d beeman - industrial power systems handbook

Industrial Power Systems HandbookD O N A L D BEEMAN, EditorManager, 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?r 1

by Donald Beeman, Alan Graeme Darling,and

R. H. Kaufmann

Short-circuit-current CalculatingProceduresFUNDAMENTALS OF A-C SHORT-CIRCUIT CURRENTSThe determination of short-circuit currents in power distribution systems 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 i t will deliver. This is done so that circuit breakers or fuses may he selected with adequate interrupting capacity (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 orI

2

SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES

capacity of t,he power source. The larger the apparatus which supplies electric power t o 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 a t 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 breakers, a 15- or 20-amp circnit, breaker wnuld he specified. However, the size of t,he power 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. A simple mathematical example is shown in Fig. 1.1. These numbersMUST BE CAPABLE OF INTERRUPTING1000 AMPERES

El

MOTORIOOV 100 A

LOAD

~ ~ 1 0O. MS H1

CURRENT 5 AMP APPARENT IMPEDANCE 20 OHMS E ZT:

SHORT

CIRCUIT CURRENT =

I00 - = 1000- AMPERES 0.1

MUST

BE CAPABLE OF INTERRUPTING 10,000 AMPERES

w MOTOR LOAD CURRENT 5 AMP

I000 A 2 1 = 0.01 OHMS

FIG. 1.1

Illustrotion showing that copocity of power source has more effect on rhortcircuit-current magnitude than load.


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 t o 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 a t 1000 amp, is substituted for the 100-amp unit. A short circuit a t 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 shortcircuit current will now he 10,000 amp, and circuit breaker A must be able t o 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 t o 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 t o the flow of current through a short circuit in the distribution 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 SHORT-CIRCUIT CURRENTS F

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

4


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 t o produce voltage because the field excitation is maintained and the prime mover drives the generator at substantially normal speed. The generated voltage produces a shortcircuit 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 a t the terminals of the generator, the current from the generator is limited only by its own impedance.

FIG. 1.2

Normal load and short-circuit currents are analogous to the conditions shown in

the hydroelectric plant.

SHORT-CIRCUIT-CURRENT ULCULATlNG PROCEDURES

5

METAL CLAD SWITCHGEAR

SHORT CIRCUIT

CURRENT FROMINDUCTION MOTOR

FIG. 1.3current.

Generators, synchronous motors, and induction motors all produce short-circuit

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 current 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 slowing down. In other words, the rotating energy of the load and rotor drives the synchronous motor just as the prime mover drives a generator.

6


The synchronous motor then becomes a generator and delivers shortcircuit current for many cycles after the short circuit occurs on the system. . Figure 1 5 shows 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

F G 1.4 I.UlILITY SYSTEM

Normally motors draw load current from the source or utility system but produce rhortcircuit current when a short cirw i t occurs in the d a d .

SYNCHRONOUS MOTOR

-tSHORT CIRCUIT CURRENT FROM MOTOR

, -

\

. .-.. .SYSTEM

SYNCMOYOUS

'

YorollSHORT CIRCUIT

'.

-

I

FIG 1._ IBmlowl. l.r o c e o. 0s. . . 5 , .., . ... . . f . cillogrclm of short-circuit current produced by a synchronous motor

-.

__

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. However, 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 woundrotor 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 a t a point in the circuit close to its input (stator) terminals a t time TI. The current shown is measured on the motor side of the short circuit; so the shortcircuit 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 a t full load show that the short-circuit current produced

T.

FIG. 1.6

Tracer of oxillograms of short-circuit currents produced running a t light load.,

by an induction motor

8


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 current 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 a t the time of short circuit, cmesponds closely with the impedance a t 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 delivered 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 i n Fig. 1.7. The current starts out a t 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,ance of 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


9

occurrence of any short circuit requires a complicated formula involving time as one of the variables. For the sake of simplification in short-circuit 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 current 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 determines the current flow when a steady-state condition is reached. It is not effective until 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

OCCURS A T THIS TIME.

OF TOTAL OSCILLOGRAM

ONLY TWO ENDS SHOWN HERE. THIS REPRESENTS THE BREAK BETWEEN THE TWO PARTS.

FIG. 1.7 Trace of orcillograrn of hart-circuit current produced by a generator.

