842
IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS. VOL. IA-21, NO. 4. JULY/AUGUST 1985
Understanding Asymmetry
CRAIG N. HARTMAN,
MEMBER, IEEE
Abstract-When calculated fault levels lie very close to circuit breaker
interrupting ratings, a thorough evaluation of the asymmetrical currents
involved may become the deciding factor between breakers of varied
capabilities and costs. The increasing emphasis on efficiency tends to
exacerbate concerns in this area. An attempt is made to help the reader
visualize the events of the first few cycles after fault inception as well as
providing a rigorous quantitative analysis of the magnitude of currents
involved. Equations for sizing both molded-case and power circuit
breakers are developed.
INTRODUCTION
WHENEVER a short-circuit occurs on a conductor there
Time (Cycles)
exists a transient response time during which the circuit
(a)
circuits
In
ac
attempts to reach a steady-state fault condition.
Magnitude
(P.U.)
of
those
excess
in
currents
produce
may
response
transient
this
2.0
encountered either before the fault occurred or after the
DC Componn
Asymmetria
steady-state condition is achieved. These high currents must be
taken into consideration when specifying power system
equipment.
It is instructive to note the effect which high efficiency
equipment is having on this type of analysis. Generally
transformers constitute the bulk of the systems impedance, and
thus have a predominate effect on the system X/R ratio. A
00 0 Q1 0.2 0.3 0 4 0.5 0.6 0.7 0 8 0.9 1 0
cursory look at this situation would indicate that increasing the
Time (Cycles)
efficiency of a transformer from 98-percent to 99-percent
(b)
would have the effect of doubling the X/R ratio. This effect is
Fig. 1. Fully offset wave.
seen in virtually all high efficiency equipment. The X/R ratio
3) the fault impedance is purely inductive;
of modem distribution systems commonly exceeds the estab4)
no current was flowing prior to fault inception.
lished X/R ratio of the protective equipment, thus requiring
derating.
Referring to Fig. 1(a) we note that at time zero both current
The author feels that it is important not only to take these and voltage are zero. Recall that the rate of change of current
factors into account when sizing the equipment, but also to in an inductor is proportional to the voltage across that device.
have some visual conception of what actually takes place At time 0.05 cycles the voltage is beginning to rise but is still
during those first few cycles following inception of a fault. It is rather low. The current in the inductor is also beginning to
felt that a good visualization of the situation would clarify a rise, but the slope of current change is small in response to the
number of misconceptions which exist relative to this subject. low voltage value. As the voltage increases, the rate of change
Accordingly this article shall begin by "walking through" a (slope) of the current wave increases in response. At time 0.25
hypothetical fault.
cycles the voltage reaches a maximnum, and the slope of
current is at its steepest value. NoW the voltage begins to drop,
FAULT VISUALIZATION
but since it is still positive the current continues to increase. At
A worst-case condition is assumed in which
time 0.45 cycles the current is still increasing, but due to the
1) the fault occurs at an instant of time when the source low voltage level the current increases at a very slow rate. At
time 0.5 cycles the current finally reaches its maximum value.
voltage is equal to zero;
2) the source voltage can supply an infinite amount of Not until the voltage goes negative can the current begin to
decrease. Using the same arguments one can follow the course
current with no voltage dip;
cycles.
Paper IPSD 84-38, approved by the Power Semiconductor Committee of of the current wave back to zero magnitude at time 1.0
Using the same reasoning above, but with the fault
the IEEE Industry Applications Society for presentation at the 1984 Industrial
and Commercial Power Systems Conference, Milwaukee, WI, May 9-12, occurring at a voltage maximum, one could trace a current
1984. Manuscript released for publication November 2, 1984.
(no
The author is with the Westinghouse Electric Corporation, 3900 S. wave which is symmetrical with regard to current zero
offset). By looking at the two waves it bbecomes obvious that
Wadsworth Blvd., Lakewood, CO 80235.
