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<p>Industrial Power Systems</p><p>Handbook</p><p>D O N A L D BEEMAN, Editor</p><p>Manager, Industriaf P w e r Engineering</p><p>Industrial Engineering Seclwn</p><p>General Electric Company, Schenectady, New Yorlc</p><p>FIRST EDIT ION</p><p>McGRAW-HILL BOOK COMPANY, INC.</p><p>1955 New York Toronto London</p><p>Ch.UPh?r 1 by Donald Beeman, Alan Graeme Darling,</p><p>and R. H. Kaufmann</p><p>Short-circuit-current Calculating</p><p>Procedures</p><p>FUNDAMENTALS OF A-C SHORT-CIRCUIT CURRENTS</p><p>The determination of short-circuit currents in power distribution sys-</p><p>tems is just as basic and important as the determination of load currents</p><p>for the purpose of applying circuit breakers, fuses, and motor starters.</p><p>The magnitude of the shoncircuit current is often easier to determine</p><p>than the magnitude of the load current.</p><p>Calculating procedures have been so greatly simplified compared with</p><p>the very complicated procedures previously used that now only simple</p><p>arithmetic is required to determine the short-circuit currents in even the</p><p>most complicated power systems.</p><p>SHORT-CIRCUIT CURRENTS AND THEIR EFFECTS</p><p>If adequate protection is to he provided for a plant electric system, the</p><p>size of the electric power system must also be considered to determine</p><p>how much short-circuit current i t will deliver. This is done so that cir-</p><p>cuit breakers or fuses may he selected with adequate interrupting capac-</p><p>ity (IC). This interrupting capacity should be high enough to open</p><p>safely the maximum short-circuit current which the power system can</p><p>cause to flow through a circuit breaker if a short circuit occurs in the</p><p>feeder or equipment which it protects.</p><p>The magnitude of the load current is determined by the amount Of</p><p>work that is being done and hears little relation to the size of the system</p><p>supplying the load. However, the magnitude of the short-circuit current</p><p>is somewhat independent of the load and is directly related to the size or</p><p>I</p><p>2 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>capacity of t,he power source. The larger the apparatus which supplies</p><p>electric power to the system, the greater the short-circuit current will be.</p><p>Take a simple case: A 440-volt three-phase lo-lip motor draws about</p><p>13 amp of current a t full load and will draw only this amount whether</p><p>supplied by a 25-kva or a 2500-kva transformer bank. So, if only thc</p><p>load currcnts arc considered when selecting motor branch circuit break-</p><p>ers, a 15- or 20-amp circnit, breaker wnuld he specified. However, the</p><p>size of t,he power system back of the circuit breaker has a real bearing on</p><p>the amount of the short,-circuit, current. which can flow as a result of a</p><p>short circuit on the load side of the circuit breaker. Hence, a much</p><p>larger circuit breaker would be required to handle the short-circuit current</p><p>from a 2500-kva bank than from a 25-kva bank of transformers.</p><p>These numbers A simple mathematical example is shown in Fig. 1.1.</p><p>MUST BE CAPABLE OF INTERRUPTING 1000 AMPERES</p><p>El I O O V</p><p>100 A</p><p>~ ~ 1 0 . 1 OHMS</p><p>MOTOR LOAD</p><p>CURRENT</p><p>5 AMP</p><p>APPARENT</p><p>IMPEDANCE</p><p>20 OHMS</p><p>E I 00 SHORT CIRCUIT CURRENT = - : - = 1000- AMPERES</p><p>Z T 0.1</p><p>MUST BE CAPABLE OF INTERRUPTING 10,000 AMPERES</p><p>w</p><p>I000 A</p><p>2 1 = 0.01 OHMS</p><p>MOTOR LOAD</p><p>CURRENT</p><p>5 AMP</p><p>FIG. 1.1</p><p>circuit-current magnitude than load.</p><p>Illustrotion showing that copocity of power source has more effect on rhort-</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 3</p><p>have been chosen for easy calculation rather than a representation of</p><p>actual system conditions.</p><p>The impedance, limiting the flow of load current, consists mainly of</p><p>the 20 ohms apparent impedance of the motor. If a short circuit occurs</p><p>at F , the only impedance t o limit the flow of short-circuit current is the</p><p>transformer impedance (0.1 ohm compared with 20 ohms for the motor);</p><p>therefore, the short-circuit current is 1000 amp, or 200 times as great as</p><p>the load current. Unless circuit breaker A can open 1000 amp, the</p><p>short-circuit current will continue to flow, doing great damage.</p><p>Suppose the plant grows and a larger transformer, one rated a t 1000</p><p>amp, is substituted for the 100-amp unit. A short circuit a t F , (bottom</p><p>in Fig. 1.1) will now be limited by only 0.01 ohm, the impedance of the</p><p>larger transformer. Although the load current is still 5 amp, the short-</p><p>circuit current will now he 10,000 amp, and circuit breaker A must be</p><p>able t o open that amount. Consequently it is necessary to coiisider the</p><p>size of the system supplying the plant as well as the load current, to be</p><p>sure that circuit breakers or fuses are selected which have adequate</p><p>interrupting rating for stopping the flow of the short-circuit current.</p><p>Short-circuit and load currents are analogous t o the flow of xvater in a</p><p>hydroelectric plant, shoivn in Fig. 1.2. The amount of water that flows</p><p>under normal conditions is determined by the load on the turbines.</p><p>Within limits, i t makes little difference whether the reservoir behiiid the</p><p>dam is large or small. This flow of water is comparable to the flow of load</p><p>current in the distribution system in a factory.</p><p>On the other hand, if the dam breaks, the amount of water that will</p><p>flow will depend upon the capacity of the reservoir and will bear little</p><p>relation to the load on the turbines. Whether the reservoir is large or</p><p>small will make a great difference in this case. This flow of water is</p><p>comparable t o the flow of current through a short circuit in the distribu-</p><p>tion system. The load currents do useful work, like the water that flows</p><p>down the penstock through the turbine water wheel. The short-circuit</p><p>currents produce unwanted effects, like the torrent that rushes madly</p><p>downstream when the dam breaks.</p><p>SOURCES OF SHORT-CIRCUIT CURRENTS</p><p>When determining the magnitude of short-circuit currents, it is</p><p>extremely important that all sources of short-circuit current he considered</p><p>and that the reactance characteristics of these sources be known.</p><p>There are three basic sources of short-circuit current:</p><p>1. Generators</p><p>2. Synchronous motors and synchronous condensers</p><p>3. Induction motors</p><p>4 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>All these can feed shorecircuit current into a short circuit (Fig. 1.3).</p><p>Generators are driven by turbines, diesel engines, water wheels, or</p><p>other types of prime movers. When a short circuit occurs on the circuit</p><p>fed by a generatar, the generator continues to produce voltage because the</p><p>field excitation is maintained and the prime mover drives the generator</p><p>at substantially normal speed. The generated voltage produces a short-</p><p>circuit current of a large magnitude which flows from the generator (or</p><p>generators) to the short circuit. This flow of short-circuit current is</p><p>limited only by the impedance of the generator and of the circuit between</p><p>the generator and the short circuit. For a short circuit a t the terminals</p><p>of the generator, the current from the generator is limited only by its own</p><p>impedance.</p><p>FIG. 1.2</p><p>the hydroelectric plant.</p><p>Normal load and short-circuit currents are analogous to the conditions shown in</p><p>SHORT-CIRCUIT-CURRENT ULCULATlNG PROCEDURES 5</p><p>METAL CLAD SWITCHGEAR</p><p>SHORT CIRCUIT</p><p>CURRENT FROM</p><p>INDUCTION</p><p>MOTOR</p><p>FIG. 1.3 Generators, synchronous motors, and induction motors all produce short-circuit</p><p>current.</p><p>HOW SYNCHRONOUS MOTORS PRODUCE SHORT-CIRCUIT CURRENT</p><p>Synchronous motors are constructed substantially like generators; i.e.,</p><p>they have a field excited by direct current and a stator winding in which</p><p>alternating current flows. Normally, synchronous motors draw a-c</p><p>power from the line and convert electric energy to mechanical energy.</p><p>However, the design of a synchronous motor is so much like that of a</p><p>generator that electric energy can be produced just as in a generator, by</p><p>driving the synchronous motor with a prime mover. Actually, during a</p><p>system short circuit the synchronous motor acts like a generator and</p><p>delivers shortcircuit current to the system instead of drawing load cur-</p><p>rent from it (Fig.</p><p>or line-to-line short circuits than for the</p><p>three-phase short circuits. Thus, the simple three-phase short-circuit-</p><p>current calculations will suffice for application of short-circuit protective</p><p>devices in most industrial systems.</p><p>In some very</p><p>large systems where the high-voltage-system neutral is solidly grounded,</p><p>maximum short-circuit current flows for a single phase-to-ground short</p><p>rircuit. Such a system might be served from a large delta-Y trans-</p><p>former bank or directly from the plant generators.</p><p>Hence the only time that single-phase short-circuit-current calculations</p><p>need be made is on large high-voltage systems (2400 volts and above)</p><p>with solidly grounded generator neutrals or where main transformers</p><p>that supply a plant from a utility are ronnected in delta on the high-</p><p>voltage side (incoming line) and in Y with solidly grounded neutrals</p><p>on the low-voltage (load) side.</p><p>The calculations of unbalanced short-circuit currents in large power</p><p>systems can best be done by symmetrical components, see Chap. 2.</p><p>Normally, generator and large delta-Y transformer secondaries are</p><p>grounded through a reactor or resistor to limit the short-circuit current</p><p>for a single line-to-ground short circuit on the system to letis than the</p><p>value of short-circuit current for a three-phase short circuit.</p><p>Several tests have been</p><p>made to evaluate the effect of arc drop at the point of short circuit in</p><p>reducing the short-circuit-current magnitude. It was felt by some</p><p>engineers that the current-limiting effect of the arc was pronounced.</p><p>These tests showed, however, that for circuit voltages as low as 300 volts</p><p>Unbalanced Short Circuits in Large Power Systems.</p><p>Bolted Short Circuits Only Are Considered.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 43</p><p>there may be no substantial difference in the current that flows for a</p><p>bolted short circuit and when there is an arc of several inches of length.</p><p>These test,s also confirmed modern calculating procedure as an accurate</p><p>method of estimating the short-circuit-current magnitude in systems</p><p>of 600 volts and less.</p><p>.4rcs cannot be counted on to limit the flow of short-circuit currents</p><p>even in louvoltage circuits; so short-circuit-current calculations for</p><p>all circuit voltages are made on the basis of zero impedance at the point</p><p>of short circuit, or, in other words, a bolted short circuit. This materially</p><p>simplifies calculation because all other circuit impedances are linear in</p><p>magnitude, whereas arcs have a nonlinear impedance characteristic.</p><p>At What Point in the System Should the Short Circuit Be Considered</p><p>to Occur? The maximum short-circuit current will flow through a cir-</p><p>cuit breaker, fuse, or motor starter when the short circuit occurs at the</p><p>4160V.</p><p>I I I</p><p>$? MAX.SHORT CIRCUIT DUTY ON $- $EW:RS FOR SHORT CIRCUIT</p><p>BREAKERS ON THIS BUS</p><p>1</p><p>T</p><p>?;</p><p>A&??</p><p>Y T T - 3</p><p>& + * + r y r-x</p><p>MAX. DUTY FOR</p><p>THESE BREAKERS</p><p>OCCURS FOR</p><p>SHORT CIRCUIT</p><p>HERE</p><p>FIG. 1.30 Location of faults for maximum Short-circuit duty on circuit breakers.</p><p>44 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>terminals of the circuit breaker, etc. (Fig. 1.30). These devices, if</p><p>properly applied, should be capable of opening the maximum short-</p><p>circuit current that can flow through them. Therefore, only one short-</p><p>circuit location (at the terminal of the device) need be considered for</p><p>checking the duty on a given circuit breaker, fuse, or motor starter.</p><p>DETERMINING REACTANCES AND RESISTANCES OF CIRCUITS AND MACHINES</p><p>Typical reactances of circuit elements and machines are given at the</p><p>end of this chapter. These</p><p>tables may be used as a basis for assigning values to the various elements</p><p>of the impedance diagram. The reactances and resistances are all line-</p><p>to-neutral values for one phase of a three-phase circuit. Where the</p><p>reactances of a specific motor, generator, or transformer are known,</p><p>these values should he used in lieu of the typical reactances in this</p><p>chapter. The following is a guide to general practice in selecting and</p><p>representing reactances.</p><p>Whether or not</p><p>the reactance of a certain circuit element of a system is significant</p><p>depends upon the voltage rating of the system where the short circuit</p><p>occurs. In all cases, generator, motor, and transformer reactances</p><p>are used. In systems rated above 600 volts, the reactances of short</p><p>bus runs, current transformers, disconnecting switches, circuit breakers,</p><p>and other circuit elements of only a few feet in length are so low that</p><p>they may be neglected without significant error.</p><p>In circuits rated 600 volts or less, the reactances of low-voltage current</p><p>transformers, air circuit breakers, disconnecting switches, low-voltage bus</p><p>runs, etc., may have a significant hearing on the magnitude of total short-</p><p>circuit current.</p><p>As a general guide, the reactance of the low-voltage secondary-switch-</p><p>gear section in load-center unit substations with closely coupled trans-</p><p>formers and secondary switchgear is not significant for all voltages of</p><p>600 volts and below. However, where there are several transformers or</p><p>generators paralleled on one bus, or connections several feet long between</p><p>a single transformer and its switchgear, reactances of the bus connections</p><p>will generally be significant and should be considered in short-circuit</p><p>calculations. In systems of more than about 1000 kva on one bus a t</p><p>208Y/120 or 240 volts, reactance of all circuit components such as short</p><p>bus runs, current transformers, circuit breakers, etc., should be included</p><p>in the short-circuit study.</p><p>I n systems of more than about 3000 kva on one bus a t 480 volts or</p><p>600 volts, reactances of all components such as current transformers,</p><p>circuit breakers, short bus runs, etc., should be considered.</p><p>It should be remembered that the lower the voltage, the more effective</p><p>Resistances are included for certain items.</p><p>U s e R e a c t a n c e s of All S ign i f i can t Circuit Elements.</p><p>SHORT-CIRCUIT-CURRENl CALCUUTING PROCEDURES 45</p><p>a small impedance is in limiting the short-circuit-current magnitude.</p><p>That is why extreme care should he used to include all circuit elements</p><p>in the impedance diagram, particularly for large ZORY/lZO-volt or</p><p>240-volt systems. If care is not used, the calculations will result in a</p><p>value of current far higher than will actually be realized in practice.</p><p>See the example outlined in Figs. 1.46 and 1.47. This often results in the</p><p>adoption of low-voltage switchgear of higher interrupting rating and</p><p>higher cost than are actually required. If care is used in including all</p><p>reactances, the calculated reiults will be close to the short-circuit currents</p><p>obtained in practice. Short-circuit calculations are of most value if</p><p>they reflect accurate answers.