industrial-power-systems-handbook-donald-beeman_compress.pdf

Industrial Power Systems
Handbook
D O N A L D BEEMAN, Editor
Manager, Industriaf P w e r Engineering
Industrial Engineering Seclwn
General Electric Company, Schenectady, New Yorlc
FIRST EDITION
McGRAW-HILL BOOK COMPANY, INC.
1955 New York Toronto London

Ch.UPh?r1 by Donald Beeman, Alan Graeme Darling,
and R. H. Kaufmann
Short-circuit-current Calculating
Procedures
FUNDAMENTALS OF A-C SHORT-CIRCUIT CURRENTS
The determination of short-circuit currents in power distribution sys-
tems is just as basic and important as the determination of load currents
for the purpose of applying circuit breakers, fuses, and motor starters.
The magnitude of the shoncircuit current is often easier to determine
than the magnitude of the load current.
Calculating procedures have been so greatly simplified compared with
the very complicated procedures previously used that now only simple
arithmetic is required to determine the short-circuit currents in even the
most complicated power systems.
SHORT-CIRCUIT CURRENTS AND THEIR EFFECTS
If adequate protection is to he provided for a plant electric system, the
size of the electric power system must also be considered to determine
how much short-circuit current it will deliver. This is done so that cir-
cuit breakers or fuses may he selected with adequate interrupting capac-
ity (IC). This interrupting capacity should be high enough to open
safely the maximum short-circuit current which the power system can
cause to flow through a circuit breaker if a short circuit occurs in the
feeder or equipment which it protects.
The magnitude of the load current is determined by the amount Of
work that is being done and hears little relation to the size of the system
supplying the load. However, the magnitude of the short-circuit current
is somewhat independent of the load and is directly related to the size or
I

2 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES
capacity of t,he power source. The larger the apparatus which supplies
electric power to the system, the greater the short-circuit current will be.
Take a simple case: A 440-volt three-phase lo-lip motor draws about
13 amp of current at full load and will draw only this amount whether
supplied by a 25-kva or a 2500-kva transformer bank. So, if only thc
load currcnts arc considered when selecting motor branch circuit break-
ers, a 15- or 20-amp circnit, breaker wnuld he specified. However, the
size of t,hepower system back of the circuit breaker has a real bearing on
the amount of the short,-circuit,current. which can flow as a result of a
short circuit on the load side of the circuit breaker. Hence, a much
larger circuit breaker would be required to handle the short-circuit current
from a 2500-kva bank than from a 25-kva bank of transformers.
These numbers
A simple mathematical example is shown in Fig. 1.1.
MUST BE CAPABLE OF INTERRUPTING 1000 AMPERES
El I O O V
100 A
~ ~ 1 0 . 1
OHMS
MOTOR LOAD
CURRENT
5 AMP
APPARENT
IMPEDANCE
20 OHMS
E I 0 0
SHORT CIRCUIT CURRENT = - : - = 1000- AMPERES
Z T 0.1
MUST BE CAPABLE OF INTERRUPTING 10,000 AMPERES
w
I000 A
2 1 = 0.01 OHMS
MOTOR LOAD
CURRENT
5 AMP
FIG. 1.1
circuit-current magnitude than load.
Illustrotion showing that copocity of power source has more effect on rhort-

SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 3
have been chosen for easy calculation rather than a representation of
actual system conditions.
The impedance, limiting the flow of load current, consists mainly of
the 20 ohms apparent impedance of the motor. If a short circuit occurs
at F , the only impedance to limit the flow of short-circuit current is the
transformer impedance (0.1 ohm compared with 20 ohms for the motor);
therefore, the short-circuit current is 1000 amp, or 200 times as great as
the load current. Unless circuit breaker A can open 1000 amp, the
short-circuit current will continue to flow, doing great damage.
Suppose the plant grows and a larger transformer, one rated at 1000
amp, is substituted for the 100-amp unit. A short circuit at F , (bottom
in Fig. 1.1) will now be limited by only 0.01 ohm, the impedance of the
larger transformer. Although the load current is still 5 amp, the short-
circuit current will now he 10,000 amp, and circuit breaker A must be
able to open that amount. Consequently it is necessary to coiisider the
size of the system supplying the plant as well as the load current, to be
sure that circuit breakers or fuses are selected which have adequate
interrupting rating for stopping the flow of the short-circuit current.
Short-circuit and load currents are analogous to the flow of xvater in a
hydroelectric plant, shoivn in Fig. 1.2. The amount of water that flows
under normal conditions is determined by the load on the turbines.
Within limits, it makes little difference whether the reservoir behiiid the
dam is large or small. This flow of water is comparable to the flow of load
current in the distribution system in a factory.
On the other hand, if the dam breaks, the amount of water that will
flow will depend upon the capacity of the reservoir and will bear little
relation to the load on the turbines. Whether the reservoir is large or
small will make a great difference in this case. This flow of water is
comparable to the flow of current through a short circuit in the distribu-
tion system. The load currents do useful work, like the water that flows
down the penstock through the turbine water wheel. The short-circuit
currents produce unwanted effects, like the torrent that rushes madly
downstream when the dam breaks.
SOURCES O
F SHORT-CIRCUIT CURRENTS
When determining the magnitude of short-circuit currents, it is
extremely important that all sources of short-circuit current he considered
and that the reactance characteristics of these sources be known.
There are three basic sources of short-circuit current:
1. Generators
2. Synchronous motors and synchronous condensers
3. Induction motors

All these can feed shorecircuit current into a short circuit (Fig. 1.3).
Generators are driven by turbines, diesel engines, water wheels, or
other types of prime movers. When a short circuit occurs on the circuit
fed by a generatar,the generator continues to produce voltage because the
field excitation is maintained and the prime mover drives the generator
at substantially normal speed. The generated voltage produces a short-
circuit current of a large magnitude which flows from the generator (or
generators) to the short circuit. This flow of short-circuit current is
limited only by the impedance of the generator and of the circuit between
the generator and the short circuit. For a short circuit at the terminals
of the generator, the current from the generator is limited only by its own
impedance.
FIG. 1.2
the hydroelectric plant.
Normal load and short-circuit currents are analogous to the conditions shown in

SHORT-CIRCUIT-CURRENT ULCULATlNG PROCEDURES 5
METAL CLAD SWITCHGEAR
SHORT CIRCUIT
CURRENT FROM
INDUCTION
MOTOR
FIG. 1.3 Generators, synchronous motors, and induction motors all produce short-circuit
current.
HOW SYNCHRONOUS MOTORS PRODUCE SHORT-CIRCUIT CURRENT
Synchronous motors are constructed substantially like generators; i.e.,
they have a field excited by direct current and a stator winding in which
alternating current flows. Normally, synchronous motors draw a-c
power from the line and convert electric energy to mechanical energy.
However, the design of a synchronous motor is so much like that of a
generator that electric energy can be produced just as in a generator, by
driving the synchronous motor with a prime mover. Actually, during a
system short circuit the synchronous motor acts like a generator and
delivers shortcircuit current to the system instead of drawing load cur-
rent from it (Fig. 1
.
4
)
.
As soon as a short circuit is established, the voltage on the system is
reduced to a very low value. Consequently, the motor stops delivering
energy to the mechanical load and starts slowing down. However, the
inertia of the load and motor rotor tends to prevent the motor from slow-
ing down. In other words, the rotating energy of the load and rotor
drives the synchronous motor just as the prime mover drives a generator.

The synchronous motor then becomes a generator and delivers short-
circuit current for many cyclesafter the short circuit occurson the system.
Figure 1
.
5shows an oscillogram of the current delivered by a synchronous
motor during a system short circuit. The amount of current depends
upon the horsepower, voltage rating, and reactance of the synchronous
motor and the reactance of the system to the point of short circuit.
LOAD CURRENT
SYNCHRONOUS
MOTOR
-€t
F
I
G
.1.4 Normally motors draw
load current from the source or
utility system but produce rhort-
circuit current when a short cir-
wit occurs in the d a d .
UlILITY
SYSTEM
,
-
SHORT CIRCUIT
CURRENT FROM
MOTOR
..-.. .
SYSTEM
SYNCMOYOUS '
FIG 1 5 IBmlowl lroce o
f 0s-
Yoroll
..-. .._ ,
. __..,
. . .
.
..
. .
.
cillogrclm of short-circuit current
produced by a synchronous
motor
SHORT
' . - I
CIRCUIT
SHORT CIRCUIT CURRENT DELIVERED BY
A SYNCHRONOUS MOTOR.

SHORT.CIRCUIT-CURRENT CALCULATING PROCEDURES 7
HOW INDUCTION MOTORS PRODUCE SHORT-CIRCUIT CURRENT
The inertia of the load and rotor of an induction motor has exactly the
same effect on an induction motor as on a synchronous motor; i.e., it
drives the motor after the system short circuit occurs. There is one
major difference. The induction motor has no d-c field winding, but
there is a flux in the induction motor during normal operation. This flux
acts like flux produced by the d-c field winding in the synchronous motor.
The field of the induction motor is produced by induction from the
stator rather than from the d-c winding. The rotor flux remains normal
as long as voltage is applied to the stator from an external source. How-
ever, if the external source of voltage is removed suddenly, as it is when a
short circuit occurs on the system, the flux in the rotor cannot change
instantly. Since the rotor flux cannot decay instantly and the inertia
drives the induction motor, a voltage is generated in the stator winding
causing a short-circuit current to flow to the short circuit until the rotor
flux decays to zero. To illustrate the short-circuit current from an
induction motor in a practical case, oscillograms were taken on a wound-
rotor induction motor rated 150 hp, 440 volts, 60 cycles, three phase, ten
poles, 720 rpm. The external rotor resistance was short-circuited in each
case, in order that the effect might he similar to that which would he
obtained with a low-resistance squirrel-cage induction motor.
Figure 1.6 shows the primary current when the machine is initially
running light and a solid three-phase short circuit is applied at a point in
the circuit close to its input (stator) terminals at time TI. The current
shown is measured on the motor side of the short circuit; so the short-
circuit current contribution from the source of power does not appear, but
only that contributed by the motor. Similar tests made with the machine
initially running at fullload show that the short-circuit current produced
T.
FIG. 1.6 , Tracer of oxillograms of short-circuit currents produced by an induction motor
running at light load.

