to operate, thus the time for contact
separation (under the action of the
instantaneous setting of a relay) is
thus a minimum value of 65 ms. Thus
Top = 65 ms in Eqn. 4. It is advisable
to check these assumptions with the
energize - Jan/Feb 2008 - Page 75
breaker manufacturer. In all cases try to
take the shortest time for Top.
• For 50 Hz systems, half a cycle corresponds
to 10 ms for the Tt value in Eqn. 4.
• Using the value of Top = 65 ms,
Tt = 10 ms and the system time constant
calculated above, enter these values
in Eqn. 4 to give the % DC-offset at the
instant of breaker opening.
APPLICATION
50 Hz systems
τ (milliseconds)
X/R
APF
45
60
90
120
133
14,1
2,55
18,8
2,61
28,3
2,68
37,7
2,72
41,8
2,73
Table 2: Relationship between X/R ratio and Asymmetrical Current Peak Factor (APF).
a severe case but it demonstrates the
argument effectively. A similar effect
(albeit with faster AC-decay) could also
be achieved by only having a large
amount of induction machines feeding
into the fault but with no generators.
This example case is selected as the
AC-decay is slower thus emphasizing
the effects under discussion.
• Adding 20 MVA of induction machine
load to case number 2 and removing
the generators gives us case number 3.
In these three examples the following is
observed:
Fig. 7: Two of the cases used in this article.
• Check with the manufacturer whether
t h e i r b r e a ke r c a n w i t h s t a n d t h e
calculated asymmetrical breaking
current (Ib(asym)).
(4)
• Repeat points 3 and 4 above but take
the current values at the point of contact
separation from Fig. 4 and Table 1 for
induction machines and generators,
these are the AC-decay components.
Add these values to the value found (Iutility)
to give the total symmetrical current at
contact separation. To differentiate this
value from the initial symmetrical fault
current Ik’’ we will call this value Ik’ (also
from IEC60909).
• Check with the breaker manufacturer
whether their breaker can withstand Iutility
for a second or longer as calculated
This may be as clear as mud, and leave
the reader saying: “this is too complicated”.
Well these values are critical to the correct
selection of circuit breakers for fault ratings,
especially for the asymmetrical breaking
value and the peak making current.
• Taking the Ik’ value calculated above,
select a breaker whose rated short-time
withstand current (Ik is usually specified for
1 s or 3 s) and is greater or equal to
this calculated Ik’ value. Ik’ is called the
symmetrical breaking current denoted by
Ib. The value the manufacturer will supply is
given the symbol Isc from (IEC62271-100).
Rating using simulation tools
We have now looked at the hand calculation
method, and it is probably much easier to do
these studies using simulation tools.
As an example three cases are taken (see
Fig. 7) where the breaker in question is fed
from Bus 1:
• Next calculate the DC-offset current value
by taking Ik’ from above and multiplying it
by the value in Eqn. 4 (for example 30%
or 0,3 and not 30). This then gives the
value Idc in amps.
• An 11kV system with a three-phase
symmetrical source fault level of
150 MVA and an X/R = 18,4. There is
only a 2 MVA constant impedance load
attached (such as residential loads)
to the bus from which the breaker is
supplied.
• To c a l c u l a t e t h e a s y m m e t r i c a l
breaking current (I b(asum) ) substitute
I b above and I k’ from into equation
Eqn. 5:
• Case 2 but with a significant amount
of generation (3 off 8,5 MW units)
connected supplying mostly constant
impedance loads (not shown) and a
2 MVA, 90% motive load. This is quite
(5)
Peak or rms
value
Symbols and
units
Case 1
Case 2
Case 3
rms
Ik, kA
7.87
11.28
7.87
Peak current or making onto
a fault current
peak
Ip or IMA, kA
20.65
72.94*
37.15
Short-circuit breaking current
(asymmetrical)
rms
Ib(asum), kA
8.59
28.44*
13.68
Short-time short-circuit
withstand current
Table 3: Results from switchgear rating simulations.
* A 20kA circuit breaker will not suffice.
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An engineer who isn’t familiar with IEC62271-100
(and AC-decay and DC-offset components)
and who hasn’t applied the methodology
described above will only look at Ik or Iutility (as
this is very readily calculated by hand) and
would rate the breakers in all three cases at
20 kA (1 s value) from “fault calculations”.