</p><p>The resistance of all generators,</p><p>transformers, reactors, motors, and high-capacity buses (above about</p><p>1000-amp rating) is so low, compared with their reactance, that their</p><p>resistance is not considered, regardless of their voltage rating. The</p><p>resistance of all other circuit elements of the high-voltage system (above</p><p>600 volts) is usually neglected, because the resistance of these parts has</p><p>no significant bearing on the total magnitude of short-circuit currents.</p><p>In systems of 600 volts and less the error of omitting resistances of</p><p>all parts of the circuit except cables and small ampere rating buses is</p><p>usually less than 5 per cent. However, the resistance of cable circuits is</p><p>often the predominant part of the total impedance of a cable. When</p><p>appreciable lengths of cable are involved in the circuit through which</p><p>short-circuit current flows in a system of GOO volts or less, the resistance</p><p>as well as the reactance of the cable circuits should be included in the</p><p>When Is Resistance Considered?</p><p>GENERATOR</p><p>_ _ _ -. . . -.</p><p>1100 FT. 101</p><p>AND RESISTANCE OF THESE</p><p>IN GENERAL USF REACTANCE</p><p>OF-THESE CIRCUIT ELEMENTS.</p><p>SHORT CIRCUIT CURRENT CONSIDERING</p><p>REACTANCE ONLY : 20800 AMPERES</p><p>SHORT CIRCUIT CURRENT CONSIDERING</p><p>REACTANCE OF ALL PARTS PLUS</p><p>RESISTANCE OF COW VOLTAGE</p><p>CABLE = 11500 4MPERES.</p><p>----</p><p>(20 FT</p><p>FIG. 1.31 One-line diagram showing effect of resistance in cable circuits.</p><p>46 SHORT-CIRCUIT.CURRENT CALCULATING PROCEDURES</p><p>impedance diagram. The example of Fig. 1.31 shows the error that</p><p>might result in neglecting cable resistance.</p><p>In secondary network systems of 600 volts and less, the resistance as</p><p>well as the reactance of the tie-cable circuits between substation buses</p><p>should be included in the impedance diagram. The example of Fig. 1.32</p><p>shows the effect of cable resistance in reducing short-circuit current in a</p><p>typical industrial network.</p><p>n n</p><p>SHORT CIRCUIT CURRENT USING</p><p>REACTANCE ONLY = 51000 AMPERES,</p><p>SHORT CIRCUIT CURRENT USING</p><p>REACTANCE PLUS RESISTANCE OF</p><p>T IE CIRCUIT= 41000 AMPERES.</p><p>T I E CIRCUITS</p><p>208 Y/ lZOVOLTS.</p><p>200 FT</p><p>2- 250 M,CM</p><p>3 CONO. CABLES ~~~~~T</p><p>I N PARALLEL</p><p>200 F T</p><p>FIG. 1.32</p><p>resistance of cable tie circuits.</p><p>One-line diogrtlm of low-voltage secondary network system showing effect of</p><p>Where to Use Exact Multiplying Factors. In low-voltage systems</p><p>having considerable lengths of cahle, the X / R ratio may be so low that</p><p>the 1.25 multiplying factor would be considerably in error. Hence in</p><p>these systems where resistance is considered, determine the correct</p><p>X / R ratio and then use minimum multiplying factor.</p><p>GUIDE FOR REPRESENTING THE REACTANCE O F A GROUP O F MOTORS</p><p>A group of motors fed from one substation or from one generating</p><p>station bus may range in rating from fractional to several thousand horse-</p><p>power per motor. All motors that are running at the time a short circuit</p><p>occurs in the power system contribute short-circuit current and therefore</p><p>should be taken into consideration.</p><p>In that portion of the power sys-</p><p>tem operating at 600 volts or less, there are generally numerous small</p><p>Motors Roted 600 Volts and Below.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES A?</p><p>motors, i.e., under about 50 hp. I t becomes impractical to represent</p><p>each small motor in the impedance diagram. These motors are con-</p><p>stantly being turned off and on; so it is practically impossible to predict</p><p>which ones will be on the line when a short circuit occurs. Furthermore,</p><p>it would be impractical to obtain the characteristics of each small motor</p><p>and to account for the effect of the impedance of their leads.</p><p>Where more accurate data are not available, the following procedure</p><p>may be used with satisfactory results for representing the combined</p><p>reactance of a group of miscellaneous motors operating a t 600 volts or</p><p>less.</p><p>1. In systems rated 240, 480, or 600 volts a t each generator and/or</p><p>transformer bus, assume that the maximum horsepower of motors</p><p>runniug a t any one time is equal to the combined kva rating of the step-</p><p>down transformer and/or generators supplying that one bus (see Figs.</p><p>1.33 and 1.34).</p><p>2. 10 systems rated 208Y/120 volts, a substantial portion of the load</p><p>usually consists of lights and a lesser proportion of motor load than in</p><p>240-, 480-, or 000-volt systems. Hence in 208Y/120-volt systems where</p><p>more accurate data are not available, assume a t each generator and/or</p><p>transformer bus that the maximum horsepower of motors running a t</p><p>REbCTbNCE QOW,</p><p>TO UTILITY SYSTEM OF UTILITY OR5.,s 0.25% OR</p><p>25 %</p><p>REbCTbNCE OF</p><p>EQUIVALENT</p><p>MOTOR</p><p>SYSTEM</p><p>REbCTbNCE</p><p>OF 750 K V b</p><p>TRbNSF. 5.5%</p><p>IMPEObNCE OIbGRbM</p><p>750 KVb BASE</p><p>SHORT EQUIVALENT MOTOR</p><p>CIRCUIT 7 5 0 K V b</p><p>240, 480, 600 VOLT SYSTEMS</p><p>TO UTILITY SYSTEM</p><p>0.50% OR</p><p>50 %</p><p>REACTbNCE OF</p><p>EQUIVALENT El MOTOR REbCTbNCE</p><p>OF 750 KVb</p><p>TRbNSF.</p><p>REbCTbNCE</p><p>OF UTILITY</p><p>SYSTEM</p><p>EQUIVILENT MOTOR SHORT</p><p>CIRCUIT</p><p>375 K V b IMPEObNCE OIbGRbM hKVA 750 KVb BASE</p><p>208Y/120 VOLT SYSTEMS</p><p>FIG. 1.33 Oiagromr illustrating how to include motors in low-voltage radial systems.</p><p>40 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>any one time is equal t,o 50 per cent of the combined rating of all step-</p><p>down trausformers and/or generators supplying power to that one bus,</p><p>Fig. 1.33. For large commercial buildings the 50 per cent figure may</p><p>be too low. Check carefully the mot,or load on all large 208Y/120-volt</p><p>systems.</p><p>In the generalized rases referred t o in paragraphs 1 and 2 , no specific</p><p>ratio of induction to synchronous motors or no specific number of motors</p><p>which prcduce unusually high short-circuit current,s has been set fort,h.</p><p>T o account for these variables, an average motor reitctance ihcluding</p><p>leads is assumed to be 25 per cent for the purpose of preparing application</p><p>tables like Table 1.5 and in making short-circuit st,udies where no more</p><p>accurat,e data are available. It will he noted that the average motor</p><p>reactance of 25 per cent is based on the transformer or supply-generator</p><p>kva rating. This figure is between the values of 28 per cent for induc-</p><p>tion mot,ors and 21 per cent for synchronous motors given in Table 1.14.</p><p>Where the division between synchronous and iuduction motors is known,</p><p>then more accurate calculations can be made by using the assumed motor</p><p>reactances of Table 1.14. The reactances given in Table 1.14 are based</p><p>on motor kva ratings and not supply transformer or generator ratings.</p><p>T 750 KVA</p><p>A 500 KVA 750 KVA</p><p>500 KVA</p><p>v</p><p>EQUIVALENT MOTORS WOULD BE 250 KVA AND</p><p>FOR 280Y/120 VOLT SECONDARY SYSTEM</p><p>375 K VA</p><p>- 480 VOLTS</p><p>FIG. 1.34</p><p>rvrternr.</p><p>Diagram illustrating how lo include motors in lowvoltage secondary network</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 49</p><p>Although a portion of the load connected to a bus rated GOO voks or</p><p>less may be heaters, lights, a-c welders, solderitig irons, appliances, arid</p><p>other devices which produce no short-circuit curreiit, the total installed</p><p>horsepower of motors connected t,o such a bus is geiierally much greater</p><p>than the kva rating of the supply transformers and generators. Hov-</p><p>ever, allowing for diversity, generally the total comhitied horsepower</p><p>rat,ing of all mot,ors running a t one time ix-ould trot produce short-cir-</p><p>cuit currents in excess of the values obtained when using the ahore</p><p>assumptions.</p><p>In systems of 000 volts or Icss, the large motors (i,e., mot,ors 011 t,he</p><p>order of several hundred horsepomerj are usually few i n number and</p><p>represent only a small portion of the tot,al connected horsepower; there-</p><p>fore, these larger motors are generally lumped in with the smaller motors</p><p>and the complete group is represented as one equivalent motor i t i the</p><p>impedance diagram.</p><p>Synchrouous and induction motors need not be segregated when com-</p><p>bining the motors in these low-voltage systems, because lorn-voltage air</p><p>circuit breakers operr so fast that only the current flow duritig the first half</p><p>cycle is considered; i.e., only suhtraiisient reactances ( X y ) of marhiiies</p><p>are considered.</p><p>Motors Rated above 600 Volts. High-voltage motors (rated 2200</p><p>volts and ahove) are generally larger in horsepower rating thau motors</p><p>on systems operating under 600 volts. These largcr motors may have</p><p>a much more significant hearing on short-circuit-current magnitudes</p><p>than smaller motors, and, therefore, more exact determinatiou of the</p><p>reactances of the larger motors is in order. Therefore, it is often foutid</p><p>convetiient to represent each large high-voltage motor individually in</p><p>the impedance diagram.</p><p>However, in large plants like steel mills, paper mills, etc., where there</p><p>are numerous motors of several huridred horsepower each, it is often found</p><p>desirable to group these larger motors iii one group arid represent them</p><p>by one reartaiire in the impedance diagram. Individual motors of</p><p>several thousand horsepoitrer should be coiisidered individually and</p><p>their reactances accurately determined hefore starting the short-circuit</p><p>Whether considering motors individually or in groups, regardless of</p><p>voltage rating of the motors, it is necessary to obtain an equivalent kva</p><p>rating of the individual or group of motors. This can be done precisely</p><p>for large motors by Eq. (1.9) or can be approximated hy Eq. (l.lO),</p><p>( l , l l ) , or (1.12), when the full-load current is not known. The latter</p><p>equations are used when considering a single reactance to represent a</p><p>group of miscellaneous motors.</p><p>study.</p><p>50 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>I n high-voltage systems, complete motor data may not be available.</p><p>Lacking these data, the connected horsepower is assumed to he equal</p><p>t o the generator and/or transformer capacity supplying a given high-</p><p>voltage bus.</p><p>If the reactance of the leads between the transformer and/or gen-</p><p>erator bus and the motors is significant, the reactanre of these leads</p><p>should be included.</p><p>MAKING THE IMPEDANCE DIAGRAM</p><p>After it has been decided what elements of the one-line diagram are</p><p>to be considered in the impedance diagram, the mechanirs of making</p><p>the impedance diagram and of determining the short-circuit-current</p><p>magnitude are as follows.</p><p>Treatment of Sources of Short-circuit</p><p>Current. The generators and motors</p><p>are treated as if they comprised a gen-</p><p>erator of zero reactance plus an external</p><p>reactor to represent the reactance of the</p><p>EXTERNAL TO machine windings, Fig. 1.35. The first</p><p>REPRESENT IMPEDANCE OF step in making an impedance diagram</p><p>GENERATOR OR MOTOR. is torepresent every generator and motor</p><p>or groups of motors and utility supply</p><p>FIG. 1.35 One-line representation by a reactance connected to a zero im-</p><p>of generator or motor in impedance pedance bus or so-called “infinite bus,”</p><p>Fig. 1.36. This bus represents the in- diogmm.</p><p>ternal voltage of the generators and motors.</p><p>The second step is to add the</p><p>reactance of cables, buses, transformers, current transformers, circuit</p><p>GENERATOR OR MOTOR OF</p><p>ZERO IMPEDANCE 7</p><p>Completing the Impedance Diagram.</p><p>flG. 1.36</p><p>system shown in one-line diagram form in Fig. 1.28.</p><p>Representation of reactances of generators, motors, and utility supply of</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 51</p><p>breakers, switches, etc., in their proper location to complete the imped-</p><p>ance diagram, top of Fig. 1.37.</p><p>The</p><p>next step is to decide whether to use ohms, per cent ohms, or per-unit</p><p>ohms to represent the various circuit impedances in the impedance</p><p>diagram.</p><p>Choice of Ohms, Per Cent Ohms, or Per-unit Ohms Method.</p><p>INFINITE BUS</p><p>SHORT 6.04V</p><p>INFINITE BUS CIRCUIT</p><p>STEP NO i COMBINE SERIES REACTANCES</p><p>C + D = 0 . 0 4 + 0 . 1 5 ~ 0 . 1 9 %</p><p>H + 1 + J = 2.Ot ~0.0+0.10~ 12.10%</p><p>STEP NO.:! COMBINE PARALLEL REACTANCES</p><p>F,G AND I H + I + J )</p><p>XI F G H + I + J</p><p>- ' = _ ' + L + I</p><p>I I - _ -o,'40t 3 + p-j=j 2.5+0.2+0.083</p><p>I -= 2.783 X =0.3698</p><p>XI</p><p>STEP N0 .3 COMBINE SERIES REACTANCES</p><p>X,,AND E</p><p>X t = XI + E = 0.36+0.04'0.40%</p><p>STEP NO. 4 COMBINE PARALLEL REACTANCES</p><p>X o , A . B . AND IC+D)</p><p>XR XI A B C+D '0.40 025 2.0 0.19</p><p>I - 1 + 1 + L +- 1 - +- +- +- l i I I</p><p>2.5+ 4 + 0 . 5 +5.3=12.3</p><p>X ,</p><p>RESULTANT SINGLE REACTANCE</p><p>I 0.0805 % X ~ O ~ z</p><p>FIG. 1.37</p><p>bining reactances into o single resultant value.</p><p>Complete reaclomce diagram for system shown in Fig. 1.28. Steps for com-</p><p>52 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>Ohms are generally not used because of the difficulty of converting</p><p>ohms from one voltage base to another without error and because of</p><p>the very small numbers, which make accurate and easy calculation more</p><p>difficult than the per cent or per-unit system.</p><p>In many of the examples in this book, the assumed or given impedance</p><p>or reactance data are listed in per cent, hut in the reactance dia, -rams</p><p>these are converted to per-unit. No notation will he made when that is</p><p>done as it will be obvious.</p><p>Equations (1.1) to (1.4) show how to convert ohms to per cent ohms,</p><p>ohms to per-unit ohms.</p><p>The Per-unit System for Electrical Calculations.* A per-unit system</p><p>is a means of expressing numbers for ease in comparing them. A per-unit</p><p>value is a ratio:</p><p>a number Per-unit = ~~ base number (1.21)</p><p>The base number is also called unit value since in the per-unit system</p><p>Thus, base voltage is also called unit</p><p>For</p><p>i t has a value of 1, or unity.</p><p>voltage.</p><p>example, for the columns below, a base of 560 is used:</p><p>Any convenient number may be selected for the base number.</p><p>Per-unit Volue</p><p>with 560 as a Base Number</p><p>93 0.17</p><p>125 0.22</p><p>560 1 .oo</p><p>2053 3.65</p><p>Each number in the second column is a per-unit part of the base</p><p>number. In the first column, to compare the numbers, first mentally</p><p>determine the ratio of one to the other. In the second column this is</p><p>already accomplished.</p><p>The comparison can be aided by selection of the base number which</p><p>will illustrate the comparison best. In the foregoing example, if i t is</p><p>desired to show how much larger each uumber is when compared with</p><p>the smallest number, the number 93 might have been selected as the</p><p>base. This would then be obtained as follows:</p><p>Per-""it Valve</p><p>Number with 93 (I, (I Base</p><p>93 I .oo</p><p>125 I .35</p><p>560 6.00</p><p>2053 22.20</p><p>The value of a per-unit system is particularly useful when comparing</p><p>* From material originally prepared by H. J. Finison. iormrrly of General Ekctrir</p><p>Company.</p><p>SHORT-CIRCUIT.CURRENT CALCULATING PROCEDURES 53</p><p>For numbers that are similarly related to two different base numbers.</p><p>example :</p><p>Core A Cole B</p><p>Norm01 "0th 2300 460</p><p>Volts during motor starting 2020 420</p><p>The above figures in themselves have little significance until they are</p><p>compared each with its normal condition as follows:</p><p>Vollr during starting per-unit of normal 0.88 0.91</p><p>Per Cent. Obviously per cent and per-unit systems are similar. The</p><p>per cent system is obtained by multiplying the per-unit value arbitrarily</p><p>by 100 to keep many frequently used per-unit values expressed as whole</p><p>integers. By definition,</p><p>x 100 a number</p><p>base number Per cent = (1.22)</p><p>Thus to change per cent to per-unit, divide by 100. For example, a</p><p>transformer which has an impedance of 6 per cent has an impedance of</p><p>0.06 per-unit.