by the motor when short-circuited is substantially the same, regardless of
initial loading on the motor. Note that the maximum current occurs in
the lowest trace on the oscillogram and is about ten times rated full-load
current. The current vanishes almost completely in four cycles, since
there is no sustained field current in the rotor to provide flux, as in the
case of a synchronous machine.
The flux does last long enough to prodnce enough short-circuit current
to affect the momentary duty on circuit breakers and the interrupting
duty on devices which open within one or two cycles after a short circuit.
Hence, the short-circuit current produced by induction motors must he
considered in certain calculations. The magnitude of short-circuit cur-
rent produced by the induction motor depends upon the horsepower,
voltage rating, reactance of the motor, and the reactance of the system to
the point of short c. "cuit. The machine impedance, effective at the time
of short circuit, cmesponds closely with the impedance at standstill.
Consequently, the i iitial symmetrical value of Short-circuit current is
approximately equnl to the full-voltage starting current of the motor.
TRANSFORMERS
Transformers are often spoken of as a source of short-circuit current.
Strictly speaking, this is not correct, for the transformer merely delivers
the short-circuit current generated by generators or motors ahead of the
transformer. Transformers merely change the system voltage and mag;
nitude of current but generate neither. The short-circuit current deliv-
ered by a transformer is determined by its secondary voltage rating and
reactance, the reactance of the generators and system to the terminals of
the transformer, and the reactance of the circuit from the transformer to
the short circuit.
ROTATING-MACHINE REACTANCE
The reactance of a rotating machine is not one simple value as it is for a
transformer or a piece of cable, but is complex and variable with time.
For example, if a short circuit is applied to the terminals of a generator,
the short-circuit current behaves as shown in Fig. 1.7. The current starts
out at a high value and decays to a steady state after some time has
elapsed from the inception of the short cirroit. Since the field excitation
voltage and speed have remained snbstantially constant within the short
interval of time considered, a change of apparent react,anceof the machine
may he assumed, to explain the change in the magnitude of short-circuit
current with time.
The expression of such variable reactance at any instant after the

occurrence of any short circuit requires a complicated formula involving
time as one of the variables. For the sake of simplification in short-cir-
cuit calculating procedures for circuit-breaker and relay applications,
three values of reactance are assigned to generators and motors, viz.,
subtransient reactance, transient reactance, and synrhronous reactance.
The three reactances can be briefly described as follows:
1. Subtransient reactance X y is the apparent reactance of the stator
winding at the instant short circuit occurs, and it determines the current
Row during the first few cycles of a short circuit.
2. Transient reactance X i is the apparent initial reactance of the
stator winding, if the effect of all amortisseur windings is ignored and
only the field winding considered. This reactance determines the cur-
rent following the period when subtransient reactance is the controlling
value. Transient reactance is effective up to 45 see or longer, depending
upon the design of the machine.
3. Synchronous reactance X d is the apparent reactance that deter-
mines the current flow when a steady-state condition is reached. It is not
effectiveuntil several seconds after the short circuit occurs; consequently,
it has no value in short-circuit calculations for the application of circuit
breakers, fuses, and contactors but is useful for relay-setting studies.
Figure 1.8 shows the variation of current with time and associates the
various reactances mentioned above with the time and current scale.
Previous loading has an effect on the total magnitude of short-circuit
CURRENT DETERMINED
BY SYNCHRONOUS
OF TOTAL OSCILLOGRAM
ONLY TWO ENDS SHOWN
HERE. THIS REPRESENTS
THE BREAK BETWEEN
THE TWO PARTS.
OCCURS A
T
THIS TIME.
FIG. 1.7 Trace of orcillograrn of hart-circuit current produced by a generator.

MAX, SUBTRANSIENT CURRENT- USE SUBTRANSIENT REACTANCE X"d
/-
T
I
M
E
-
( 8 )
FIG 1.8 Variation of generotor short-circuit current wilh time.
current delivered by a generator.
given by the machine designer is the lowest value obtainable.
use will show maximum short-circuit current.
before a system analysis can he made.
The value of X i or X y generally
Hence, its
Certain characteristics of short-circuit currents must he understood
SYMMETRICAL AND ASYMMETRICAL SHORT-CIRCUIT CURRENTS
These terms are used to describe the symmetry of the a-c waves about
the zero axis. If the envelopes of the peaks of the current waves are
symmetrical about the zero axis, the current is called symmetrical current
(Figs. 1.9 and 1.10). If the envelopes of the peaks of the current waves
are not symmetrical about the zero axis, the current is called asymmetrical
ENVEWPES OF PEAKS
OF SINE WAVE ARE
SYMMETRIGAL ABOUT
THE ZERO AXIS.
ZERO
AXIS
FIG. 1.9 Symmelrical a-c wove.

SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES
THE ENVELOPES OF PEAKS
ARE SVHHETRICAL ABOUT
ZERO AXIS
FIG, 1.10 Symmetrical dternating current from a short-circuited generotor.
11
ENVELOPES OF PEAKS ARE NOT
SYMMETRICAL ABOUT ZERO AXIS
AX1S
TOTALLY 0FFSET
PARTIALLY OFFSEl
FIG. 1.11
for the purpose of illustration only.
circuits.
Asymmetrical (I-c waver. The conditions shown here ore theoreticol a n d ore
D-C component will rapidly decay to zero in actual
FIG. 1.12 Trace of orcillogram of a typical short-circuit current

current (Fig. 1.11). The envelope is a line drawn through the peaks of
the waves, as shown in Figs. 1.9 to 1.12.
For the sake of explanation, many of the illustrations, such as Figs.
1.11, 1.15 to 1.19, show sine waves o
f current uniformly offset for several
cycles. It should be noted that in practical circuits the amount of asym-
metry decreases rapidly after the occurrence of the short circuit in the
system. This decrease of asymmetry is shown qualitatively in illustra-
tions such as Figs. 1.12, 1.20, 1.23, and 1.24.
Oscillogramsshow that short-circuit currents are nearly always asym-
metrical during the first few cycles after the short circuit occurs. They
also show that the asymmetry is maximum at the instant the short circuit
occurs and that the current gradually becomes symmetrical a few cycles
after the occurrence of the short circuit. The trace of an oscillogram of a
typical short-circuit current is shown in Fig. 1.12.
WHY SHORT-CIRCUIT CURRENTS ARE ASYMMETRICAL
In the usual industrial power systems the applied or generated voltages
are of sine-wave form. When a short circuit occurs, substantially sine
wave short-circuit currents result. For simplicity, the following discus-
sion assumes sine-wave voltages and currents.
In ordinary power circuits the resistance of the circuit is negligible com-
pared with the reactance of the circuit. The short-circuit-current power
factor is determined by the ratio of resistance and reactance of the circuit
only (not of the load). Therefore the short-circuit current in most power
circuits lags the internal generator voltage by approximately 90" (see
Fig. 1.13). The internal generator voltage is the voltage generated in
the stator coils by the field flux.
If in a circuit mainly containing reactance a short circuit occurs at the
peak of the voltage wave, the short-circuit current would start at zero
and trace a sine wave which would be symmetrical ahout the zero axis
(Fig. 1.14).
If in the same circuit (i.e., one containing a large ratio of reactance to
resistance) a short circuit occurs at the zero point of the voltage wave, the
current will start at zero but cannot follow a sine wave symmetrically
about the zero axis because such a current would be in phase with the
voltage. The wave shape must be the same as that of voltage hut 90'
behind. That can occur only if the current is displaced from the zero
axis, as shown in Fig. 1.15. In this illustration the current is a sine wave
and is displaced 90' from the voltage wave and also is displaced from the
zero axis. The two cases shown in Figs. 1.14 and 1.15 are extremes.
One shows a symmetrical current and the other a completely asym-
metricd current.
This is known as a symmetrical short-circuit current.

WORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 13
GENERATOR TRANSFORMER
INTERNAL VOLTAGE OF GENERATOR APPLIED HERE
ONE LINE IMPEDANCE
ioxazx 7
x 0.m
REACTANCE, X = 19%
RESISTANCE. R = 1.4%
I
RESISTANCE IS LESS THAN OF THE REACTANCE HENCE MAY
BE NEGLECTED WITHOUT AN APPRECIABLE ERROR
INTERNAL VOLTAGE OF GENERATOR
NEARLY 90'
-SHORT CIRCUIT CURRENT
DIAGRAM
SHOWING
SINE WAVES
CORRESPONDING
TO VECTOR
DIAGRAM
FOR ABOVE
CIRCUIT
FIG. 1.13 Diagrams Illustrating the phase relations of voltage and short-circuit current.

14 SHORT-CIRCUll-CURRENT CALCULATING PROCEDURES
GENERATED VOLTAGE
SHORT CIRCUIT CURRENT
ZERO
AXIS
SHORT CIRCUIT OCCURRED AT THIS POINT
FIG. 1.14
cirwit.
Symmetric01short-circuit current and generoted voltage for zero-power-factor
-SHORT
CIRCUIT
CURRENT
F
I
G
.1.15 Asymmetrical short-circuit current and generated voltage in zero-power-factor
circuit. Condition is theoretical and is shown for illustration purposes only.

SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES I S
If,in a circuit containing only reactance, the short circuit occurs at any
point except at the peak of the voltage wave, there will be some offset of
the current (Fig. 1.16). The amount of offset depends upon the point on
the voltage wave at which the short circuit occurs. It may vary from
zero (shown in Fig. 1.14) to a maximum (shown in Fig. 1.15).
In circuits containing both
reactance and resistance, the s~,?&&,R&!~~
amount of offset of the short-
circuit current may vary be-
tween the same limits as for
circuits containing only react-
ance. However, the point on
the voltage wave at which the
short circuit must occur topro-
duce maximum asymmetry
dependsupon theratioof react-
ance to resistance of the cir-
cuit. Maximum asymmetry
is obtained when the short cir-
cuit occurs at a time angle
equal to 90" +0 (measured
forward in degrees from the
zero point of the voltage wave)
CURRENT
where tangent 0 equals there- ASYMMETRICAL
actance-to-resistance ratio of
FIG. 1.16 Short-circuit current and generated
the circuit' The short-circuit voltage in zero-Dower-factor circuit. Short cir-
current will be symmetrical cuit occurred between the
when the faultoccurs 90"from
that point onthevoltage wave.
point and peak of the generated voltctge wove.
This condition is theoretical and for illustration
an example, assumeacir- purporer only. The short-circuit current will
gradually become symmetrical in practical
cuit that has equal resistance
and reactance, i.e., the react-
ance-to-resistance ratio is 1. The tangent of 45" is I ;hence, maximum
offset is obtained when the short circuit occurs at 135' from the zero
point of the voltage wave (Fig. 1.17).
CiTCUit.,
D-C COMPONENT OF ASYMMETRICAL SHORT-CIRCUIT CURRENTS
Asymmetrical alternating currents when treatedas a singlecurrent wave
are difficult to interpret for circuit-breaker application and relay-setting
purposes. Complicated formulas are also required to calculate their
magnitude unless resolved into components. The asymmetrical alter-
nating currents are, for circuit-breaker applications and relay-setting

16 SHORT-CIRCUIT-CURRENT CALCUUTING PROCEDURES
MAXIMUM OFFSET
Short-circuit current and generated voltage in circuit with equal reactance and
The
FIG. 1.17
resistance.
short-circuit current will gradually become symmetrical in practical circuits.
purposes, arbitrarily divided into simple components, which makes it
easy to calculate the short-circuit magnitude at certain significant times
after the short circuit occurs.
The asymmetrical alternating current behaves exactly as if there were
two component currents flowing simultaneously. One is a symmetrical
a-c component and the other a d-c component. The sum of those two
components at any instant is equal to the magnitude of the total asym-
metrical a-c wave at the same instant.
The d-c component referred to here is generated within the a-c system
with no external source of direct current being considered. In some cases,
particularly in the neighborhood of the d-c railways, direct current from
the railways flows through neighboring a-c systems. This type of d-c
current is not considered in this discussion or in the calculating procedures
which follow.
As an example of the resolution of asymmetrical alternating currents
into components, refer to Fig. 1.15 which shows an asymmetrical short-
circuit current which is resolved into a symmetrical a-c and a d-c compo-
nent in Fig. 1.18. If the instantaneous values of the two components
(dashed lines) are added at any instant, the resultant will be that of the
asymmetrical current wave.
This condition is theoretical and is shown for illustration purposes only.

