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Fault Calculations
2.7 Applications to Sizing Circuit Breakers (ANSI C37) The
application of fault calculations to power systems is considered,
including the simplified method of short-circuit calculation (the E
/ X method), as utilized in the IEEE and ANSI standards for rating
circuit breakers and fuses. However, the presentation is not a
complete discussion of sizing and rating breakers and fuses.
Short-circuit calculations for protective relaying are also briefly
discussed.
2.7.1 Introduction There are several applications for
short-circuit calculations in power systems, including: fuse and
circuit breaker sizing, protective relay setting, and calculation
of maximum values of electromagnetic forces applied to conductors.
The classical theory of rotating machinery gives an accurate
mathematical expression for the worst three-phase short circuit
current for an initially unloaded synchronous generator with an
external reactance of Xe.
isc(t) = 1.414 Iac cos( t + ) + Idc exp(-t / Ta) Idc = 1.414 E /
(Xd" + Xe)
Iac = (I" - I') exp(-t / Td") + (I' - I) exp(-t / Td') + I
I = E / (Xd + Xe) = rms steady-state short-circuit current
I' = E / (Xd' + Xe) = rms transient short-circuit current
I" = E / (Xd" + Xe) = rms subtransient short-circuit current
E = prefault voltage (usually 1.00 per unit)
Xd = synchronous reactance
Xd' = transient reactance
Xd" = subtransient reactance
Td' = Tdo' (Xd' + Xe) / (Xd + Xe)
Td" = Tdo" (Xd" + Xe) / (Xd' + Xe)
Tdo' = open-circuit transient time constant
Tdo" = open-circuit subtransient time constant
Usually, the rms short-circuit current is the most interesting
quantity to compute:
Irms(t) = (Iac2 + Idc2)1/2
Notice that the rms current decreases as time proceeds, due to
decrements in both the AC and the DC components. This expression
can be used for any short-circuit calculation, for example the
momentary duty on a circuit breaker would be Irms at t = 1/2 cycle
and the circuit breaker interrupting duty would be Irms at t =
contact parting time.
This is a rigorous approach, but it is so laborious for systems
with many generators in an extensive network that simpler methods
are generally used in most applications.
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C. W. Brice August 2002
2.7.2 Simplified methods The original ANSI (American National
Standards Institute) standards for circuit breakers C37.4-1953,
C37.5-1953, and C37.6-1953 used an interrupting rating that was
based on the "... rms value including the DC component at the
instant of contact separation as determined from the envelope of
the current wave." This is known as the total current basis of
rating a circuit breaker. A simplified method was included in the
standards to allow the symmetrical short-circuit current
calculation to be multiplied by a factor (depending on system
characteristics and breaker speed) to obtain the worst-case total
rms current at contact separation.
In the early 1950's the AIEE (one of the IEEE's predecessors)
Switchgear Committee began work on revising the C37 standards and
developing circuit breaker ratings based on symmetrical
interrupting currents. This effort eventually led to the standards
labeled ANSI C37.04, C37.06, C37.010, etc. These standards utilize
the Symmetrical Current Basis of Rating contrasted to the previous
Total Current Basis of Rating (which are labeled C37.4, C37.5,
C37.6, etc.). At present both rating structures are still in use
although the understanding is that new circuit breaker developments
will be directed toward the symmetrical standards.
The simplified method of calculating short-circuit currents is
to calculate the symmetrical short-circuit current produced by
modeling each generator as a voltage source behind an appropriate
reactance. Use of the subtransient reactance gives the initial
symmetrical short-circuit current, and use of the transient
reactance gives the short-circuit current a few cycles later. The
DC component is ignored. The methods discussed previously may be
used to perform this calculation using a digital computer program.
Alternately, manual circuit analysis methods may be used for small
systems.
The calculation of short-circuit currents by the simplified (or
E / X) method is an acceptable approximation for most purposes,
except for faults electrically close to large generating units.
Most high-voltage lines and transformers have reactance values that
are much larger than their resistance values. In cases where the
resistance is significant (usually, but not always, low-voltage
devices) the magnitude of the impedance Z may be substituted for
the reactance X.
If the value of E / X does not exceed 80% (70% for a
line-to-ground fault) of the symmetrical interrupting capability of
a circuit breaker, the simplified method may always be used to size
circuit breakers. This may result in excessively conservative
design, though, and ANSI C37.010-1979 gives a set of factors, whose
values depend on the system X / R ratio and the breaker speed, that
may be used to adjust for AC and DC decrements. This will be
discussed in more detail later.
It has been tacitly assumed that the simplified method of
calculating the short-circuit current was to be used. This is
consistent with usual practice, since multiplying factors can
usually be used to estimate the total (or asymmetrical) current if
needed.