</p><p>The per cent system is somewhat more difficult to work with and more</p><p>subject to possible error since it must always be remembered that the</p><p>numbers have been arbitrarily multiplied by 100. For a simple example,</p><p>money may draw interest a t the rate of 4 per cent per year. Early in</p><p>arithmetic one learns to determine the interest by multiplying the princi-</p><p>pal by 0.04. It is thus necessary to remember to convert to the per-unit</p><p>value before using the figure. In a complex calculation, this repeated</p><p>conversion may invite errors. In effect it is safer and more convenient</p><p>to say that interest is a t the rate of 0.04 per-unit.</p><p>I t is</p><p>usually convenient to convert these figures immediately to per-unit by</p><p>dividing by 100 and thereafter do all calculating in terms of per-unit</p><p>rather than attempt to remember always during the calculations whether</p><p>a number should or should not be multiplied or divided by 100 to obtain</p><p>the true value.</p><p>Just as the per cent system has a symbol (%) to desig-</p><p>nate that a given number is expressed in terms of per cent (as 6%) so also</p><p>does the per-unit system have a symbol. The symbol for per-unit is</p><p>(%).</p><p>In a per-unit system as used for expressing</p><p>electrical quantities of voltage, current, and impedance, it is necessary to</p><p>select numbers arbitrarily for the following:</p><p>Impedances of electric apparatus are usually given in per cent.</p><p>Symbol.</p><p>Thus 0.06 per-unit is written as 0.06 91.</p><p>Selection of Base Number.</p><p>Base volts</p><p>Base amperes</p><p>54 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>Do not then in addition arbitrarily select base ohms since it has already</p><p>been fixed by the first two selections because of Ohm’s law.</p><p>E z = -</p><p>I</p><p>base volts</p><p>base a m p z s Base ohms = (1.23)</p><p>Using the selected base values, all parts of an electric circuit or system</p><p>may be expressed in per-unit terms as follows:</p><p>(1.24)</p><p>(1.25)</p><p>(1.26)</p><p>volts</p><p>base volts</p><p>amperes</p><p>base amperes</p><p>ohms</p><p>base ohms</p><p>Per-unit volts =</p><p>Per-unit amperes =</p><p>Per-unit ohms =</p><p>In practice it is</p><p>more convenient to select:</p><p>Base volts</p><p>Base kva</p><p>The base values of other quant.ities are thus automatically fixed.</p><p>for a single-phase system,</p><p>Hence,</p><p>(1.27)</p><p>(1.28)</p><p>base kva X 1000</p><p>base volts</p><p>base kva</p><p>base kv</p><p>Base amperes =</p><p>Base amperes =</p><p>Base ohms = (1.23)</p><p>where base kva is single-phase kva and base volts is single-phase volts.</p><p>base volts</p><p>base amperes</p><p>For a three-phase system:</p><p>(1.29) base kva X 1000</p><p>X base voks</p><p>Bme amperes =</p><p>base kva</p><p>4 X base kv</p><p>Base amperes = (1.30)</p><p>(1.31)</p><p>where base kva is three-phase kva, base volts is line-to-line, and hase ohms</p><p>is per phase.</p><p>Per-unit Ohms. In practice i t is desirable to convert directly from</p><p>ohms to per-unit ohms, without first determining base ohms. By Ohm’s</p><p>law,</p><p>hase volts</p><p>4 X base amperes</p><p>Base ohms =</p><p>base volts</p><p>base amperes Base ohms = (1.23)</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 55</p><p>Substitute Eq. (1.27) (which gives the base amperes) into Eq. (1.23), to</p><p>obtain</p><p>base volts</p><p>(base kva X 1000)/base volts</p><p>(base volts)P</p><p>bsse kva x 1000</p><p>Base ohms =</p><p>Base ohms = (1.32)</p><p>By definition:</p><p>ohms</p><p>base ohms Per-unit ohms = (1.26)</p><p>Substitute Eq. (1.32) into Eq. (1.26) to obtain</p><p>ohms</p><p>(base volts)e/(base kva X 1000)</p><p>ohms X base kva X 1000</p><p>(base voltd2</p><p>Per-unit ohms =</p><p>Per-unit ohms = (1.33)</p><p>ohms X base kva</p><p>(base kv)2 X 1000 Per-unit ohms = (1.34)</p><p>where base kva is single-phase kva and base kv is single-phase kv.</p><p>When dealing with a three-phase system, i t is usual to select three-phase</p><p>kva and line-to-line volts for the base values. Convert the above expres-</p><p>sions to these bases to obtain</p><p>ohms X base kva X 1000 X 3</p><p>(base volts X d3)z Per-unit ohms =</p><p>. ,</p><p>ohms'X base kva X 1000</p><p>(base volts)2</p><p>ohms X base kva</p><p>(base kv)* X 1000</p><p>Per-unit ohms =</p><p>Per-unit ohms = (1.35)</p><p>where ohms are per phase, kva is three-phase kva, and kv is line-to-line</p><p>voltage.</p><p>Usual Base Numbers for System Studies. If per cent or per-unit ohms</p><p>reactance is used, the next step is to choose a kva base.</p><p>In system studies it is usually desirable to select as the base voltage the</p><p>nominal-system voltage or the voltage rating of the generators and supply</p><p>transformers. Base kva will usually be selected as the kva rating of one</p><p>of the machines or transformers in the system, or a convenient round</p><p>number such as 1000, 10,000, or 100,OOO kva. After choosing the kva</p><p>base, convert ohmic reactance of cables, wires, current transformers,</p><p>etc., to per cent or per-unit ohms reactance on the chosen base, using</p><p>Eq. (1.1) or (1.2) or Table 1.3.</p><p>If ohms reactance is used, convert all per cent reactances to ohms by</p><p>Eq. (1.3).</p><p>Where two systems of differing voltage are interconnected through a</p><p>56 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>transformer, select a common kva base for both systems and the rated</p><p>voltage of each system as its own base voltage. (These base voltages</p><p>must have the same ratio to each other as the turn ratio of the transformer</p><p>connecting the two systems.) Base ohms and base amperes for the two</p><p>systems will thus he correspondingly different. Figure 1.38 shows a</p><p>typical example.</p><p>Once the system values are expressed as per-unit values, the two inter-</p><p>connected systems may be treated as a single system and any calculations</p><p>necessary carried out. Only in reconverting the per-unit values of the</p><p>results to actual voltage and current values is i t necessary to remember</p><p>tha t two different voltages actually existed in the system.</p><p>Frequently the impedance of a circuit ele-</p><p>ment may be expressed in terms of a particuiar base kva, and it may be</p><p>desirable to express it in terms of a different base kva. For example, the</p><p>reactance of devices like transformers, generators, and motors is given in</p><p>per cent on their own kva rating, and their reactances must be converted</p><p>to the common base, chosen for the study by means of Eq. (1.5) or (1.36).</p><p>Per-unit ohms on kva base 2</p><p>Change of Base Number.</p><p>- - base kva x (per-unit ohms on kva base 1) (1.36) base kva 1</p><p>Similarly, a machine rated a t one voltage may actually be used i n a</p><p>Its per-unit impedance must thus be circuit a t a different voltage.</p><p>changed to a new base voltage.</p><p>GENERATOR MOTOR</p><p>1000 KVA I0;YKVA o(lOOO KVA)</p><p>13800 2300</p><p>VOLTS VOLTS</p><p>PRIMARY SECONDARY</p><p>RATING RATING</p><p>13200 2400</p><p>VOLTS VOLTS</p><p>TRANSFORMER RATIO= 13 200/2400=5.5</p><p>(A1 RATIO -</p><p>(El</p><p>(8 ) LOW VOLTAGE SYSTEM (A)HIGH VOLTAGE SYSTEM</p><p>13 800 BASE VOLTS 2500 5.5</p><p>I000 BASE KVA 1000 I .o</p><p>41.6 EASE AMPS 233 115 5</p><p>190 BASE OHMS 6.2 5 (5.5?</p><p>FIG. 1.38</p><p>onother.</p><p>Method of converting bore volts, kva, amperes, and ohms from one value to</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES n</p><p>Reference to Eq. (1.35) shows that per-unit ohms is inversely propor-</p><p>tional to the square of base volts. Thus:</p><p>(1.37)</p><p>Per-unit ohms on new base volts - (old base volt.s)*</p><p>Per-unit ohms on old base volts (new base volts)*</p><p>-</p><p>and</p><p>Per-unit ohms on new base volts = per-unit ohms on old base volts</p><p>(1.38)</p><p>Equations (1.37) and (1.38) may be used for per cent ohms as well as per-</p><p>unit ohms.</p><p>When using ohms</p><p>instead of per cent or per-unit in the impedance diagram, it is important</p><p>to convert the ohmic values to a common voltage base by Eq. (1.13).</p><p>For example, if the short-circuit current is being calculated in a 480-volt</p><p>system (supplied by transformers rated 480-volt secondary) fed through a</p><p>cable and a transformer from a 2400-volt system, the ohms impedance of</p><p>the cable in the 2400-volt circuit must be multiplied by 48O2/24OO2 to</p><p>convert it to ohms on a 480-volt base. The transformer ratings, i.e., 480,</p><p>240, etc., and not system ratings, if different from transformer rating, are</p><p>used as the voltage base for short-circuit-current calculations.</p><p>The utility system must be</p><p>represented by a reactance in the impedance diagram. Sometimes this</p><p>utility-system reactance is available in per cent on a certain base. If so,</p><p>it is merely necessary to convert this value to the common base used in</p><p>the impedance diagram. In some cases the</p><p>utility engineers will give the short-circuit kva or current that the utility</p><p>system will deliver a t the plant site. In otker cases, only the interrupting</p><p>capacity of the incoming-line circuit breaker is known. In these cases to</p><p>convert short-circuit kva, current, or incoming-line breaker interrupting</p><p>rating to per cent reactance on the kva base used in the reactance diagram,</p><p>proceed as follows:</p><p>If given short-circuit kva, convert to per cent by using Eq. (1.6).</p><p>If per-unit is desired, use also Eq. (1.4).</p><p>If given short-circuit amperes (rms symmetrical), convert to per cent</p><p>by Eq. (1.7) and to per-unit by Eqs. (1.7) and (1.4).</p><p>If only the kva interrupting rating of the incoming line circuit breaker</p><p>is known, convert to per cent by Eq. (1.8) and to per-unit by Eqs. (1.8)</p><p>and (1.4).</p><p>(old base volts)2</p><p>(new base volts)2</p><p>Converting Ohms to a Common Voltage Base.</p><p>Representing the Utility Supply System.</p><p>To do this, use Eq. (1.5).</p><p>DETERMINING THE EQUIVALENT SYSTEM IMPEDANCE OR REACTANCE</p><p>After completing the impedance diagram and inserting the values of</p><p>reactance or impedance for each part of the diagram, it is necessary to</p><p>reduce this network to one equivalent value. This can be done either by</p><p>58 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>longhand calculation or with the aid of a calculating board. Since so few</p><p>engineers have access to calculating hoards and must use longhand meth-</p><p>ods, this method will be covered in sufficient detail to enable solving the</p><p>short-circuit problems commonly encountered.</p><p>A d-c calculating board will permit</p><p>accurate solution of all short-circuit problems where reactance only is</p><p>considered. In most cases where resistance is a significant factor and</p><p>must be considered, the d-c calculating board cannot be used readily.</p><p>However, in some problems involving resistance, certain approximations</p><p>can be made to obtain reasonably accurate</p><p>answers on d-c calculating</p><p>boards. For exact calculating-board solutions of problems factoring</p><p>resistance and reactance, the a-c calculating board may he employed.</p><p>A-c calculating boards have boxes to represent both the resistance and</p><p>reactance of a circuit. The procedure for using calculating boards is</p><p>beyond the scope of this book.</p><p>Longhand methods of</p><p>combining reactances vary in some respects. To illustrate the principles</p><p>involved, refer to Figs. 1.37 and 1.39.</p><p>Arbitrary values of reactance have been assigned to the various</p><p>branches. Combining the various branches of the diagram is merely a</p><p>question of reducing two or more series reactances to one value and</p><p>reducing two or more parallel reactances to one value until one single</p><p>equivalent value is obtained.</p><p>Use of Calculating Boards.</p><p>Longhand Method of Combining Reactances.</p><p>The following shows how to combine reactances and resistances.</p><p>1. Combining reactance and resistance to determine impedance,</p><p>z = r + j z (1.39)</p><p>2. Adding series reactance of circuits where resistance is neglected add</p><p>z = m</p><p>wherej = 47</p><p>reactances arithmetically, i.e.,</p><p>x, + x2 + xa = x. = equivalent reactance</p><p>z,, z2, and x3 = reactances of circuit components</p><p>zs = equivalent reactance</p><p>3. Combining parallel reactances,</p><p>zo = equivalent reactance</p><p>For two reactances only x, and z2</p><p>(1.40) ( d ( z 2 )</p><p>21 + 2 2</p><p>XI = -</p><p>For combining several parallel reactances</p><p>(1.41) 1 1 1 1 1 1</p><p>2. 2, 2 2 X I 2,</p><p>- = _ + - + - + - + E</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 59</p><p>ONE-LINE DIAGRAW</p><p>INFINITE</p><p>c.</p><p>REACTANCE DIAGRAM OF CIRCUIT</p><p>SHOWN IN ONE LINE DIAGRAM TO</p><p>THE LEFT.</p><p>P ~ T & T( $*,</p><p>mP.T* EQUIVALENT Y</p><p>CONVERT PITI , PITI e c , TO</p><p>c. EQUIVALENT Y.</p><p>S T E P # I</p><p>COMBINE SERIES REACTANCES</p><p>P I ~ T I , R B T ~ , E T C .</p><p>STEP x z</p><p>--&- I</p><p>&Pa. C t a"' 3 + c *</p><p>L COMBINE 2 + Ct , 3+ C+ AND</p><p>UNTIL ONE EOUIVALENT</p><p>cs c4</p><p>THEN REPEAT STEPS 2.3 e 4</p><p>DRAW NEW DIAGRAM REACTANCE IS OBTAINED.</p><p>STEP-* 3 STEP tt4</p><p>FIG. 1.39</p><p>into a single resultant value.</p><p>Example of the method of combining remtmces of a network-type system</p><p>MI SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>Some systems are such that they cannot he reduced by merely com-</p><p>bining series and parallel rgactances. For example, take the one-line</p><p>diagram of a circuit as show in upper left-hand corner of Fig. 1.39. The</p><p>reactance diagram is shown the ypper righehand corner of Fig. 1.39.</p><p>In addition to combining serieszind parallel reactances, it is necessary to</p><p>convert a triangle of reactances such as PI, TI, Pzr T , and C1 to an equiva-</p><p>lent Y of reactances by the formulas of Fig. 1.40. By these conversions,</p><p>\</p><p>I</p><p>ob + ac + be</p><p>b</p><p>a b + a c + bc</p><p>B =</p><p>c =</p><p>A = o b + o c + b c</p><p>a</p><p>0c</p><p>A + B + C</p><p>a=-</p><p>b : "</p><p>A + B + C</p><p>A 8</p><p>A + B + C C :</p><p>FIG. 1.40 Formula for converting a triangle or delta of three impedances to a Y of three</p><p>equivalent impedances, and vice verso.</p><p>any commonly encountered system reactance diagram can be reduced to</p><p>one equivalent reactance.</p><p>Sometimes i t is desirable to consider the</p><p>resistance and reactance of a circuit. This involves combining imped-</p><p>ances. The procedure for combining impedances is outlined here. The</p><p>combining of parallel impedances necessitates multiplication and division</p><p>of impedances (complex quantities) and is outlined here.</p><p>When two or more impedances are in</p><p>series, the resistance and reactance components are added separately to</p><p>combine the series into one equivalent value.</p><p>Combining Impedances.</p><p>Adding Series Impedances.</p><p>Refer to Fig. 1.41. The three series impedances are</p><p>z1 = TI + jzl</p><p>za = 72 i- jxa</p><p>zz = Tp + ja</p><p>SHORT-ClRCUIT+CURRENT CALCULATING PROCEDURES 61</p><p>3 SERIES EQUIVALENT</p><p>IMPEDANCES IMPEDANCE</p><p>FIG. 1.41 Example illustrating the combining of series impedances.</p><p>The equivalent impedance</p><p>2 % = rl + V Z + 73 + j(z1 + zz + 4 (1.42)</p><p>Using the numerical values of Fig. 1.41,</p><p>2 , = 1 + j 2</p><p>22 = 2 + j 3</p><p>= 0.5 +jl</p><p>21 = (1 + 2 + 0.5) + j ( 2 + 3 + 1) = 3.5 + j G</p><p>The above is applicable when impedances are expressed in ohms, per-</p><p>Combining Parallel Impedances. Parallel impedances may be</p><p>unit or per cent.</p><p>reduced to one equivalent impedance as follows (see Fig. 1.42):</p><p>TWO PARALLEL IMPEDANCES EQUIVALENT IMPEDANCE</p><p>FIG. 1.42 Example illustrating Le combining of parallel impedances.