10

SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES MAX, SUBTRANSIENT CURRENT- USE SUBTRANSIENT REACTANCE X"d

/-

TM I E (8)

FIG 1.8

Variation of generotor short-circuit current wilh time.

current delivered by a generator. The value of X i or X y generally given by the machine designer is the lowest value obtainable. Hence, its use will show maximum short-circuit current. Certain characteristics of short-circuit currents must he understood before a system analysis can he made.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 asymmetricalENVEWPES 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

11

ZERO AXIS

FIG, 1.10

Symmetrical

d t e r n a t i n g current f r o m a short-circuited generotor.

ENVELOPES OF PEAKS ARE NOT SYMMETRICAL ABOUT ZERO AXIS

AX1 S TOTALLY 0 F F SET PARTIALLY O F F S E l

FIG. 1.11 Asymmetrical (I-c waver. The conditions shown here ore theoreticol a n d ore for the purpose of illustration only. D-C component will r a p i d l y d e c a y to zero i n a c t u a lcircuits.

FIG. 1.12

Trace of o r c i l l o g r a m of a t y p i c a l short-circuit current

12


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 current uniformly offset for several f cycles. It should be noted that in practical circuits the amount of asymmetry decreases rapidly after the occurrence of the short circuit in the system. This decrease of asymmetry is shown qualitatively in illustrations such as Figs. 1.12, 1.20, 1.23, and 1.24. Oscillograms show that short-circuit currents are nearly always asymmetrical 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 s i n e wave short-circuit currents result. For simplicity, the following discussion assumes sine-wave voltages and currents. In ordinary power circuits the resistance of the circuit is negligible compared 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). This is known as a symmetrical short-circuit current. 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 a t 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 asymmetricd current.

WORT-CIRCUIT-CURRENT CALCULATING PROCEDURES

13

GENERATOR TRANSFORMER INTERNAL VOLTAGE OF GENERATOR APPLIED HERE

ioxazx

ONE LINE IMPEDANCE

7 0.m x

REACTANCE, X = 19% RESISTANCE. R = 1.4%

RESISTANCE I S LESS THAN OF THE REACTANCE BE NEGLECTED WITHOUT AN APPRECIABLE ERROR

I

HENCE MAY

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. 1.14cirwit.

Symmetric01 short-circuit current and generoted voltage for zero-power-factor

-SHORT CIRCUIT CURRENT

F G 1.15 I.circuit.

Asymmetrical short-circuit current and generated voltage in zero-power-factor Condition i s theoretical and is shown for illustration purposes only.


IS

If,in a circuit containing only reactance, the short circuit occurs a t any point except a t 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). I n circuits containing both reactance and resistance, the s~,?&&,R&!~~ amount of offset of the shortCURRENT circuit current may vary between the same limits as for circuits containing only reactance. However, the point on the voltage wave a t which the short circuit must occur to produce maximum asymmetry dependsupon the ratioof reactance to resistance of the circuit. Maximum asymmetry is obtained when the short circuit occurs a t a time angle equal to 90" 0 (measured forward in degrees from the zero point of the voltage wave) where tangent 0 equals thereASYMMETRICAL 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 circurrent will be symmetrical cuit occurred between the when the fault occurs 90"from point and peak of the generated voltctge wove. that point onthe voltage wave. This condition i s theoretical and for illustration an example, assumeacir- purporer only. The short-circuit current will gradually become symmetrical in practical cuit that has equal resistance CiTCUit., and reactance, i.e., the reactance-to-resistance ratio is 1. The tangent of 45" is I ; hence, maximum offset is obtained when the short circuit occurs a t 135' from the zero point of the voltage wave (Fig. 1.17).

+

D-C COMPONENT OF ASYMMETRICAL SHORT-CIRCUIT CURRENTS

Asymmetrical alternating currents when treatedas a single current 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 alternating currents are, for circuit-breaker applications and relay-setting

16

SHORT-CIRCUIT-CURRENT CALCUUTING PROCEDURES

MAXIMUM OFFSET

FIG. 1.17 Short-circuit current and generated voltage in circuit with equal reactance and resistance. This condition i s theoretical and is shown for illustration purposes only. The 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 a t 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 a t any instant is equal t o the magnitude of the total asymmetrical a-c wave a t 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. I n 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 shortcircuit current which is resolved into a symmetrical a-c and a d-c component in Fig. 1.18. If the instantaneous values of the two components (dashed lines) are added a t any instant, the resultant will be that of the asymmetrical current wave.

SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES F I N S T A N T AT WHICH SHORT CIRCUIT OCCURS

17

ASYMMETRICAL

AC COMPONENT

FIG. 1.18current.

Theoretical Ihort-circuit-cvrrent wove illustrating components of asymmetrical In practical circuits, d-c component would decay to zero in o few cycler.

INSTANT

OF SHORT CIRCUIT

TOTAL CURRENT

DC COMPONENT AC COMPONENT

ZERO A X I S

a = b = D C COMPONENTFIG. 1.19 Components of asymmetrical short-circuit current in which short circuit occurred at some point between the zero point and p e a k of the generated voltage wave. This is a lhsoretical condition similar to that shown in Fig. 1.18.

I8


As mentioned previously, the examples shown in Figs. 1.13 and 1.18 are for purposes of illustration only. In practical circuits the d-c component decays very rapidly, as shown in Fig. 1.20.INITIAL M A G N I T U D E OF D-C C O M P O N E N T

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 a t thc instant of short circuit. The above limit,s hold true for the initial magiiitudc of d-c eomporient in a system regardless of the reactance and resistance. Ilowever, the d-c componeut does not continue to flo~v t a constant value, as a shown i n Figs. 1.18 and 1.19, unless there is zero resistauce i i i the circuit.DECREMENT

There is uo d-c voltage in the system t o 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 CURRENTC

COMPONENT AC COMPONENT

FIG. 1.20 Trace of orcillogrom showing decay of d-c component and how orymmetricd short-circuit currenl gradually becomes symmetrical when d-c component diroppearr.


19

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 components gives combination of the decaying of d-c and symmetriral a-(* 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 circuit. The decrement or rate of decay of the d-c component is proportional 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. 1.18 and 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 definite time to decrease to substantially zero, as shown in Fig. 1.20. ! I n generators the ratio of subtransient reactance to resistance may be as ?much as 7 0 : l ; so it takes several cycles for the d-c component to disappear. 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 dissipated sooner.D-C TIME CONSTANT

Often it is said that generators, motors, or circuits have a certain d-c time constant. This refers again to the rate of decay of the d-c compoO C COMPONENT

a

= 37Y. OF b (APPROX

)

C -

TIMEOF D C COMPONENT

CONSTANT I N SECONDSFIG. 1.21

Graphic illustration of time constant.

20


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 a t the instant of short circuit. I t 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 t o decay a t 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. I n 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 offset is expressed by the equation

where C

=

a b

= =

rrns value of offset or asymmetrical current wave over one cycle rrns value of a-c component value of d-c component at one-half cycle

MULTIPLYING FACTOR

Calculation of the precise rrns value of an asymmetrical current a t any time after the inception of a short circuit may be very involved. Accurate decrement factors to account for the d-c component a t 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 a t 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.


21

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,o the 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 circuit a multiplying factor to calculate the total current at the first half cycle of 1.5 is used. For some calculations, rms current evaluations a t 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 Charts showing multiplying factors to account for decoy of d-c component for various X / R ratio of circuits.

22

SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES

short-circuit protective devices. These factors range from 1 t o 1.6, depending upon whether the short-circuit calculation is being made t o determine the interrupting or momentary duty on the short-circuit protective device.SHORT-CIRCUIT RATIO OF GENERATORS

This term is referred t o frequently in short-circuit discussions. With present AIEE procedures of short-rircuit ralrulations, it has become a n accessory with no practical significance from this standpoint. For the sake of completeness, a definition is given here. Short-circuit ratio field current t o produce rated voltage a t no load -~ field current t o 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 shortcircuit current from the utility. This act,ually comes from ut,ility generators, 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, however, a slight decrement, as indicated in Fig. 1.23. The other sources are synchronous motors which act something like plant generators, except that t,hey have a higher rate of decay of the symmetriral component, and induction motors whirh have a very rapid rate of dccay of the symmetrical component of current. When all these currents are added, the total symmetrical short-circuit rurrent is typical of that shown a t the bottom of Fig. 1.23. The magnitude of the first few cycles of the t,otal symmetrical shortcircuit, 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 t o 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 appliration. The problem of doing this has been simplified by standardized procedures to a poiut xhere t o determine the rms asymmetriral current one need only divide t,he line-to-neutral roltage by the proper reactance


23

RG. 1.23 Tracer of orcillogramr of rymmetrical short-circuit currents from utility, panerator, synchronous motors, and induclion motors. The shape of the total combined currents is illurtmted by the bottom hace.