2.7.3 System X / R ratio calculation and significance To
calculate the system X / R ratio for a fault at a given location,
there are two different (but equivalent) methods:
-2.7.2-
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Fault Calculations
The first involves network reductions to find the Thevenin
equivalent reactance and resistance. The proper procedure is to
construct a network of reactances (neglecting all resistances
regardless of value), then reduce the network to find the Thevenin
reactance at the point of the fault. Next, construct a network of
resistances (neglecting all reactances regardless of value), and
then reduce this network to find the Thevenin resistance at the
point of the fault. The X / R ratio for a fault in this location is
the ratio of the Thevenin reactance to the Thevenin resistance
The second method is to use the Zbus matrix building algorithm
that was discussed earlier. The program must be run twice, first
for the reactances neglecting all resistance (call this result
Xbus), then for the resistances neglecting all reactances (call
this result Rbus). The X / R ratio for a fault at bus number i is
then found by the ratio of the i-th diagonal element of Xbus over
the i-th diagonal element of Rbus.
(X / R)i = Xbusii / Rbusii
The advantage of this approach is that X / R ratios for faults
at every bus are determined from running the Zbus building
algorithm only twice.
Note that the Zbus algorithm can handle the full R + jX
impedance data, but the procedure outlined above should be used
always for calculation of X / R. This will ensure that the X / R
ratio is always over-estimated, thereby giving conservative
results. If we were to use complex impedance data in the Zbus
building algorithm there is a possibility of non-conservative
results. Note also that the ANSI standards that we will use specify
the separate network reductions for X's and then for R's, which is
equivalent to running the Zbus building algorithms twice, first
with only X's, then with only R's.
Practice Problem: For the system shown in Figure 2.7.1,
calculate the X / R ratio for a three-phase short circuit at bus
B1. Be sure to omit all resistances in the reactance calculation
and to omit all reactances in the resistance calculation.
50 MILE LINE
0.60 + j 0.80OHMS PER MILE
G1: 100 MVA X = 15% R = 0.50% T1: 100 MVA X = 7.0% R = 0.35%
T2: 75 MVA X = 7.0% R = 0.35%G2: 75 MVA X = 10% R = 0.50%
T1
13.8 kV
G1
115 kV
B1 B2
115 kV 13.8 kV
T2
G2
Figure 2.7.1 Practice problem for calculation of X/R ratio.
Solution: Convert data to per unit on 100 MVA base:
G1: Xd" = .15, R = .005
G2: Xd" = .10 100/75 = .1333, R = .005 100/75 = .00667
-2.7.3-
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C. W. Brice August 2002
T1: X = .070, R = .0035
T2: X = .070 100/75 = .0933, R = .0035 100/75 = .00467
115 kV line: X = .80 x 50 x 100/1152 = .302, R = .60 x 50 x
100/1152 = .227
For a fault at B1
Xthev = .220 x .529/(.220 + .529) = 0.1554
Rthev = .0085 x .238/(.0085 + .238) = 0.00821
X / R = 0.1554/.00821 = 18.9
For the sake of comparison, the calculation using full 60 Hz
impedance data gives Zthev = 0.162/81.8o, and X / R = tan 81.8o =
6.94, which is smaller than the value computed above. Note that the
first method is the correct one, in the sense that it is the more
conservative approach and is consistent with the ANSI C37 standard.
The differences between the two calculations are not always this
great.
The reason for our interest in the X / R ratio is twofold: 1) we
need it get appropriate multiplying factors in the ANSI C37
standards, and 2) it has fundamental physical significance. To
illustrate the latter point, consider a simple series RL circuit
excited by an ideal sinusoidal voltage source (an infinite bus). At
t = 0, a switch closes connecting the source to the RL circuit.
e(t) = 1.414 E sin(t + ) = R i + L di/dt i(0) = 0
The complete solution (transient plus steady-state) is
i(t) = 1.414 I { sin(t + - ) - sin( - ) exp[-(R / X) t] } where
= tan-1(X / R) and I = E / Z = (E / X) / [ (R / X)2 + 1 ]1/2 Note
that in a real system with multiple generators and a network of
lines, there is no single value of X / R. Our calculations are
simply good engineering approximations.
2.7.4 First-cycle duties for fuses and low-voltage circuit
breakers Usually, the first-cycle duty is the only value of
short-circuit current needed for fuses and low-voltage breakers.
The subtransient reactances are to be used for all generators and
for all motors (assuming that data is available for the motors,
which is sometimes not true for groups of low-voltage motors).
A group of low-voltage motors is often fed from a low-voltage
substation. If the substation supplies only this group of motors,
then the transformer kVA rating (self-cooled) is usually
approximately equal to the total motor horsepower rating. If this
is so, then the entire group of motors may be represented by a
single reactance of 25% on a kVA base equal to the self-cooled
rating of the transformer. If the total motor horsepower rating is
not equal to the transformer kVA rating then the reactance value
should be adjusted appropriately (as the motor HP rating
-2.7.4-
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Fault Calculations
decreases, the contribution to fault current decreases, and the
reactance should increase accordingly).