F I N S T A N T AT WHICH SHORT CIRCUIT OCCURS
17
ASYMMETRICAL
AC COMPONENT
FIG. 1.18
current.
Theoretical Ihort-circuit-cvrrentwove illustrating components of asymmetrical
In practical circuits, d-c component would decay to zero in o few cycler.
INSTANT OF SHORT CIRCUIT
T O T A L C U R R E N T
DC COMPONENT
AC COMPONENT
ZERO AXIS
a = b = D C COMPONENT
FIG. 1.19
at some point between the zero point and peak of the generated voltage wave.
lhsoretical condition similar to that shown in Fig. 1.18.
Components of asymmetrical short-circuit current in which short circuit occurred
This is a

I8 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES
As mentioned previously, the examples shown in Figs. 1.13and 1.18are
In practical circuits the d-c component
for purposes of illustration only.
decays very rapidly, as shown in Fig. 1.20.
INITIAL MAGNITUDE OF D-C COMPONENT
The magnitude of the d-c component depends upon the iustant, the
short circuit occurs and may vary from zero, as in Fig. 1.14, to a maximum
initial value equal to the peak of the a-c symmetrical compoiieiit, as i n
Figs. 1.15 and 1.18. When the short circuit occurs at any other point,
such as shown in Fig. 1.19, the initial magnitude of the d-c componciit is
equal to the value of the a-c symmct,riral component at thc instant of
short circuit. The above limit,shold true for the initial magiiitudc of d-c
eomporient in a system regardless of the reactance and resistance. Ilow-
ever, the d-c componeut does not continue to flo~v
at a constant value, as
shown in Figs. 1.18 and 1.19, unless there is zero resistauce iii the circuit.
DECREMENT
There is uo d-c voltage in the system to sustaiu the flax of direct
current; therefore the energy represeuted by the dirert. component of
current will be dissipated as ZZR loss from the direct current flowiug
through the resistance of the circuit. If the circuit had zero resistance,
the direct current would flow at a constant value (Figs. 1.18 and 1.19)
TOTAL ASYMMETRICAL CURRENT
C COMPONENT
AC COMPONENT
FIG. 1.20
short-circuit currenl gradually becomes symmetrical when d-c component diroppearr.
Trace of orcillogrom showing decay of d-c component and how orymmetricd

until the circuit was interrupted. However, all practical circuits have
some resistance; so the d-c romponent decays as shown in Fig. 1.20. The
combination of the decaying of d-c and symmetriral a-(*
components gives
an asymmetrical wave that changes to a symmetriral wave whcti the
d-c component has disappeared. The rate of decay of the currents is
called the decrement.
X / R RATIO
The X / R ratio is the ratio of the reactance to the resistance of the cir-
cuit. The decrement or rate of decay of the d-c component is propor-
tional to the ratio of reactance to resistance of the complete circuit from
generator to short circuit. The theory is the same as opening the circuit
of a battery and an inductive coil.
If the ratio of reactance to resistance is infinite (i.e., zero resistance),
the d-c component never decays, as shown in Figs. 1.18and 1.19. On the
other hand, if the ratio is zero (all resistance, no reartance), it decays
instantly. FOFany ratio of reactarice to resistance in between these
limits, the d-c component takes a definitetime to decrease to substantially
zero, as shown in Fig. 1.20.
! In generators the ratio of subtransient reactance to resistance may be as
?much as 70:l; so it takes several cycles for the d-c component to dis-
appear. In circuits remote from generators, the ratio of reactance to
resistance is lower, and the d-c component decays more rapidly. The
higher the resistance in proportion to the reactance, the more IaRloss
from the d-c c.omponent, and the energy of the direct current is dis-
sipated sooner.
D-C TIME CONSTANT
Often it is said that generators, motors, or circuits have a certain d-c
This refers again to the rate of decay of the d-c compo-
time constant.
O C COMPONENT
a = 37Y. OF b (APPROX )
C
- TIME
CONSTANT IN OF D C COMPONENT
SECONDS
FIG. 1.21 Graphic illustration of time constant.

nent. The d-c time constant is the time, in seconds, required by the d-c
component to reduce to about 37 per cent of its original value at the
instant of short circuit. It is the ratio of the inductance in henrys to the
resistance in ohms of the machine or circuit. This is merely a guide to
how fast the d-c component decays.
Stated in other terms, it is the time in seconds for the d-c component to
reach zero if it continued to decay at the same rate it does initially
(Fig. 1.21).
RMS VALUE INCLUDING D-C COMPONENT
The rms values of a-c waves are significant since circuit breakers, fuses,
and motor starters are rated in terms of rrns current or equivalent kva.
The maximum rrns value of short-circuit current occurs at a time of about
one cycle after short circuit, as shown in Fig. 1.20. If there were no
decay in the d-c component, as in Fig. 1.18, the rrns value of the first
cycle of current would be j.732 times the rrns value of the a-c component.
In practical circuits there is always some d-c decay during the first cycle.
An approximate rrns value of one cycle of an offset wave whether it is
partially or totally offsetis expressed by the equation
where C = rrns value of offset or asymmetrical current wave over one
cycle
a = rrns value of a-c component
b = value of d-c component at one-half cycle
MULTIPLYING FACTOR
Calculation of the precise rrns value of an asymmetrical current at any
time after the inception of a short circuit may be very involved. Accu-
rate decrement factors to account for the d-c component at any time are
required, as well as accurate factors for the rate of change of the apparent
reactance of the generators. This precise method may he used if desired,
but simplified methods have been evolved whereby the d-c component is
accounted for by simple multiplying factors. The multiplying factor
converts the rrns value of the symmetrical a-c wave into rms amperes of
the asymmetrical wave including a d-c component.
The magnitude of the d-c component depends upon the point on the
voltage wave at which the short circuit occurs. For protective-device
application, only the maximum d-c component is considered, since the
circuit breaker must be applied to handle the maximum short-circuit
current that can occur in a system.

In the general case for circuits rated above 600 volts, the multiplying
factor to account for d-c component is 1.6 times the rms value of the a-c
symmetrical component at the first half cycle.
For circuits rated 5000 volts or less where there is no local generation,
that is, where the supply t,othe bus is through transformers or long lines,
the multiplying factor to ralculate the total current at the first half cycle
may be reduced to 1.5. For circuits 600 volts and less, t,he multiplying
factor to calculate the total current at the first half cycle is 1.25 when the
circuit breaker is applied on the average current in three phases. Where
single-phase conditions must be considered in circuits GOO volts and less,
then to account for the d-c component in one phase of a three-phase cir-
cuit a multiplying factor to calculate the total current at the first half
cycle of 1.5 is used.
For some calculations, rms current evaluations at longer time intervals
than the first half cycle, such as three to eight cycles corresponding to the
interrupting time of circuit breakers, are required. Multiplying factors
for this purpose may be taken from the curve in Fig. 1.22.
Table 1.2 gives the multiplying factors commonly used for applying
e
FIG. 1.22
various X / R ratio of circuits.
Charts showing multiplying factors to account for decoy of d-c component for

22 SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES
short-circuit protective devices. These factors range from 1 to 1.6,
depending upon whether the short-circuit calculation is being made to
determine the interrupting or momentary duty on the short-circuit pro-
tective device.
SHORT-CIRCUIT RATIO OF GENERATORS
This term is referred to frequently in short-circuit discussions. With
present AIEE procedures of short-rircuit ralrulations, it has become an
accessory with no practical significance from this standpoint. For the
sake of completeness, a definition is given here.
Short-circuit ratio
field current to produce rated voltage a t no load
field current to produce rated current at sustained short circuit
-~
-
No further mention will he made of short-circuit ratio.
TOTAL SHORT-CIRCUIT CURRENT
The total symmetrical short-rirruit current is made up of currents from
several sourves, Fig. 1.23. At the top of the figure is shown the short-
circuit current from the utility. This act,ually comes from ut,ility gener-
ators, but generally the industrial system is small and remote electrically
from the utility generators so that the Symmetrical short-rircuit current
is substant,ially constant,. If there are generators in the indust,rial plant,
then they cont,ribute a symmet,rical short-circuit rurreiit which for all
practical purposes is constant over the first few cycles. There is, how-
ever, a slight decrement, as indicated in Fig. 1.23.
The other sources are synchronous motors which act something like
plant generators, except that t,heyhave a higher rate of decay of the sym-
metriral component, and induction motors whirh have a very rapid rate
of dccay of the symmetrical component of current. When all these cur-
rents are added, the total symmetrical short-circuit rurrent is typical of
that shown at the bottom of Fig. 1.23.
The magnitude of the first few cycles of the t,otal symmetrical short-
circuit, current is further increased by the presence of a d-c compouent,
Fig. 1.24. The d-c component,offsets the a-c ware and, therefore, makes
it asymmetrical. The d-c component decays to zero within a few cycles
in most indust,rial power systems.
It is this total rms asymmetrical short-circuit current, as shown in Fig.
1.24, that must he determilied for short-circuit protective-derice applira-
tion. The problem of doing this has been simplified by standardized
procedures to a poiut xhere to determine the rms asymmetriral current
one need only divide t,he line-to-neutral roltage by the proper reactance

RG. 1.23 Tracer of orcillogramr of rym- FIG. 1.24 Arymmelrical short-circuit current
metrical short-circuit currents from utility, from dl sources illustrated in Fig. 1.23 plus
panerator, synchronous motors, and induc- d-c component.
lion motors. The shape of the total com-
bined currents is illurtmted by the bottom
hace.

24 SHORT.CIRCUIT-CURRENT U L C U U l l N G PROCEDURES
or impedance and then multiply by the proper multiplying factor from
Table 1.2.
BASIS OF RATING A-C SHORT-CIRCUIT PROTECTIVE DEVICES
The background of the circuit-breaker rating structure as well as the
basic characteristics of short-circuit currents must be understood to
enable the engineer to select the proper rotating-machine reactances and
multiplying factors for the d-c component to determine the sbort-circuit-
current magnitude for checking the duty on a particular circuit breaker,
such as momentary duty or interrupting duty.
The rating structure of circuit breakers, fuses, and motor starters is
designed to tell the application engineer how circuit breakers, fuses, or
motor starters will perform under conditions where the short-circuit cur-
rent varies with time. In discussing these rating bases, and for the sake
of clarity, they will be arbitrarily divided into two sections, i.e., the rating
basis of high-voltage short-circuit protective devices above 600 volts and
the rating basis of low-voltage Short-circuit protective devices 600 volts
and below.
HIGH-VOLTAGE SHORT-CIRCUIT PROTECTIVE DEVICES (ABOVE 600 VOLTS)
Power-circuit-breaker Rating Basis. Thestandard indoor oillesspower
circuit breakers as used in metal-clad switchgear will be used here to
explain power circuit-breaker ratings. The same fundamental principles
apply to all other high-voltage power circuit breakers.
The circuit-breaker rating structure is complicated because of the time
of operation of the circuit breakers after a short circuit occurs.
The few cycles needed for the power circuit breaker to open the circuit
and stop the flow of short-circuit current consist of the time required for
(1) the protective relays to close their contacts, (2) the circuit-breaker
trip coil to move its plunger to release the breaker operating mechanism,
(3) the circuit-breaker contacts to part, and (4)the circuit breaker to
interrupt the short-circuit current in its arc chamber. During this time,
the short-circuit current produces high mechanical stresses in the circuit
breaker and in other parts of the circuit. These stresses are produced
almost instantaneously in phase with the current and vary as the square
of the current. Therefore, they are greatest when maximum current is
flowing. The foregoing discussion showed that t,he short-circuit current
is maximum during the first cycle or loop, because of the presence of the
d-c component and because the motors contribute the most short-circuit
current at that time. Thus, the short-circuit stresses on the circuit
breakers and other parts of the circuit are maximum during the first loop
of short-circuit current.
During the time from the inception of the short circuit until the circuit-
breaker contacts part, the current decreases in magnitude because of the