-
0093-9994/85/0700-0842$01.00 (© 1985 IEEE
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843
HARTMAN: UNDERSTANDING ASYMMETRY
the fully offset current wave is simply obtained by adding a dc
current equal to the peak magnitude of the symmetrical wave
to the wave itself (Fig. 1(b)). The actual value of the dc
component would depend on the time at which the short-circuit
in the voltage wave and is quantitatively equal and
opposite to the value of the steady-state symmetrical current
wave at time zero. When the dc current assumes a value equal
to the peak value of the symmetrical current, we say that the
wave is fully offset (maximum asymmetry). It is possible
under certain circumstances to get more than 100-percent
offset, but these special circumstances are beyond the scope of
this paper.
occurs
QUANTIFYING THE CURRENT VALUES
In real life, of course, perfect inductors do not exist any
more than do perfect voltage sources. In order to obtain more
realistic actual values the hypothetical fault circuit discussed
above will be modified to include a resistance. This circuit is
now as shown in Fig. 2.
If the switch is closed at time zero to simulate initiation of
the fault and if current values are per-unitized according to the
peak symmetrical current (IO) then the current plot is a
function of three independent variables (0, X/R, and time).
Equation (1) results from using Kirchhoffs voltage law to sum
the voltages around the circuit of Fig. 2 and then solving the
resulting nonhomogeneous first order differential equation for
current.
I= IO{sin [arctan (X/R)
exp [- 27rt/(X/R)1 (dc component)
+ sin [2rxt
arctan (X/R)I} (ac component) (1)
where
Io symmetrical peak current
X reactance value
R resistance value
t
time (cycles).
A few comments concerning (1) might be expedient at this
point. The voltage and current waves will be displaced from
each other by an angle corresponding to the amount of
reactance in the circuit compared to the amount of resistance in
the circuit. This angle is equal to arctan (X/R). When the
circuit is purely inductive (which it is to a large degree during
faults on high power circuits) the current wave will be
displaced from the voltage wave by 90 degrees. As resistance
is added to the circuit this angle will go from 90 degrees to
zero degrees so that in the purely resistive circuit the voltage
and current will be completely in-phase. Also note that at time
zero the dc component is exactly equal in magnitude to the
value of the ac component but opposite in sign. This condition
must exist due to the fact that current cannot change
instantaneously in an inductive circuit. The argument of the
exponential indicates that the dc component will decay at a rate
dependent on the X/R ratio.
MAXIMUM ASYMMETRY VERSUS MAXIMUM PEAK
CURRENT
to
It is Interesting note that the maximum asymmetrical peak
currents do not generally occur during conditions of maximum asymmetry. It has been shown [I] that the maximum
peak asymmetrical current will be produced for any X/R ratio
when the fault occurs at voltage zero (O = 0). As shown in
(I), however, maximum asymmetry exists when the fault
occurs at a symmetrical current component maximum (0 =
arctan (X/R) - 90°). (Maximum asymmetry is defined as
being that state which produces the maximum dc component
value.) Thus the fault angle for maximum peak current is
always zero whereas, the fault angle for maximum asymmetry
ranges from 0-90 degrees.
Table I shows the maximum actual peak current for several
X/R ratios, along with the dc component, ac component, and
the time in cycles at which the current peaks. Table II
TABLE I
MAXIMUM FAULT CURRENTS
-
X/R
I
Ii
I.
I
Time
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
2
3
4
5
10
15
20
25
30
35
40
45
50
100
1000
1.00
1.00
5E-9
3E-5
6E-4
3E-3
0.01
0.02
0.03
0.04
0.06
0.07
0.25
0.39
0.49
0.56
0.74
0.82
0.86
0.88
0.90
0.92
0.93
0.93
0.94
0.97
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
0.26
0.28
0.29
0.31
0.32
0.33
0.34
0.35
0.36
0.36
0.41
0.43
0.44
0.45
0.47
0.48
0.48
0.49
0.49
0.49
0.49
0.49
0.49
-
V Ldi
VL=L-Ll
dit
(J'
v(t) =Vo sin
v
st
VR=Ri(t)
(wt +o)
I.,,
id
I.
In
T
Fig.
2.
Asymmetrical circuit model.