If motor contributions to low-voltage system faults cannot be
calculated due to lack of data, the following "rule of thumb" may
be used:
Imotor = 4 X (Sum of the Rated Current of Motors Connected to a
Faulted Bus)
This is usually a satisfactory approximation, but actual
calculation of motor currents is more accurate.
According to ANSI C37.13, short-circuit calculations for sizing
low-voltage power circuit breakers usually do not need to consider
the X / R ratio. The initial symmetrical short-circuit current I"
can be used directly, since ample margins are built in to the
breaker ratings for most purposes. There are some exceptions,
however, which are listed below: 1) Local generation at circuit
breaker voltage in sizes greater than 500 kVA, 2) Gas-filled or
dry-type transformers in sizes greater than 1000 kVA, 3) Any
transformer in sizes 2500 kVA and larger, 4) Network systems, 5)
Transformers with impedances higher than those specified in ANSI
C57 standard, 6) Current-limiting reactors at circuit breaker
voltage on the source side, 7) Current-limiting busway at circuit
breaker voltage on source side, 8) Any other application where
available short-circuit current approaches 80% of the breaker
short-circuit current rating (note: this is unlikely).
For any of these exceptions, use this table from the ANSI C37.13
standard:
Power Factor System X / R Multiplying Factor Unfused
Breaker Fused
Breaker 20% 4.90 1.00 1.00 15% 6.60 1.00 1.07 12% 8.27 1.04 1.11
10% 9.95 1.07 1.15 8.5% 11.72 1.09 1.18 7.0% 14.25 1.11 1.21 5.0%
20.00 1.15 1.26
Note that most low-voltage short-circuit duty calculations use
the initial symmetrical short-circuit current.
Note also that for many low-voltage systems, the resistance
should not be neglected. Instead of E/X, use E/Z, where
Z = (R2 + X2)1/2
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C. W. Brice August 2002
Other relevant standards are NEMA AB 1 Molded-Case Circuit
Breakers and NEMA SG 3 Low-Voltage Power Circuit Breakers, IEEE 141
Recommended Practice for Electric Power Distribution for Industrial
Plants (The Red Book), IEEE 242 Recommended Practice for Protection
and Coordination of Industrial and Commercial Power Systems (The
Buff Book).
The C37 series contains several standards related to fuses,
including ANSI C37.41, C37.42, C37.44, C37.45, and C37.46. Fuses
may have either symmetrical or total current ratings. Use the
subtransient reactances for rotating machines. For fuses with
symmetrical current interrupting ratings, use the initial
symmetrical short-circuit current I" to calculate the duty. If the
fuse is rated on a total current basis, a multiplying factor is
specified in ANSI C37.41, as follows:
Multiplying factor = 1.55 in most cases. Multiplying factor =
1.20 for special cases.
Special cases are (a) all distribution fuse cutouts and (b)
power fuses (but not current-limiting fuses) that are remote from
stations. The special case applies only if remote from generation,
the system X / R ratio is less than 4, and the fuse is applied at
15 kV or less.
Example: X / R = 3 I" = 5000 A
Distribution fuse cutout with a total current rating. System is
remote from all generation. Interrupting duty = 1.20 x 5000 A =
6000 A
Now replace the fuse cutout with one having a symmetrical
current rating: Interrupting duty = 5000 A
Now replace with a current-limiting power fuse with a total
current rating: Interrupting duty = 1.55 X 5000 A = 7750 A
If the current-limiting fuse has a symmetrical current rating:
Interrupting duty = 5000 A
The point of this example is that the basis for the rating
(symmetrical current or total current) must be known for a proper
application.
Practice Problem: In the previous example, a current-limiting
fuse is to be applied on the low-voltage side of the transformer,
which serves an induction motor. The transformer has an X / R ratio
of 8.00, and a reactance of 5.00% on a 100 kVA base. Calculate the
three-phase short-circuit current through the fuse for a fault on
the motor terminals. Calculate the X / R ratio and the interrupting
duty (both symmetrical and total current duties).
Solution: 4000 V / 1.732 = 2309 V L-N Xsource = 2309 V / 5000 A
= 0.462 ohms per phase Xtransf = .05 (4.00 kV)2 / 100 kVA = 8.00
ohms per phase Xtot = Xsource + Xtransf = 8.46 ohms per phase
-2.7.6-
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Fault Calculations
I" = 2309 V / 8.46 ohms = 273 A on high-voltage side I" = (4000
/ 208) 273 A = 5250 A on low-voltage side R = 0.462 / 3 + 8.00 / 8
= 1.154 ohms per phase X / R = 8.46/1.154 = 7.33 Symmetrical: 5250
A. Asymmetrical: 1.55 x 5250 = 8140 A.
2.7.5 First-cycle duties for medium and high voltage circuit
breakers This section applies to circuit breakers for applications
with nominal voltage greater than 1000 V. The first-cycle duty is
for comparison with the breaker momentary rating (or closing and
latching capability).