</p><p>61 WORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>(1) Reduce the per cent values of resistance and reactance in each of</p><p>the given parallel circuits to a per-unit basis by dividing per cent figures</p><p>by 100 or convert the per cent values to ohms. Per cent values can be</p><p>used in the following if the multiplier 100 is applied properly, e.g.,</p><p>T X</p><p>(Branch 1) 0.05 0.15</p><p>(Branch 2) 0.008 0.108</p><p>(2) Calculate the impedance squared z2 of each circuit</p><p>1 2% = r' + 2'</p><p>(Branch 1) rlz + 21' = ZI', e.g., 0 .052 + 0.1547-0>25</p><p>(Branch 2) r 2 + zz2 = zz2, e.g., 0.008z + 0.108* = 0.0117</p><p>(3) Obtain the ratios of r/z' of each circuit</p><p>T l 0.05</p><p>21 0.025</p><p>rz</p><p>z '2 0.0117</p><p>(Branch 1) -', e.g., - = 2.0</p><p>0.0°8 - 0.683 (Branch 2) -, e.g., ~ -</p><p>(4) Add the foregoing</p><p>r /zz = Ga = 2.683</p><p>(5) Obtain the ratios of x/z* for each circuit</p><p>2 1 0.15</p><p>21 0.025 (Branch 1) -2 e.g., - = 6</p><p>XP 0.108</p><p>2 2 0.0117 (Branch 2) 7 j e.g., - = 9.2</p><p>(6) Add the foregoing</p><p>X/L' = Ba = 15.2</p><p>(7) Ya2 = 02 + Ba2, e.g., = 2.683' + 15.24 = 238.2</p><p>(8) ra = - 9 Ga e.g., = 22;; ~ - - 0.0112</p><p>Y3'</p><p>BJ</p><p>Ya2 238.2 15" 0.0642 (9) xa = -2 e.g., = __ =</p><p>The foregoing may be tabulated for convenience in solving a number of</p><p>parallel pairs of circuits:</p><p>r z z4 = r' + z2 r/z' 2 / 2 2</p><p>(Branch 1) 00 0 00</p><p>(Branch 2) 00 0 0 0</p><p>(Branch 3, etc.) ( ) ( ) ( ) ()()</p><p>By addition- Go( )Bo( )</p><p>The combination of the circuits results in</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 63</p><p>Any number of parallel circuits may be accommodated by additional</p><p>horizontal columns as fo branch 1 and branch 2, etc., their resultant</p><p>(r /z2) ’s and (x/z2)’s heling added to obtain GO and Bo.</p><p>Multiplying and Dividing Impedances. Two impedmces may be</p><p>multiplied as per the following equations:</p><p>1</p><p>(21) (22) = 23</p><p>21 = T I + j X l</p><p>ZP = T S + j x ,</p><p>23 = r8 + j x a</p><p>2 3 = (T I + j X l ) ( T t + jZ2)</p><p>= (TIT2 - 2 1 2 2 ) + j ( T I X 2 + TBZL)</p><p>(1.44)</p><p>Two impedances may be divided according to the following equations:</p><p>1 13 = (nrz - XIXZ)</p><p>j a = j(r1zz + rczJ</p><p>T I + j x , T Z - jxt =-x-</p><p>r2 + j x 2 T Z - jxt</p><p>(1.45)</p><p>DETERMINING THE SHORT-CIRCUIT-CURRENT MAGNITUDE</p><p>After the reactance diagram has been reduced to a single value, the</p><p>value of symmetrical short-circuit kva can be determined by Eq. (1.14),</p><p>(1.15), or (1.16). To determine the symmetrical short-circuit current, use</p><p>Eq. (1.17), ( l . lS) , or (1.19).</p><p>Table</p><p>1.4 gives figures for converting kva to amperes.</p><p>The final step is to apply the</p><p>proper multiplying factor from Table 1.2. To determine the total rms</p><p>short-circuit current or kva, use Eq. (1.20).</p><p>Equations (1.14) to (1.19) do not allow for any d-c component.</p><p>Apply Proper Multiplying Factor.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 64</p><p>Three phore</p><p>volh</p><p>line-to-line,</p><p>110</p><p>115</p><p>120</p><p>180</p><p>I99</p><p>208</p><p>220</p><p>230</p><p>240</p><p>440</p><p>460</p><p>480</p><p>550</p><p>575</p><p>600</p><p>1,100</p><p>1,150</p><p>1,200</p><p>2,200</p><p>2,300</p><p>2,400</p><p>3.300</p><p>3.450</p><p>3,600</p><p>3,800</p><p>4,000</p><p>4.160</p><p>6,600</p><p>6.900</p><p>7.200</p><p>11,000</p><p>11,500</p><p>12,000</p><p>TABLE 1.4 Amperes per Kva</p><p>Amperes</p><p>ier phase</p><p>per kva</p><p>5.25</p><p>5.02</p><p>4.81</p><p>3.21</p><p>2.90</p><p>2.78</p><p>2.63</p><p>2.51</p><p>2.41</p><p>1.31</p><p>I .25</p><p>1.20</p><p>1.05</p><p>1 .oo</p><p>0.962</p><p>0.525</p><p>0.502</p><p>0.481</p><p>0.263</p><p>0.251</p><p>0.241</p><p>0.175</p><p>0.167</p><p>0.160</p><p>0.152</p><p>0.144</p><p>0.138</p><p>0.0875</p><p>0.0838</p><p>0.0803</p><p>0.0525</p><p>0.0502</p><p>0.0481</p><p>-</p><p>13.200</p><p>13,800</p><p>14,400</p><p>22,000</p><p>23,000</p><p>24,000</p><p>33,000</p><p>34,500</p><p>36,000</p><p>44,000</p><p>46,000</p><p>48,000</p><p>66,000</p><p>69,000</p><p>72,000</p><p>I I0.000</p><p>1 I5.000</p><p>120,000</p><p>132,000</p><p>138,000</p><p>144,000</p><p>154,000</p><p>161,000</p><p>168,000</p><p>220,000</p><p>230,000</p><p>240,000</p><p>330,000</p><p>345,000</p><p>360,000</p><p>Amperes</p><p>rer phase</p><p>per kvo</p><p>0.0437</p><p>0.0419</p><p>0.0401</p><p>0.0263</p><p>0.0251</p><p>0.0241</p><p>0.0175</p><p>0.0167</p><p>0.0160</p><p>0.0131</p><p>0.0125</p><p>0.0120</p><p>0.00875</p><p>0.00838</p><p>0.00803</p><p>0.00525</p><p>0.00502</p><p>0.00481</p><p>0.00437</p><p>0.0041 9</p><p>0.00401</p><p>0.00375</p><p>0.00359</p><p>0.00344</p><p>0.00263</p><p>0.00251</p><p>0.00241</p><p>0.00175</p><p>0.00167</p><p>0.00160</p><p>"0 wire</p><p>V d h</p><p>c or d-l</p><p>24</p><p>48</p><p>110</p><p>115</p><p>120</p><p>125</p><p>220</p><p>230</p><p>240</p><p>250</p><p>275</p><p>300</p><p>440</p><p>460</p><p>480</p><p>550</p><p>575</p><p>600</p><p>650</p><p>750</p><p>1,200</p><p>1,500</p><p>2,200</p><p>2,300</p><p>2,400</p><p>3,000</p><p>Amperes</p><p>er kro or</p><p>d-c kw</p><p>41.7</p><p>20.8</p><p>9.10</p><p>8.70</p><p>8.33</p><p>8.00</p><p>4.55</p><p>4.35</p><p>4.17</p><p>4.00</p><p>3.64</p><p>3.33</p><p>2.27</p><p>2.17</p><p>2.08</p><p>I .82</p><p>I .74</p><p>1.67</p><p>1.54</p><p>1.33</p><p>0.833</p><p>0.666</p><p>0.455</p><p>0.435</p><p>0.417</p><p>0.333</p><p>/</p><p>~</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 65</p><p>EQUIVALENT CIRCUITS</p><p>The redurtion of impedance diagrams to a single value of impedance</p><p>can he greatly simplified by using equivalent circuits for duplex reactors</p><p>and three-winding transformers.</p><p>Equivalent Circuit for Duplex Reactors. The duplex reactor consists</p><p>of two sections of winding per phase on the same core, with a t ap brought</p><p>out from the junction point. The current ratings and reactances of the</p><p>two sections arc generally equal.</p><p>Aside from the midtap connections, whirh necessitate a total of nine</p><p>leads, the construction is similar to that of the series reactor.</p><p>If 1, and l2 are the self-inductances ( X , and X , are the corresponding</p><p>reactances) of the individual sections, and f c is the “coupling factor” of</p><p>the mutual inductance betmeen sections, then the simplified equivalent</p><p>LEX REACTOR</p><p>ONE LINE DlAGRPlM</p><p>I</p><p>J</p><p>k GENERATOR</p><p>- X I f c</p><p>FIG. 1.43 One-line diagram and equivalent circuit for duplex reactor.</p><p>66 SHORT-CIRCUIT-CURRENT CALCUUllNG PROCEDURES</p><p>circuit for the duplex reactor is as shown in Fig. 1.43. For preliminary</p><p>calculations, an average figure off. = 0.5 should give results of sufficient</p><p>accuracy.</p><p>When making</p><p>short-circuit calculations of power systems which include three-winding</p><p>transformers, there is a question on how to use the designer's reactance</p><p>values.</p><p>Figure 1.44A shows a three-winding transformer, and Fig. 1.44B shows</p><p>its equivalent circuit. The following equations are easily derived and are</p><p>the proper ones to use in short-circuit studies:</p><p>Equivalent Circuit of Three-winding Transformer.</p><p>Designers give reactance values between pairs of windings.</p><p>x,. + X A C - X B C</p><p>2 x. =</p><p>(1.46) XIB + Xec - X A C</p><p>2</p><p>X B C + X d C - X d B</p><p>2</p><p>xs =</p><p>x, =</p><p>All reactance6 must be on same kva base.</p><p>NOTE: The equivalent circuit and equations for a four-winding trans-</p><p>former are more complicated and will not he evident by simple analogy</p><p>from Eq. (1.46). , A</p><p>(A1 mi</p><p>FIG. 1.44</p><p>transformer.</p><p>(A1 One-line diagram and (61 equivalent circuit diagram of three-winding</p><p>EXAMPLES OF SHORT-CIRCUIT-CURRENT CALCULATIONS'</p><p>The following examples are indicative of methods of applying the short-</p><p>circuit-current calculating procedures outlined in the foregoing.</p><p>Systems 600 Volts and Below. The system shown in Fig. 1.45</p><p>involves one source of supply through a transformer from a primary sys-</p><p>tem. The kva base for the short-circuit calculations is taken as the kva</p><p>*NOTE: Numbers in parentheses in Figs. 1.45 and 1.47 to 1.50 refer to numbers</p><p>of formulas used.</p><p>SHORT-CIRCUIT-CURRENT CALCUVITING PROCEDURES</p><p>INCOMING LINE</p><p>A A</p><p>67</p><p>0.25 Yt</p><p>MOTORS</p><p>SOURCE</p><p>TRANSFORMER</p><p>I</p><p>750 KVA</p><p>5.5 x x</p><p>(0.055%)</p><p>REACTANCE DIAGRAM</p><p>USE 750 KVA BASE ? T ? ? FOR CALCULATIONS</p><p>480 VOLTS</p><p>M$</p><p>( 0 )</p><p>SOURCE REACTANCE ON 750 KVA BASE : loo,ooo 750 -0.0075% (1.61</p><p>0.0625 1 2 5 v 1 x=-- XI XI% +x2-0.0625t025 0.0625XC125-0,05% T 5 %</p><p>( d ) I 750</p><p>X yj;xo,4& 18,000 AMPERES SYMMETRICAL [ 1.18)</p><p>18,000 X 1.25"22.500 AMPERES ASYMMETRICAL (1.201</p><p>-</p><p>o,050</p><p>( e l</p><p>FIG. 1.45</p><p>center system.</p><p>Illustration of procedure for calculation of short-circuit currents in radial load-</p><p>68 SHORT-CIRCUIT-CURREHI CALCULATING PROCEDURES</p><p>rating of the transformer. The kva of the connected motors is assumed</p><p>to be 750 with an equivalent reactance of 25 per cent. Only reactances</p><p>are used in these calculations. This problem is the type on which</p><p>Table 1.5 is based.</p><p>Large 208Y/120-volt Systems. Problems, particularly those involving</p><p>secondary-network systems in the downtown area of the large cities or in</p><p>large buildings, require the determination of the short-circuit current on a</p><p>208Y/120-volt basis. In these systems it is particularly important that</p><p>the reactance of all circuit elements, however small, be taken into account,</p><p>as they have a much more significant effect in reducing the short-circuit</p><p>current a t 208Y/120 volts than a t 480 or 600 volts.</p><p>PLAN</p><p>FEEDERS</p><p>BREAKERS</p><p>CHANNEL B U $ - 4 0 0 0 A - 150'</p><p>I w</p><p>NETWORK</p><p>PROTECTOR</p><p>NETWORK TRANSFORMER</p><p>13200-216~ / I25 VOLTS Z 5 0 0 A n KvA I I</p><p>INCOMING LINE</p><p>y Z 5 O YVA SC L- CD"Yl</p><p>nus o'</p><p>CIRCUIT</p><p>BREAKER</p><p>4000</p><p>rnY"l . ""...</p><p>ELEVATION</p><p>FIG. 1.46 Arrmgement of equipment for large 208Y/120-volt spot network system.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 69</p><p>The</p><p>one-line diagram is shown in Fig. 1.47.4 which iurludes the hayir reartanre</p><p>data on the circuit elemenk. The impedauce diagram is shown in Fig.</p><p>1.47B. Figure 1.47C shows the condensed diagram to illustrate t,he rela-</p><p>tive distribution of reactance in the system. It will be noted t,hat the</p><p>overhead bus R has 70 per cent as much impedance as the romhinatiotr of</p><p>all the transformers an8,huses ahead of it,. Elimiiiatiug this item would</p><p>cause a serious error in th magnitude of short-circuit, curretit.</p><p>The intermediate steps etween Figs. 1.47H and 1.47C can be worked</p><p>out by followiug the fa J oing text.</p><p>The short circuit is located just ahead of the maiii 4000-amp circuit</p><p>breaker as this determiires the available short-circuit, curreut, which this</p><p>circuit breaker must interrupt. As pointed out previously, air circuit</p><p>breakers are applied 011 the basis of availahle rurreiit, and therefore \\.heir</p><p>calculat,ing the short,-rirruit duty oil them, t,he impedalire of t,he rirciiit</p><p>breaker is not included.</p><p>The examplc shown in Fig. 1.48</p><p>is typical of what might, be eucouritered i n a steel mill. The kva base</p><p>chosen is 100,000 kva. Precise data are available 011 large motors and are</p><p>used in the short,-circuit, st,udy. Since the large mot,ors roiistitute only</p><p>part of the motor load, the remaining motor load is estimated. For short</p><p>circuits on the 22-kv system t,he motor load is assumed to be equal to the</p><p>capacity supplying each 22-kv bus, or 62,500 k r a aiid 20,000 kva.</p><p>Should more precise data be available regarding ronnevted mot,or load,</p><p>these data should be used for simulating motor ront,ribution for faults on</p><p>the 22-kv system. In t,his example, the connected horsepower 011 the</p><p>6.0-kv bus mas known t,o be as shown in t,he diagram.</p><p>To check the momentary dut,y at F , 011 the KY-kv bus, the primary sys-</p><p>tem should be represented by its equivalrut, subt,raiisieiit reartaure nf</p><p>12.2 per cent. For interrupting duty on the 6.9-kv bus, t,he primary</p><p>syst,em should be represented by a reartanre equivalent to the iirterrupt-</p><p>irig duty on t,he 22-kv system, or 17.5 per cent.</p><p>These large complicated syst,ems should he set up 011 a calculating</p><p>board to enable accurate ausivers t,o he obtained easily.</p><p>The equipment for this example is arrauged as shoirn i ir Fig. 1.46.</p><p>Large High-voltage Power System.</p><p>SHORT CIRCUITS IN SINGLE-PHASE LIGHTING AND</p><p>WELDING POWER SYSTEMS (600 VOLTS A N D LESS)</p><p>A common p r a h c e is to use single-phase trausformers roiiuected to</p><p>three-phase primary systems t,o supply single-phase loiv-voltage power for</p><p>welders and for lightirrg rircuits in some of the older syst,ems.</p><p>When determining the short-circuit current a t the serondaries of these</p><p>transformers, it, is necessary to use the proper impedance to represerrt the</p><p>primary system. In three-phase short-circuit calculations, the reactance</p><p>70 SHORT-CIRCUIT-CURRENT</p><p>CALCUATING PROCEDURES</p><p>FIG. 1.47 One-line diogram, reactance diagram,</p><p>SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES 71</p><p>and short-circuit-current calculation procedure for spot network system show in Fig. 1.46.</p><p>72 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>c:</p><p>74 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>of a conductor is the reactance from the center of the condurtor to the</p><p>theoretical neutral. Assume that for eaeh phase the rurrent leaves on the</p><p>phase conductor and returus through the neutral. In a three-phase short</p><p>circuit, the three currents balance; so there is no rurrent flowing in the</p><p>neutral. With single-phase line-to-line short cirruits, the eurreut leaves</p><p>on one phase conductor and returns ou the other. Therefore this rurrent</p><p>sees the reactance of two condurtors as beiug in series. Heure, for siugle-</p><p>phase tramformers conuected line-to-hie on the primary, twire the</p><p>primary system impedance must be used to represent i t in a true relation</p><p>to the rest of the circuit. The remaining calculatious are essentially the</p><p>same as for three-phase circuits using the transformer and loiv-voltage-</p><p>circuit reactances.</p><p>Single-phase tramformers used for supplying 120/240-volt single-phase</p><p>lighting circuits usually have the midtap available for ronnerting to three-</p><p>mire neutral and ground by the user and are usually relatively low iu kva.</p><p>These small transformers have a relatively high resistatire-t~reactance</p><p>ratio compared with three-phase trausformers of a higher seroridary-</p><p>voltage rating and of larger kva rating.</p><p>100,000 KVA 3 PHASE -7 SHORT CIRCUIT OUTY</p><p>BASE 500 KVA</p><p>4tL</p><p>PRIMARY SYSTEM REACTANCE ON 3-PHASE BASIS.</p><p>PRIMARY SYSTEM REACTANCE ON SINGLE PWSE</p><p>BASIS = 0.005X 2 * 0.01 Va</p><p>PRIMARY SYSTEM X : 0.01%</p><p>TRANSFORMER X =O.O3Ym i</p><p>TOTAL X ~ 0 . 0 4 %</p><p>1%' o , 0 4 ~ ~ , , e o : % : 26000 AMP SYMMETRICAL 0.0192</p><p>11.18 MODIFIED)</p><p>1.25 X 26000 = 32500 A M P ASYMMETRICIL K2Ol</p><p>FIG. 1.49</p><p>system.</p><p>Short-circuit-current calculating procedure for single-phase two-wire 480-volt</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 7.5</p><p>The most severe short-circuit condition in this case is a line-to-neutral</p><p>short circuit because it involves a much higher primary-to-secondary turn</p><p>ratio than does a line-to-line short circuit. Hence, this is the basis on</p><p>which protective equipment should be selected.</p><p>Since the reactance and resistance of the transformers are given on the</p><p>basis of a full winding, it is necessary to convert to the proper values when</p><p>only one-half the secondary winding is involved as is the case when a</p><p>line-to-line neutral short circuit occurs. The reactance is increased by a</p><p>factor of 1.2 and the resistance by a factor of 1.41. Therefore, the pub-</p><p>lished reactances and resistances of these transformers are multiplied by</p><p>those figures.