FIG. 1.24 Arymmelrical short-circuit current from dl sources illustrated in Fig. 1.23 plus d-c component.

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-circuitcurrent 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 current 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. The standard indoor oilless power circuit breakers as used in metal-clad switchgear will be used here t o 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 a t 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 circuitbreaker contacts part, the current decreases in magnitude because of the


25

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 circuit breakers: (1) the momentary rating or its ability to withstand mechanical stresses due to high short-circuit current and (2) the interrupting rating or its ability t,o interrupt the flow of short-circuit current within its interrupting element. What Comprises the Circuit-breaker-rating Structure. Circuitbreaker-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,rical Engineers (AIEE) standards he referred to. To illustrate the various factors that comprise the circuit-breakerrating 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 sake of completeness the General Electric Company nomenclature is used in this column. 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 4800volt circuits) 250 = interrupting rating in mva a t 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 a t which the circuit 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% 2500.) Some utility syst.ems operate a t 1330 X volts near the substation. 4. Minimum operating kv a t 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-amperesi.

(see Appendix).

t

16


I !

I

I (a/

t


27

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 and other surges. This is applied with an impulse generator as a 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 current that a circuit breaker will withstand including short-circuit cnrrents 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 mechanical stresses produced by the very large offset first cycle of the shortcircuit 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 proportioned 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 4 sec 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 t o 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 a t which the circuit 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 a t any operating voltage. 11. Amperes a t rated voltage (35,000): the maximum total rms amperes which the circuit breaker will interrupt a t 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. In this example, this current is 37,500 amp. At minimum operating voltage, 3.85 kv, this corresponds to 250 mva, and, for example, a t a voltage of 2.3 kv this corresponds to 150mva. The circuit breaker will not interrupt this much current a t all voltages, i.e., i t 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 ) a t 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 proportional to the reduction in operating voltage below the value of column 4. 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 interrupting-and Momentary-duty Basis? In so far as applying power cir-

cuit breakers on an interrupting-duty basis is concerned i t 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 a t any voltage. This limit is significant only 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 interrupting amperes limit the mva to values less than the rnva rating. 3. Maximum interrupting rating amperes should never be exceeded

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 a t rated mva). 4 Momentary current should never be exceeded a t 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 contribution 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. How to Check Momentary Duty on Power Circuit Breakers. Siuce the short-circuit current is maximum a t the first half cycle, the short-circuit current must be determined a t the first half cycle to determine the maximum momentary duty on a circuit breaker. To determine the short-circuit current a t the first half cycle, it is necessary 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 a t 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 outlined 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. How to Check Interrupting Duty on Power Circuit Breakers. To check the interrupting duty on a power circuit breaker, the short-circuit current should be determined a t 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 circuitbreaker operating speed. The fewer cycles required for the circuitbreaker 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 shortcircuit calculations, for the purpose of determining interrupting duties, the relays are assumed to operate instantaneously. To account for

SEPES-DIVEN SEN-RIO-EIELI', tCA30

1HIGH VOLTAGE INCOMING LINE


2400 4160 4800 VOLT INCOMING L I N E FROM UTILITY

$

o,:4600 V BUS A6,0

(0)

T O P L A N T LOAD NO GENERATION IN THE P L A N T

TO P LANT L O AD NO GENERATION IN THE P L A N T

(b)

13.6 KVU U

u.-L

USE 1.6 MULTIPLYING FACTOR NO GENERATION ON THIS BUS NO GENERATION

2400, 4160 OR

(C)

TO LOAD

FIG. 1.25 One-line diogrom of carer where the multiplying factor 1.5 may be used on circuits rated less than 5 h.

c

,,

,..

.:

.. ..