The rotating machine reactances to be used in this study are in
the following table (from C37.5 and C37.010):
Turbo-generators, hydro-generators with damper
windings, and all condensers
1.00 Xd"
Hydro-generators with no damper windings
0.75 Xd'
Synchronous motors 1.00 Xd"
Induction motors
above 1000 hp at 1800 rpm or less
above 250 hp at 3600 rpm 1.00 Xd"
50 - 1000 hp at 1800 rpm or less
50 - 250 hp at 3600 rpm 1.20 Xd"
Below 50 hp Neglect
For induction motors Xd" is one divided by the per-unit
locked-rotor current at rated voltage. Induction motors and small
synchronous motors are often neglected in utility short-circuit
calculations, but should be included on station service systems and
on substations supplying large industrial customers. Industrial
distribution systems must include motor contributions. These
multiplying factors account for the fact that the motor current
will decay with a subtransient time constant, and more accurate
multipliers may be obtained from the manufacturer for specific
large motors that contribute significantly to the fault
current.
With these machine reactances, compute the initial symmetrical
short-circuit current I". The first-cycle duty is 1.60 I". This
quantity approximates the duty on the breaker during the first
-2.7.7-
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C. W. Brice August 2002
half cycle of short-circuit current. The breaker must be able to
close and latch, while withstanding the mechanical forces produced
by this large current. This is particularly severe on systems with
large motor contributions.
Breakers that are rated on a total current basis have a
momentary rating, which must exceed the calculated first-cycle
duty. Breakers that are rated on a symmetrical current basis have
this same rating, but it is called the closing and latching
capability.
Example: A circuit breaker on a 115 kV system has a rated
maximum voltage of 121 kV, a rated continuous current of 1200 A, a
rated interrupting time of 3 cycles, a maximum symmetrical
interrupting capability of 20 kA, and a closing and latching
capability of 32 kA. If I" = 15 kA then the first-cycle duty is 1.6
I" = 24 kA. Since the first-cycle duty is less than the closing and
latching capability, the breaker is adequate in this regard. If
this example were repeated for I" = 25 kA the first cycle duty
would be 40 kA, and the breaker would not have a sufficient closing
and latching capability. Note that the first-cycle duty is a total
(or asymmetrical) current quantity, even though the breaker has a
symmetrical interrupting rating.
Practice Problem: A circuit breaker on a 13.8 kV system has a
maximum symmetrical interrupting capability of 48 kA, and a closing
and latching capability of 77 kA. If the initial symmetrical
short-circuit current is I" = 45 kA, calculate the first-cycle duty
and evaluate the breaker's closing and latching capability.
Solution: The first-cycle duty is 1.60 x 45 kA = 72 kA. The
closing and latching capability of 77 kA is adequate, but if future
short-circuit currents are expected to increase significantly the
breaker will not be adequate.
2.7.6 Interrupting duties for medium and high voltage circuit
breakers For the interrupting duty calculation, we must know the
contact parting time. Figure 2.7.2, which comes from ANSI C37.010,
shows that the contact parting time is the sum of the tripping
delay and the opening time. This is the total time between the
initiation of the short circuit and the parting of the primary
arcing contacts.
-2.7.8-
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Fault Calculations
PRIMARYCONTACTS
PART
CONTACT PARTING TIME
INTERRUPTING TIME
INITIATIONOF SHORTCIRCUIT
TRIPCIRCUIT
ENERGIZED
TRIPPINGDELAY
OPENINGTIME
ARC ONPRIMARYCONTACTS
EXTINGUISHED
TIMEARCING
TIME
Figure 2.7.2 Events in the clearing of a fault by a circuit
breaker.
The breaker contacts part, which initiates an arc that is
subsequently extinguished by the breaker. The arcing time plus the
opening time is the interrupting time. There is a relationship
between the rated interrupting time for a given breaker and its
minimum contact parting time. This is shown in the table below.
Note that the actual contact parting time (but not the interrupting
time) can always be increased by time delays in the protective
relaying system.