1.00
1.00
1.01
1.02
1.03
1.04
1.05
1.07
1.24
1.38
1.48
1.55
1.74
1.81
1.86
1.88
1.90
1.92
1.93
1.93
1.94
1.96
2.00
0.99
0.99
0.99
0.99
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.01
1.06
1.14
1.21
1.27
1.45
1.53
1.57
1.60
1.62
1.64
1.65
1.66
1.66
1.70
1.73
0.50
0.50
maximum current in per-unit of peak symmetrical
dc component in per-unit of peak symmetrical
ac component in per-unit of peak symmetrical
rms current in per-unit of symmetrical rms
time to I, in cycles.
Note: Maximum peak always occurs when fault is initiated at zero voltage (not
maximum asymmetry).
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IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS, VOL.
844
IA-21. NO. 4. JULY/AUGUST 1985
TABLE II
DIFFERENCES IN CALCULATED FAULT LEVELS ACCORDING TO METHOD OF CALCULAFION
.I
Ipeak
maximum
X/R
asymmetry
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
2
1.000
1.000
1.000
1.000
3
4
5
10
15
20
25
30
35
40
45
50
100
1000
Inf.
1.002
1.005
1.011
1.020
1.030
1.043
1.208
1.351
1.456
1.533
1.730
1.811
1.855
1.882
1.901
1.914
1.924
1.933
1.939
1.969
1.997
2.000
1p,ek (maximum asymmetry)
I'k (maximum peak)
'ems (maximum asymmetry)
,e,s (maximum peak)
rm*
'peak
maximum
peak
1.000
1.000
1.001
1.003
1.008
1.016
1.026
1.039
1.053
1.069
1.242
1.379
1.477
1.550
1.737
1.814
1.857
1.883
1.902
1.915
1.925
1.933
1.939
1.696
1.997
2.000
Difference
(percent)
0
0
0.10
0.26
0.61
1.05
1.44
1.86
2.14
2.41
2.75
2.04
1.43
1.07
0.38
0.16
0.13
0.06
0.05
0.04
0.03
0.02
0.01
0
0
0
frms
I,ms
maximum
asymmetry
maximum
peak
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Difference
1.002
1.042
1.116
1.190
1.253
1.438
1.522
1.569
1.599
1.619
1.634
1.646
1.655
1.662
1.273
1.477
1.526
1.572
1.601
1.621
1.636
1.647
1.656
1.663
1.697
1.728
1.732
1.732
1.001
1.004
1.009
1.016
1.025
1.036
0.09
0.16
0.21
0.31
1.76
2.07
1.99
1.60
0.64
0.28
0.21
0.15
0.11
0.10
0.06
0.05
0.03
0.02
1.214
1.627
1.728
1.000
0
0
0
0
0
0
1.000
1.000
1.000
1.000
1.001
1.002
1.003
1.005
1.061
1.140
1.001
I.*
(percent)
1.048
1.062
1.076
1.2 19
1.324
1.395
1.446
1.571
1.620
1.646
1.662
1.673
1.681
1.687
1.692
1.695
1.713
1.729
1.732
0.02
0
peak current under conditions of maximum asymmetry
maximum peak current possible
rms current under conditions of maximum asymmetry
rms current under conditions of maximum peak current
actual calculated rms current over the first half-cycle assuming conditions of maximum asymmetry.
compares the actual maximum peak current to the peak current
under conditions of maximum asymmetry.
Since finding maximum current magnitudes using (1) is
rather unwieldy, the traditional method for calculating currents has been to assume a fault of maximum asymmetry and
then calculate the peak current at 0.5 cycles. Substituting these
assumptions into (1) yields the following:
(2)
Ipeak = 'o exp [ -7r/(X/R)] + 1.
As shown in Table II the maximum error one can expect from
using this simplification will be less than 3-percent. For high
X/R ratios, where asymmetrical current is most important,
the errors are negligible.
In addition to the maximum peak value of current, one may
also be interested in the value of peak or root mean square
(rms) current at other times (e.g., at the time the circuit
breaker interrupts the circuit). Fig. 3 shows a typical plot of
current for an X/R ratio of 15. Either peak or rms current
values may be obtained at any instant of time by simply
knowing the values of the dc and ac components.