...

,


31

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 a t 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 subtransient to 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 interrupting 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 shortcircuit value reaches 500 mva. High-voltage Fuses. High-voltage fuses are either of the currentlimiting 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 a t 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-

u w

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

1 Generators. 1

I0

IEight cycle or slower (general case). Rva cycle..

IAbove 600 volt, Any ploee where symmetricmi short-circuit kva i s loss than 500 mva

1I .O1.1

frequency changers

IInterrupting duty

w

a

C

2i i

..............................

.......... Above 600 wlh

Subtransient Subtransient

Momentary duty

Generol GOSO.. Lr than 5 k.. a

........................... ..........................

s z

s

Above 600 volt) 601 to 5000 volh

Near generoting station Remote from generating dolion (X/R rotio l e u thon I1 0 High-voltaqe Fuses

1.6 1.5


5Three-phose I n o interrupting duly

All typos, including dl wrront-limiting fuses.

.... Above 600 wih... Above 600 volt'

Anywhere in system

I .O

Subhqndent

1

Transient

1

Neglect

All types, including dl current-limiting fuses.. Non-current-limiting lypes only..

............. 601 to 15,000 wlh

1

I

Maximum rms ampere interrupting duty Anywhere in system Remote from generoting %to. tion ( X / R mtio leu lhm 41 1.6 1 .?

i

Subtronsient Svbtronrient Svbwmrient Subwoniiont Subhmrient Subtransient

i

i

All h e p o w e r ratings..

....................

2400 and 4i60YWlh

Anywhere in system

1.0

All horsepower rotingr..

....................

2400 and 4160YYolh

Anywhere in system

I .6

CIrmit breaker w conladm l y p e . .

...........

601 10 5000 volts

Cirwit b r w b r or contocto~ lype. Clrcvit b r e e b r or contartor type..

............ 601 to MOO volts ........... 601 lo 5000 volts

0bywhere in system temote from gener.ting 1 . 1 lion lX/R ratio leis than 101

1.6 1.5

Subtransient Subtrmdent Subtransient Subtrmdent


8R 0m

Apparatus. 600 Volts and Below Interrupting or momentary duty Air circuit breakers or breaker-contactor combino. lion motor stoners.. Low-voltacp furas or fused combination motor

z

.................... Slarte" ...............................

600 volts and below Anywhere in system600 volt* and below Anywhere in system

I .251 .25

Subtransient Subtianrient

Svbtronrienl

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 application in systems with power circuit breakers, some fuses are given interrupting 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 I S REACHED

1 FIG. 1.26 Grophic sxplonotion of the current-limiting action of current-limiting fuses. See Fig. 1.27 for method o determining available short-circuit current. f

SHORT-CIRCUIT-CURRENT CAKULATING PROCEDURES

35

These three-phase mva ratings have been selected so they will line u p 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-voltoge M o t o r Starters. 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.LOW-VOLTAGE CIRCUIT PROTECTIVE EQUIPMENT (600 VOLTS A N D BELOW)

low-voltage Air Circuit Breokers. The present designs of low-voltage air circuit breakers differ from those of high-voltage power circuit breakers because they are substantially instantaneous in operation a t 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 a t the first cycle after short circuit and withstanding the mechanical forces of that rurrent. It is necessary to calculate the current a t 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 lowvoltage circuits as in some high-voltage circuits. This is due to the generally 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 maximum phase. The generally lower ( X / R ) ratio and the use of an average d-c component 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 t o 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 systems 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-voltage air 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-circuitbreaker 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 M o t o r Starters. Low-voltage motor starters are of two types: those using fuses and those using air circuit breakers for shortcircuit 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 same basisas fuses in so far as short-circuit current is concerned.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 shortcircuit current is determined by assuming that the protective device is shorted out by a bar of zero impedance (Fig. 1.27). The short-circuit/


37

current which flows in such a circuit is commonly called available shortcircuit 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 shortcircuit 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. I t 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.

0MOTORS

GENERATOR

TRANSFORMER

CABLE

SHORT ClRCUlTED 8 1 J UMPER OF Z E R O IMPEDANCE

CABLE SHORT

CIRCUIT

FIG. 1.27 Connections

for determining available short-circuit current for testing rhort-

circuit protective devices.