Table showing the relationship between the interrupting time and
the contact parting time:
Rated Interrupting Time Minimum Contact Parting Time 2 Cycles
1.5 Cycles 3 Cycles 2 Cycles 5 Cycles 3 Cycles 8 Cycles 4
Cycles
The rotating machine reactances to be used in interrupting duty
calculations are in the following table (from C37.5 and
C37.010):
-2.7.9-
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C. W. Brice August 2002
Turbo-generators, hydro-generators with damper
windings, and all condensers
1.00 Xd"
Hydro-generators with no damper windings
0.75 Xd'
Synchronous motors 1.50 Xd"
Induction motors
above 1000 hp at 1800 rpm or less
above 250 hp at 3600 rpm 1.50 Xd"
50 - 1000 hp at 1800 rpm or less
50 - 250 hp at 3600 rpm 3.00 Xd"
Below 50 hp Neglect
Perform a short-circuit study using these machine reactances,
and find the initial symmetrical short-circuit current I".
Determine the X / R ratio. Remember to determine X, neglecting R's;
then determine R, neglecting X's.
The procedure now depends on whether the breaker is rated on a
total current basis or on a symmetrical current basis. We consider
each case in turn.
A. Total current basis: If the fault is electrically close to a
generator, both AC and DC decrements need to be considered. If the
fault is remote from generation, then only DC decrements need to be
considered. In this context, electrically close means that there is
at most one transformer between the fault and the nearest
generator, or that the external impedance between the fault and the
generator is less than 1.50 times the Xd" value of the
generator.
Using ANSI C37.5, look up the appropriate multiplying factor
from one of three curves, which are reproduced in Figure 2.7.4. The
total interrupting duty is the calculated value of the initial
symmetrical short-circuit current I" times the multiplying factor.
The first curve is for three-phase faults electrically close to
generation. The second curve is for single line to ground faults
electrically close to generation. The third curve is for either
type of fault remote from generation.
Multiplying the appropriate factor times the initial symmetrical
short-circuit current I" gives the total current interrupting duty.
This is to be compared to the circuit breaker interrupting
rating.
-2.7.10-
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Fault Calculations
Older breakers (1960's and before) were often rated in MVA. The
example below is taken, in part, from Wagner [see bibliography at
end of document].
Example: Breaker rating 5000 MVA at 69 kV. The breaker has a
rated-voltage interrupting capability of
5000 MVA / (1.732 x 69 kV) = 42 kA
The maximum interrupting capability of the breaker is 44 kA. If
the operating voltage is 67 kV, the breaker can interrupt a current
proportionately larger:
42 kA x 69 kV / 67 kV = 43 kA
If the operating voltage is 64 kV, a proportionate increase in
the interrupting capability would be 42 kA x 69 kV / 64 kV = 45 kA,
but that would exceed the maximum capability of 44 kA, so the
breaker interrupting capability is 44 kA at an operating voltage of
64 kV.
Now consider applying this breaker on a system with a Thevenin
reactance of 0.05 per unit (on 100 MVA and 69 kV bases) and an X /
R = 20, generation remote. Let the breaker have a rated
interrupting time of 5 cycles, which corresponds to a contact
parting time of 3 cycles. This gives a multiplying factor of 1.13
(from the third curve), and a breaker duty of
I" = 1 / 0.05 = 20 per unit
Ibase = 100 MVA / (1.732 x 69 kV) = 0.837 kA
I" = 20 x 0.837 kA = 16.7 kA
Breaker duty = 1.13 x 16.7 kA = 18.9 kA
Since this is well below the 69 kV rating of 42 kA, the
application is satisfactory.
-2.7.11-
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C. W. Brice August 2002
Figure 2.7.3 (a) Multiplying factors for circuit breakers rated
on symmetrical basis. Three-phase short circuits including both AC
and DC decrements, to be used when fault is electrically close to
generators (no more than one transformation away).
-2.7.12-
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Fault Calculations
Figure 2.7.3 (b) Multiplying factors for circuit breakers rated
on symmetrical basis. Single phase to ground faults near
generators.
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C. W. Brice August 2002
Figure 2.7.3 (c) Multiplying factors for circuit breakers rated
on symmetrical basis. All faults remote from generators.
-2.7.14-
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Fault Calculations
Figure 2.7.4 Multiplying factors for circuit breakers rated on
asymmetrical (total) short-circuit current basis.
-2.7.15-
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C. W. Brice August 2002
Practice Problem: A 13.8 kV, 8 cycle (4 cycle contact parting
time) circuit breaker has an interrupting rating of 42 kA (total
current) and a momentary rating of 80 kA. If applied on a 13.8 kV
system with X / R = 30, and X = 11% (on 100 MVA, 13.8 kV base),
evaluate the breaker interrupting rating.
Solution:
I" = 1 / 0.11 = 9.09 pu
Ibase = 100 MVA / (1.732 x 13.8 kV) = 4.18 kA
I" = 9.09 x 4.18 kA = 38 kA
If the breaker is fed predominantly from nearby generation, the
multiplying factor is 1.04, but the factor is 1.16 if it is remote
from generation.