A common misconception is that high asymmetrical currents are the result of stored energy which is released from the
system during the first few cycles. Actually, exactly the
opposite is true. Any energy stored in the system inductances
tends to reduce the amount of asymmetry which occurs. Also,
asymmetry is maximum when the power factor of the fault is
close to zero indicating that the energy loss is only a small
Magnitude (P.U.)
20
DC Components
1.8
g
1.6 -
14
_Actual Current
-
1.21.0
0.8p
A
tu
l
XI
-02
0.2
04
- 0 6
- 0.8
=06. -,i 11olI I I I 7 I AI I I I X I I
00 05 1.0 1.5 20 25 3.0 3.5 40 45 5.0 5.5 60 6.5 70
Time (Cycles)
Fig. 3. Current
wave
for
an
X/R of 15.
fraction of the kilovolt ampere figure.
Thus far we have discussed only a simple series inductanceresistance (LR) circuit. It is obvious that, were two such
circuits to feed a fault, the asymmetrical current would be the
sum of the current contributions from each leg of the circuit.
While combining the two impedances into a Thevenin's
equivalent impedance would yield accurate values of symmetrical current, the asymmetric current would contain an error
dependent on the mismatch of X/R ratios of the two legs.
Modern distribution systems contain many series/parallel
elements making rigorous analysis unfeasible. In addition, the
power system data is rarely characterized with better than 10percent accuracy when all is told. One should therefore not be
caught up in assuming an accuracy of calculation greatly
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845
HARTMAN: UNDERSTANDING ASYMMETRY
beyond the accuracy of the assumptions. All calculation
methods presently employed are the result of judicious
compromise, generally containing both conservative and
nonconservative elements.
A final note to consider is that when generation is close to
the fault or there is a substantial motor load the ac component
of the wave will decay along with the dc component. Neither
ac decrement nor methods of computing fault levels in power
systems will be covered in this discussion. Although brief
mention is made of ac decriment curves in the high voltage
circuit breaker section of this paper one should refer to a more
extensive treatise on circuit breaker application [2] for this
information.
APPLICATIONS TO AC CIRCUIT BREAKERS
Since asymmetrical currents play an important role in the
rating of ac circuit breakers it may be appropriate to address
the subject of circuit breaker ratings in connection with this
discussion.
rms and Peak Currents
Since circuit breakers are generally rated with reference to
rms amperes it is important to discuss exactly what this term
means and how it relates to peak currents and circuit breaker
ratings. Mathematically an rms value is computed by taking
the square root of the mean (arithmetic average) of the square
of the current wave oVf r a certain period of time. Thus the
term rms is inseparably ,onnected with the value of time over
which it is compared. The rms value bears no relationship to
the peak value unless the shape of the current wave is known.
When an ac periodic nondecaying sinusoidal wave is superimposed on a dc wave the rms value will be the square root of the
sum of the squares of the ac symmetrical rms value and the dc
value.
In circuit breaker ratings rms values are computed in this
manner by taking a "snapshot" of the dc current component at
a certain instant of time and calculating the rms value as if this
dc component were constant over the entire cycle. This value
might be referred to an an "instantaneous rms" (which almost
seems to be a contradiction of terms) and is much simpler to
compute than a real rms value over a period of time.
According to ANSI C37.04-1979 [7].
...the required asymmetrical interrupting capability of a
circuit breaker is the highest value of the total short-circuit
current rms amperes at the instant of primary arcing contact
separation...
All rms currents mentioned henceforth is this article will be
computed using the instantaneous rms concept. (Table II
shows the actual rms values over the first half-cycle of the
fault current wave, assuming maximum asymmetry, for
comparison.)
A Circuit Breaker Primer
When fault current flows through a breaker, magnetic
forces proportional to the square of the instantaneous current
are created which tend to blow the contacts apart or otherwise
cause mechanical damage. These considerations are most
important in establishing the momentary rating of a circuit
Magnitude (P.U.)