38


HOW TO MAKE A SHORT-CIRCUIT STUDY FOR DETERMINING SHORT-CIRCUIT CURRENTFORMULAS FOR SHORT-CIRCUIT STUDY'

1. Changing ohms to per cent ohms, etc.:

Per cent (%) ohms reactance Per-unit

= =

(90 ohms reactance

(ohms reactance) (kva.base) (1.1) (kvt)*(lO) (ohms reactance)(kva base) (kv)*(1000) (1.2)

[see Eq. (1.34)] Ohms reactance=

( % reactance)(kv)2(10)=

Per-unit ohms reactance

kva base per cent ohms reactance 100

(1.3) (1.4)

2. Changing per cent or per-unit ohms reactance from one kva base to another:

% ohms reactance on kva base 2 - kva base 2 X (% ohms reactance on base 1) (1.5) kva base 1 9f reactance on kva base 2 - kva base 2 X (% ohms reactance on kva base 1) (1.36) kva base 13. Converting utility-system reactance to per cent or per-unit ohms reactance on kva base being used in study: a. If given in per cent ohms reactance on a kva base different than that used in the study, convert according to Eq. (1.5). b. If given in short-circuit kva, convert to per-unit ohms thus:

kva base used in reactance diagram (1.6) short-circuit kva of utility system c. If given in short-circuit amperes (rms symmetrical), convert t o perunit ohms thus:

9i reactance

=

Yi reactance =

kva base used in reactance diagram (short-circuit current) ( d $ ) ( k v rating of system)

(1.7)

d. If only the kva interrupting rating of the incoming line breaker is known,* See pp. 54 to 57 for more prr-unit formulas

1 kv

= line-to-line kilovolts.

SHORTT-CIRCUIT.CURRENTCALCULATING PROCEDURES

39

9f ohms reactance-

kva base used in reactance diagram kva interrupting rating of incoming line breaker The exact kva base of a motor=

(1.8)

4. Determining kva base of motors:

EI 4 3

(1.9)

where E = name-plate voltage rating I = name-plate full-load current rating When motor full-load currents are not known, use the following kva bases: Induction motors: kva base = horsepower rating (1.10) 0.8-power factor synchronous motor: (1.11) kva base = 1.0 (horsepower rating) 1.0-power factor synchronous motors: (1.12) kva base = 0.80 (horsepower rating) 5. Changing voltage base when ohms are used: Ohms on basis of voltage 1-

')* X (ohms on basis of voltage 2) (voltage 2)2

(1.13)

In Eqs. (1.1) to (1.4), ohms impedance or ohms resistance may be substituted for ohms reactance. The final product is then per-unit or per cent ohms impedance or resistance, respectively. 6 . Determining the symmetrical short-circuit kva: Symmetrical short-circuit kva=~

% X*

(kva base)

(1.14) (1.15)(1.16)

- y? -~

'& base) (kva

(line-to-neutral voltage)2 ohms reactance X 1000 kv2 X lo00 ohms reactance 7. Determining the symmetrical short-circuit current: (100) (kva base) Symmetrical short-circuit current = (% X*)(v%(kvt) kva base (% X*)(&)(kvt) k v t X lo00 ( d ) ( o h m s reactance) * X = reactance or impedanoe. t kv = line-&line kilovolts.= 3

(1.16a)

.

(1.17) (1.18) (1.19)

TABLE 1.3 Factor ( K ) to Convert Ohms to Per Cent or Per-unit Ohms for Three-phase Circuits*Base kvo

0 L

lootP r .

1 50Per-""it Per centPer-""it

200Per cant Por-un1t Per cent

300

- __Per-""it Per cent

500Per-""itv)

c*nt

216Y/125 240 480

'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

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

321.5 260.4 65.21 4.166 2.604 0.808 0.651

3.215 2.604 0.6521 0.4166 0.02604 0.00808

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

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

t3 4

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

6.43 5.21 1.302 0.833 0.0511 0.0172 0.01302 0.0063 0.00579 0.00247 0.00226 0.002080.001 92

071 868 217 I38 8.68 2.88

- 2 0.718.68 2.17 1.38 0.0868 0.0288 0.0217 0.0105 0.00965 0.00413 0.00377 0.00347 0.0032 0.00286 0.00262 0.00045 0.000355 0.000236 0.000105