decay of the d-c component and the change in motor reactance, as
explained previously. Consequently, the current that the circuit breaker
must interrupt, four or five cycles after the inception of t.he short circuit,
is generally of less magnitude than the maximum value of the first loop.
The fact that the current changes in magnitude with time has led to the
establishment of two bases of short-circuit-current ratings on power cir-
cuit breakers: (1) the momentary rating or its ability to withstand
mechanical stresses due to high short-circuit current and (2) the inter-
rupting rating or its ability t,ointerrupt the flow of short-circuit current
within its interrupting element.
What Comprises the Circuit-breaker-rating Structure. Circuit-
breaker-rating structures are revised and changed from time to time. It
is suggested that where specific problems require the latest information on
circuit-breaker ratings the applicahlc American Standards Association
(ASA), National Electrical Manufacturers Association (XEMA), or
American Instituteof Elect,ricalEngineers (AIEE)standardshe referred to.
To illustrate the various factors that comprise the circuit-breaker-
rating structure, an oilless power circuit breaker for metal-clad switchgear
rated 4.16 kv 250 mva* has been chosen. The complete rating is shown
on line 5, Table 1.1. The following will explain the meaning of the several
columns of Table 1.1, starting at the left. The rircuit-breaker-type
designation, column 1,varies among manufacturers. For the sakeof com-
pleteness the General Electric Company nomenclature is used in this col-
umn. The remainder of the items are uniform throughout the industry.
1. Type of Circuit Breaker (AM-4.16-250)
AM = magne-blast circuit breaker
4.16 = for 4.16-kv class of circuits (not applicable to 4800- and 4800-
volt circuits)
250 = interrupting rating in mva at 4.16 kv
2-4. Voltage Rating
2. Rated kv (4.16): the nominal voltage class or classes in which the
circuit breaker is rated.
3. Maximum design kv (4.76): the maximum voltage at which the cir-
cuit breaker is designed to operate. The 4.16-kv circuit breakers,
for example, are suitable for a 1330-volt system plus 10 per cent for
voltage regulation or 4.76 kv.
(Note: 4330 is 4%
X 2500.) Some utility syst.ems operate at 1330
volts near the substation.
4. Minimum operating kv at rated mva (3.85) :the minimum voltage a t
which the circuit breaker will interrupt its rated mva or in this case
it is 3.85 kv. At any voltages below this value, the circuit breaker
* blegavalt-amperes (see Appendix).
t
i.

I !
I (
t
a
/
I

is not designed to interrupt the rated mva but will interrupt some
value less than rated mva.
This is very significant in the rating of power circuit breakers for,
as poiuted out later, the circuit hreaker will interrupt a maximum of
only so many amperes regardless of voltage. At any voltage less
than the minimum operating voltage the product of the maximum
kiloampere interrupting rating times the kv times the square root of
3 is less than the mva interrupting rating of the circuit breaker.
5-6. Insulation Level (Withstand Test)
5. Low-frequency rrns kv (19): the 60-cycle high-potential test.
6. Impulse crest kv (60):a measure of its ability to withstand lightning
This is applied with an impulse generator as a
and other surges.
design test.
7-9. Current Ratings in Amperes
7. Continuous 60 cycles (1200 or 2000): the amount of load current
which the circuit breaker will carry continuously without exceeding
the allowable temperature rise.
8-9. Short-time Rating
8. Momentary amperes (60,000):the maximum rms asymmetrical cur-
rent that a circuit breaker will withstand including short-circuit cnr-
rents from all sources and motors (induction and synchronous) and
the d-c component. This rating is independent of operating voltage
for a given circuit breaker.
This is just as significant a limitation as mva interrupting rating.
It defines the ability of the circuit breaker to withstand the mechani-
cal stresses produced by the very large offset first cycle of the short-
circuit current. This rating is nnusually significant because the
mechanical stresses in the circuit hreaker vary as the square of the
current. It is the only rating that is affected by the square law, and
therefore is one of the most critical in the application of the circuit
breakers. The rating schedules of power circuit breakers are so pro-
portioned that the momentary rating is about 1.6 times the maximum
interrupting rating amperes.
9. Four-second (37,500): the maximum current that the circuit breaker
will withstand in the closed position for a period of 4sec to allow for
relaying operating time. This value is the same as the maximum
interrupting rating amperes.
10-13. Interrupting Ratings
10. Three-phase rated mva (250):the three-phase mva which the circuit
breaker will interrupt over a range of voltages from the maximum
design kv down to the minimum operating kv. In this case the

28 SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES
interrupting rating is 250 rnva between 4.76 and 3.85 kv. The mva
to be interrupted is obtained by multiplying the kv at which the cir-
cuit breaker operates times the symmetrical current in kiloamperes to
be interrupted times the square root of 3. The product of these must
not exceed the rnva interrupting rating at any operating voltage.
11. Amperes at rated voltage (35,000): the maximum total rms amperes
which the circuit breaker will interrupt at rated voltage, i.e., in the
case of the example used above 35,000 at 4.16 kv (4.16 X 35.000 x
fi= 250 mva). These figures are rounded. This figure is given
for information only and does not have a limiting significance of
particular interest to the application engineer.
12. Maximum amperes interrupting rating (37,500):the maximum total
rms amperes that the circuit breaker will interrupt regardless of how
low the voltage is. At
minimum operating voltage, 3.85 kv, this corresponds to 250 mva,
and, for example, at a voltage of 2.3 kv this corresponds to 150mva.
The circuit breaker will not interrupt this much current at all volt-
ages, i.e., it will not interrupt this much current if the product of
current, voltage, and the square root of 3 is greater than the mva
interrupting rating. This current limit determines the minimum kv
at which the circuit breaker will interrupt rated mva (column 4
)
. At
any voltage lower than that given in column 4,this maximum rms
total interrupting current determines how much the circuit breaker
will interrupt in mva. Therefore, when the voltage goes below the
limit of column 4, the mva which the circuit breaker will interrupt is
lower than the rnva rating given in column 10 by an amount propor-
tional to thereduction in operating voltage belowthe valueof column4.
13. Rated interrupting time (8 cycles on 60-cycle basis): the maximum
total time of operation from the instant the trip coil is energized until
the circuit breaker has cleared the short circuit.
What limits the Application of Power Circuit Breakers an on inter-
rupting-and Momentary-duty Basis? In so far as applying power cir-
cuit breakers on an interrupting-duty basis is concerned it can be seen
from the foregoing that there are four limits, none of which should be
exceeded. These must all be checked for any application.
1. Operating voltage should never at any time exceed the limit of
column 3, Table 1.1, i.e., the maximum design kv.
2. Interrupting rnva should never be exceeded at any voltage. This
limit is sig’nificantonly when the operating voltage is between the limits
of columns 3 and 4, Table 1.1. It is not significant when the operating
voltage is below the limit of column 4, Table 1.1,because maximum inter-
rupting amperes limit the mva to values less than the rnva rating.
3. Maximum interrupting rating amperes should never be exceeded
In this example, this current is 37,500 amp.

SHORT-CIRCUIT.CURRENT CALCUUTING PROCEDURES 29
even though the product of this current times the voltages times the
square root of 3 is less than the interrupting rating in mva. This figure
is the controlling one in so far as interrupting duty is involved when the
voltage is below that of column 4, Table 1.1 (minimum operating voltage
at rated mva).
4
. Momentary current should never be exceeded at any operating
voltage. Modern power circuit breakers generally have a momeutary
rating in rms amperes of 1.6 times the maximum interrupting rating in
rms amperes. As a result, where there is no short-circuit-current contri-
bution from motors, a check of the interrupting duty only is necessary.
If this is within the circuit-breaker interrupting rating then the maximum
Short-circuit current, including the d-c component, mill be within the
momentary rating of the circuit breaker.
Where there is short-circuit contribution from motors, the momentary
rating of the circuit breaker may be exceeded, before the interrupting
rating is exceeded in a given cirruit. Whenever there are motors to be
considered in the short-circuit calculations, the momentary duty and the
interrupting duty should both be checked.
Siuce the
short-circuit current is maximum at the first half cycle, the short-circuit
current must be determined at the first half cycle to determine the maxi-
mum momentary duty on a circuit breaker.
To determine the short-circuit current at the first half cycle, it is neces-
sary to consider all sources of short-circuit current, that is, the generators,
synchronous motors, induction motors, and utility connections. The
subtransient reactances of generators, synchronous motors, and inductiou
motors are employed in the reactance diagram. Since the d-r component
is present at this time, it is necessary to account for it by the use of a
multiplying factor. This multiplying factor is either 1.5 or l.G, as out-
lined in Table 1.2. Typical circuits where the 1.5 multiplying factor can
be used are shown in Fig. 1.25. The procedure is the same, regardless of
the type of power circuit breaker involved.
To check
the interrupting duty on a power circuit breaker, the short-circuit current
should be determined at the time that the circuit-breaker contacts part.
The time required for the circuit-breaker contacts to part will vary over a
considerable range, because of variation in relay time and in circuit-
breaker operating speed. The fewer cycles required for the circuit-
breaker contacts to part, the greater will be the curreut to interrupt.
Therefore, the maximum interrupting duty is imposed upon the circuit
breaker when the tripping relays operate instantaneously. In all short-
circuit calculations, for the purpose of determining interrupting duties,
the relays are assumed to operate instantaneously. To account for
How to Check Momentary Duty on Power Circuit Breakers.
How to Check Interrupting Duty on Power Circuit Breakers.

SEPES-DIVEN
SEN-RIO-EIELI', tCA
1
2400
4160
4800 VOLT
INCOMING
LINE FROM
UTILITY
TO PLANT LOAD
( 0 ) NO GENERATION
IN THE PLANT
u.-LNO GENERATION
ON THIS BUS
2400, 4160 OR
( C ) TO LOAD
13.6 KV
HIGH VOLTAGE
INCOMING LINE
$:,oA,,60
4600 V BUS
TO PLANT LOAD
NO GENERATION (b)
IN THE PLANT
U U
USE 1.6
MULTIPLYING
FACTOR
NO GENERATION
FIG. 1.25
circuits rated less than 5 h.
One-line diogrom of carer where the multiplying factor 1.5 may be used on

. . ,
c ..
.: .
,, ,.. . .
variation in the circuit-breaker operating speed, power circuit breakers
have been grouped into classes, such as eight-cycle, five-cycle, three-cycle
circuit breakers, etc. It is assumed for short-circuit-calculation purposes
that circuit breakers of all manufacturers, in any one speed grouping,
operate substantially the same with regard to contact parting time.
Instead of specifying a time at which the short-circuit current is to he
calculated, it is determined by the simpler approach of specifying the
generator and motor reactances and using multiplying factors. These
factors are listed in Table 1.2.
In industrial plants, eight-cycle circuit breakers are generally used.
Normally, the induction-motor contribution has disappeared, and that of
the synchronous motors has changed from the subtransientto the transient
condition before the contacts of these circuit breakers part. Therefore,
in calculating the interrupting duty on commonly used power circuit
breakers, generator subtransient reactance and synchronous-motor
transient reactance are used and induction motors are neglected. The
elapsed time is so long that usually all the d-c component has disappeared.
What d-c component is left is more than offset by the reduction in a-c
component due to the increase in reactance of the generators. Hence, a
multiplying factor of one (1) is used.
In very large power systems, when symmetrical short-circuit interrupt-
ing duty is 500 mva or greater, there is an exception to this rule. In such
large power systems, the ratio of reactance to resistance is usually so high
that there may be considerable d-c component left when the contacts of
the standard eight-cycle circuit breaker part. To account for this, the
multiplying factor of 1.1is used in determining the total rms short-circuit
mva that a circuit breaker may have to interrupt in these large systems.
The multiplying factor of 1.1 is not applied until the symmetrical short-
circuit value reaches 500 mva.
High-voltage fuses are either of the current-
limiting type, Fig. 1.26, which open the circuit before the first current
peak, or of the non-current-limiting type, which open the circuit within
one or two cycles after the inception of the short circuit. For the sake of
standardization, all fuse-interrupting ratings are on the basis of maximum
rms current that will flow in the first cycle after the short circuit occurs.
This is the current that will flow if the fuse did not open the circuit
previously, i.e., fuses are rated in terms of “available short-circuit
current.”
To determine the available short-circuit current at the first cycle for
the application of high-voltage fuses, use the subtransient reactances of
all generators, induction motors, synchronous motors, and utility sources
and allow for the maximum d-c component. The multiplying factor for
allowing for d-c component is 1.6, the same as for allowing for d-c compo-
High-voltage Fuses.