1.04 x 38 kA = 39.5 kA
1.16 x 38 kA = 44 kA
The application is satisfactory only if it is fed predominantly
from nearby generation. Note that some judgment may be required in
interpreting the word "predominantly". The remote assumption is
always the conservative one.
B. Symmetrical current method: The newer ANSI standards have the
multiplying factors built in to the circuit breaker ratings, so
that the breaker has a symmetrical current interrupting rating
(ANSI C37.010). The new standard is based on X / R = 15. If X / R
exceeds 15, or if it is unknown, then the Simplified Method (E / X
Method) can not be used. The exception is that if the symmetrical
short-circuit current is < 80% of the breaker three-phase fault
interrupting capability (70% for line-to-ground fault) then the
application is satisfactory regardless of the X / R ratio.
If X / R < 15, the symmetrical short-circuit current can
reach 100% of the breaker interrupting capability. If X / R >
15, the standard C37.010 contains curves of multiplying factors,
which are to be multiplied by the symmetrical short-circuit current
I" to obtain the breaker interrupting duty. These curves are
duplicated in Figure 2.7.3. The use of these curves is called the E
/ X Method with Adjustments for Decrements.
The new breaker ratings are contained in ANSI C37.06, which
defines 1) rated maximum voltage, 2) rated voltage range factor K,
3) rated short-circuit current, among others. The rated maximum
voltage is the highest voltage for which the breaker is designed.
For example, a system with nominal voltage of 115 kV would require
a breaker with a maximum voltage of 121 kV, while a nominal voltage
of 138 kV would require 145 kV maximum voltage.
Circuit breakers with maximum voltages above 72.5 kV have
voltage range factors of K = 1.0, which means that their
interrupting rating is independent of the operating voltage. In
other words, the interrupting rating is a current rating. Breakers
with maximum voltages of 72.5 kV and below, have voltage range
factors 1.0 < K < 3.75. For these, the interrupting
capability is inversely proportional to the operating voltage, up
to a limit of K times the rated short-circuit current.
-2.7.16-
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Fault Calculations
Example: A circuit breaker has a rated maximum voltage of 15 kV,
a voltage range factor of K = 2.27, and a rated short-circuit
current of 19.0 kA. Apply on a system with a maximum operating
voltage of 13 kV. The interrupting capability is 19 kA X 15 kV / 13
kV = 21.9 kA. But K times the rated short-circuit current = the
maximum interrupting capability = 2.27 X 19.0 kA = 43.1 kA. The
actual interrupting capability is 21.9 kA.
Apply the same breaker on a 6 kV system. 19.0 kA X 15 / 6 = 47.5
kA. The maximum interrupting capability is 43 kA. The actual
capability is 43 kA.
Note also that the close and latch capability is 1.6 K times the
rated short-circuit current.
Practice Problem: A 13.8 kV system has a circuit breaker rated
(on a symmetrical basis) as follows: rated maximum voltage = 15 kV,
voltage range factor K = 1.30, rated interrupting time = 5 cycles
(contact parting time = 3 cycles), maximum symmetrical interrupting
capability = 48 kA, rated short-circuit current (at rated maximum
voltage) = 37 kA, closing and latching capability = 77 kA. If the
system X / R is unknown, and X = 14% (on 100 MVA, 13.8 kV base),
evaluate the breaker interrupting capability.
Solution:
I" = 1 / 0.14 = 7.14 pu
Ibase = 4.18 kA
I" = 7.14 x 4.18 kA = 30 kA
The interrupting capability at 13.8 kV is 37 kA (15/13.8) = 40
kA. 80% of 40 kA is 32 kA, which is greater than the 30 kA
interrupting duty. Thus, the breaker is satisfactory, regardless of
the X / R ratio.
Example: A 5 cycle breaker has ratings as follows: maximum
voltage = 38 kV, K = 1.65, rated short-circuit current = 22 kA. It
is operated at 33 kV. The three-phase symmetrical current
capability is 22 kA X 38 / 33 = 25.3 kA. Max. capability = 36.3 kA.
Actual capability = 25.3 kA.
Note that the single line-to-ground capability is 1.15 times the
three-phase capability, but not to exceed the maximum symmetrical
capability.
Going back to the previous example, the line-to-ground fault
capability is 1.15 X 25.3 kA = 29.1 kA, and the maximum capability
is 36 kA. The actual capability is 29.1 kA.
If there is a need to calculate the asymmetrical (total) current
capability, ANSI C37.010 gives S-factors to accomplish this. For
example, a 5 cycle breaker (3 cycle contact parting time) has S =
1.10, which would give in the last example the following
capabilities:
three phase fault: 1.1 X 25.3 = 27.8 kA
line to ground fault: 1.1 X 29.1 = 32 kA.
Another example: A 46 kV system, a 5 cycle breaker with maximum
voltage of 48.3 kV, K = 1.21, rated short-circuit current = 17 kA.