I

600 2,400 41 60 .4,800 6.900 7,200 l1,OOO 11.500 12,000 12,500 13.200 13,800 23,000 37.4M) 46,00069,OCU

2 K E 2

0.3150.289 0.123 0.113 0.104 0.096 0.086 0.0787 0.0283 0.0107 0.00708

0.00651 0.0031 5 0.002890.00123 0.00113 0.00104 0.00096 0.00086 0.000787 0.000283 0.000106 0.0000708

2.171.05 0.965 0.413 0.377 0.347 0.32 0.286 0.262 0.045 0.0355 0.0236 0.0105

Bf

I

n

250

0,00172 0.001 57 0.000547 0.00021 3 0.0001 41 0.000063

6 c

R v,

0.0021 2 - 0.0000212=

-

0.0031 5

0.000031 5

* For per-unit, K

kva base , kva base For per cent, K = kv' X 1wO kv' X 10

kv = line-to-line kilovolts

t To determine multiplying factors far any other base use figures under 100-kvs base columns multiplied by new base in kva,100


41

8. Determining the asymmetrical short-circuit current:Asymmetrical short-circuit current = (symmetrical current) (multiplying factor) Asymmetrical short-circuit kva = (symmetrical kva) (multiplying factor)DIAGRAMS

(1.20)

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, synchronous condensers, rotary converters, etc., and all significant circuit elements, such as transformers, cables, circuit breakers, etc. (Fig. 1.28). M a k e an Impedance or Reactance Diagram. The second step is to make an impedance or reactance diagram showing all significant reactances and resistances (Pig. 1.29). In the following pages this will beGENERATOR C

I

UTILITY SYSTEM TRANS

D

GENERATOR

CABLE E SHORT CIRCUIT LARGE MOTOR CABLE J

480 VOLT MOTORS

FIG. 1.28

e diagram c

, typical large industrial power system.

H

INFINITE BUSES

-SHORT

CIRCUIT CURRENT GOES THROUGH HERE

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. I n most industrial 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-circuitcurrent calculations will suffice for application of short-circuit protective devices in most industrial systems. Unbalanced Short Circuits in Large Power 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 transformer 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 highvoltage 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. Bolted Short Circuits Only Are Considered. 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


43

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 BREAKERS ON THIS BUS $ WR E : S FOR SHORT CIRCUIT

1 TA&?? Y T T - 3&

?;+

r

y

MAX. DUTY FOR THESE BREAKERS OCCURS FOR SHORT CIRCUITHERE

rx -

* +

FIG. 1.30

Location of faults for maximum Short-circuit duty on circuit breakers.

44


terminals of the circuit breaker, etc. (Fig. 1.30). These devices, if properly applied, should be capable of opening the maximum shortcircuit current that can flow through them. Therefore, only one shortcircuit 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. Resistances are included for certain items. 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 lineto-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. U s e R e a c t a n c e s of All S i g n i f i c a n t Circuit E l e m e n t s . 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-voltage bus runs, etc., may have a significant hearing on the magnitude of total shortcircuit current. As a general guide, the reactance of the low-voltage secondary-switchgear section in load-center unit substations with closely coupled transformers 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. I n systems of more than about 1000 kva on one bus a t 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. I n systems of more than about 3000 kva on one bus a t 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

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 care is not used, the calculations will result in a f 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 care is used in including all f 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. When Is Resistance Considered? 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 theGENERATOR

OF-THESE CIRCUIT ELEMENTS. IN GENERAL USF REACTANCE AND RESISTANCE OF THESE

___

-. . . -. 1100 FT. 101

---(20 FT

SHORT CIRCUIT CURRENT CONSIDERING REACTANCE ONLY :20800 AMPERESSHORT CIRCUIT CURRENT CONSIDERING REACTANCE OF A LL PARTS PLUS RESISTANCE OF COW VOLTAGE CABLE = 11500 4MPERES.

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. I n 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 nSHORT CIRCUIT CURRENT USING REACTANCE ONLY = 51000 AMPERES, SHORT CIRCUIT CURRENT USING REACTANCE PLUS RESISTANCE OF T I E CIRCUIT= 41000 AMPERES.