TABLE 1.2 Condensed Table of Multiplying Factors and Rotating-machine Reactances
To Be Used for CaLdatina Swt-dreuit Cunanh for Circuit-breaker, Fuse, and Motor.rtartor Applicdons
Eight cycle or slower (general case). .......... Above 600 wlh
Rva cycle.. .............................. Above 600 volt,
Any ploee where symmetricmi I .O
short-circuit kva i s loss than 1.1
500 mva
........................... 1.6
L
a
r than 5 k............................ 601 to 5000 volh Remote from generating do- 1.5
Generol GOSO.. Above 600 volt) Near generoting station
lion (X/R rotio l e u thon I
0
1
High-voltaqe Fuses
All typos, including dl wrront-limiting fuses. ....Above 600 wih Anywhere in system I .O
w
u
Interrupting duty 2
Subtransient
Subtransient
i
i
s
s
z
Momentary duty
Subtransient
Subtransient
5
Three-phose Ino interrupting duly
Subhqndent 1 Transient 1 Neglect
Maximum rms ampere interrupting duty
I I
1 Generators. 1 I
0
a
w
C
frequency
1 changers I
All types, including dl current-limiting fuses.. ... Above 600 volt'
Non-current-limiting lypes only.. .............601 to 15,000 wlh Remote from generoting %to.
IAnywhere in system
1 tion ( X / R mtio leu lhm 41
1.6 Subtronsient Svbtronrient Svbwmrient
1.? Subwoniiont Subhmrient Subtransient
i i i

All h e p o w e r ratings.. .................... Anywhere in system
2400 and 4i60Y
Wlh
All horsepower rotingr.. .................... 2400 and 4160Y Anywhere in system
Yolh
1.0
I .6
CIrmit breaker w conladm lype.. ...........
Cirwit b r w b r or contocto~
lype. ............
Clrcvit breebr or contartor type.. ...........
601 10 5000 volts
601 to MOO volts
601 lo 5000 volts
bywhere in system
lion lX/R ratio leis than 101
temote from gener.ting 1
1
.
-
m
z
Apparatus. 600 Volts and Below
1.6
1.5
Air circuit breakers or breaker-contactor combino.
lion motor stoners.. ....................
Low-voltacp furas or fused combination motor
Slarte" ...............................
Subtransient
Subtransient
600 volts and below
600 volt* and below
0
Subtrmdent Subtransient
Subtrmdent Subtransient 8
0
R
Interrupting or momentary duty
Anywhere in system
Anywhere in system
I .25 Subtransient Subtianrient Svbtronrienl
1 .25 Subtransient Subtransient Svbtraniient

34 SHORT-CIRCUIT.CURRENT CALCULATING PROCEDURES
nent when determining the momentary duty on a power circuit breaker
(see Table 1.2).
The interrupting rating of fuses in amperes is exactly parallel, in so far
as short-circuit+urent calculations are concerned, to the momentary
rating of power circuit breakers.
The ampere interrupting rating of high-voltage fuses is the only rating
that has any physical significance. For the sake of simplicity of applica-
tion in systems with power circuit breakers, some fuses are given inter-
rupting ratings in three-phase mva. The three-phase mva interrupting
rating has no physical significance, because fuses are single-phase devices,
each fuse functioning only on the current which passes through it.
WAVE OF AVAILABLE
THE FUSE ELEMENTS MELT
BEFORE PEAK VALUE OF
AVAILABLE SHORT CIRCUIT
CURRENT IS REACHED
1
FIG. 1.26
See Fig. 1.27 for method o
f determining available short-circuit current.
Grophic sxplonotion of the current-limiting action of current-limiting fuses.

SHORT-CIRCUIT-CURRENT CAKULATING PROCEDURES 35
These three-phase mva ratings have been selected so they will line up
with power-circuit-breaker ratings. For example, a high-voltage fuse
rated 150 mva and a power circuit breaker rated 150 mva can he applied
on the basis of the same short-circuit-current calculations. Of course, the
application voltage must he factored in each case.
High-voltage motor starters generally
employ for short-circuit protection either current-limiting fuses or power
circuit breakers. The short-circuit-current calculations for applying
these motor starters are the same as those for high-voltage fuses and
power circuit breakers, respectively.
High-voltoge Motor Starters.
LOW-VOLTAGE CIRCUIT PROTECTIVE EQUIPMENT (600VOLTS AND BELOW)
low-voltage Air Circuit Breokers. The present designs of low-voltage
air circuit breakers differ from those of high-voltage power circuit break-
ers because they are substantially instantaneous in operation at currents
near their interrupting rating. The contacts often begin to part during
the first cycle of current. Therefore, low-voltage air circuit breakers are
subject to interrupting the current at the first cycle after short circuit and
withstanding the mechanical forces of that rurrent. It is necessary to
calculate the current at only one time for the application of low-voltage air
circuit breakers. The current determined should be that of the first halt
cycle and should be determined on exactly the same hasis as for checking
the momentary duty of high-voltage power circuit breakers, except for a
change in the multiplying factor as discussed in the next paragraph. The
suhtransient reactances of generators, induction motors, and synrhronous
motors are used, and the d-c component is considered (see Table 1.2).
The multiplying factor for the d-c component is not so high in low-
voltage circuits as in some high-voltage circuits. This is due to the gener-
ally lower level of reactance-to-resistance ( X I R ) ratio in low-voltage
circLits, which causes the d-c component to decay faster than in some
high-voltage circuits.
In rating low-voltage air circuit breakers, the average d-c component of
the three phases is used, which is somewhat lower than that for the maxi-
mum phase.
The generally lower ( X / R )ratio and the use of an average d-c compo-
nent for the three phases result in a considerably lower multiplying factor
in low-voltage circuits. The multiplying factor has been standardized
at 1.25 for the average for the three phases. This is equivalent to a
multiplier of about 1.5 to account for the d-c component in the maximum
phase.
Application of High-voltage Oil Circuit Breokers to 600-volt Systems.
In the 192Os,5-kv oil circuit breakers were used extensively on 600-volt

36 SHORT-CIRCUIT-CURRENT CALCULAnNG PROCEDURES
systems. The procedure for determining short-circuit currents in sys-
tems of 600 volts and below is slightly modified for checking duty on oil
breakers of the 5-kv class as compared with low-voltage air circuit
breakers.
Both the momentary duty and interrupting duty must be checked for
the oil-circuit-breaker application. To check the momentary duty, use
the same procedure as for low-voltageair circuit breakers, i.e., generators,
utility sources, induction motors, and synchronous motors (subtransient
reactance). However, a multiplying factor of 1.5 is used instead of 1.25
as for low-voltage air circuit breakers. Oil-circuit-breaker momentary
ratings are based on the maximum current through any one pole, not on
the average current in the three phases which is employed in the rating of
low-voltage circuit breakers.
To determine the interrupting duty, use the generator subtransient
reactance and utility-source reactance plus the synchronous-motor
transient reactance and a multiplying factor of 1.0.
Low-voltage Fuses. Several low-voltage fuses with published a-c
interrupting ratings are appearing on the market. There are no industry
standards to follow, but most of these seem to be following air-circuit-
breaker standards, i.e., using the same rating base and same method of
determining short-circuit duty as is used for low-voltage air circuit
breakers. Hence, the procedure will not be repeated here except to
point out that the 1.25 multiplying factor is used (see Table 1.2).
So-called National Electrical Code (NEC) plug and cartridge fuses
have no established a-c interrupting ratings. Many tests have been made
to determine their a-c interrupting ability, but to date the industry has
not applied a-c interrupting ratings.
Low-voltage motor starters are of two
types: those using fuses and those using air circuit breakers for short-
circuit protection. Those using air circuit breakers for short-circuit
protection are applied 04 exactly the same basis as low-voltage air circuit
breakers in so far as short-circuit currents are concerned.
Motor starters using fuses for short-circuit protection are applied on
exactly the samebasisasfusesin sofarasshort-circuit current is concerned.
Low-voltage Motor Starters.
AVAILABLE SHORT-CIRCUIT CURRENT
In determining the short-circuit current, the impedance of the circuit
protective device connected in the faulty feeder is neglected. The short-
circuit current is determined by’ assuming that the protective device is
shorted out by a bar of zero impedance (Fig. 1.27). The short-circuit
/

current which flows in such a circuit is commonly called available short-
circuit cumat. The procedure for determining the available short-circuit
current is based on setting up impedance or reactance diagrams. The
impedance of the short-circuit protective device that is nearest the short
circuit (electrically) is omitted from the impedance diagram.
Practically all protective devices are so rated and tested for short-
circuit interrupting ability; hence this procedure may be followed in
short-circuit calculations. This greatly simplifies the calculations and
removes the effect of impedance variations between different types and
makes of devices having the same interrupting rating. It means that
one set of short-circuit-current calculations for a given set of conditions
is all that is needed for applying any type of protective device, regardless
of the impedance of the devices themselves.
GENERATOR
0
TRANSFORMER
MOTORS
CABLE
CABLE
SHORT
CIRCUIT
SHORT ClRCUlTED 81
JUMPER OF ZERO
IMPEDANCE
FIG. 1.27
circuit protective devices.
Connections for determining available short-circuit current for testing rhort-

HOW TO MAKE A SHORT-CIRCUIT STUDY
FOR DETERMINING SHORT-CIRCUIT CURRENT
FORMULAS FOR SHORT-CIRCUIT STUDY'
1. Changing ohms to per cent ohms, etc.:
(ohms reactance)(kva.base)
(kvt)*(lO)
(ohms reactance)(kva base)
(1.1)
Per cent (%) ohms reactance =
Per-unit (90ohms reactance = (kv)*(1000) (1.2)
[see Eq. (1.34)]
(1.3)
(1.4)
(% reactance)(kv)2(10)
kva base
Ohms reactance =
Per-unit ohms reactance =
per cent ohms reactance
100
2. Changing per cent or per-unit ohms reactance from one kva base to
another:
% ohms reactance on kva base 2
kva base 2
kva base 1
-
- X (% ohms reactance on base 1) (1.5)
9f reactance on kva base 2
kva base 2
kva base 1
-
- X (% ohms reactance on kva base 1) (1.36)
3. Converting utility-system reactance to per cent or per-unit ohms
a. If given in per cent ohms reactance on a kva base differentthan that
b. If given in short-circuit kva, convert to per-unit ohms thus:
reactance on kva base being used in study:
used in the study, convert according to Eq. (1.5).
(1.6)
kva base used in reactance diagram
short-circuit kva of utility system
9i reactance =
c. If given in short-circuit amperes (rms symmetrical), convert to per-
unit ohms thus:
Yi reactance =
d. If only the kva interrupting rating of the incoming line breaker is
* See pp. 54 to 57 for more prr-unit formulas
1kv = line-to-line kilovolts.
(1.7)
kva base used in reactance diagram
(short-circuit current)(d$)(kv rating of system)
known,

SHORTT-CIRCUIT.CURRENTCALCULATING PROCEDURES 39
9f ohms reactance
(1.8)
- kva base used in reactance diagram
kva interrupting rating of incoming line breaker
-
4. Determining kva base of motors:
The exact kva base of a motor = EI 4
3 (1.9)
where E = name-plate voltage rating
When motor full-loadcurrents are not known,use the followingkva bases:
Induction motors:
0.8-power factor synchronous motor:
1.0-power factor synchronous motors:
I = name-plate full-load current rating
kva base = horsepower rating
kva base = 1.0 (horsepower rating)
kva base = 0.80 (horsepower rating)
(1.10)
(1.11)
(1.12)
5. Changing voltage base when ohms are used:
Ohms on basis of voltage 1
-
- ')* X (ohms on basis of voltage 2) (1.13)
In Eqs. (1.1)to (1.4), ohms impedance or ohms resistance may be sub-
The final product is then per-unit or per
(voltage 2)2
stituted for ohms reactance.
cent ohms impedance or resistance, respectively.
6. Determining the symmetrical short-circuit kva:
Symmetrical short-circuit kva = ~ (kva base) (1.14)
% X *
-~
- y? '&
(kva base) (1.15)
(1.16)
(1.16a)
(line-to-neutral voltage)2
ohms reactance X 1000
kv2 X lo00
ohms reactance
= 3
-
-
.
7. Determining the symmetrical short-circuit current:
(1.17)
(100)(kva base)
(% X*)(v%(kvt)
(% X*)(&)(kvt)
Symmetrical short-circuit current =
(1.18)
(1.19)
- kva base
-
- kvt X lo00
-
( d ) ( o h m s reactance)
* X = reactance or impedanoe.
t kv = line-&line kilovolts.