2.8
1 = Peak value (in P.U. of sym. RMS)
2.6
2 = Peak value (in P.U. of sym. Peak)
= RMS value (in P.U. of sym. RMS)
2.2
=
=
2.0
1.8
1.6
1.4
N
2
-
1.2Z
1.0
0.0
23
L..ILI I.L
1.0
2.0 3.0 4.0
Tiime (Cycles)
5.0
6.0
7.0
Fig. 4. Peak versus rms currents.
breaker. Since these forces respond to instantaneous values of
current, the momentary rating is dependent on the maximum
peak current value. In addition to the mechanical stress,
thermal stresses proportional to the current times the arc
voltage (i.e., power input) exist which cause temperatures
within the arc to approach 50 000 K. This energy tends to
ionize particles, which then perpetuate the arc. The interrupting device must be able to deal with the heat being generated
within the arc to achieve successful interruption. It is the rms
value or heating effect which is most important here. Although
the above is oversimplified, neglecting other important criteria, it will serve to allow us to look at these two important
values (peak current and rms current).
Fig. 4 shows the curve for peak current and rms current as a
function of time for an X/R ratio of 15. Note that while the
maximum asymmetrical rms value is 1.732 times the symmetrical rms value, the peak instantaneous value may reach
double the symmetrical peak value. At approximately 0.5
cycles where the first real peak would occur in Fig. 4 the peak
value is 1.81 times that of the symmetrical wave while the rms
value is only 1.52 times that of the rms symmetrical value. A
circuit breaker seeing an asymmetrical rms fault value of 1.52
would therefore have to withstand a peak value of 1.81. (High
voltage circuit breakers have a momentary rating of 1.6 times
symmetrical, or 1.88 times peak. This is 5 percent higher than
shown in Fig. 4.)
APPLICATION TO AC HIGH-VOLTAGE CIRCUIT
BREAKERS
Most ac high-voltage circuit breakers today are rated on the
"symmetrical current basis of rating" which, simply stated,
means that when the breaker is rated to interrupt the
symmetrical value of fault current, and the X/R ratio is below
a predetermined value, then the circuit breaker will be able to
withstand and interrupt any asymmetrical current values which
occur. This makes sizing a circuit breaker easy given that the
X/R ratio is less than the predetermined value. The actual
capability of circuit breakers is stated in ANSI C37.04 and
C37.06 [7] and [8], and the test procedures are given in ANSI
C37.09 [9]. The application procedures are given in ANSI
C37.010 [10]. According to ANSI/IEEE C37.010-1979 [10],
A circuit breaker having adequate symmetrical interruption
capability will have adequate capability to meet all of the
related short-circuit requirements unless there is a significant
contribution from motor load or unless the X/R ratio is greater
than approximately 15.
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IEEE TRANSACTIONS ON INDUJSTRY
846
TABLE I1I
CIRCUIT BREAKER OPERATION TIMES
Contact
Rated
Interrupting
Time
(cycles)
Opening
(cycles)
(cycles)
Time
Capability
Factor
2
3
5
8
1.0
1.5
2.5
3.5
1.5
2.0
3.0
4.0
1.3
1.2
1.1
1.0
Time
Parting
APPLICATIONS, VOL. IA-21
NO. 4, JULY/AUGUST 1985
While (6) gives the estimated capability factor, it should be
understood that for each circuit breaker a certain capability
factor is mandated by standards and that circuit breakers are
tested to those stated values. These values were noted for each
breaker in Table III.
It would appear from the above that a circuit breaker with a
large capability factor could be applied on systems with low
X/R ratios at a symmetrical current higher than its rating.
Current interruption, however, is a function of the transient
recovery voltage (TRV) as well as the thermal energy in the
arc. TRV is a function of di/dt which relates to the
symmetrical component of the current, so the symmetrical
value of the fault current is just as important as the
asymmetrical value. Circuit breakers should therefore never
be applied where the symmetrical fault current exceeds their
symmetrical rating, no matter what the X/R ratio.
For systems with larger X/R ratios, however, the capability
factor can be put to good use. Equation (7) may be used to
determine a multiplying factor for X/R ratios higher than 15
by which the symmetrical fault current must be multiplied.
(The reciprocal of (7) could also be used as a derating factor
for the breaker.)