At 46 kV the three-phase capability is 17.9 kA, the line
-2.7.17-
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C. W. Brice August 2002
to-ground capability is 20.5 kA, and the maximum capability is
20.6 kA. If the three-phase fault current is
I" = E / X = 16.5 kA
and the line-to-ground fault current is
I"LG = 3E / (2X1 + X0) = 17.0 kA
The three-phase capability is (16.5/17.9) 100% = 92% >
80%
The line-to-ground capability is (17/20.5) 100% = 83% >
70%
The multiplying factors must be used. Assume the breaker is
remote from generation and the system X / R ratio is 25 for
three-phase faults, and X / R = 30 for line-to-ground faults. Using
minimum contact parting time of 3 cycles, the three-phase factor is
1.10 and the line-ground factor is 1.14 (from the third set of
curves).
The three-phase duty is 1.10 x 16.5 kA = 18.2 kA.
The line-ground duty is 1.14 x 17.0 kA = 19.4 kA.
Since the three-phase capability is only 17.9 kA, this
application is unsatisfactory. The line-ground capability of 20.5
kA is marginally satisfactory.
Practice Problem: Same as the previous practice problem, but X /
R = 30, and X1 = 11%, X0 = 10%, a) electrically close to
generation, b) remote from generation. Evaluate interrupting
capability for three-phase and for line-to-ground short
circuits.
Solution:
Three-phase short circuits:
I" = (1 / 0.11) x 4.18 kA = 38 kA
a) close to generation: multiplying factor is 1.03
Interrupting duty = 38 kA x 1.03 = 39 kA. Interrupting
capability at 13.8 kV = 40 kA; application is satisfactory.
b) remote from generation: multiplying factor is 1.13
Interrupting duty = 38 kA x 1.13 = 43 kA.
Application is not satisfactory.
For a single line-to-ground short circuit:
I" = 3 x 1.00 / (.22 + .10) = 9.375 per unit
9.375 x 4.18 kA = 39.2 kA
Interrupting capability = 1.15 x 40.2 kA = 46.2 kA
a) close to generation:
Interrupting duty = 1.10 x 39.2 kA = 43.1 kA. Application is
satisfactory.
-2.7.18-
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Fault Calculations
b) remote from generation:
Interrupting duty = 1.13 x 39.2 kA = 44.3 kA. Application is
satisfactory.
Although the application is satisfactory with regard to
line-to-ground faults, it is satisfactory with regard to
three-phase faults only if it is fed predominantly from
generation.
A serious problem in sizing circuit breakers is unanticipated
growth in the available short-circuit MVA. Any new construction of
generating plants and lines tends to decrease the reactance of the
system, which will result in an increase in the available
short-circuit MVA. The only defenses against this problem are to
size the breaker somewhat larger than the present system would
require, or to install current limiting apparatus such as
reactors.
Automatic reclosing of circuit breakers is a useful method of
decreasing the severity of system faults. This is true since the
majority of faults on many systems are temporary. Some breakers
(especially oil circuit breakers) with more than two operations, or
less than 15 second delay before reclosing, will require derating
of the breaker. SF6 breakers usually do not require derating. This
method of derating the breaker is given in ANSI C37.06.
Note that this is not a complete discussion of circuit breaker
application. In particular, we have not mentioned transient
recovery voltage (the voltage across the open breaker contacts
after operation), or switching capacitors. See ANSI standards and
Wagner for more information.
The last practice problem is an industrial power system with
utility feed and in-plant generation. This problem is very similar
to an illustrative example in IEEE 141 (the Red Book) but is not as
extensive.
Practice Problem: For the system of Figure 2.7.5, perform a
short-circuit study for a three-phase fault on the 13.8 kV bus.
Calculate the first-cycle duty and the interrupting duty on the
13.8 kV breakers. Note that different reactances are used for the
first-cycle duty than for the interrupting duty. This requires two
short-circuit studies. Note also that the duty on all breakers
connected to a bus is the total fault current into a bus fault.
This is because the actual fault may be on either side of a given
breaker, so that each one must be able to handle the total fault
current at the bus. Select one of the circuit breakers listed
below. All have a nominal voltage class of 13.8 kV rms, a rated
maximum voltage of 15 kV rms, and a rated interrupting time of 5
cycles.