L
0
TABLE 1.3 Factor ( K ) to Convert Ohms to Per Cent or Per-unit Ohms for Three-phase Circuits*
216Y/125
240
480
600
2,400
4
.
1 60
4,800
6.900
7,200
l1,OOO
11.500
12,000
12,500
13.200
13,800
23,000
37.4M)
46,000
69,OCU
loot
P
.
r c*nt
'14
73
43.4
27.7
1.73
0.56
0.435
0.210
0.193
0.0825
0.0755
0.0695
0.064
0.0574
0.0525
0.0187
0.00711
0.00471
0.00212
-
Per-""it
2.14
1.73
0.434
0.277
0.0173
0.00576
0.00435
0.0021
0.001 93
0.000825
0.000755
0.000695
0.00064
0.000574
0.000525
0,000187
0.000071 I
0.0000471
0.0000212
Per cent
321.5
260.4
65.21
4.166
2.604
0.808
0.651
0.315
0.289
0.123
0.113
0.104
0.096
0.086
0.0787
0.0283
0.0107
0.00708
0.00315
-
1 50
Per-""it
3.215
2.604
0.6521
0.4166
0.02604
0.00808
0.00651
0.00315
0.00289
0.00123
0.00113
0.00104
0.00096
0.00086
0.000787
0.000283
0.000106
0.0000708
0.0000315
Base kvo
200
Per cant
128
147
86.8
55.5
3.47
1.15
0.868
0.42
0.386
0.165
0.151
0.138
0.127
0.114
0.105
0.0378
0.0142
0.00945
0.0042
Por-un1t
4.28
3.47
0.868
0.555
0.0347
0.0115
0.00868
0.0042
0.00386
0.00165
0.00151
0.00138
0.00127
0.00114
0.00105
0.000378
0.000142
0.0000945
0.000042
300
Per cent
t
4
3
a1
30.2
83.3
5.21
1.72
1.302
0.63
0.579
0.247
0.226
0.208
0.192
0.172
0.157
0.0567
0.0213
0.0141
0.0063
kva base , kva base
kv' X 10
* For per-unit, K = For per cent, K = kv = line-to-line kilovolts
kv' X 1wO
-
Per-""it
6.43
5.21
1.302
0.833
0.0511
0.0172
0.01302
0.0063
0.00579
0.00247
0.00226
0.00208
0.001 92
0,00172
0.001 57
0.000547
0.00021 3
0.00014
1
0.000063
500
__
Per cent
071
868
217
I38
8.68
2.88
2.17
1.05
0.965
0.413
0.377
0.347
0.32
0.286
0.262
0.045
0.0355
0.0236
0.0105
Per-""it v)
I
- 2
2
2.17 K
0.71
8.68
E
1.38 2
0.0868
0.0288
0.0105 n
0.00965
0.00413
0.00377 5
0.00347 f
0
0.0032
0.00286
0.00262
0.00045 c
0.000355 R
0.000236 v,
0.000105
0.0217 B
-
I
2
6
new base i
n kva,
100
t To determine multiplying factors far any other base use figures under 100-kvs base columns multiplied by

8. Determining the asymmetrical short-circuit current:
INFINITE
H BUSES
-SHORT CIRCUIT CURRENT GOES THROUGH HERE
41
(1.20)
Asymmetrical short-circuit current
Asymmetrical short-circuit kva
= (symmetrical current) (multiplying factor)
= (symmetrical kva)(multiplying factor)
DIAGRAMS
One-line Diagram. The first step in making a short-circuit study is to
prepare a one-line diagram showing all sources of short-circuit current,
i.e., utility ties, generators, synchronous motors, induction motors, syn-
chronous condensers, rotary converters, etc., and all significant circuit
elements, such as transformers, cables, circuit breakers, etc. (Fig. 1.28).
The second step is to
make an impedance or reactance diagram showing all significant react-
ances and resistances (Pig. 1.29). In the following pages this will be
Make an Impedance or Reactance Diagram.
C UTILITY SYSTEM
I
TRANS D GENERATOR
GENERATOR
CABLE E
SHORT
CIRCUIT
LARGE CABLE J
MOTOR
FIG. 1.28 e diagram c
480 VOLT
MOTORS
, typical large industrial power system.
FIG. 1.29 Reactonce diagram of system shown in Fig. 1.28.

42 SHORT-ClRCUIT.CURRENT CALCULAltNG PROCEDURES
referred to as an impedance diagram, recognizing of course that only
reactances will be used in many diagrams. The circuit element,s and
machines considered in the impedance diagram depend upon many
factors, i.e., circuit voltage, whether momentary or interrupting duty are
to be checked, etc.
The foregoing discussion and Table 1.2 explain when motors are to be
considered and what motor reactances are to he used for checking
the dut,y on a given circuit breaker or fuses of a given voltage class.
There are other problems, i.e., (1) selecting the type and location of the
short circuit in the system, (2) determining the specific reactance of a
given circuit element or machine, and (3) deciding whether or not circuit
resistance should be convidered.
SELECTION OF TYPE AND LOCATION OF SHORT CIRCUIT
Three-phase Short Circuits Generally Considered. In most indus-
trial systems, the maximum short-circuit current is obtained when a
three-phase short circuit occurs. Short-rircuit-current magnitudes are
generally less for line-to-neutral or line-to-line short circuits than for the
three-phase short circuits. Thus, the simple three-phase short-circuit-
current calculations will suffice for application of short-circuit protective
devices in most industrial systems.
In some very
large systems where the high-voltage-system neutral is solidly grounded,
maximum short-circuit current flows for a single phase-to-ground short
rircuit. Such a system might be served from a large delta-Y trans-
former bank or directly from the plant generators.
Hence the only time that single-phase short-circuit-current calculations
need be made is on large high-voltage systems (2400 volts and above)
with solidly grounded generator neutrals or where main transformers
that supply a plant from a utility are ronnected in delta on the high-
voltage side (incoming line) and in Y with solidly grounded neutrals
on the low-voltage (load) side.
The calculations of unbalanced short-circuit currents in large power
systems can best be done by symmetrical components, see Chap. 2.
Normally, generator and large delta-Y transformer secondaries are
grounded through a reactor or resistor to limit the short-circuit current
for a single line-to-ground short circuit on the system to letis than the
value of short-circuit current for a three-phase short circuit.
Several tests have been
made to evaluate the effect of arc drop at the point of short circuit in
reducing the short-circuit-current magnitude. It was felt by some
engineers that the current-limiting effect of the arc was pronounced.
These tests showed, however, that for circuit voltages as low as 300 volts
Unbalanced Short Circuits in Large Power Systems.
Bolted Short Circuits Only Are Considered.

there may be no substantial difference in the current that flows for a
bolted short circuit and when there is an arc of several inches of length.
These test,s also confirmed modern calculating procedure as an accurate
method of estimating the short-circuit-current magnitude in systems
of 600 volts and less.
.4rcs cannot be counted on to limit the flow of short-circuit currents
even in louvoltage circuits; so short-circuit-current calculations for
all circuit voltages are made on the basis of zero impedance at the point
of short circuit, or, in other words, a bolted short circuit. This materially
simplifies calculation because all other circuit impedances are linear in
magnitude, whereas arcs have a nonlinear impedance characteristic.
At What Point in the System Should the Short Circuit Be Considered
to Occur? The maximum short-circuit current will flow through a cir-
cuit breaker, fuse, or motor starter when the short circuit occurs at the
4160V.
I I I
$? MAX.SHORT CIRCUIT DUTY ON
$- $
E
W
:
R
S FOR SHORT CIRCUIT
BREAKERS ON THIS BUS
1
T
?;
A&??
Y T T - 3
& + * +
r y r
-
x
MAX. DUTY FOR
THESE BREAKERS
OCCURS FOR
SHORT CIRCUIT
HERE
FIG. 1.30 Location of faults for maximum Short-circuit duty on circuit breakers.

terminals of the circuit breaker, etc. (Fig. 1.30). These devices, if
properly applied, should be capable of opening the maximum short-
circuit current that can flow through them. Therefore, only one short-
circuit location (at the terminal of the device) need be considered for
checking the duty on a given circuit breaker, fuse, or motor starter.
DETERMINING REACTANCES AND RESISTANCES OF CIRCUITS AND MACHINES
Typical reactances of circuit elements and machines are given at the
end of this chapter. These
tables may be used as a basis for assigning values to the various elements
of the impedance diagram. The reactances and resistances are all line-
to-neutral values for one phase of a three-phase circuit. Where the
reactances of a specific motor, generator, or transformer are known,
these values should he used in lieu of the typical reactances in this
chapter. The following is a guide to general practice in selecting and
representing reactances.
Whether or not
the reactance of a certain circuit element of a system is significant
depends upon the voltage rating of the system where the short circuit
occurs. In all cases, generator, motor, and transformer reactances
are used. In systems rated above 600 volts, the reactances of short
bus runs, current transformers, disconnecting switches, circuit breakers,
and other circuit elements of only a few feet in length are so low that
they may be neglected without significant error.
In circuits rated 600 volts or less, the reactances of low-voltage current
transformers, air circuit breakers, disconnecting switches, low-voltagebus
runs, etc., may have a significant hearing on the magnitude of total short-
circuit current.
As a general guide, the reactance of the low-voltage secondary-switch-
gear section in load-center unit substations with closely coupled trans-
formers and secondary switchgear is not significant for all voltages of
600 volts and below. However, where there are several transformers or
generators paralleled on one bus, or connections several feet long between
a single transformer and its switchgear, reactances of the bus connections
will generally be significant and should be considered in short-circuit
calculations. In systems of more than about 1000 kva on one bus at
208Y/120 or 240 volts, reactance of all circuit components such as short
bus runs, current transformers, circuit breakers, etc., should be included
in the short-circuit study.
In systems of more than about 3000 kva on one bus at 480 volts or
600 volts, reactances of all components such as current transformers,
circuit breakers, short bus runs, etc., should be considered.
It should be remembered that the lower the voltage, the more effective
Resistances are included for certain items.
U s e Reactances of All Significant Circuit Elements.