</p><p>Figure 1.49 shows a typical example where reactance only is used, as</p><p>would be the case for a relatively large 480-volt transformer supplying a</p><p>welder circuit. In these calculations it is necessary to use twire the line-</p><p>to-neutral reactance of the primary system. In the example of Fig. 1.50</p><p>use twice the line-to-neutral reactance of the primary. Use the proper</p><p>100 000 KV4 3 PH4SE 1 SH& ClRCUlT DUTY F+ 120,240-V )IR X = : 3 % 1.2 X ON ,. FULL ,. WINDING 84SIs</p><p>B4SE 50 KV4</p><p>PRIM4RY SYSTEM RE4CT4NCE ON 3 PH4K 84%</p><p>0.00198</p><p>II 61 PRlM4R"X - : 0 0 0 0 5 ~ ~</p><p>0036%</p><p>TR4NS X</p><p>PRIM4RI SYSTEM RE4Cl4NGE ON 4 SINGLE PH45E</p><p>B 4 5 1 S ~ 0 0 a ) 5 X 2 i O O 0 1 9 ~</p><p>H4LF WlNDlNG RE4CT4NCE OF TRMIYORMER42 XO0310036X</p><p>T W W R</p><p>00172%</p><p>'I RESIST4NCE " .' ~144X0012~001720/1</p><p>1.25 X 10300 i I2900 4MPS ASYMHETRICbL 11.201</p><p>FIG. 1 .SO Short-circuit-current colculating procedure for single-phase three-wire</p><p>120/24Q.volt system.</p><p>76 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>reactance and resistance for line-to-neutral short circuit a t the secondary</p><p>of the transformer. In both cases there is assumed to be no motor</p><p>feedback.</p><p>TABLES AND CURVES FOR ESTIMATING</p><p>SHORT-CIRCUIT CURRENTS</p><p>To make short-circuit protective equipment application easier, par-</p><p>ticularly in circuits of 60 volts or less, many charts, tables, and curves</p><p>have been prepared to eliminate the necessity for detailed calculations.</p><p>Some of the more usef 1 ones are presented here. :o.</p><p>UNIT SUBSTATIONS</p><p>Standard low-voltage unit substations so widely used have standard</p><p>transformer section impedance and voltage ratings. Hence, the second-</p><p>ary short-circuit currents available can be easily tabulated, as shown in</p><p>Tables 1.5 and 1.6. The available short-circuit duty may be read directly</p><p>from the table as a function of transformer kva, secondary voltage, and</p><p>available primary short-circuit kva.</p><p>Example of Use of Table 1.5. Assume a lonn-kva unit substation for</p><p>480-volt power service having an available</p><p>SHORT CIRCUIT primary short-circuit capacity of 150,000 "']?%- kva.</p><p>"2 x, See 480-volt application table. Follow</p><p>FIG. 1.51 0 n e - k diagram the vertical column under the 1000-kva suh-</p><p>station rating down to the 150,000-kva avail-</p><p>able primary three-phase short-circuit kva</p><p>line in thetable. The availableshort-circuit</p><p>showing location of short circuit</p><p>for determinotion of short-circuit</p><p>currents shown in Table 1.5.</p><p>current a t the 480-volt bus is indicated as 30,400 amp.</p><p>REDUCTION OF SHORT-CIRCUIT CURRENT DUE TO FEEDER IMPEDANCE</p><p>The unit substation application Tables 1.5 and 1.6 make it easy to</p><p>determine the short-circuit current a t the main unit substation bus. By</p><p>the use of the simple estimating curves the short-circuit, current at the end</p><p>of the secondary feeders can he easily determined too. Henre these tables</p><p>and the curves shown in Figs. 1.52 and 1.53 make it easy quickly to esti-</p><p>mate the short-circuit current a t any point in a secondary system 600</p><p>volts and less fed by standard load-center unit substations.</p><p>The curves are for 60-cycle operation. Figure 1.52 is for cable cirruits</p><p>and Fig. 1.53 for bus feeders.</p><p>The results are in terms of the three-phase average asymmetrical rm</p><p>value during the first cycle corresponding with the basis of rating for low-</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 77</p><p>voltage air circuit breakers. The effect of circuit resistance both in</p><p>increasing the impedanre and speeding the decay of the d-c component</p><p>'has been included.</p><p>The range of operat,ing conditions encompassed is as follows:</p><p>System operating voltage (nominal) :</p><p>ZOSY/lZO volts, three phase, four wire; or 208 volts, three phase,</p><p>480 volts, three phase, three wire; or 480Y/27' volts, three phase,</p><p>600 volts, three phase, three wire</p><p>three wire</p><p>four wire</p><p>Short-circuit-current magnitudes:</p><p>10,000 to 100,000 amp</p><p>Feeder-circuit construction :</p><p>Three-conductor cable, No. 4 Awg to 500 MCM</p><p>Busway, plug-in bus of representative designs in current ratings from</p><p>Interlared loiv-reactance feeder bus (LVD) rated 2,000 amp, t,hrec</p><p>225 to 800 amp.</p><p>phase (four bars per phase).</p><p>y/</p><p>3</p><p>.</p><p>CABLE FLLOER LENCTM- FEET</p><p>FIG. 1.52</p><p>three-conductor cable in conduit or interlocked-armor cable (60 cycler).</p><p>Chart for determining short-circuit current a t end of cable circuit consisting of</p><p>TA</p><p>BL</p><p>E</p><p>1.</p><p>5</p><p>A</p><p>va</p><p>ila</p><p>bl</p><p>e</p><p>S</p><p>h</p><p>o</p><p>rt</p><p>-c</p><p>ir</p><p>cu</p><p>it</p><p>C</p><p>u</p><p>rr</p><p>e</p><p>n</p><p>t</p><p>fr</p><p>o</p><p>m</p><p>A</p><p>va</p><p>ila</p><p>bl</p><p>e</p><p>P</p><p>ri</p><p>m</p><p>ar</p><p>y</p><p>th</p><p>re</p><p>e-</p><p>ph</p><p>as</p><p>e</p><p>sh</p><p>or</p><p>t-</p><p>Fi.</p><p>C"</p><p>it</p><p>k</p><p>w</p><p>SE</p><p>C</p><p>O</p><p>N</p><p>D</p><p>A</p><p>R</p><p>Y</p><p>R</p><p>A</p><p>TI</p><p>N</p><p>G</p><p>:</p><p>20</p><p>8Y</p><p>/l</p><p>20</p><p>VO</p><p>LT</p><p>S,</p><p>T</p><p>HR</p><p>EE</p><p>P</p><p>H</p><p>A</p><p>SE</p><p>Su</p><p>bs</p><p>ta</p><p>tio</p><p>n</p><p>kv</p><p>a</p><p>ra</p><p>ti</p><p>n</p><p>g</p><p>11</p><p>2.</p><p>5</p><p>1 150</p><p>1 225</p><p>I 300</p><p>I 500</p><p>I 750</p><p>1 1000</p><p>1 I500</p><p>N</p><p>or</p><p>m</p><p>ol</p><p>c</p><p>ur</p><p>re</p><p>nt</p><p>,</p><p>am</p><p>p</p><p>31</p><p>3</p><p>41</p><p>7</p><p>1 625</p><p>1 834</p><p>1 138</p><p>8</p><p>1 208</p><p>0</p><p>1 278</p><p>0</p><p>1 417</p><p>0</p><p>10</p><p>.0</p><p>10</p><p>.3</p><p>10</p><p>.4</p><p>10</p><p>.4</p><p>10</p><p>.5</p><p>10</p><p>.5</p><p>.</p><p>11</p><p>.9</p><p>15</p><p>.9</p><p>20</p><p>.7</p><p>32</p><p>.4</p><p>42</p><p>.3</p><p>53</p><p>.3</p><p>48</p><p>.7</p><p>9.</p><p>4</p><p>11</p><p>.2</p><p>15</p><p>.1</p><p>12</p><p>.2</p><p>16</p><p>.5</p><p>21</p><p>.7</p><p>35</p><p>.0</p><p>46</p><p>.8</p><p>60</p><p>.4</p><p>61</p><p>.3</p><p>9.</p><p>6</p><p>11</p><p>.5</p><p>15</p><p>,6</p><p>12</p><p>.3</p><p>16</p><p>.7</p><p>22</p><p>.1</p><p>36</p><p>.0</p><p>48</p><p>.5</p><p>63</p><p>.3</p><p>74</p><p>.5</p><p>9.</p><p>7</p><p>11</p><p>.6</p><p>15</p><p>.8</p><p>12</p><p>.4</p><p>16</p><p>.9</p><p>22</p><p>.4</p><p>36</p><p>.8</p><p>50</p><p>.0</p><p>65</p><p>.9</p><p>80</p><p>.0</p><p>9.</p><p>7</p><p>11</p><p>.7</p><p>16</p><p>.0</p><p>12</p><p>.5</p><p>17</p><p>.1</p><p>22</p><p>.6</p><p>37</p><p>.5</p><p>51</p><p>.3</p><p>67</p><p>.9</p><p>85</p><p>.5</p><p>9.</p><p>8</p><p>11</p><p>.8</p><p>16</p><p>.1</p><p>12</p><p>.6</p><p>17</p><p>.2</p><p>22</p><p>.9</p><p>38</p><p>.1</p><p>52</p><p>.5</p><p>70</p><p>.2</p><p>90</p><p>.0</p><p>9.</p><p>8</p><p>11</p><p>.8</p><p>16</p><p>.2</p><p>50</p><p>.0</p><p>00</p><p>10</p><p>0.</p><p>00</p><p>0</p><p>15</p><p>0.</p><p>00</p><p>0</p><p>25</p><p>0,</p><p>00</p><p>0</p><p>50</p><p>0.</p><p>00</p><p>0</p><p>or</p><p>d</p><p>iff</p><p>er</p><p>en</p><p>t</p><p>vo</p><p>lt</p><p>og</p><p>e</p><p>ba</p><p>re</p><p>, m</p><p>ul</p><p>tip</p><p>ly</p><p>s</p><p>ho</p><p>rt</p><p>-c</p><p>irc</p><p>ui</p><p>t c</p><p>ur</p><p>re</p><p>nt</p><p>v</p><p>al</p><p>ue</p><p>s</p><p>in</p><p>ta</p><p>bl</p><p>e</p><p>by</p><p>20</p><p>8</p><p>th</p><p>e</p><p>ra</p><p>ti</p><p>o</p><p>na</p><p>w</p><p>v</p><p>ol</p><p>to</p><p>ge</p><p>U</p><p>nl</p><p>im</p><p>ite</p><p>d</p><p>N</p><p>O</p><p>TE</p><p>:</p><p>3.</p><p>F</p><p>or</p><p>di</p><p>ff</p><p>er</p><p>ed</p><p>w</p><p>ltm</p><p>g</p><p>e</p><p>ho</p><p>se</p><p>.</p><p>I</p><p>24</p><p>0</p><p>n</p><p>o</p><p>r</p><p>*o</p><p>lt.</p><p>*e</p><p>va</p><p>lu</p><p>es</p><p>in</p><p>to</p><p>bl</p><p>e</p><p>by</p><p>th</p><p>e</p><p>ra</p><p>lio</p><p>~</p><p>N</p><p>O</p><p>TE</p><p>:</p><p>52</p><p>.2</p><p>58</p><p>.3</p><p>60</p><p>.8</p><p>63</p><p>.0</p><p>64</p><p>.8</p><p>66</p><p>.7</p><p>2</p><p>'t</p><p>o</p><p>n</p><p>d</p><p>a</p><p>rd</p><p>T</p><p>h</p><p>re</p><p>e</p><p>-p</p><p>h</p><p>a</p><p>s</p><p>e</p><p>U</p><p>ni</p><p>t</p><p>S</p><p>u</p><p>b</p><p>s</p><p>to</p><p>ti</p><p>o</p><p>n</p><p>s</p><p>SE</p><p>C</p><p>O</p><p>N</p><p>D</p><p>A</p><p>R</p><p>Y</p><p>R</p><p>A</p><p>TI</p><p>N</p><p>G</p><p>:</p><p>24</p><p>0</p><p>VO</p><p>LT</p><p>S,</p><p>T</p><p>HR</p><p>EE</p><p>P</p><p>H</p><p>A</p><p>SE</p><p>71</p><p>.2</p><p>'</p><p>87</p><p>.5</p><p>2</p><p>92</p><p>.0</p><p>5</p><p>95</p><p>.9</p><p>2</p><p>10</p><p>0.</p><p>0</p><p>$</p><p>82</p><p>.5</p><p>$</p><p>S</p><p>ub</p><p>st</p><p>ot</p><p>io</p><p>n</p><p>kr</p><p>a</p><p>rc</p><p>lti</p><p>ng</p><p>27</p><p>0</p><p>36</p><p>1</p><p>1 542</p><p>To</p><p>ta</p><p>l</p><p>lo</p><p>w</p><p>-v</p><p>o</p><p>lt</p><p>o</p><p>g</p><p>e</p><p>sh</p><p>or</p><p>t-c</p><p>irc</p><p>ui</p><p>t C</p><p>u</p><p>rl</p><p>en</p><p>h</p><p>,</p><p>th</p><p>ou</p><p>sa</p><p>nd</p><p>s</p><p>of</p><p>a</p><p>m</p><p>pe</p><p>re</p><p>s</p><p>~</p><p>~</p><p>rm</p><p>al</p><p>c</p><p>ur</p><p>re</p><p>nt</p><p>, e</p><p>n</p><p>72</p><p>2</p><p>1 1203</p><p>19</p><p>.7</p><p>31</p><p>.1</p><p>20</p><p>.6</p><p>33</p><p>.3</p><p>21</p><p>.0</p><p>34</p><p>.2</p><p>21</p><p>.2</p><p>I</p><p>34</p><p>.9</p><p>21</p><p>.5</p><p>35</p><p>.5</p><p>21</p><p>.7</p><p>I 36.1</p><p>-</p><p>18</p><p>04</p><p>-</p><p>-</p><p>41</p><p>.3</p><p>45</p><p>.1</p><p>46</p><p>.6</p><p>48</p><p>.0</p><p>49</p><p>.0</p><p>50</p><p>.</p><p>I</p><p>-</p><p>tip</p><p>iy</p><p>9</p><p>N</p><p>O</p><p>TE</p><p>: 2</p><p>.</p><p>M</p><p>o</p><p>to</p><p>r</p><p>sh</p><p>or</p><p>t-c</p><p>irc</p><p>ui</p><p>t</p><p>cu</p><p>rr</p><p>en</p><p>t</p><p>co</p><p>nt</p><p>rib</p><p>ut</p><p>io</p><p>n</p><p>is</p><p>2.</p><p>5</p><p>tim</p><p>es</p><p>t</p><p>he</p><p>t</p><p>ra</p><p>ns</p><p>fo</p><p>rm</p><p>er</p><p>n</p><p>or</p><p>m</p><p>al</p><p>I N</p><p>O</p><p>TE</p><p>:</p><p>4.</p><p>M</p><p>ot</p><p>or</p><p>s</p><p>ho</p><p>rt-</p><p>ci</p><p>rc</p><p>ui</p><p>t</p><p>cu</p><p>rr</p><p>en</p><p>t-</p><p>co</p><p>nt</p><p>rib</p><p>ut</p><p>io</p><p>n</p><p>is</p><p>5.</p><p>0</p><p>ti</p><p>m</p><p>n</p><p>lh</p><p>e</p><p>tm</p><p>n</p><p>r</p><p>a</p><p>cu</p><p>rr</p><p>en</p><p>t</p><p>fo</p><p>r</p><p>50</p><p>%</p><p>co</p><p>nn</p><p>ec</p><p>te</p><p>d</p><p>m</p><p>ot</p><p>or</p><p>s.</p><p>fo</p><p>rm</p><p>er</p><p>n</p><p>or</p><p>m</p><p>01</p><p>c</p><p>ur</p><p>re</p><p>nt</p><p>f</p><p>or</p><p>1</p><p>00</p><p>%</p><p>c</p><p>on</p><p>ne</p><p>ct</p><p>ed</p><p>m</p><p>ot</p><p>on</p><p>.</p><p>4.</p><p>0</p><p>4.</p><p>5</p><p>5.</p><p>0</p><p>5.</p><p>0</p><p>5.</p><p>0</p><p>5.</p><p>5</p><p>5.</p><p>5</p><p>5.</p><p>5</p><p>4.</p><p>0</p><p>4.</p><p>5</p><p>5.</p><p>0</p><p>5.</p><p>0</p><p>5.</p><p>0</p><p>5.</p><p>5</p><p>5.</p><p>5</p><p>5.</p><p>5</p><p>fo</p><p>rm</p><p>er</p><p>im</p><p>pe</p><p>da</p><p>nc</p><p>e,</p><p>%</p><p>80 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>FIG. 1.53</p><p>designations refer to General Electric Company bus I60 cycles).</p><p>Chart for determining short-circuit current (it end of feeder bur. The type</p><p>Required Data. The basic data needed to enable the use of Figs. 1.52</p><p>1. System operating voltage</p><p>2. Available short-circuit current at the source bus (average asym-</p><p>3. Length and construction of the feeder circuit</p><p>4. Connected motor load at the feeder terminal</p><p>Procedure for Use of Figs. 1.52 and 1.53.</p><p>and 1.53 are the following:</p><p>metrical)</p><p>The evaluation of feeder</p><p>terminal short-circuit current involves only four simple steps (see Fig.</p><p>1.54):</p><p>1. Locate the magnitude of source-end short-circuit current on the</p><p>proper left-hand operating voltage scale.</p><p>2. From this starting point move along to the right following along a</p><p>curve or an interpolation between adjacent curves until the desired length</p><p>of specific feeder construction (horizontal scales) is reached.</p><p>3. Project the latter point horizontally to the left and read the short-</p><p>circuit current contributed by the feeder on the same scale as used in 1.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 81</p><p>4. Add the feeder terminal connected motor-current contribution (five</p><p>times the sum total of the motor full-load current).</p><p>MODIFICATIONS FOR SPECIAL CONDITIONS</p><p>Parallel Circuit Feeders. A feeder circuit composed of two or more</p><p>identical circuits in parallel can be readily treated by making a correction</p><p>in the apparent length. The impedance presented by a feeder consisting</p><p>of two circuits in parallel will be identical to that of a sing16 circuit of half</p><p>the length; that of three circuits in parallel will be identical to that of a</p><p>single circuit of one-third the length; etc.</p><p>In the case of parallel circuit feeders, divide the true feeder length by</p><p>the number of circuits in parallel and proceed on the basis of single-circuit</p><p>data.</p><p>. I</p><p>Y</p><p>t</p><p>k-- H I 1</p><p>.</p><p>i</p><p>I</p><p>I l l I</p><p>I_ 850</p><p>, / C P ) O I C I I</p><p>FEEDER L m m " I</p><p>FIG. 1.54 Example rhowing how to use the charts of Fig. 1.52 and 1.53.</p><p>Available primary</p><p>lhree-phose</p><p>hoil-circull kro</p><p>25 137.5 I 50 I 75 1 100 150 1 200 1 250 I 333 1 500</p><p>Normal eurrenl. ornperes a1 240 volts</p><p>104 I 156 1 208 I 313 1 417 I 625 1 833 1042 I 1388 12083</p><p>12.6</p><p>13.3</p><p>13.6</p><p>13.7</p><p>13.8</p><p>13.9</p><p>14.0 _ _</p><p>1.2</p><p>3.0</p><p>Tolo1 lox-vollage shw-circuil c ~ ~ r e n l , lhousandr of rms omperes for m e 120-volt winding</p><p>short-circuited, lhe olhei opon-circuiled</p><p>15.9</p><p>16.9</p><p>17.5</p><p>17.8</p><p>17.9</p><p>18.0</p><p>18.1</p><p>1.2</p><p>3.5</p><p>25,000</p><p>50,000</p><p>100,000</p><p>150.000</p><p>250,000</p><p>500,000</p><p>Unlimited</p><p>full-winding</p><p>impadançe:</p><p>Per cent R . . . . .</p><p>Per cent 2.. . . . .</p><p>Transformei</p><p>20.4</p><p>22.1</p><p>25 .1</p><p>23.5</p><p>-</p><p>6.5</p><p>6.7</p><p>6.8</p><p>6.8</p><p>6.9</p><p>6.9</p><p>6.9 -</p><p>1.4</p><p>3.0</p><p>-</p><p>28.4 35.2 32.3 37.3 48.9</p><p>31.8 40.8 36.9 43.5 60.2</p><p>33.9 44.2 39.7 47.4 68.1</p><p>34.7 45.5 41.0 49.0 71.1</p><p>-</p><p>9.6</p><p>10.0</p><p>10.2</p><p>10.2</p><p>10.3</p><p>10.3</p><p>10.4 -</p><p>1.4</p><p>3.0</p><p>-</p><p>I I I I I</p><p>23.7 35.3 46.6 41.7 50.1 73.9</p><p>24.0 135 .8 147 .5 142 .3 151 .2 176.0</p><p>1 . 2 1 1 . 2 1 1 . 2 1 1.0</p><p>3.5 3.5 3.5 5 .0 5 .5 5 .5</p><p>A short circuit invalving one of the secondrtry half windings (terminals Xi to X 2 ai</p><p>terminals X, to X, ) , Fig. 1.51, allows eansiderahly more short-çireuit current to flow</p><p>than a short circuit involving the full seeondary minding (terminals X i to X d . Con-</p><p>sequently, the circuit-hreaker seleetions are based on the half-winding value of short-</p><p>circuit current.</p><p>The eonditions on whieh the tables are hased are summsrizcd below:</p><p>1. A salid half-winding short cireuit at the tcrminals (scc Fig. 1.51).</p><p>2. Primary three-phase short-eircuit capacities vsrying from 25,000 kva to unlimited</p><p>kva. For the worst case, the single-phasr short-cireuit capaeity is me-half the three-</p><p>phase primsry short-circuit capacity, and this value has bem used in thc celculations.</p><p>This worst csse involves the assumption that the primary of the transformer is con-</p><p>nected line-to-line on the high-voltage system, not line-to-neutral.</p><p>3. The full-winding per cent impedance and per cent resistances m e given in</p><p>Table 1.6.</p><p>4. The half-winding reactance was taken as 1.2 times the full-winding reactance,</p><p>while the half-winding resistance was taken as 1.44 times the full-winding resistance,</p><p>on full kva base.