-2.7.19-
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C. W. Brice August 2002
3 IND.MOTORS
69 kV SYSTEM: 1000 MVA SHORT CIRCUIT AVAILABLE
G1: 20 MVA GENERATOR Xd" = 0.09 X/R = 40
T1: 20 MVA X = 0.07 X/R = 21
EACH T2: 1.5 MVA X = 0.055 X/R = 10
EACH LV SUBSTATION HAS MOTOR LOAD:
0.5 MVA TOTAL MOTORS EACH 50 TO 150 HP
1.0 MVA TOTAL MOTORS EACH < 50 HP
Xd" = 0.25 X/R = 9
1750 HP IND. MOTORS, EACH:
1800 RPM Xd" = 0.17 X/R = 30
500 HP IND. MOTORS, EACH:
1800 RPM Xd" = 0.18 X/R = 20
EACH 1750 HP
SUBSTATIONS
69 kV SYSTEM
LOAD
STATIC
T2
T1
T2
UNIT
2 LV
T2 T3
G1
4 IND.EACH 500 HP
4.16 kV BUS
13.8 kV BUS
MOTORS
Figure 2.7.5 Example system for short-circuit calculations.
-2.7.20-
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Fault Calculations
Nominal 3-phase MVA
class
Rated voltage range factor
K
Rated symmetrical short-circuit current
(at rated max. voltage) kA, rms
Maximum symmetrical short-circuit interrupting
capability kA, rms
Closing & latching
capability kA, rms
250 2.27 9.3 21 34
500 1.30 18 23 37
750 1.30 28 36 58
1000 1.30 37 48 77
Solution:
Transformer reactances in per unit on 10 MVA base:
T1: X = .07 (10/20) = .035
T2: X = .055 (10/1.5) = .367
T3: X = .055 (10/7.5) = .0733
Rotating machine subtransient reactances in per unit on 10 MVA
base:
69 kV system: I" = 100 per unit, X = 1/100 = .01
Generator G1: X" = .09 (10/20) = .045
1750 HP Induction Motor (kVA rating = HP rating): X" = .17
(10/1.75) = .971
500 HP Induction Motor (kVA rating = HP rating): X" = .18
(10/0.50) = 3.60
LV motor group of 1.0 MVA, less than 50 HP each: X" = .25 (10/1)
= 2.50
LV motor group of 0.5 MVA, 50 to 150 HP each: X" = .25 (10/0.50)
= 5.00
Reactances for AC high-voltage circuit breaker first-cycle
duties. Use subtransient reactances, except for:
500 HP motor, use 3.60 x 1.20 = 4.32 LV motor group, 50 to 150
HP, use 5.00 x 1.2 LV motor group, less than 50 HP, omit.
Reactances for AC high-voltage circuit breaker interrupting
duties. Use subtransient reactances, except for:
1750 HP motor, use .971 x 1.50 = 1.457 500 HP motor, use 3.60 x
3.00 = 10.8 LV motor group, 50 to 150 HP, use 5.00 x 3.00 = 15.0 LV
motor group, less than 50 HP, omit.
-2.7.21-
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C. W. Brice August 2002
X / R ratio and resistances for interrupting duties. T1: X / R =
21, X = .0350, R = .001667 T2: X / R = 10, X = .367, R = .0367 T3:
X / R = 14, X = .0733, R = .00524 69 kV system: X / R = 22, X =
.010, R = .000455 G1: X / R = 40, X = .045, R = .001125 1750 HP IM:
X / R = 30, X = 1.457, R = .0486 500 HP IM: X / R = 20, X = 10.8, R
= .540 LV motor group: X / R = 9, X = 15.0, R = 1.667 1 / .045 =
22.2 1 / .322 = 3.11 1 / 3.18 = .314
For a three-phase short circuit on the 13.8 kV bus, I" = 22.2 +
22.2 + 3.11 + .314 = 47.8 per unit
First-cycle duty = 47.8 x 1.6 x 10 MVA / (1.732 x 13.8 kV) = 32
kA 1 / .045 = 22.2 1 / .485 = 2.06 1 / 7.68 = .13
For a three-phase short circuit on the 13.8 kV bus, I" = 22.2 +
22.2 + 2.06 + 0.13 = 46.6 per unit
Symmetrical interrupting duty = 46.6 x 10 MVA / (1.732 x 13.8
kV) = 19.5 kA
The 750 MVA, 15 kV breaker has an interrupting capability of 28
kA (15/13.8) = 30 kA at 13.8 kV. 80% of this value is 24 kA, which
is sufficient for this application. The closing and latching
capability is 58 kA, which is sufficient (compare to only 32 kA
first-cycle duty).
To see if this is excessively conservative, calculate the X / R
ratio:
Resistance network for interrupting duty. 1/R = 1/.00212 +
1/.001125 + 1/.0197 + 1/.852 R = .000708 per unit X / R =
.02144/.000708 = 30.3
If remote from generation, interrupting duty = 1.13 x 19.5 = 22
kA. If close to generation, 1.03 x 19.5 = 20 kA.
The 500 MVA breaker has a symmetrical interrupting capability of
18 (15/13.8) = 19.6 kA, which is not adequate. Select the 750 MVA
breaker. The 1000 MVA breaker might be justified if short-circuit
current levels are expected to rise in the near future.
-2.7.22-