V I + 2 exp [ - 4xt/(X/R)]
multiplying factor
(7)
S
In ac high-voltage circuit breakers, the predetermined X/R
ratio value has been established at 15. Thus, all ac highvoltage breakers must be able to withstand the 1.52 per-unit
rms current and 1.81 per-unit peak current (2.55 times the
symmetrical rms value) as shown in Fig. 4 at 0.5 cycles. As
stated previously they are actually rated at 1.6 per-unit rms or
2.7 per-unit peak.)
The interrupting rating is somewhat more difficult to assess
since different circuit breakers will interrupt the fault at
different times. As a result these circuit breakers are classified
according to their rated interrupting times. Table III shows
circuit breaker operating times according to ANSI/IEEE
C37.04-1979 [7].
The rated interrupting time is the maximum permissible where
interval between the energizing of the trip circuit at rated
t
contact parting time in cycles
control voltage and the interruption of the main circuit on all
X/R
actual circuit X/R
poles. A breaker should be rated to interrupt the amount of
S
factor.
capability
current present at the contact parting time. Contact parting
time is determined by adding 1/2 cycle relay time to the
A plot of (7) will correspond to the graphs shown in ANSI/
breaker opening time. The curves of Fig. 4 may be drawn for IEEE C37.010-1979 [10, fig. 10, p. 36]. The equation is not
any X/R ratio using (3), (4), and (5).
meant to imply greater accuracy than the curve it represents,
but only greater convenience should the user wish to use some
dc
t in cycles
27rt/(X/R)]
exp [
sort of computerized selection procedure for large distribution
systems.
(4)
peak current dc + 1
So far we have discussed derating only with regard to the
rms current- l2dc2+l .
(5) rms values. Should we not have equal concern for the peak
A five-cycle breaker with a standard three-cycle contact current values? Table IV shows the peak values of current
parting time will be used as an example. As seen in Fig. 4 the along with the multiplying factors for rms current from (7).
Note that under certain conditions the peak ratio multiplying
rms current which the breaker must interrupt is 1.08 times the
symmetrical value. This value occurs at the contact parting factor exceeds that of the rms by about 2 percent. Since the
time of 3.0 cycles. The ability of a circuit breaker to interrupt actual momentary tested value of 1.6 discussed earlier exceeds
the calculated value of 1.52 by 5 percent it would appear that
a higher rms value of asymmetrical current than its symmetrithe
multiplying factors provided by (7) would provide the
cal value is called "capability factor." The breaker just
necessary
protection for peak currents as well as rms.
discussed would need a capability factor of about 1.08. The
to ANSI/IEEE C37.010-1979 [10], if the fault is
According
actual capability factor of this breaker is 1.1. Therefore a fivefrom generators through a) not more than
fed
predominantly
cycle circuit breaker with a contact parting time of three cycles
or b) a per-unit reactance external to the
transformation
one
can interrupt 10 percent more asymmetrical rms current that
is
which
less
than 1.5 times the generator per-unit
its rated symmetrical value. The approximate capability factor generator
on
a common system megavolt-ampere
subtransient
reactance
for any circuit breaker may be obtained from (6)
base, then one should use Figs. 8 and 9 from [10] for derating.
These figures include the effect of ac decrement and have been
(6)
S = 1+2 exp[-4irt/(X/R)]
determined
by empirical methods.
where
The IEEE working group has determined that the standards
S
are slightly "pessimistic" as far as handling the dc component
capability factor
t
is concerned, but "optimistic" as far as handling the ac
contact parting time in cycles
X/R
15.