SHORT-CIRCUIT-CURRENl CALCUUTING PROCEDURES 45
a small impedance is in limiting the short-circuit-current magnitude.
That is why extreme care should he used to include all circuit elements
in the impedance diagram, particularly for large ZORY/lZO-volt or
240-volt systems. I
f care is not used, the calculations will result in a
value of current far higher than will actually be realized in practice.
See the example outlined in Figs. 1.46 and 1.47. This often results in the
adoption of low-voltage switchgear of higher interrupting rating and
higher cost than are actually required. I
f care is used in including all
reactances, the calculated reiults will be close to the short-circuit currents
obtained in practice. Short-circuit calculations are of most value if
they reflect accurate answers.
The resistance of all generators,
transformers, reactors, motors, and high-capacity buses (above about
1000-amp rating) is so low, compared with their reactance, that their
resistance is not considered, regardless of their voltage rating. The
resistance of all other circuit elements of the high-voltage system (above
600 volts) is usually neglected, because the resistance of these parts has
no significant bearing on the total magnitude of short-circuit currents.
In systems of 600 volts and less the error of omitting resistances of
all parts of the circuit except cables and small ampere rating buses is
usually less than 5 per cent. However, the resistance of cable circuits is
often the predominant part of the total impedance of a cable. When
appreciable lengths of cable are involved in the circuit through which
short-circuit current flows in a system of GOO volts or less, the resistance
as well as the reactance of the cable circuits should be included in the
When Is Resistance Considered?
GENERATOR
___-. .. -.
1100 FT. 101
AND RESISTANCE OF THESE
IN GENERAL USF REACTANCE
OF-THESE CIRCUIT ELEMENTS.
SHORT CIRCUIT CURRENT CONSIDERING
REACTANCE ONLY :
20800 AMPERES
SHORT CIRCUIT CURRENT CONSIDERING
REACTANCE OF ALL PARTS PLUS
RESISTANCE OF COW VOLTAGE
CABLE = 11500 4MPERES.
----
(20 FT
FIG. 1.31 One-line diagram showing effect of resistance in cable circuits.

46 SHORT-CIRCUIT.CURRENT CALCULATING PROCEDURES
impedance diagram. The example of Fig. 1.31 shows the error that
might result in neglecting cable resistance.
In secondary network systems of 600 volts and less, the resistance as
well as the reactance of the tie-cable circuits between substation buses
should be included in the impedance diagram. The example of Fig. 1.32
shows the effect of cable resistance in reducing short-circuit current in a
typical industrial network.
n n
SHORT CIRCUIT CURRENT USING
REACTANCE ONLY = 51000 AMPERES,
SHORT CIRCUIT CURRENT USING
REACTANCE PLUS RESISTANCE OF
TIE CIRCUIT= 41000 AMPERES.
T I E CIRCUITS
208 Y/lZOVOLTS.
200 FT
2- 250 M,CM
3 CONO. CABLES ~~~~~T
I N PARALLEL
200 F T
FIG. 1.32
resistance o
f cable tie circuits.
One-line diogrtlm of low-voltage secondary network system showing effect of
Where to Use Exact Multiplying Factors. In low-voltage systems
having considerable lengths of cahle, the X / R ratio may be so low that
the 1.25 multiplying factor would be considerably in error. Hence in
these systems where resistance is considered, determine the correct
X / R ratio and then use minimum multiplying factor.
GUIDE FOR REPRESENTING THE REACTANCE O F A GROUP O F MOTORS
A group of motors fed from one substation or from one generating
station bus may range in rating from fractional to several thousand horse-
power per motor. All motors that are running at the time a short circuit
occurs in the power system contribute short-circuit current and therefore
should be taken into consideration.
In that portion of the power sys-
tem operating at 600 volts or less, there are generally numerous small
Motors Roted 600 Volts and Below.

SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES A?
motors, i.e., under about 50 hp. It becomes impractical to represent
each small motor in the impedance diagram. These motors are con-
stantly being turned off and on; so it is practically impossible to predict
which ones will be on the line when a short circuit occurs. Furthermore,
it would be impractical to obtain the characteristics of each small motor
and to account for the effect of the impedance of their leads.
Where more accurate data are not available, the following procedure
may be used with satisfactory results for representing the combined
reactance of a group of miscellaneous motors operating at 600 volts or
less.
1. In systems rated 240, 480, or 600 volts at each generator and/or
transformer bus, assume that the maximum horsepower of motors
runniug at any one time is equal to the combined kva rating of the step-
down transformer and/or generators supplying that one bus (see Figs.
1.33 and 1.34).
2. 10 systems rated 208Y/120 volts, a substantial portion of the load
usually consists of lights and a lesser proportion of motor load than in
240-, 480-, or 000-volt systems. Hence in 208Y/120-volt systems where
more accurate data are not available, assume at each generator and/or
transformer bus that the maximum horsepower of motors running at
REbCTbNCE QOW,
TO UTILITY SYSTEM OF UTILITY OR5.,s 0.25% OR
25 %
REbCTbNCE OF
EQUIVALENT
MOTOR
SYSTEM
REbCTbNCE
OF 750 KVb
TRbNSF. 5.5%
IMPEObNCE OIbGRbM
750 KVb BASE
SHORT EQUIVALENT MOTOR
CIRCUIT 7 5 0 KVb
240, 480, 600 VOLT SYSTEMS
TO UTILITY SYSTEM
0.50%OR
50 %
REACTbNCE OF
EQUIVALENT
ElMOTOR
REbCTbNCE
OF 750 KVb
TRbNSF.
REbCTbNCE
OF UTILITY
SYSTEM
EQUIVILENT MOTOR
SHORT
CIRCUIT
375 K V b IMPEObNCE OIbGRbM
hKVA
750 KVb BASE
208Y/120 VOLT SYSTEMS
FIG. 1.33 Oiagromr illustrating how to include motors in low-voltage radial systems.

any one time is equal t,o 50 per cent of the combined rating of all step-
down trausformers and/or generators supplying power to that one bus,
Fig. 1.33. For large commercial buildings the 50 per cent figure may
be too low. Check carefully the mot,or load on all large 208Y/120-volt
systems.
In the generalized rases referred to in paragraphs 1 and 2, no specific
ratio of induction to synchronous motors or no specific number of motors
which prcduce unusually high short-circuit current,s has been set fort,h.
To account for these variables, an average motor reitctance ihcluding
leads is assumed to be 25 per cent for the purpose of preparing application
tables like Table 1.5 and in making short-circuit st,udies where no more
accurat,e data are available. It will he noted that the average motor
reactance of 25 per cent is based on the transformer or supply-generator
kva rating. This figure is between the values of 28 per cent for induc-
tion mot,orsand 21 per cent for synchronous motors given in Table 1.14.
Where the division between synchronous and iuduction motors is known,
then more accurate calculations can be made by using the assumed motor
reactances of Table 1.14. The reactances given in Table 1.14 are based
on motor kva ratings and not supply transformer or generator ratings.
T
750 KVA
A 500 KVA 750 KVA
500 KVA
v
EQUIVALENT MOTORS WOULD BE 250 KVA AND
FOR 280Y/120 VOLT SECONDARY SYSTEM
375 KVA
-480 VOLTS
FIG. 1.34
rvrternr.
Diagram illustrating how lo include motors in lowvoltage secondary network

Although a portion of the load connected to a bus rated GOO voks or
less may be heaters, lights, a-c welders, solderitig irons, appliances, arid
other devices which produce no short-circuit curreiit, the total installed
horsepower of motors connected t,osuch a bus is geiierally much greater
than the kva rating of the supply transformers and generators. Hov-
ever, allowing for diversity, generally the total comhitied horsepower
rat,ing of all mot,ors running at one time ix-ould trot produce short-cir-
cuit currents in excess of the values obtained when using the ahore
assumptions.
In systems of 000 volts or Icss, the large motors (i,e., mot,ors 011 t,he
order of several hundred horsepomerj are usually few in number and
represent only a small portion of the tot,al connected horsepower; there-
fore, these larger motors are generally lumped in with the smaller motors
and the complete group is represented as one equivalent motor iti the
impedance diagram.
Synchrouous and induction motors need not be segregated when com-
bining the motors in these low-voltage systems, because lorn-voltage air
circuit breakers operr so fast that only the current flowduritig the first half
cycle is considered; i.e., only suhtraiisient reactances ( X y ) of marhiiies
are considered.
Motors Rated above 600 Volts. High-voltage motors (rated 2200
volts and ahove) are generally larger in horsepower rating thau motors
on systems operating under 600 volts. These largcr motors may have
a much more significant hearing on short-circuit-current magnitudes
than smaller motors, and, therefore, more exact determinatiou of the
reactances of the larger motors is in order. Therefore, it is often foutid
convetiient to represent each large high-voltage motor individually in
the impedance diagram.
However, in large plants like steel mills, paper mills, etc., where there
are numerous motors of several huridred horsepower each, it is often found
desirable to group these larger motors iii one group arid represent them
by one reartaiire in the impedance diagram. Individual motors of
several thousand horsepoitrer should be coiisidered individually and
their reactances accurately determined hefore starting the short-circuit
Whether considering motors individually or in groups, regardless of
voltage rating of the motors, it is necessary to obtain an equivalent kva
rating of the individual or group of motors. This can be done precisely
for large motors by Eq. (1.9) or can be approximated hy Eq. (l.lO),
(l,ll), or (1.12), when the full-load current is not known. The latter
equations are used when considering a single reactance to represent a
group of miscellaneous motors.
study.

In high-voltage systems, complete motor data may not be available.
Lacking these data, the connected horsepower is assumed to he equal
to the generator and/or transformer capacity supplying a given high-
voltage bus.
If the reactance of the leads between the transformer and/or gen-
erator bus and the motors is significant, the reactanre of these leads
should be included.
MAKING THE IMPEDANCE DIAGRAM
After it has been decided what elements of the one-line diagram are
to be considered in the impedance diagram, the mechanirs of making
the impedance diagram and of determining the short-circuit-current
magnitude are as follows.
Treatment of Sources of Short-circuit
Current. The generators and motors
are treated as if they comprised a gen-
erator of zero reactance plus an external
reactor to represent the reactance of the
EXTERNAL TO machine windings, Fig. 1.35. The first
REPRESENT IMPEDANCE OF step in making an impedance diagram
GENERATOR OR MOTOR. is torepresent everygeneratorand motor
or groups of motors and utility supply
FIG. 1.35 One-line representation by a reactance connected to a zero im-
of generator or motor in impedance pedance bus or so-called “infinite bus,”
Fig. 1.36. This bus represents the in-
diogmm.
ternal voltage of the generators and motors.
The second step is to add the
reactance of cables, buses, transformers, current transformers, circuit
GENERATOR OR MOTOR OF
ZERO IMPEDANCE
7
Completing the Impedance Diagram.
flG. 1.36
system shown in one-line diagram form in Fig. 1.28.
Representation of reactances of generators, motors, and utility supply of

breakers, switches, etc., in their proper location to complete the imped-
ance diagram, top of Fig. 1.37.
The
next step is to decide whether to use ohms, per cent ohms, or per-unit
ohms to represent the various circuit impedances in the impedance
diagram.
Choice of Ohms, Per Cent Ohms, or Per-unit Ohms Method.
INFINITE BUS
SHORT 6.04V
INFINITE BUS CIRCUIT
STEP NO i COMBINE SERIES REACTANCES
C + D = 0 . 0 4 + 0 . 1 5 ~ 0 . 1 9 %
H + 1 + J = 2.Ot ~0.0+0.10~
12.10%
STEP NO.:! COMBINE PARALLEL REACTANCES
F,G AND I H + I + J)
XI F G H + I + J
-
' = _' + L + I
I I
- _
-o,'40t
3 + p-j=j 2.5+0.2+0.083
I
-= 2.783 X =0.3698
XI
STEP N0.3 COMBINE SERIES REACTANCES
X,,AND E
X t = XI + E = 0.36+0.04'0.40%
STEP NO. 4 COMBINE PARALLEL REACTANCES
X o , A . B . AND IC+D)
XR XI A B C+D '
0
.
4
0 025 2.0 0.19
I - 1 + 1 + L+- 1 - +- +- +-
l i I I
2.5+ 4 + 0 . 5 +5.3=12.3
X,
RESULTANT SINGLE REACTANCE
I 0.0805 %
X ~ O ~ z
FIG. 1.37
bining reactances intoo single resultantvalue.
Complete reaclomce diagram for system shown in Fig. 1.28. Steps for com-

