top of page # Group

Public·29 members

# Industrial Power System Handbook Donald Beeman Pdf 27

Short-circuit-current CalculatingProceduresFUNDAMENTALS OF A-CSHORT-CIRCUIT CURRENTSThe determination of short-circuit currentsin power distribution systems is just as basic and important as thedetermination of load currents for the purpose of applying circuitbreakers, fuses, and motor starters. The magnitude of theshoncircuit current is often easier to determine than the magnitudeof the load current. Calculating procedures have been so greatlysimplified compared with the very complicated procedures previouslyused that now only simple arithmetic is required to determine theshort-circuit currents in even the most complicated powersystems.SHORT-CIRCUIT CURRENTS AND THEIR EFFECTS

## Industrial Power System Handbook Donald Beeman Pdf 27

If adequate protection is to he provided for a plant electricsystem, the size of the electric power system must also beconsidered to determine how much short-circuit current i t willdeliver. This is done so that circuit breakers or fuses may heselected with adequate interrupting capacity (IC). Thisinterrupting capacity should be high enough to open safely themaximum short-circuit current which the power system can cause toflow through a circuit breaker if a short circuit occurs in thefeeder or equipment which it protects. The magnitude of the loadcurrent is determined by the amount Of work that is being done andhears little relation to the size of the system supplying the load.However, the magnitude of the short-circuit current is somewhatindependent of the load and is directly related to the size orI

capacity of t,he power source. The larger the apparatus whichsupplies electric power t o the system, the greater theshort-circuit current will be. Take a simple case: A 440-voltthree-phase lo-lip motor draws about 13 amp of current a t fullload and will draw only this amount whether supplied by a 25-kva ora 2500-kva transformer bank. So, if only thc load currcnts arcconsidered when selecting motor branch circuit breakers, a 15- or20-amp circnit, breaker wnuld he specified. However, the size oft,he power system back of the circuit breaker has a real bearing onthe amount of the short,-circuit,current. which can flow as aresult of a short circuit on the load side of the circuit breaker.Hence, a much larger circuit breaker would be required to handlethe short-circuit current from a 2500-kva bank than from a 25-kvabank of transformers. A simple mathematical example is shown inFig. 1.1. These numbersMUST BE CAPABLE OF INTERRUPTING1000AMPERES

Synchronous motors are constructed substantially likegenerators; i.e., they have a field excited by direct current and astator winding in which alternating current flows. Normally,synchronous motors draw a-c power from the line and convertelectric energy to mechanical energy. However, the design of asynchronous motor is so much like that of a generator that electricenergy can be produced just as in a generator, by driving thesynchronous motor with a prime mover. Actually, during a systemshort circuit the synchronous motor acts like a generator anddelivers shortcircuit current to the system instead of drawing loadcurrent from it (Fig. 1 4 . .) As soon as a short circuit isestablished, the voltage on the system is reduced to a very lowvalue. Consequently, the motor stops delivering energy to themechanical load and starts slowing down. However, the inertia ofthe load and motor rotor tends to prevent the motor from slowingdown. In other words, the rotating energy of the load and rotordrives the synchronous motor just as the prime mover drives agenerator.

The synchronous motor then becomes a generator and deliversshortcircuit current for many cycles after the short circuit occurson the system. . Figure 1 5 shows an oscillogram of the currentdelivered by a synchronous motor during a system short circuit. Theamount of current depends upon the horsepower, voltage rating, andreactance of the synchronous motor and the reactance of the systemto the point of short circuit.

The inertia of the load and rotor of an induction motor hasexactly the same effect on an induction motor as on a synchronousmotor; i.e., it drives the motor after the system short circuitoccurs. There is one major difference. The induction motor has nod-c field winding, but there is a flux in the induction motorduring normal operation. This flux acts like flux produced by thed-c field winding in the synchronous motor. The field of theinduction motor is produced by induction from the stator ratherthan from the d-c winding. The rotor flux remains normal as long asvoltage is applied to the stator from an external source. However,if the external source of voltage is removed suddenly, as it iswhen a short circuit occurs on the system, the flux in the rotorcannot change instantly. Since the rotor flux cannot decayinstantly and the inertia drives the induction motor, a voltage isgenerated in the stator winding causing a short-circuit current toflow to the short circuit until the rotor flux decays to zero. Toillustrate the short-circuit current from an induction motor in apractical case, oscillograms were taken on a woundrotor inductionmotor rated 150 hp, 440 volts, 60 cycles, three phase, ten poles,720 rpm. The external rotor resistance was short-circuited in eachcase, in order that the effect might he similar to that which wouldhe obtained with a low-resistance squirrel-cage induction motor.Figure 1.6 shows the primary current when the machine is initiallyrunning light and a solid three-phase short circuit is applied a ta point in the circuit close to its input (stator) terminals a ttime TI. The current shown is measured on the motor side of theshort circuit; so the shortcircuit current contribution from thesource of power does not appear, but only that contributed by themotor. Similar tests made with the machine initially running a tfull load show that the short-circuit current produced