component is concerned. Combined, the results are quite
=
=
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847
HARTMAN: UNDERSTANDING ASYMMETRY
TABLE IV
COMPARISON OF DERATING HIGH-VOLTAGE BREAKERS BY PEAK
VERSUS DERATING BY RMS
X/R
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Peak
ratio
1
1.006
1.011
1.016
1.020
1.024
1.028
1.031
1.034
1.037
1.039
1.041
1.044
1.046
1.048
1.049
-rms (by maximum asymmetry)Two cycle Three cycle Five cycle Eight cycle
0.964
0.978
0.991
1.004
1.015
1.026
1.036
1.046
1.055
1.064
1.072
1.079
1.087
1.093
1.100
1.106
0.977
0.992
1.006
1.019
1.032
1.044
1.056
1.067
1.077
1.087
1.097
1.106
1.114
1.123
1.131
1.138
0.980
0.992
1.003
1.015
1.027
1.038
1.049
1.060
1.071
1.082
1.092
1.102
1.112
1.121
1.130
1.139
1.034
1.042
1.051
1.060
1.069
1.078
1.087
1.097
1.107
1.116
1.126
1.135
1.145
1.154
1.163
1.172
Peak ratio: peak current in per-unit of peak current with X/R = 15.
rms (by maximum asymmetry): derating factor by (7) [6-10] for each
breaker interrupting time.
TABLE V
POWER FACTOR OF TEST CIRCUIT FOR MOLDED CASE BREAKERS
Rated Interrupting Current
Amperes
Power Factor
10000 and less
10001 to 2000
Above 20000
0.45-0.50
0.25-0.30
0.15-0.20
TABLE VI
CAPABILITIES OF MOLDED CASE CIRCUIT BREAKERS
Rated
Interrupting
Current
(amperes)
X/R
(rms)
iaym
lsym
(peak)
10000 and less
10001 and 20000
Above 20000
1.7-2.0
3.2-3.9
4.9-6.6
1.03-1.04
1.13-1.18
1.25-1.33
1.16-1.21
1.37-1.44
1.53-1.62
Iaym (rms) is in per-unit of symmetrical rms.
I8,y. (peak) is in per-unit of symmetrical peak.
Using (3), (4), and (5) as before we can calculate the
close. It should be stated, also, that long experience with asymmetrical currents shown in Table VI. For values of Iasym
applying circuit breakers by these methods has seemed to in per-unit of symmetrical rms current, simply multiply by the
square root of 2. These values are usually taken to be 1.7, 2.0,
verify their adequacy.
and 2.2, respectively, for the three breaker ratings. For system
X/R ratios higher than those in Table VI, the multiplying
LOW-VOLTAGE POWER CIRCUIT BREAKERS
Low-voltage power circuit breakers are assumed to have a factor to use for modifying the calculated fault values is
arrived at using (9)
contact parting time of 1/2 cycle unless otherwise stated. This
means that their momentary and interrupting ratings should be
1 + exp [ - 7r/(X/R)]
multiplying factor =
identical. An X/R ratio of 6.6 is also used rather than the 15
Iasym (peak)
used for high-voltage breakers. Finally considering the characteristics of low-voltage breakers (both power and molded case)
(see Table VI). (9)
it has been decided that peak currents rather than rms currents
will determine the interrupting capability. From (3) and (4),
OTHER METHODS OF BREAKER RATINGS
using a time of 0.5 cycles and an X/R ratio of 6.6, it can be
Occasionally circuit breakers (especially older models) may
seen that these breakers must interrupt a peak current of 1.62
be
rated according to asymmetrical single-phase amperes or
times that of the symmetrical peak (2.3 times that of the
asymmetrical
three-phase amperes. In the asymmetrical sinsymmetrical rms). For X/R ratios higher than 6.6 one may
gle-phase
case
one may use (3) and (5) to find the nns current
use (8) to modify the fault current prior to selecting a breaker
and match this to the interrupting rating. In the asymmetrical
three-phase case, (10) should be used and matched to the
1 + exp [-rl(XIR)]
multiplying factor
(8)
interrupting rating. A convenient table to use for this purpose
1.62
is included in NEMA standard publication No. AB1-1975 [4,
Equation 8 would correspond to table 3, ANSI/IEEE C37. 13- pt. 2, p. 121.
1981. Again, (8) should be not used for X/R ratios lower than
6.6.
Iave (asym)- {{Il + 2 exp [- 4it/(X/R)
MOLDED CASE CIRCUIT BREAKERS
+ 2 V11+ 1/2 exp [-47rt/(X/R)] }
(10)
Due to the lack of standards, the following procedures are
suggested for applying molded case circuit breakers.
t in cyles.