</p><p>5. The d-e offset multiplier for the first half eycle was taken as 1.25.</p><p>6. It is sssumed that the 120/240-volt units will supply lighting loads only, i.e., no</p><p>7. The only source of power connected to the secondary bus is one transformer of the</p><p>motor feedbaek.</p><p>capaeity indicated.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 83</p><p>Make</p><p>an independent evaluation of each common circuit construction starting</p><p>at the source end.</p><p>1. Evaluate the short-circuit current a t the end of the first section of</p><p>common feeder construction in the standard manner.</p><p>2. Using the answer derived from 1 as the source short-circuit-current</p><p>value for section 2, proceed in the standard manner to evaluate the short-</p><p>circuit current a t the end of the second section.</p><p>3. Using the answer derived from 2 as the source short-circuit-current</p><p>value for the third section, proceed in the standard manner to evaluate</p><p>the short-circuit current a t the end of the third section.</p><p>Results obtained for sections beyond the first will be somewhat on the</p><p>conservative side (higher than the true short-circuit-current value). This</p><p>follows from the fact that the basic analysis assumes an X / R ratio of 12</p><p>a t the source end of the feeder. The true X / R ratio at the source termi-</p><p>nals of any feeder section beyond the first will necessarily be less than 12</p><p>since no feeder construction exhibits an X / R ratio as high as 12.</p><p>Specific cable</p><p>feeder length scales have been inscribed for conductor sizes of 500 MCM,</p><p>250 MCM, No. 2/0 Awg and No. 4 Awg. For intermediate valuesof cable</p><p>size locate the horizontal scale points for the desired length of adjacent</p><p>cable sizes which</p><p>are charted, and interpolate between these values. For</p><p>example, a No. 3/O-Awg conductor is about midway between a No. 2/0-</p><p>Awg and a 250-MCM. To evaluate the effect of a 100- f t run of No. 3/0-</p><p>Awg cable based on Fig. 1.52, locate the 100-ft point on the No. 2/0-scale</p><p>and on the 250-MCM scale. A point midway between these two points</p><p>will closely represent 100 ft of No. 3/O-Awg conductor.</p><p>Results obtained from the</p><p>estimating curves without correction can be safely used to select protec-</p><p>tive interrupters.</p><p>If desired, a closer approximation of the actual value can be obtained</p><p>by increasing the apparent feeder length to account for the higher imped-</p><p>ance of single-conductor feeder circuits.</p><p>Feeders Consisting of Different Circuit Construction in Series.</p><p>Interpolation for Intermediate Cable Conductor Sizes.</p><p>Three Single-conductor Cables in Conduit.</p><p>Conductor Sirs Use an Appored Lenglh of</p><p>500 MCM.. ........ 130% of lhe acluol feeder Imglh</p><p>250 MCM.. ........ 120% of the o c h d feeder lenglh</p><p>No. 2/0 A r g . . ..... 110% of lhe amal feeder lmglh</p><p>No. 4 Awg ......... No correction</p><p>Both the 60-cycle resistance and reactance of a three-single-conductor</p><p>cable feeder in conduit are greater than those of a three-conductor cable</p><p>feeder in conduit or steel armor in the ratios reflected in the accompanying</p><p>table:</p><p>84 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>Conductor sire Residence, 7% Reactance. Yo</p><p>I I</p><p>500MCM .............. i 25 1-50</p><p>No .4Awg ............. I 102 I 150</p><p>No. 2/0 Awg. . ......... 106 150</p><p>NOTE: Spaced open-wire circuits should be treated by conventional</p><p>calculation procedures; a suitable one is given under Circuit Analysis-</p><p>General Case.</p><p>Results obtained from the curves, Figs. 1.52</p><p>and 1.53, may be used with safety for the selection of protective inter-</p><p>rupters.</p><p>The true short-circuit-current value for a two-wire single-phase circuit</p><p>operating at line-to-line voltage will be about 87 per cent of the three</p><p>phase evaluation.</p><p>Frequency. The curves, Figs. 1.52 and 1.53, are restricted to 60-cycle</p><p>operation. For operating frequencies other than 60 cycles, conventional</p><p>calculations should be used, such as outlined under Circuit Analysis-</p><p>General Case. Note that feeder circuit resistance is not appreciably</p><p>affected by frequency, while reactance varies directly with frequency.</p><p>Single-phase Circuits.</p><p>UhIN SOURCE BUS 48O"OLTS ,.P"**E</p><p>6 0 C I C L E S SHORT CIRCUIT C W l R E N T i</p><p>4CCOOAYP</p><p>CABLES IN PARALLEL</p><p>250 YCY 3IC INTERLOCKED ARMOR</p><p>FIG. 1.55 System diagram used as on</p><p>example to illustrate the determination</p><p>of short-circuit currenk a t the end of</p><p>feeder circuits.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 85</p><p>Short-circuit current at bus A? Example of Application-Fig. 1.55.</p><p>Source short-cirruit current = 40,000 amp</p><p>Equivalent single cable feeder length = 1595 = 75 ft</p><p>From curve Fig. 1.52 (4GO-volt short-circuit current scale; 250-MCM</p><p>feeder Irngt,h scale) :</p><p>Contribut,ion via feeder cable = 23,000 amp</p><p>Motor contribution, bus A = 5 X 310 = 1,550</p><p>24,550 amp</p><p>Motor contribution, bus R = 5 X 03 = 315</p><p>Short-circuit current bus A = 24,8G5 amp</p><p>~</p><p>~</p><p>Short-circuit current a t bus B?</p><p>Source short-circuit current for section 2 = 24,550 amp (say 25,000)</p><p>Feeder lengt,h = 75 ft</p><p>From curve (4GO-volt short-circuit current scale) interpolate between the</p><p>7 5 f t point on ;To. 210 and KO. 4 feeder length scales-Ko. 2 about one-</p><p>third of the way from Xo. 4 to No. 2/0.</p><p>Contribution via feeder cable</p><p>Motor ront,ribution, bus R = 5 X G3 =</p><p>Short-circuit current bus R</p><p>= 11,000 amp</p><p>315</p><p>= 11,315 amp</p><p>~</p><p>CIRCUIT ANALYSIS-GENERAL CASE</p><p>The circuit, problem involved in resolving short-circuit-current magni-</p><p>tudes in low-voltage feeder systems is outlined in Fig. 1.56.</p><p>I n general, low-voltage short-circuit current,s are expressed in terms of</p><p>three-phase average asymmetrical rms amperes during the first cycle of</p><p>currcnt flwv. Since main low-voltage source systems exhibit an X / R</p><p>rat,io of about, 10, it, is standard convention t o multiply the symmet,rical</p><p>short,-rirruit, current, by 1.25 to obtain the short-circuit current a t the</p><p>main buses (this corresponds with an X / R ratio of 12) (see Table 1.2).</p><p>Therefore, at the main bus</p><p>1.25 E Short-circuit current = 1.25 X I symm = - X - v5 z*</p><p>1.25 E</p><p>& short-circuit current z , = - x</p><p>Considering the source system X / R ratio = 12</p><p>1.25 E</p><p>z . = - x 4 short-circuit current (A + jl) = R. + j X .</p><p>86 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>2, (obtained from reference tables) = R, + j X ,</p><p>2, (impedance to end of feeder) = R, + R, + j ( X . + X I )</p><p>X, /R, ratio a t end of feeder = x. + XI - x, _ -</p><p>R. + Izi Rt</p><p>M is the factor to account for d-c offset and is a direct function of</p><p>the X , / R , ratio</p><p>XdRt ratio.. ......... 2</p><p>K ................... I 1;;s 1 I:* I 1:s I l!l I I i 6 1 1.02</p><p>I, is the local motor contribution, and the three-phase average assym-</p><p>metrical rms value may be taken as five times the motor full-load rated</p><p>amperes.</p><p>Available short-circuit current at X = I, (three-phase avg assym-</p><p>metrical rms) + I,</p><p>61</p><p>2 -SOURCE SVSTEY IMPEDANCE '1 Rg+ j X s OHMS/PHASE</p><p>I MAIN LOW-VOLTaGE BUS</p><p>FEE0ER:Zf:Rf t j x f OHMSIPHASE</p><p>IFROH TABLES)</p><p>VAIL4ELE</p><p>SHORT ClRCUlT CURRENT DESIRED HERE FIG. 1.56 One-line diagram for rhort-</p><p>circuit-current calculation ot the end of</p><p>IS'CURRENT CONTRIBUTION FROM</p><p>ly*CURRENT CONTRlBUTlON FROM</p><p>SOURCE *"STEM</p><p>LCCAL YOTORS feeder circuits-genernl core.</p><p>: \J 4 'I</p><p>LOAD</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 07</p><p>TABLES SHOWING EFFECl OF CABLE LENGTH</p><p>Another useful way of showing the effect of length of cable in reducing</p><p>short-circuit currents is given in the Tables 1.7 to 1.10. These show how</p><p>much cable length is required to reduce the short-circuit current from one</p><p>protective-device rating level to another for circuits GOO volts and less.</p><p>Standard protective-device rating levels are:</p><p>100,000 amp rms asymmetrical</p><p>75,000 amp rms asymmetrical</p><p>50,000 amp rms asymmetrical</p><p>25,000 amp rms asymmetrical</p><p>15,000 amp rms asymmetrical</p><p>5000 amp rms asymmetrical</p><p>The tables show how long a cable with a given con</p><p>necessary</p><p>a t various values of I, to keep the cable from being damaged before the</p><p>protective circuit breaker operates. Referring to Chap. 3, i t will be</p><p>noted, for example, that a t 50,000 amp (I,, Tables 1.7 to 1.10) the mini-</p><p>mum size cable which a 50,000-amp interrupting rating low-voltage air</p><p>circuit breaker will protect is No. 4/0 Awg. Hence, the only values in</p><p>the right-hand column of Tables 1.7 to 1.10 that have any practical signifi-</p><p>cance are the two at the bottom of the column. The values above that</p><p>are of academic interest only.</p><p>TABLE 1.7 Lirnitina Effect of Cable on Short-circuit Currents at 400 Volts.</p><p>27.5</p><p>43.6</p><p>69.5</p><p>110.0</p><p>171.5</p><p>265.5</p><p>407.0</p><p>497.0</p><p>606.0</p><p>723.0</p><p>852.0</p><p>990.0</p><p>1072.0</p><p>Conductor</p><p>size</p><p>109.4</p><p>170.3</p><p>263.0</p><p>402.0</p><p>No. 14 A x g . .</p><p>No. 12 Awg . .</p><p>No. 10 Awg. .</p><p>No. 8 Awg . . .</p><p>No. 6 Awg . . .</p><p>N e . 4 A w g ...</p><p>No. 2 Awg. . .</p><p>No. I Awg , , .</p><p>No. 110 Awg .</p><p>No. 2 '0 A x g .</p><p>No. 3/0 Awg .</p><p>No. 4/0 Awg .</p><p>250MCM.. . .</p><p>34.2</p><p>53.0</p><p>81.0</p><p>122.1</p><p>I , = avail</p><p>488.0</p><p>592.0</p><p>706.0</p><p>827.0</p><p>960.0</p><p>038.0</p><p>Coble length 1, ft</p><p>146.8</p><p>175.0</p><p>206.0</p><p>237.5</p><p>271.0</p><p>290.5</p><p>-</p><p>26.9</p><p>42.6</p><p>67.5</p><p>106.5</p><p>165.0</p><p>254.0</p><p>384.5</p><p>468.0</p><p>564.0</p><p>664.0</p><p>775.0</p><p>890.0</p><p>962.0</p><p>-</p><p>7.3</p><p>11.4</p><p>17.9</p><p>28.0</p><p>42.6</p><p>63.7</p><p>91 .O</p><p>111.0</p><p>126.8</p><p>144.8</p><p>162.0</p><p>180.0</p><p>190.5</p><p>le short-circu:</p><p>I f = short-circuit current</p><p>:::: I 1::;</p><p>69.0 21.8</p><p>9.1</p><p>14.4</p><p>22.8</p><p>36.0</p><p>56.1</p><p>86.3</p><p>131 .O</p><p>159.1</p><p>192.8</p><p>228.5</p><p>267.0</p><p>308.0</p><p>333.0</p><p>-</p><p>5.3</p><p>8.5</p><p>13.4</p><p>21.1</p><p>32.7</p><p>43.8</p><p>75.8</p><p>91 .4</p><p>110.7</p><p>128.8</p><p>149.1</p><p>171.2</p><p>184.1</p><p>-</p><p>__</p><p>2.4</p><p>3.8</p><p>5.9</p><p>9.3</p><p>14.2</p><p>21.4</p><p>31.8</p><p>38. I</p><p>44.0</p><p>50.9</p><p>55.4</p><p>64.7</p><p>69.0</p><p>-</p><p>urrent in kiloamperes a t source end of cable</p><p>kiloamperes ior short circuit a t end of cable of length L</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 89</p><p>TABLE 1.8 limiting Effect of Cable on Short-circuit Currents at 480 Volts.</p><p>Three Phase</p><p>Three Single-conductor Cables in a Mmnetic Dud</p><p>5.8</p><p>9.2</p><p>14.3</p><p>22.4</p><p>34.1</p><p>I.</p><p>I ,</p><p>21.9</p><p>34.6</p><p>55.2</p><p>87.6</p><p>136.5</p><p>7.0</p><p>11.0</p><p>17.4</p><p>27.4</p><p>42.4</p><p>64.8</p><p>97.8</p><p>117.5</p><p>140.1</p><p>164.8</p><p>190.0</p><p>216.8</p><p>232.8</p><p>3.8</p><p>6.0</p><p>9.3</p><p>14.6</p><p>22.4</p><p>34.1</p><p>50.4</p><p>60.2</p><p>69.8</p><p>80.6</p><p>91.4</p><p>102.2</p><p>108.6</p><p>51.0</p><p>73.0</p><p>89.0</p><p>101.5</p><p>115.8</p><p>210.5</p><p>321.5</p><p>391.0</p><p>474.0</p><p>566.0</p><p>No. 1 4 A x g ....__..</p><p>N o . 1 2 A x g ........</p><p>No. 1 0 A w g ........</p><p>No. 8 Awg ..... ....</p><p>No.6 A x g ....__.._</p><p>No. 4 A x g .... .....</p><p>No. 2 Awg. . . . . . . . .</p><p>No. 1 Awg ........_</p><p>No. 110 A x g ... ....</p><p>No. 2/0 Awg. . . . . . .</p><p>No. 3/0 A x g .._....</p><p>No. 4/0 Axg.. . . . . .</p><p>250MCM ........_.</p><p>10.8</p><p>17.1</p><p>27.0</p><p>42.7</p><p>66.3</p><p>101.5</p><p>153.8</p><p>187.0</p><p>226.0</p><p>266.0</p><p>310.0</p><p>356.0</p><p>385.0</p><p>25 5 / 1 5 / 25 50 5 i l 5 ~ 2 5 / 50</p><p>50 100 5~ 100 I 5</p><p>Cable length 1, fl</p><p>Conductor</p><p>sire</p><p>-</p><p>21.5</p><p>34.1</p><p>54.1</p><p>85.4</p><p>132.5</p><p>203.3</p><p>308.0</p><p>374.0</p><p>452.0</p><p>532.0</p><p>621.0</p><p>713.0</p><p>771 . O</p><p>-</p><p>short</p><p>~</p><p>7 .3</p><p>11.5</p><p>18.3</p><p>28.8</p><p>44.8</p><p>69.0</p><p>105.0</p><p>127.4</p><p>154.2</p><p>183.1</p><p>214.0</p><p>246.5</p><p>266.8</p><p>~</p><p>4.3</p><p>6.8</p><p>10.7</p><p>16.9</p><p>26.2</p><p>39.9</p><p>60.7</p><p>73.2</p><p>88.7</p><p>03.0</p><p>19.3</p><p>37.0</p><p>47.7</p><p>-</p><p>No. I 4 Awg. . . . ,</p><p>No. 12 Awg.. . . ,</p><p>N o . l O A w g .....</p><p>No. 8 A x g . . . .. ,</p><p>No. 6 Axg.. . . . .</p><p>No. 4 Axg.. . . . .</p><p>No. 2 Axg.. . . . ,</p><p>No. I A x g ......</p><p>No. 1/0 Awg . . . .</p><p>No. 210 Awg . . . .</p><p>No. 3/0 Awg . . . .</p><p>No. 4/0 Awg , . . .</p><p>250 M C M . . . . . . .</p><p>1.9</p><p>3.0</p><p>4.7</p><p>7.4</p><p>11.4</p><p>17.2</p><p>25.5</p><p>30.5</p><p>35.2</p><p>40.7</p><p>44.4</p><p>51.8</p><p>55.2</p><p>-</p><p>22.0</p><p>34.9</p><p>55.7</p><p>88.1</p><p>485.0</p><p>579.0</p><p>682.0</p><p>793.0</p><p>860.0</p><p>-</p><p>29.6</p><p>144.1</p><p>52.5</p><p>663.0</p><p>768.0</p><p>832.0</p><p>-</p><p>rcuit current kiloampcrcs i S O I I I ~ P cnd of</p><p>uit current in kiloarnperPs for short circuit at m i l of c</p><p>~</p><p>I . = availab</p><p>I1 = short-c</p><p>TABLE 1.9 limiting Effect of Cable on Short-circuit Currents a t 240 Volts,</p><p>Three Phase</p><p>Three Single-conductor Cables in a Mmgnetic Duct -</p><p>100</p><p>50</p><p>-</p><p>__</p><p>I .o</p><p>I . 5</p><p>2.4</p><p>3.7</p><p>5.7</p><p>8.6</p><p>2.8</p><p>5.3</p><p>7.6</p><p>0.4</p><p>2.2</p><p>5.9</p><p>7.6</p><p>-</p><p>-</p><p>100</p><p>25</p><p>~</p><p>~</p><p>2.2</p><p>3.4</p><p>5.4</p><p>8.5</p><p>3.1</p><p>'0.0</p><p>0.4</p><p>6.6</p><p>4.4</p><p>I .5</p><p>9.7</p><p>8.5</p><p>3.9</p><p>-</p><p>Conduclor</p><p>*i*e I Cable length 1, ft</p><p>-</p><p>3.7</p><p>5.8</p><p>9.2</p><p>14.4</p><p>22.4</p><p>34.5</p><p>52.5</p><p>63.7</p><p>77. I</p><p>91.6</p><p>107.0</p><p>123.3</p><p>!33.4</p><p>-</p><p>-</p><p>2.9</p><p>4.6</p><p>7.2</p><p>11.2</p><p>17.1</p><p>25.5</p><p>36.5</p><p>44.5</p><p>50.8</p><p>57.9</p><p>64.8</p><p>72.1</p><p>76.3</p><p>-</p><p>__</p><p>11.0</p><p>17.3</p><p>27.6</p><p>43.8</p><p>68.3</p><p>105.3</p><p>160.8</p><p>195.5</p><p>237.0</p><p>283.0</p><p>331.5</p><p>384.0</p><p>416.0</p><p>-</p><p>-</p><p>3.5</p><p>5.5</p><p>8.7</p><p>13.7</p><p>21.2</p><p>32.4</p><p>48.9</p><p>58.8</p><p>70. I</p><p>82.4</p><p>85.0</p><p>108.4</p><p>116.4</p><p>-</p><p>-</p><p>1.9</p><p>3.0</p><p>4.7</p><p>7.3</p><p>11.2</p><p>17.1</p><p>25.2</p><p>30.1</p><p>34.9</p><p>40.3</p><p>45.7</p><p>51.1</p><p>54.3</p><p>-</p><p>__</p><p>11.0</p><p>17.5</p><p>27.9</p><p>44. I</p><p>68.7</p><p>106.4</p><p>163.0</p><p>199.3</p><p>242.5</p><p>289.5</p><p>341 .o</p><p>396.5</p><p>430.0</p><p>-</p><p>I . = available short-circuit currcnt in kiloampercs at source end of eahle</p><p>I , = short-circuit current in kiloamperrs for short circuit a t end of cahlc of length L</p><p>PO SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>TABLE 1.10 Limiting Effect of Cable an Short-circuit Currents ot 208 Volts,</p><p>Three Phase</p><p>Three Single-condudor Cables in a Magnetic Duct</p><p>9.5</p><p>15.0</p><p>23.9</p><p>38.0</p><p>59.2</p><p>Conductor</p><p>sire</p><p>3.0</p><p>4.8</p><p>7.5</p><p>11.9</p><p>18.4</p><p>No. 14 Awg.. . . .. .</p><p>No. 12 Awg.. . . . . .</p><p>No. 10 Awg.. . . . . .</p><p>No. 8 Awg.. . . . . . . ,</p><p>No. 6 Awg.. . . . . . .</p><p>No. 4 Awg.. . . . . . .</p><p>No. 2 Awg.. . . . . . .</p><p>No. 1 Awg ........</p><p>No. 110 Awg .. . . . .