by the motor when short-circuited is substantially the same,regardless of initial loading on the motor. Note that the maximumcurrent occurs in the lowest trace on the oscillogram and is aboutten times rated full-load current. The current vanishes almostcompletely in four cycles, since there is no sustained fieldcurrent in the rotor to provide flux, as in the case of asynchronous machine. The flux does last long enough to prodnceenough short-circuit current to affect the momentary duty oncircuit breakers and the interrupting duty on devices which openwithin one or two cycles after a short circuit. Hence, theshort-circuit current produced by induction motors must heconsidered in certain calculations. The magnitude of short-circuitcurrent produced by the induction motor depends upon thehorsepower, voltage rating, reactance of the motor, and thereactance of the system to the point of short c. "cuit. The machineimpedance, effective a t the time of short circuit, cmespondsclosely with the impedance a t standstill. Consequently, the iiitial symmetrical value of Short-circuit current is approximatelyequnl to the full-voltage starting current of themotor.TRANSFORMERS

In the usual industrial power systems the applied or generatedvoltages are of sine-wave form. When a short circuit occurs,substantially s i n e wave short-circuit currents result. Forsimplicity, the following discussion assumes sine-wave voltages andcurrents. In ordinary power circuits the resistance of the circuitis negligible compared with the reactance of the circuit. Theshort-circuit-current power factor is determined by the ratio ofresistance and reactance of the circuit only (not of the load).Therefore the short-circuit current in most power circuits lags theinternal generator voltage by approximately 90" (see Fig. 1.13).The internal generator voltage is the voltage generated in thestator coils by the field flux. If in a circuit mainly containingreactance a short circuit occurs at the peak of the voltage wave,the short-circuit current would start at zero and trace a sine wavewhich would be symmetrical ahout the zero axis (Fig. 1.14). This isknown as a symmetrical short-circuit current. If in the samecircuit (i.e., one containing a large ratio of reactance toresistance) a short circuit occurs at the zero point of the voltagewave, the current will start a t zero but cannot follow a sine wavesymmetrically about the zero axis because such a current would bein phase with the voltage. The wave shape must be the same as thatof voltage hut 90' behind. That can occur only if the current isdisplaced from the zero axis, as shown in Fig. 1.15. In thisillustration the current is a sine wave and is displaced 90' fromthe voltage wave and also is displaced from the zero axis. The twocases shown in Figs. 1.14 and 1.15 are extremes. One shows asymmetrical current and the other a completely asymmetricdcurrent.

The total symmetrical short-rirruit current is made up ofcurrents from several sourves, Fig. 1.23. At the top of the figureis shown the shortcircuit current from the utility. This act,uallycomes from ut,ility generators, but generally the industrial systemis small and remote electrically from the utility generators sothat the Symmetrical short-rircuit current is substant,iallyconstant,. If there are generators in the indust,rial plant, thenthey cont,ribute a symmet,rical short-circuit rurreiit which forall practical purposes is constant over the first few cycles. Thereis, however, a slight decrement, as indicated in Fig. 1.23. Theother sources are synchronous motors which act something like plantgenerators, except that t,hey have a higher rate of decay of thesymmetriral component, and induction motors whirh have a very rapidrate of dccay of the symmetrical component of current. When allthese currents are added, the total symmetrical short-circuitrurrent is typical of that shown a t the bottom of Fig. 1.23. Themagnitude of the first few cycles of the t,otal symmetricalshortcircuit, current is further increased by the presence of a d-ccompouent, Fig. 1.24. The d-c component, offsets the a-c ware and,therefore, makes it asymmetrical. The d-c component decays t o zerowithin a few cycles in most indust,rial power systems. It is thistotal rms asymmetrical short-circuit current, as shown in Fig.1.24, that must he determilied for short-circuit protective-dericeappliration. The problem of doing this has been simplified bystandardized procedures to a poiut xhere t o determine the rmsasymmetriral current one need only divide t,he line-to-neutralroltage by the proper reactance