Unless otherwise stated, molded case circuit breakers are
APPLICATIONS TO OTHER DEVICES
assumed to have contact parting times of 1/2 cycle. The rated
X/R ratio, however, depends on the size of the breaker. Table
Fuses will not be covered in this paper, but the point should
V, taken from NEMA publication No. ABI-1975 141, shows be made that (3), (4), and (5) of this paper are much preferred
the power factor of the test circuit under which each of these to the "up-over-down" method used on most fuse curves, the
breakers must be tested.
"up-over-down" method assumes an X/R for you.
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848
IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS, VOL. IA-21, NO. 4, JULY/AUGUST 1985
Solid-state overcurrent protective devices usually operate on
peak current and are thus prone to operate sooner than [1]
expected for asymmetrical faults. Coordination by peak
current value is sometimes helpful. A similar condition occurs
when using adjustable magnetic-trip-only breakers for motor [2]
protection. The National Electric Code Article 430, Sections [3]
52 and 152, prohibits magnetic-only trips to be set above 1300 [4]
percent. It is not uncommon today for high-efficiency motors
to have asymmetrical peak currents more than 20 times their [51
rated value. The result is sporadic "nuisance" tripping.
[6]
Many other applications can be found where asymmetrical
currents play an important part in the engineering solution to a [7]
problem. A sound understanding of the principle of asymme- [8]
try will greatly enhance the ability of the engineer to find an
acceptable solution. While not covered in this paper, an
understanding of iron saturation is also essential in many of [9]
these applications.
[10]
CONCLUSION
A familiarization with the mechanics of asymmetry can be a
valuable tool in the analyzation of disturbances on a power
system. At times the asymmetry involved may become the
deciding factor between breakers of different interrupting
ratings with considerably varied price tags. In addition, a good
visual understanding of the "phenomenon" of asymmetry
makes one much more comfortable in responding to the
problem. The traditional methods of calculating fault currents
were shown to produce values very close to the actual
maximum currents attained. Finally, in an age where computers are readily available to most practicing engineers the
equations for circuit breaker derating may be programmed for
convenience. This is particularly useful on modem large
distribution systems where X/R ratios are high and copious
numbers of interrupting devices are to be sized.
REFERENCES
H. W. Reichenstein and J. C. Gomez, "Relationship of X/R, I, and
IRMs' to asymmetry in resistance/reactance circuits," presented at the
IEEE/IAS conference, Mexico City, Mexico, 1983.
T. E. Brown, Jr., Theory and Techniques of Circuit Interruption.
Marcel Dekkar, Inc., 1984.
R. E. Friedrich, Application of Power Circuit Breakers, an IEEE
Tutorial Course. New York: IEEE, 1975.
Molded Case Circuit Breakers, National Electrical Manufacturers
Association Standard AB1-1975.
Low- Voltage Power Circuit Breakers, National Electrical Manufacturers Association Standard SG3-1975.
Low-VoltageACPower Circuit Breakers Used in Enclosures, ANSI
Standard C37.13-1981.
Rating StructureforAC High-Voltage Circuit Breakers Rated on a
Symmetrical Current Basis, ANSI Standard C37.04-1979.
Preferred Ratings and Related Required Capabilitiesfor AC HighVoltage Circuit Breakers Rated on a Symmetrical Current Basis,
ANSI Standard C37.06-1979.
Test Procedure for AC High- Voltage Circuit Breakers Rated on a
Symmetrical Current Basis, ANSI Standard C37.09-1979.
Application Guidefor AC High- Voltage Circuit Breakers Rated on
a Symmetrical Current Basis, ANSI Standard C7.010-1979.
Craig N. Hartman (S'78-M'80) received the
B.S.E.E. degree from the University of Utah, Salt
Lake City, in 1980.
He joined the Westinghouse Electric Corporation, San Francisco, CA, as an Industrial Application Engineer in 1980. Since 1983 he has worked as
a District Engineer for Westinghouse in Denver,
CO, responsible for technical support of the corporation with reference to industrial business in that
area.
Mr. Hartman currently serves as the Editor of the
IEEE Newsletter for Colorado, and he is a Registered Professional Engineer
in the State of California.
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