</p><p>No. 2/0 Awg.. . . . . ,</p><p>No. 310 Awg.. . . . . ,</p><p>No. 4/0 Awg.. . . . . ,</p><p>250 MCM.. . . . . . . . ,</p><p>9.5</p><p>15.1</p><p>24.2</p><p>38.2</p><p>59.6</p><p>25</p><p>5</p><p>3.1</p><p>5.0</p><p>7.9</p><p>12.5</p><p>19.5</p><p>25</p><p>50 I lo: I lG</p><p>91.3</p><p>139.4</p><p>169.3</p><p>205.3</p><p>245.0</p><p>287.0</p><p>132.5</p><p>160.0</p><p>100</p><p>25</p><p>28.1</p><p>42.4</p><p>50.9</p><p>60.7</p><p>71.5</p><p>82.4</p><p>94.0</p><p>100.8</p><p>100</p><p>50</p><p>92.2</p><p>141.4</p><p>172.5</p><p>210.0</p><p>L51.0</p><p>295.5</p><p>143.5</p><p>172.0</p><p>-</p><p>9.3</p><p>14.8</p><p>23.4</p><p>37.0</p><p>57.4</p><p>88.0</p><p>133.4</p><p>162.1</p><p>196.0</p><p>230.5</p><p>269.0</p><p>308.5</p><p>334.0</p><p>29.9</p><p>45.5</p><p>55.3</p><p>67.0</p><p>79.4</p><p>92.7</p><p>106.8</p><p>115.5</p><p>-</p><p>2.5</p><p>3.9</p><p>6.2</p><p>9.7</p><p>14.8</p><p>22.1</p><p>31.6</p><p>38.6</p><p>44.0</p><p>50.2</p><p>56.2</p><p>62.4</p><p>66.0</p><p>Cable length 1, h</p><p>-</p><p>I .6</p><p>2.6</p><p>4.1</p><p>6.3</p><p>9.7</p><p>14.8</p><p>21 .9</p><p>26.1</p><p>10.3</p><p>14.9</p><p>19.6</p><p>i4.3</p><p>17. I</p><p>-</p><p>1.9</p><p>2.9</p><p>4.6</p><p>7.3</p><p>11.3</p><p>17.3</p><p>26.3</p><p>31.7</p><p>38.4</p><p>44.7</p><p>51.8</p><p>59.5</p><p>64.0</p><p>-</p><p>0.8</p><p>1.3</p><p>2.0</p><p>3.2</p><p>4.9</p><p>7.5</p><p>11.0</p><p>13.2</p><p>15.3</p><p>17.7</p><p>19.2</p><p>22.4</p><p>23.9</p><p>-</p><p>REACTANCE AND RESISTANCE DATA</p><p>FOR MACHINES AND CIRCUITS</p><p>When making short-circuit calculations, the most accurate reactance</p><p>data available should always be used. In particular, reactauce of specific</p><p>generators, larger motors, and transformers should be obtained from the</p><p>manufacturer.</p><p>Many short-circuit studies must he made without such specific data</p><p>available, as for a proposed plant or in many older plants where the time</p><p>and work required to obtain such data from the manufacturers make it</p><p>impractical to do so. Since a great many short-circuit calculations fall in</p><p>this category, it is desirable to use approximate reactance data. Such</p><p>approximate data as are commonly used are given in Tables 1.11 to 1.31.</p><p>The most applicable reactances should be selected from these tables.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 91</p><p>10-25</p><p>10-35</p><p>18-35</p><p>17-22</p><p>28-38</p><p>10-20</p><p>15-25</p><p>25-45</p><p>APPROXIMATE MACHINE REACTANCESdO CYCLES</p><p>Large Induction Motors. The approximate short-circuit reactance of</p><p>an induction motor (or induction generator) in per cent on its own kva</p><p>base may be taken as</p><p>18</p><p>14</p><p>27</p><p>10</p><p>33</p><p>15</p><p>20</p><p>30</p><p>100</p><p>times normal stalled rotor current* Per cent X : =</p><p>15-30</p><p>20-40</p><p>25-60</p><p>The reactance of such a machine will generally be approximately as</p><p>given in Table 1.11 (in per cent on own kva base).</p><p>TABLE 1.1 1</p><p>Range M-t Common</p><p>15-25 20</p><p>TABLE 1.12 Approximate Reactances of 60-cycle Synchronous Machines</p><p>Per Cent Vdues on Moshino Kva Roting</p><p>I I</p><p>23</p><p>30</p><p>40</p><p>Salient-polo cpnerotors (without amortirre,url:</p><p>12 poles 0, leu. .....................</p><p>1 4 p o l n o i m n e .....................</p><p>12 pole* or In.. ..................... Salient-pole ganomton~ (with amortiiseur):</p><p>14 poles or more.. ...................</p><p>Synchmnoui condenrers. .................</p><p>Synchronwi converterd</p><p>600 v d h dc.. .......................</p><p>250 d t s dc.. .......................</p><p>1.4).</p><p>As soon as a short circuit is established, the voltage on the system is</p><p>reduced to a very low value. Consequently, the motor stops delivering</p><p>energy to the mechanical load and starts slowing down. However, the</p><p>inertia of the load and motor rotor tends to prevent the motor from slow-</p><p>ing down. In other words, the rotating energy of the load and rotor</p><p>drives the synchronous motor just as the prime mover drives a generator.</p><p>6 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>The synchronous motor then becomes a generator and delivers short-</p><p>circuit current for many cycles after the short circuit occurs on the system.</p><p>Figure 1.5 shows an oscillogram of the current delivered by a synchronous</p><p>motor during a system short circuit. The amount of current depends</p><p>upon the horsepower, voltage rating, and reactance of the synchronous</p><p>motor and the reactance of the system to the point of short circuit.</p><p>LOAD CURRENT</p><p>SYNCHRONOUS</p><p>MOTOR</p><p>-€t</p><p>FIG. 1.4 Normally motors draw</p><p>load current from the source or</p><p>utility system but produce rhort-</p><p>circuit current when a short cir-</p><p>wit occurs in the d a d .</p><p>U l I L I T Y</p><p>SYSTEM</p><p>,- \</p><p>SHORT CIRCUIT</p><p>CURRENT FROM</p><p>MOTOR</p><p>. . -. . .</p><p>SYSTEM</p><p>SYNCMOYOUS '</p><p>FIG 1 5 IBmlowl l roce of 0s-</p><p>Yoroll</p><p>. . - . . ._ ,. _ _ .. ,. . . .. . . ..</p><p>cillogrclm of short-circuit current</p><p>produced by a synchronous</p><p>motor</p><p>SHORT ' . - I</p><p>CIRCUIT</p><p>SHORT CIRCUIT CURRENT DELIVERED BY</p><p>A SYNCHRONOUS MOTOR.</p><p>SHORT.CIRCUIT-CURRENT CALCULATING PROCEDURES 7</p><p>HOW INDUCTION MOTORS PRODUCE SHORT-CIRCUIT CURRENT</p><p>The inertia of the load and rotor of an induction motor has exactly the</p><p>same effect on an induction motor as on a synchronous motor; i.e., it</p><p>drives the motor after the system short circuit occurs. There is one</p><p>major difference. The induction motor has no d-c field winding, but</p><p>there is a flux in the induction motor during normal operation. This flux</p><p>acts like flux produced by the d-c field winding in the synchronous motor.</p><p>The field of the induction motor is produced by induction from the</p><p>stator rather than from the d-c winding. The rotor flux remains normal</p><p>as long as voltage is applied to the stator from an external source. How-</p><p>ever, if the external source of voltage is removed suddenly, as it is when a</p><p>short circuit occurs on the system, the flux in the rotor cannot change</p><p>instantly. Since the rotor flux cannot decay instantly and the inertia</p><p>drives the induction motor, a voltage is generated in the stator winding</p><p>causing a short-circuit current to flow to the short circuit until the rotor</p><p>flux decays to zero. To illustrate the short-circuit current from an</p><p>induction motor in a practical case, oscillograms were taken on a wound-</p><p>rotor induction motor rated 150 hp, 440 volts, 60 cycles, three phase, ten</p><p>poles, 720 rpm. The external rotor resistance was short-circuited in each</p><p>case, in order that the effect might he similar to that which would he</p><p>obtained with a low-resistance squirrel-cage induction motor.</p><p>Figure 1.6 shows the primary current when the machine is initially</p><p>running light and a solid three-phase short circuit is applied a t a point in</p><p>the circuit close to its input (stator) terminals a t time TI. The current</p><p>shown is measured on the motor side of the short circuit; so the short-</p><p>circuit current contribution from the source of power does not appear, but</p><p>only that contributed by the motor. Similar tests made with the machine</p><p>initially running a t full load show that the short-circuit current produced</p><p>T.</p><p>FIG. 1.6 , Tracer of oxillograms of short-circuit currents produced by an induction motor</p><p>running at light load.</p><p>8 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>by the motor when short-circuited is substantially the same, regardless of</p><p>initial loading on the motor. Note that the maximum current occurs in</p><p>the lowest trace on the oscillogram and is about ten times rated full-load</p><p>current. The current vanishes almost completely in four cycles, since</p><p>there is no sustained field current in the rotor to provide flux, as in the</p><p>case of a synchronous machine.</p><p>The flux does last long enough to prodnce enough short-circuit current</p><p>to affect the momentary duty on circuit breakers and the interrupting</p><p>duty on devices which open within one or two cycles after a short circuit.</p><p>Hence, the short-circuit current produced by induction motors must he</p><p>considered in certain calculations. The magnitude of short-circuit cur-</p><p>rent produced by the induction motor depends upon the horsepower,</p><p>voltage rating, reactance of the motor, and the reactance of the system to</p><p>the point of short c. "cuit. The machine impedance, effective a t the time</p><p>of short circuit, cmesponds closely with the impedance a t standstill.</p><p>Consequently, the i iitial symmetrical value of Short-circuit current is</p><p>approximately equnl to the full-voltage starting current of the motor.</p><p>TRANSFORMERS</p><p>Transformers are often spoken of as a source of short-circuit current.</p><p>Strictly speaking, this is not correct, for the transformer merely delivers</p><p>the short-circuit current generated by generators or motors ahead of the</p><p>transformer. Transformers merely change the system voltage and mag;</p><p>nitude of current but generate neither. The short-circuit current deliv-</p><p>ered by a transformer is determined by its secondary voltage rating and</p><p>reactance, the reactance of the generators and system to the terminals of</p><p>the transformer, and the reactance of the circuit from the transformer to</p><p>the short circuit.</p><p>ROTATING-MACHINE REACTANCE</p><p>The reactance of a rotating machine is not one simple value as it is for a</p><p>transformer or a piece of cable, but is complex and variable with time.</p><p>For example, if a short circuit is applied to the terminals of a generator,</p><p>the short-circuit current behaves as shown i n Fig. 1.7. The current starts</p><p>out a t a high value and decays to a steady state after some time has</p><p>elapsed from the inception of the short cirroit. Since the field excitation</p><p>voltage and speed have remained snbstantially constant within the short</p><p>interval of time considered, a change of apparent react,ance of the machine</p><p>may he assumed, to explain the change in the magnitude of short-circuit</p><p>current with time.</p><p>The expression of such variable reactance at any instant after the</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 9</p><p>occurrence of any short circuit requires a complicated formula involving</p><p>time as one of the variables. For the sake of simplification in short-cir-</p><p>cuit calculating procedures for circuit-breaker and relay applications,</p><p>three values of reactance are assigned to generators and motors, viz.,</p><p>subtransient reactance, transient reactance, and synrhronous reactance.</p><p>The three reactances can be briefly described as follows:</p><p>1. Subtransient reactance X y is the apparent reactance of the stator</p><p>winding at the instant short circuit occurs, and it determines the current</p><p>Row during the first few cycles of a short circuit.</p><p>2. Transient reactance X i is the apparent initial reactance of the</p><p>stator winding, if the effect of all amortisseur windings is ignored and</p><p>only the field winding considered. This reactance determines the cur-</p><p>rent following the period when subtransient reactance is the controlling</p><p>value. Transient reactance is effective up to 45 see or longer, depending</p><p>upon the design of the machine.</p><p>3. Synchronous reactance X d is the apparent reactance that deter-</p><p>mines the current flow when a steady-state condition is reached. It is not</p><p>effective until several seconds after the short circuit occurs; consequently,</p><p>it has no value in short-circuit calculations for the application of circuit</p><p>breakers, fuses, and contactors but is useful for relay-setting studies.</p><p>Figure 1.8 shows the variation of current with time and associates the</p><p>various reactances mentioned above with the time and current scale.</p><p>Previous loading has an effect on the total magnitude of short-circuit</p><p>CURRENT DETERMINED</p><p>6 pole ..............................</p><p>8-14 pole ...........................</p><p>I 6 pole or more. .......................</p><p>Synchronous motor^'</p><p>15-35 25</p><p>25-45 35</p><p>a Nearly all salient-pole generators built by General Electric Company since 1935 have</p><p>amortisseur windings.</p><p>Add transformer reactance:</p><p>For compound-wound converters add 12 per cent.</p><p>For shunt-wound converters add 7 per cent.</p><p>These data are useful for estimating reactances of individual large motors of</p><p>several hundred or several thoumnd horsepower.</p><p>* With rated voltage and frequency applied.</p><p>92 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>TABLE 1.13 Approximate Reactance of General Electric Company</p><p>Turbine Generators, 625 to 18,750 Kva</p><p>K w ..ling</p><p>0.8 power facer</p><p>625</p><p>781</p><p>875</p><p>937</p><p>1,250</p><p>1,562</p><p>1,857</p><p>w rating</p><p>500</p><p>625</p><p>700</p><p>750</p><p>1,000</p><p>1,250</p><p>1,500</p><p>Volt.g* rating</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4,160</p><p>6,900</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4.160</p><p>6,900</p><p>240-4.1 60</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4,160</p><p>6.900</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4,160</p><p>6,900</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4,160</p><p>6.900</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4.1 60</p><p>6,900</p><p>13.800</p><p>1200</p><p>rpm</p><p>14.5</p><p>14.5</p><p>14.5</p><p>14.5</p><p>14.0</p><p>....</p><p>....</p><p>....</p><p>....</p><p>.... .... ....</p><p>14.0</p><p>.... .... ....</p><p>.... ....</p><p>....</p><p>....</p><p>15.5</p><p>15.5</p><p>15.5</p><p>15.5</p><p>....</p><p>....</p><p>.... .... .... ....</p><p>....</p><p>....</p><p>16.0</p><p>16.0</p><p>16.0</p><p>16.0</p><p>16.0</p><p>16.0</p><p>__</p><p>3600</p><p>rpm</p><p>~</p><p>8.0</p><p>9.0</p><p>9.0</p><p>9.0</p><p>9.5</p><p>6.5</p><p>8.5</p><p>8.5</p><p>7.5</p><p>9.0</p><p>9.5</p><p>5.5</p><p>11.5</p><p>7.5</p><p>7.0</p><p>7.5</p><p>7.0</p><p>9.0</p><p>10.0</p><p>10.0</p><p>10.0</p><p>9.0</p><p>10.0</p><p>8.5</p><p>9.5</p><p>8.5</p><p>8.5</p><p>9.5</p><p>9.0</p><p>8.0</p><p>10.5</p><p>9.0</p><p>8.5</p><p>8.5</p><p>9.5</p><p>7.5</p><p>-</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 93</p><p>rABLE 1.13 Approximate Reactance of General Electric Company</p><p>Turhine Generators, 625 to 18,750 Kva. (Continued)</p><p>Kvn rating</p><p>0.8 power facto,</p><p>2,500</p><p>3.125</p><p>3750</p><p>4,375</p><p>5.000</p><p>6,250</p><p>:w ‘atin(</p><p>2,000</p><p>2,500</p><p>3,000</p><p>3,500</p><p>4,000</p><p>5,000</p><p>Vdtoge ‘ding</p><p>480</p><p>600</p><p>2,400</p><p>4,160</p><p>6,900</p><p>l1.500-13.800</p><p>480</p><p>600</p><p>2,400</p><p>2.400/4.160</p><p>6.900</p><p>I 1,500</p><p>13,800</p><p>480</p><p>600</p><p>2,400</p><p>6.900</p><p>13,800</p><p>480</p><p>600</p><p>2,400</p><p>2,400/4,160</p><p>6,900</p><p>11,500</p><p>13,800</p><p>480</p><p>600</p><p>2.400</p><p>2,400/4,160</p><p>11,500</p><p>2,400/4.160</p><p>6,900</p><p>1 1,500</p><p>13,800</p><p>600</p><p>2,400</p><p>2.400/4,160</p><p>6,900</p><p>I 1,500</p><p>13,800</p><p>X&‘</p><p>3600 rpm</p><p>P.5</p><p>10.5</p><p>10.0</p><p>10.0</p><p>10.0</p><p>8.0</p><p>9.0</p><p>8.5</p><p>9.5</p><p>8.5</p><p>9.0</p><p>10.0</p><p>10.5</p><p>9.0</p><p>10.5</p><p>9.5</p><p>10.0</p><p>9.5</p><p>10.5</p><p>10.5</p><p>8.0</p><p>9.0</p><p>9.0</p><p>9.0</p><p>9.0</p><p>10.0</p><p>10.0</p><p>10.5</p><p>7.5</p><p>7.0</p><p>8.5</p><p>9.0</p><p>9.5</p><p>9.5</p><p>12.0</p><p>7.5</p><p>7.5</p><p>8.0</p><p>8.5</p><p>8.5</p><p>94 SHORT.CIRCUIT~CURRENT CALCULATING PROCEDURES</p><p>TABLE 1.13 Approximate Reactance of General Electric Company</p><p>Turbine Generators. 625 to 18.750 Kva. (Continued)</p><p>Kvo rotinp</p><p>0.8 power f.Ct0.</p><p>7,500</p><p>9.375</p><p>12,500</p><p>i 8750</p><p>-</p><p>w rating</p><p>-</p><p>6,000</p><p>7.500</p><p>10,000</p><p>15,000</p><p>-</p><p>Voltage rating</p><p>2,400</p><p>2,400/4,160</p><p>6.900</p><p>1 1,500</p><p>13.800</p><p>2,400</p><p>6,900</p><p>2.400/4,160</p><p>11.500</p><p>13,800</p><p>2,400/4,160</p><p>6,900</p><p>11.500</p><p>13.800</p><p>6,900</p><p>11,500</p><p>13.800</p><p>X;</p><p>3600 rpm</p><p>9.0</p><p>10.5</p><p>9.0</p><p>9.5</p><p>9.5</p><p>9.0</p><p>10.5</p><p>9.0</p><p>9.5</p><p>9.5</p><p>10.0</p><p>8.0</p><p>9.0</p><p>8.0</p><p>11.0'</p><p>11.0'</p><p>11.0'</p><p>* 0.5 psig hydrogen pressure.