Ohms are generally not used because of the difficulty of converting
ohms from one voltage base to another without error and because of
the very small numbers, which make accurate and easy calculation more
difficultthan the per cent or per-unit system.
In many of the examples in this book, the assumed or given impedance
or reactance data are listed in per cent, hut in the reactance dia,-rams
these are converted to per-unit. No notation will he made when that is
done as it will be obvious.
Equations (1.1) to (1.4) show how to convert ohms to per cent ohms,
ohms to per-unit ohms.
The Per-unit System for Electrical Calculations.* A per-unit system
is a means of expressing numbers for ease in comparing them. A per-unit
value is a ratio:
a number
Per-unit = ~~
base number
(1.21)
The base number is also called unit value since in the per-unit system
Thus, base voltage is also called unit
For
it has a value of 1, or unity.
voltage.
example, for the columns below, a base of 560 is used:
Any convenient number may be selected for the base number.
Per-unit Volue
with 560 as a Base
Number
93 0.17
125 0.22
560 1 .oo
2053 3.65
Each number in the second column is a per-unit part of the base
number. In the first column, to compare the numbers, first mentally
determine the ratio of one to the other. In the second column this is
already accomplished.
The comparison can be aided by selection of the base number which
will illustrate the comparison best. In the foregoing example, if it is
desired to show how much larger each uumber is when compared with
the smallest number, the number 93 might have been selected as the
base. This would then be obtained as follows:
Per-""it Valve
Number with 93 (
I
, (
I Base
93 I .oo
125 I.35
560 6.00
2053 22.20
The value of a per-unit system is particularly useful when comparing
* From material originally prepared by H. J. Finison. iormrrly of General Ekctrir
Company.

SHORT-CIRCUIT.CURRENT CALCULATING PROCEDURES 53
For
numbers that are similarly related to two different base numbers.
example:
Core A Cole B
Norm01"0th 2300 460
Volts during motor starting 2020 420
The above figures in themselves have little significance until they are
compared each with its normal condition as follows:
Vollr during starting per-unit of normal 0.88 0.91
Per Cent. Obviously per cent and per-unit systems are similar. The
per cent system is obtained by multiplying the per-unit value arbitrarily
by 100 to keep many frequently used per-unit values expressed as whole
integers. By definition,
x 100
a number
base number
Per cent = (1.22)
Thus to change per cent to per-unit, divide by 100. For example, a
transformer which has an impedance of 6 per cent has an impedance of
0.06 per-unit.
The per cent system is somewhat more difficult to work with and more
subject to possible error since it must always be remembered that the
numbers have been arbitrarily multiplied by 100. For a simple example,
money may draw interest at the rate of 4 per cent per year. Early in
arithmetic one learns to determine the interest by multiplying the princi-
pal by 0.04. It is thus necessary to remember to convert to the per-unit
value before using the figure. In a complex calculation, this repeated
conversion may invite errors. In effect it is safer and more convenient
to say that interest is at the rate of 0.04 per-unit.
It is
usually convenient to convert these figures immediately to per-unit by
dividing by 100 and thereafter do all calculating in terms of per-unit
rather than attempt to remember always during the calculations whether
a number should or should not be multiplied or divided by 100 to obtain
the true value.
Just as the per cent system has a symbol (%) to desig-
nate that a given number is expressed in terms of per cent (as 6%)so also
does the per-unit system have a symbol. The symbol for per-unit is
(%).
In a per-unit system asused for expressing
electrical quantities of voltage, current, and impedance, it is necessary to
select numbers arbitrarily for the following:
Impedances of electric apparatus are usually given in per cent.
Symbol.
Thus 0.06 per-unit is written as 0.06 91.
Selectionof Base Number.
Base volts
Base amperes

Do not then in addition arbitrarily select base ohms since it has already
been fixed by the first two selections because of Ohm’s law.
E
z = -
I
base volts
base a m p z s
Base ohms = (1.23)
Using the selected base values, all parts of an electric circuit or system
may be expressed in per-unit terms as follows:
(1.24)
(1.25)
(1.26)
volts
base volts
amperes
base amperes
ohms
base ohms
Per-unit volts =
Per-unit amperes =
Per-unit ohms =
In practice it is more convenient to select:
Base volts
Base kva
The base values of other quant.ities are thus automatically fixed.
for a single-phase system,
Hence,
(1.27)
(1.28)
base kva X 1000
base volts
base kva
base kv
Base amperes =
Base amperes =
Base ohms = (1.23)
where base kva is single-phase kva and base volts is single-phase volts.
base volts
base amperes
For a three-phase system:
(1.29)
base kva X 1000
X base voks
Bme amperes =
base kva
4 X base kv
Base amperes = (1.30)
(1.31)
where base kva is three-phase kva, base volts is line-to-line, and hase ohms
is per phase.
Per-unit Ohms. In practice it is desirable to convert directly from
ohms to per-unit ohms, without first determining base ohms. By Ohm’s
law,
hase volts
4
X base amperes
Base ohms =
base volts
base amperes
Base ohms = (1.23)

Substitute Eq. (1.27) (which gives the base amperes) into Eq. (1.23), to
obtain
base volts
(base kva X 1000)/base volts
(base volts)P
bsse kva x 1000
Base ohms =
B
a
s
e ohms = (1.32)
By definition:
ohms
base ohms
Per-unit ohms = (1.26)
Substitute Eq. (1.32) into Eq. (1.26) to obtain
ohms
(base volts)e/(base kva X 1000)
ohms X base kva X 1000
(base voltd2
Per-unit ohms =
ohms X base kva
(base kv)2X 1000
where base kva is single-phasekva and base kv is single-phase kv.
When dealing with a three-phase system, it is usual to select three-phase
kva and line-to-line volts for the base values. Convert the above expres-
sions to these bases to obtain
ohms X base kva X 1000 X 3
(base volts X d
3
)
z
Per-unit ohms =
. ,
ohms'X base kva X 1000
(base volts)2
ohms X base kva
(base kv)* X 1000
Per-unit ohms =
where ohms are per phase, kva is three-phase kva, and kv is line-to-line
voltage.
Usual Base Numbers for System Studies. If per cent or per-unit ohms
reactance is used, the next step is to choose a kva base.
In system studies it is usually desirable to select as the base voltage the
nominal-system voltage or the voltage rating of the generators and supply
transformers. Base kva will usually be selected as the kva rating of one
of the machines or transformers in the system, or a convenient round
number such as 1000, 10,000, or 100,OOOkva. After choosing the kva
base, convert ohmic reactance of cables, wires, current transformers,
etc., to per cent or per-unit ohms reactance on the chosen base, using
Eq. (1.1) or (1.2) or Table 1.3.
If ohms reactance is used, convert all per cent reactances to ohms by
Eq. (1.3).
Where two systems of differing voltage are interconnected through a

transformer, select a common kva base for both systems and the rated
voltage of each system as its own base voltage. (These base voltages
must have the same ratio to each other as the turn ratio of the transformer
connecting the two systems.) Base ohms and base amperes for the two
systems will thus he correspondingly different. Figure 1.38 shows a
typical example.
Once the system values are expressed as per-unit values, the two inter-
connected systems may be treated as a single system and any calculations
necessary carried out. Only in reconverting the per-unit values of the
results to actual voltage and current values is it necessary to remember
that two different voltages actually existed in the system.
Frequently the impedance of a circuit ele-
ment may be expressed in terms of a particuiar base kva, and it may be
desirable to express it in terms of a different base kva. For example, the
reactance of devices like transformers, generators, and motors is given in
per cent on their own kva rating, and their reactances must be converted
to the common base, chosen for the study by means of Eq. (1.5) or (1.36).
Per-unit ohms on kva base 2
Change of Base Number.
-
- base kva x (per-unit ohms on kva base 1) (1.36)
base kva 1
Similarly, a machine rated at one voltage may actually be used in a
Its per-unit impedance must thus be
circuit a t a different voltage.
changed to a new base voltage.
GENERATOR MOTOR
1000KVA I0;YKVA o(lOOO KVA)
13800 2300
VOLTS VOLTS
PRIMARY SECONDARY
RATING RATING
13200 2400
VOLTS VOLTS
TRANSFORMER RATIO=1
3200/2400=5.5
(A1
RATIO -
(El
(8)LOW VOLTAGE SYSTEM
(A)HIGH VOLTAGE SYSTEM
13 800 BASE VOLTS 2500 5.5
I000 BASE KVA 1
0
0
0 I.o
41.6 EASE AMPS 233 115 5
190 BASE OHMS 6
.
2 5 (5.5?
FIG. 1.38
onother.
Method of converting bore volts, kva, amperes, and ohms from one value to

SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES n
Reference to Eq. (1.35) shows that per-unit ohms is inversely propor-
tional to the square of base volts. Thus:
(1.37)
Per-unit ohms on new base volts - (old base volt.s)*
Per-unit ohms on old base volts (new base volts)*
-
and
Per-unit ohms on new base volts = per-unit ohms on old base volts
(1.38)
Equations (1.37) and (1.38) may be used for per cent ohms as well as per-
unit ohms.
When using ohms
instead of per cent or per-unit in the impedance diagram, it is important
to convert the ohmic values to a common voltage base by Eq. (1.13).
For example, if the short-circuit current is being calculated in a 480-volt
system (supplied by transformersrated 480-volt secondary) fed through a
cable and a transformer from a 2400-volt system, the ohms impedance of
the cable in the 2400-volt circuit must be multiplied by 48O2/24OO2to
convert it to ohms on a 480-volt base. The transformer ratings, i.e., 480,
240, etc., and not system ratings, if differentfrom transformer rating, are
used as the voltage base for short-circuit-current calculations.
The utility system must be
represented by a reactance in the impedance diagram. Sometimes this
utility-system reactance is available in per cent on a certain base. If so,
it is merely necessary to convert this value to the common base used in
the impedance diagram. In some cases the
utility engineers will give the short-circuit kva or current that the utility
system will deliver at the plant site. In otker cases, only the interrupting
capacity of the incoming-linecircuit breaker is known. In these cases to
convert short-circuit kva, current, or incoming-line breaker interrupting
rating to per cent reactance on the kva base used in the reactance diagram,
proceed as follows:
If given short-circuit kva, convert to per cent by using Eq. (1.6).
I
f per-unit is desired, use also Eq. (1.4).
If given short-circuit amperes (rms symmetrical), convert to per cent
by Eq. (1.7) and to per-unit by Eqs. (1.7) and (1.4).
If only the kva interrupting rating of the incoming line circuit breaker
is known, convert to per cent by Eq. (1.8)and to per-unit by Eqs. (1.8)
and (1.4).
(old base volts)2
(new base volts)2
Converting Ohms to a Common Voltage Base.
Representing the Utility Supply System.
To do this, use Eq. (1.5).
DETERMINING THE EQUIVALENT SYSTEM IMPEDANCE O
R REACTANCE
After completing the impedance diagram and inserting the values of
reactance or impedance for each part of the diagram, it is necessary to
reduce this network to one equivalent value. This can be done either by

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