This would be an error as demonstrated by
the red values for peak making current and
asymmetrical breaking current in Table 3.
Many 20 kA (Ik 1 s value) circuit-breakers have
a peak making current rating specified as
50 kA and the asymmetrical breaking current
rating specified as 20 kA. The simulation plots
below indicate that the breaker is adequate
for the green and the blue curves below but
not for the red-curve.
In reality the actual effects within the breaker
are a little more complicated than just the
effects of the fault current components (AC
and DC-offset) at the instant of contact
separation as shown in Fig. 8. There is a
period of arcing after the time that the
contacts separate, the duration of which is
directly dependent on the DC-component
of fault current at the instant of contact
separation. This can be seen in Fig. 9. It is
strongly advised that the DC-component
at the instant of contact separation also be
supplied to the circuit-breaker manufacturer
to ensure that the arcing time will not be
excessive for the prospective circuit breaker
to be used in the particular application. A
further issue is that of the recovery-voltage
(also seen in Fig. 9 this is the voltage across
the opening interrupter contacts at the
instant that the current stops flowing). This
is seen by the blue voltage trace which
rises from the arc-voltage up to the system
voltage. This voltage also has a higher
frequency (higher than 50 Hz) oscillation
component super-imposed onto this rising
voltage before settling down to the network
voltage. This rising voltage is commonly
called the transient recovery voltage (TRV).
The circuit breaker has to withstand this
increasing voltage across its interrupter
contacts upon opening. If this voltage rises
too quickly the interrupter may fail. This is
a serious concern and should be taken
account of when specifying switchgear.
It is however a discussion which is left for
APPLICATION
Fig. 9: Arcing time and transient recovery voltage (TRV)
of a vacuum interrupter.
[3] IEC Switchgear specifications: IEC60056,
IEC60059, IEC60298, IEC60694 and
IEC62271-100
[4] “ S p e c i f i c T i m e C o n s t a n t f o r Te s t i n g
Asymmetric Current Capability of
Switchgear”, Task Force of WG13.04 Electra
No. 173 August 1997.
Fig. 8: Fault current effects on a 20 kA (1 s) circuit-breaker under different network conditions.
Appendix A: Calculation of the
“system time constant”
The equations which describe this behavior
of the waveform seen in Fig. 1 are given
in Eqn. 6
(6)
Equation and solution to a fault
current with a DC offset
Fig. 10: Relationships between peak (making), DC component and
short-circuit ratings of a breaker.
Part 2 of this article (to be published in a
subsequent issue of Energize).
Conclusion
This article has shown that the rating of
medium voltage switchgear for fault
currents requires knowledge of AC-decays,
DC-offsets, asymmetrical peak factors (APF)
and the switchgear rating parameters that
are derived from these such as: the peak
making current (IMA) and the asymmetrical
breaking current (Ib(asym)). These parameters
are required over and above the usual
value for the short-circuit withstand current
(Ik for 1 s or 3 s).
If these are not calculated or simulated
properly it is not possible to correctly specify
switchgear for a particular application. The
switchgear manufacturer is only obliged to
supply switchgear for the ratings specified
by the user as per IEC62271-100. This thus
puts a large responsibility on the user of the
switchgear to specify these parameters
in an enquir y to the manufacturer. It is
thus advisable to contact the supplier of
switchgear if this information cannot be
determined by the prospective user or
companies who specialize in the calculation
or simulation of such, to perform a “fit-forpurpose” audit of designs. This is especially
relevant now with the very real possibility of
an inordinate amount of generators being
applied to networks because of Eskom’s
Load Shedding activities.
The “system time constant” is then found
by differentiating (taking the slope) of
the transient portion of equation 6 and
setting the time of this Eqn. to zero. This
gives us the following expression:
(7)
Differentiation of fault current at time
=0
The change in X portion of Eqn. 7 is
on the time axis and is thus called the
“system time constant” given the greek
symbol (τ) The value of this constant is
thus:
References
[1] ETAP Training Course, www.etapsa.com
[2]
Smeets, R P P, Lathouwers A G A, “Economy
Motivated Increase of DC Time Constants
in Power Systems and Cosequences for
Fault Current Interruption”, KEMA High-Power
Laboratory
energize - Jan/Feb 2008 - Page 78
(8)
Contact Dr. Ian Broake, DRPA Consulting,
Tel 011 202-8730, i.broake@drasa.co.za v