T I E CIRCUITS 208 Y / l Z O V O L T S .

200 FT2- 250 M,CM 3 CONO. CABLES ~~~~~T I N PARALLEL

200 F T

FIG. 1.32

One-line diogrtlm of low-voltage secondary network system showing effect of resistance o cable tie circuits. f

Where to Use Exact Multiplying Factors. I n 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 horsepower 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. Motors Roted 600 Volts and Below. I n that portion of the power system operating at 600 volts or less, there are generally numerous small


A?

motors, i.e., under about 50 hp. I t becomes impractical to represent each small motor in the impedance diagram. These motors are constantly 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 a t 600 volts or less. 1. In systems rated 240, 480, or 600 volts a t each generator and/or transformer bus, assume that the maximum horsepower of motors runniug a t any one time is equal to the combined kva rating of the stepdown 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 a t each generator and/or transformer bus that the maximum horsepower of motors running a tREbCTbNCE OF UTILITY SYSTEM REbCTbNCE OF 7 5 0 K V b TRbNSF.

TO UTILITY SYSTEM

QOW, OR5.,s

25 % REbCTbNCE OF EQUIVALENTMOTOR

0.25% OR

5.5%

IMPEObNCE O I b G R b M 750 K V b BASE SHORT CIRCUIT EQUIVALENT MOTOR 750 KVb

SHORT CIRCUIT

El hKVATO UTILITY SYSTEM REbCTbNCE OF UTILITY SYSTEM REbCTbNCE OF 7 5 0 KVb TRbNSF. EQUIVILENT MOTOR 375 K V b IMPEObNCE OIbGRbM 750 K V b BASE 2 0 8 Y / 1 2 0 VOLT SYSTEMS

240, 480, 600 VOLT SYSTEMS

50 % REACTbNCE OF EQUIVALENT MOTOR

0.50% OR

FIG. 1.33

Oiagromr illustrating how to include motors in low-voltage radial systems.

40


any one time is equal t,o 50 per cent of the combined rating of all stepdown 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. I n the generalized rases referred t o in paragraphs 1 and 2 , no specific ratio of induction t o synchronous motors or no specific number of motors which prcduce unusually high short-circuit current,s has been set fort,h. T o account for these variables, a n average motor reitctance ihcluding leads is assumed t o 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 induction mot,ors and 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. T h e reactances given in Table 1.14 are based on motor kva ratings and not supply transformer or generator ratings.

750 KVA

T-480 VOLTS

A 500 KVA

750 KVA

500 KVAv

EQUIVALENT MOTORS WOULD BE 250 KVA AND 375 K VA FOR 280Y/120 VOLT SECONDARY SYSTEM

FIG. 1.34 rvrternr.

Diagram illustrating how lo include motors in lowvoltage secondary network


49

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,o such a bus is geiierally much greater than the kva rating of the supply transformers and generators. Hovever, allowing for diversity, generally the total comhitied horsepower rat,ing of all mot,ors running a t one time ix-ould trot produce short-circuit currents in excess of the values obtained when using the ahore assumptions. I n systems of 000 volts or Icss, the large motors (i,e., mot,ors 011 t,he order of several hundred horsepomerj are usually few i n number and represent only a small portion of the tot,al connected horsepower; therefore, these larger motors are generally lumped in with the smaller motors and the complete group is represented as one equivalent motor i t i the impedance diagram. Synchrouous and induction motors need not be segregated when combining the motors in these low-voltage systems, because lorn-voltage air circuit breakers operr so fast that only the current flow duritig 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 t o 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 t o 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 study. Whether considering motors individually or in groups, regardless of voltage rating of the motors, it is necessary t o 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 , l l ) , or (1.12), when the full-load current is not known. The latter equations are used when considering a single reactance t o represent a group of miscellaneous motors.

50


I n high-voltage systems, complete motor data may not be available. Lacking these data, the connected horsepower is assumed to he equal t o the generator and/or transformer capacity supplying a given highvoltage bus. If the reactance of the leads between the transformer and/or generator 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.

7

are treated as if they comprised a generator 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 every generator and motor or groups of motors and utility supply FIG. 1.35 One-line representation by a rea

d beeman - industrial power systems handbook

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