</p><p>TABLE 1.14 Reactances Based on Kvo of Connected Motors</p><p>I 1</p><p>Itern Motor rotings ond connections</p><p>I per rent</p><p>I 1 -</p><p>600 "0th or less-induction</p><p>600 volts or lewynchronous l i tem 1 end 2 indude motor leads1</p><p>600 volh or leuinduct ion</p><p>600 volts or les-ynchronour litems 3 and 4 indude motor leads</p><p>Motors above 600 volt-induction</p><p>Motors above 600 volt-ynchronwr</p><p>Motors above 600 volh-indudon</p><p>Motors obwe 600 voltriynchromur litems 7 and 8 include step-</p><p>and step-down bansformen1</p><p>down transformers1</p><p>28'</p><p>21</p><p>34'</p><p>27*</p><p>20</p><p>15</p><p>26</p><p>21</p><p>I 1</p><p>* Based on AIEE Standard No. 20.</p><p>Tranrient</p><p>reactance</p><p>Xi .</p><p>per cent</p><p>29</p><p>35</p><p>25</p><p>31</p><p>-</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 95</p><p>In many</p><p>short-circuit studies, the number and size of motors, either induction or</p><p>synchronous, are not known precisely. However, the short-circuit contri-</p><p>bution from these motors must be estimated. In such cases Table 1.14 is</p><p>used to account for a large number of small induction and synchronous</p><p>motors.</p><p>The proportions of synchronous and induction motors (at all voltages)</p><p>should be known for short-circuit investigations. Some typical ratios of</p><p>total plant motor load which are usable in preliminary work are given in</p><p>Table 1.15.</p><p>The kva of the motors which are energized at one time varies also with</p><p>the type of plant and should be investigated for the more complete</p><p>studies. Approximate relations of energized to installed motors and of</p><p>energized motors to source (transformer and/or generator) capacity are</p><p>given in Table 1.16.</p><p>Assumed Motor Reactances-Group of Small Motors.</p><p>Cement ..............................</p><p>Machine shops ond IexHIe.. .............</p><p>Rubber and rolling mills.. ...............</p><p>Paper (excluding grinder mobs). ........</p><p>Commercial ond offiso.. ................</p><p>TABLE 1.15 Rotio of Induction and Synchronous Motors</p><p>40 60</p><p>85 15</p><p>50 50</p><p>67 33</p><p>50 50</p><p>Motor mio, par cent</p><p>Plant</p><p>Induction Sydrnnom</p><p>Continuous PIOLOS (cement. textile). .............</p><p>Semicontinuous (paper, reflnerier, rubberl.. .......</p><p>Rolling mills.. ...............................</p><p>Intermittent operotiom. .......................</p><p>100</p><p>90</p><p>80</p><p>75</p><p>TABLE 1.16 Rotio of Energized and Instolled Motors</p><p>PI.3</p><p>Energized</p><p>motor kva</p><p>to insbled</p><p>motor kw ,</p><p>per cent</p><p>Installed</p><p>motor k w to</p><p>source kva</p><p>(excluding</p><p>SPW").</p><p>por cent</p><p>110 110</p><p>1 67</p><p>215</p><p>400</p><p>96 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>APPROXIMATE IMPEDANCE OF TRANSFORMERS</p><p>The impedance of transformers ronsidered in a short-circuit study</p><p>should be obtained from the name platc or the manufacturer. However,</p><p>where such data cannot be obtained, the values given in Tables 1.17 to</p><p>1.19 may be used in short-rircuit studies for estimating the short-circuit</p><p>currents in the usual case.</p><p>In the usual short-circuit study, the transformer reactance and imped-</p><p>ance may be assumed to be the same without causing significant error for</p><p>transformer banks above 300 kva. This assumption is useful because</p><p>transformer name-plate data include impedance and not reactance.</p><p>Approximate Resistance, Reactance, and Impedance of TABLE 1.17</p><p>Single-phase Distribution Transformers</p><p>3</p><p>5</p><p>10</p><p>15</p><p>25</p><p>37></p><p>50</p><p>75</p><p>100</p><p>1 67</p><p>250</p><p>333</p><p>500</p><p>High voltage: 2400/416OY volts and</p><p>ow voltage; 120/240,240/480.600 volts-</p><p>2400/4800/8320Y "Olt.</p><p>Per cent</p><p>R</p><p>1.7</p><p>1.5</p><p>I .3</p><p>I .2</p><p>1 . 1</p><p>I .o</p><p>60 yclos</p><p>Per cent</p><p>X</p><p>I .5</p><p>1.7</p><p>2.2</p><p>2.3</p><p>3 . 8</p><p>4.7</p><p>Per cent</p><p>z</p><p>2.3</p><p>2.3</p><p>2.6</p><p>2.6</p><p>4.0</p><p>4.8</p><p>High voltage; 7200/12,47OY v01b</p><p>ow voltage: 120/240. 240/480,600 volts-</p><p>60 cycles</p><p>Per cent</p><p>R</p><p>2.2</p><p>I .6</p><p>1.3</p><p>1.2</p><p>1.0</p><p>1 .o</p><p>Per. cent</p><p>X</p><p>1.7</p><p>I .6</p><p>2.0</p><p>3.5</p><p>3.6</p><p>5.1</p><p>Per cent</p><p>z</p><p>~-</p><p>2.8</p><p>2.3</p><p>2.4</p><p>3 .7</p><p>3.7</p><p>5.2</p><p>APPROXIMATE REACTANCE AND RESISTANCE OF CABLES</p><p>The reactance of a cable circuit is, generally speaking, a function of the</p><p>spacing between conductor centers and the conductor diameter. Know-</p><p>ing the conductor spacing and diameter, the reactance of three-conductor</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 97</p><p>Approximate Impedance of 60-cycle Power Transformers TABLE 1.18</p><p>IAbovs 500 Kva)</p><p>High voltage</p><p>Impedance at kvo bole</p><p>equal lo 55 C rating of</p><p>largest Capacity winding for</p><p>Inwlotion doss, k r</p><p>Self-cooled or</p><p>Low voltage woter-cooled</p><p>rating, per cent</p><p>I 5 or lower. .....</p><p>25. . ...........</p><p>34.5. . .........</p><p>4 6 . 0 . . .........</p><p>6 9 . 0 . . . ........</p><p>9 2 . 0 . . .........</p><p>I 15.0. ..........</p><p>138.0 ...........</p><p>Forced-oil</p><p>cooled</p><p>rating, per Cenl</p><p>15 or lower 5 %</p><p>15 or lower 5 %</p><p>I 5 or lower 6%</p><p>I 5 or lower 7 %</p><p>15orlower 856</p><p>I5 or lower 6</p><p>I 5 o r lower 7</p><p>I 5 or lower 8</p><p>For high-voltagr insulation elassrs intermediatr of those given, use the imppdancr</p><p>For transforrncrs with a load-ratio control add 0.5 prr ccnt to the vaIu?s IistFd</p><p>Thc p ~ r c m t resistance on the hase given above rangrs from 1.0 down</p><p>to 0.06.</p><p>Approximate Reactance of Load-center-type Transformers,</p><p>of thc next higher listpd insulation class.</p><p>abovc crcrpt in those eases in which a IOWPY impedaner has heen sprrifirtl.</p><p>TABLE 1.19</p><p>60 Cycles</p><p>(Three-phase)</p><p>15-kv Maximum Primary Voltage</p><p>600-volt Maximum Secondarr Voltaoe -</p><p>Kro Range</p><p>I l256-l50 3.0</p><p>225-500 5 .0</p><p>750-2000 5 .5</p><p>Per cent Reactance on Own Kro Bore*</p><p>* Per cent resistance on own kva base is apptoiirnatcly 1.5 p ~ r ccnt for 150 kva</p><p>snd b&w and varips from approxirnatdy 1 down to 0.8 p ~ r c m t on ratings above</p><p>150 kva</p><p>cables in nonmagnetir ducts and without maglietic binders can be deter-</p><p>mined by the formula</p><p>X = 0.023 log, D + K ( 2s )</p><p>X = reactance, ohms per 1000 ft at 60 ryrlrs; S = spacing of couduc-</p><p>D = diameter of ronductors, in.; K = a rocffirient dependent upon ratio</p><p>For</p><p>tors (center to center), in.</p><p>of iriside diameter of a ronductor to outside diameter of condurtors.</p><p>standard strand construction K = 0.25</p><p>98 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>This formula does not take into account any increase of reactance due</p><p>to the spiraling of the strands. Such increase is usually negligible in</p><p>three-conductor cables and in large single-conductor cables, but it may</p><p>amount to 1 to 2 per cent in small single-conductor cables.</p><p>The effect of irregular spacing of the conductors and of magnetic</p><p>binder causes an increase of reactance of single-conductor cables, com-</p><p>pared with otherwise equivalent three-conductor cables. Cable insula-</p><p>tion thickness varies with different types of insulation for a cable of a</p><p>given voltage class. The approximate reactances of cables taking into</p><p>account these variables are shown in Tables 1.20 t o 1.22.</p><p>TABLE 1.20 Approximate Resistance, Reactance, and Impedance of</p><p>600-volt Cables in Magnetic Ducts per 100 Ft</p><p>Coble size</p><p>No. 14 Awg.</p><p>No. 12 Awg .</p><p>No. 10 Awg.</p><p>No. 8 Awg . .</p><p>No. 6 Awg . .</p><p>No. 4 Awg . .</p><p>NO. 2 Awg . .</p><p>No. I Awg..</p><p>No.l/OAwg.</p><p>No.Z/OAwg.</p><p>No. 3/0 Awg .</p><p>No.d/OAwg</p><p>25OMCM...</p><p>300 M C M . . .</p><p>350 M C M . . .</p><p>400MCM...</p><p>500MCM...</p><p>750 MCM.. .</p><p>Three single-conductor cables per dud.</p><p>ohms per 100 fi</p><p>R'</p><p>0.3135</p><p>0.1972</p><p>0.1240</p><p>0.0779</p><p>0.0498</p><p>0.0318</p><p>0.0203</p><p>0.0163</p><p>0.0131</p><p>0.0106</p><p>0.00860</p><p>0.00700</p><p>0.00608</p><p>0.00520</p><p>0.00461</p><p>0.00419</p><p>0.00359</p><p>0.00280</p><p>* Based on 75 C.</p><p>X</p><p>0.00765</p><p>0.0071 0</p><p>0.00687</p><p>0.00638</p><p>0.00598</p><p>0.00551</p><p>0.00513</p><p>0.00500</p><p>0.00495</p><p>0.00490</p><p>0.00486</p><p>0,00482</p><p>0.00480</p><p>0.00474</p><p>0.00469</p><p>0.00462</p><p>0.00450</p><p>0.00438</p><p>-</p><p>0.3135</p><p>0.1972</p><p>0.1240</p><p>0.0782</p><p>0.0500</p><p>0.0322</p><p>0.0209</p><p>0.0171</p><p>0.0140</p><p>0.0117</p><p>0.00986</p><p>0.00850</p><p>0,00778</p><p>0.00704</p><p>0.00658</p><p>0.00625</p><p>0.00575</p><p>0.00520</p><p>-</p><p>Three-conductor cable including inter-</p><p>locked armor cablo, ohms per 100 fi</p><p>R'</p><p>0.3135</p><p>0.1972</p><p>0.1240</p><p>0.0779</p><p>0.0493</p><p>0.0312</p><p>0.0197</p><p>0.0157</p><p>0.0125</p><p>0.0100</p><p>0.00800</p><p>0.00640</p><p>0.00547</p><p>0.00460</p><p>0.00400</p><p>0.00354</p><p>0.00292</p><p>0.00208</p><p>X</p><p>0.00468</p><p>0.00456</p><p>0.00448</p><p>0.00427</p><p>0.00391</p><p>0.00362</p><p>0.00344</p><p>0.00342</p><p>0.00340</p><p>0.00336</p><p>0.00333</p><p>0.00327</p><p>0.00322</p><p>0.00316</p><p>0.0031 0</p><p>0.00304</p><p>0.00295</p><p>0.00284</p><p>-</p><p>Z</p><p>0.3 1352</p><p>0.19720</p><p>0.12410</p><p>0.07460</p><p>0.04899</p><p>0.03140</p><p>0.02000</p><p>0.01606</p><p>0.01296</p><p>0.01 054</p><p>0.00866</p><p>0.00721</p><p>0.00632</p><p>0.00557</p><p>0.00510</p><p>0.00469</p><p>0.00412</p><p>0.00346</p><p>-</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 99</p><p>Approrimote Resistance. Reactance, and Impedance of TABLE 1.21</p><p>5000-volt Cables in Magnetic Ducts per 100 Ft</p><p>Fo&r for conesting</p><p>reactancn, dl rizor</p><p>of cable</p><p>Cable size</p><p>Focton for correcting redrlmces</p><p>No. l 4 t o I No.610 1 N o . 0 0 t o 1 30010 1 No. 8 A x g No. 0 Awg 250 MCM 500 MCM 750 MCM</p><p>No. 14 Awg .</p><p>No. 10 Awwg.</p><p>No. 8 Awg . .</p><p>No. 6 Awg..</p><p>N o . 4 A r g . .</p><p>No. 2 A x g . .</p><p>No. 1 Awg..</p><p>No. 110 Awg</p><p>No. 2/0 Awg .</p><p>No.3/0 A x @ .</p><p>No.4/0 Awg.</p><p>250 MCM.. .</p><p>300 MCM.. .</p><p>350MCM...</p><p>400 MCM.. .</p><p>500 MCM.. .</p><p>750 MCM. ..</p><p>Based (1</p><p>Three dngle.condudor cobler per duct,</p><p>ohms per 100 ft</p><p>R*</p><p>0.3135</p><p>0.1240</p><p>0.0779</p><p>0.0498</p><p>0.0318</p><p>0.0203</p><p>0.0163</p><p>0.0131</p><p>0.0106</p><p>0.00860</p><p>0.00700</p><p>0.00609</p><p>0.00520</p><p>0.00461</p><p>0.00419</p><p>0.00359</p><p>0.00280</p><p>75 c.</p><p>X</p><p>0.00969</p><p>O.OO8M</p><p>0.00788</p><p>0.00748</p><p>0.00681</p><p>0.00623</p><p>0.00588</p><p>0.00567</p><p>0.00545</p><p>0.00535</p><p>0.00529</p><p>0.00525</p><p>0.00519</p><p>0.00514</p><p>0.00506</p><p>0.00495</p><p>0.00474</p><p>-</p><p>___.</p><p>z</p><p>0.3135</p><p>0.1240</p><p>0.0781</p><p>0.0503</p><p>0.0325</p><p>0.0212</p><p>0.0173</p><p>0.0143</p><p>0.0119</p><p>0.0101</p><p>0.00877</p><p>0.00802</p><p>0.00735</p><p>0.00690</p><p>0.00657</p><p>0.00611</p><p>0.00551</p><p>-</p><p>Three.conductor cable including inter-</p><p>locked armor cable, ohms per 100 ft</p><p>R'</p><p>0.3135</p><p>0.1240</p><p>0.0779</p><p>0.0493</p><p>0.0312</p><p>0.0197</p><p>0.0157</p><p>0.0125</p><p>0.0100</p><p>0.00800</p><p>0.00640</p><p>0.00547</p><p>0.00460</p><p>0.00400</p><p>0.00354</p><p>0.00292</p><p>0.00208 -</p><p>X</p><p>0.006664</p><p>0.005745</p><p>0.005308</p><p>0.004941</p><p>0.004619</p><p>0.004366</p><p>0.003964</p><p>0.003792</p><p>0.003677</p><p>0.00363 1</p><p>0.003585</p><p>0.003562</p><p>0.00351 8</p><p>0.003477</p><p>0.003436</p><p>0.003344</p><p>0.003088</p><p>-</p><p>TABLE 1.22 Correction Factors for Nonmagnetic Ducts</p><p>Single-condudor able ,</p><p>z</p><p>0.3291</p><p>0.1241</p><p>0.07808</p><p>0.04944</p><p>0.03154</p><p>0.02017</p><p>0.01619</p><p>0.01304</p><p>0.01061</p><p>0.008785</p><p>0.007535</p><p>0.006527</p><p>0.005791</p><p>0.005299</p><p>0.004923</p><p>0.004439</p><p>0.003723</p><p>-</p><p>0.8 I 1.0 I 0.96 I 0.93 0.83 1 0.72</p><p>100 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>TABLE 1.22 Correction Factors for Nonmagnetic Ducts. (Continued)</p><p>Three-conductor Cables</p><p>Determine correct Z from corrected d ~ e i of X and R. No ~orreclion is required for interlocked</p><p>Three single-condudor cables in</p><p>iron conduit.. ..............</p><p>Three-conductor coble in iron con-</p><p>duit or interlocked armored</p><p>cable .....................</p><p>Three-conductor cable in nonmag-</p><p>netic duct.. ................</p><p>armor.</p><p>Factor for correcting</p><p>reoctancer. di sizes</p><p>of coble</p><p>98.3</p><p>I</p><p>71.8</p><p>58.5</p><p>0.87</p><p>15.74</p><p>11.5</p><p>9.4</p><p>Factors for correcting resistances</p><p>1.075</p><p>0.669</p><p>0.581</p><p>No. 14 to No. 00 Awg No. 0000 Awg to 750 MCM</p><p>0.222</p><p>0.194</p><p>1.0 1 0.98</p><p>I</p><p>0.11</p><p>0.0955</p><p>TABLE 1.23 Per Cent Reactance of Typical Three-phase Cable Circuits</p><p>Per Cent Reactance of 1000 Circuit Feet on o 1000-kva nose</p><p>Three single-conductor coblei in</p><p>iron conduit. ...............</p><p>Three-conductor cable in iron con-</p><p>duit or interlocked armored</p><p>cable. ....................</p><p>Three-conductor cable in nonmog-</p><p>netic duct.. ................</p><p>I</p><p>System roltoge 1 230 ! 460 1 575 ~ 2,400 j 4,160 ~ 6.900 1 13,800</p><p>I</p><p>92.5</p><p>68</p><p>54.8</p><p>Cqble sire. No. 4 to 1 Awg</p><p>0.2</p><p>0.173</p><p>0.0943</p><p>0.0818</p><p>24.6</p><p>18</p><p>14.7</p><p>Three ringlo-conductor cables in</p><p>ironconduit. ..............</p><p>Three-conductor cable in iron con-</p><p>duit or interlocked armored</p><p>coble .....................</p><p>Three-conductor cobie in nonmog-</p><p>netic duct.. ................</p><p>85 11.3</p><p>61.4 15.4</p><p>51 12.8</p><p>Cable size, No. I f 0 to No. 410 Awg</p><p>0.538 0.179</p><p>0.477 0.159</p><p>I I I</p><p>0.0796</p><p>0.07</p><p>23.2</p><p>17.1</p><p>13.72</p><p>14.85</p><p>10.9</p><p>8.8</p><p>0.955</p><p>0.6</p><p>0.52</p><p>Cable Sire, 250 to 750 MCM</p><p>13.63</p><p>9.85</p><p>8.19</p><p>-</p><p>1y 2.</p><p>0.358 ~l</p><p>0.0276</p><p>0.024</p><p>0.318 1 !</p><p>0.0237</p><p>0.0205</p><p>~</p><p>0.02</p><p>0.0176</p><p>__</p><p>SHORT.CIRCUIT-CURRENT CALCULATING PROCEDURES 101</p><p>Where more precise data are not available, the values given in Tables</p><p>1.20 t o 1.23 may be used in short-circuit-current calculations without</p><p>significant error.</p><p>Plvg-in type:</p><p>Upto600 .............</p><p>60110 1000 ...........</p><p>Upto600 .............</p><p>60110 1000 ...........</p><p>135010 1600 ..........</p><p>2000 .................</p><p>Lox-impedance type:</p><p>APPROXIMATE REACTANCE OF BUS B A R S 6 0 CYCLES</p><p>Unlike cable circuits the resistance of bus-bar circuits is so low com-</p><p>pared with the reactance that the resistances of bus bars may he neglected</p><p>in all a-c short-circuit calculations without significant error. There haye</p><p>been many papers written on the subject of bus-bar reactance calcula-</p><p>tions, and a complete bibliography is included in t,hc 1945 A I E E Tran-</p><p>sactions, Vol. 64, page 385, The Design of Bus-bar Indust,rial Distribution</p><p>System: An Epitomization of Available Data, by T . .J. Higgins. For</p><p>practical short-circuit calculations, the reactance of bus bars may</p><p>be</p><p>taken from Figs. 1.58 to 1.G2 or Tables 1.24 and 1.25.</p><p>TABLE 1.24 Reactance of Typical Three-phase Low-voltage Copper</p><p>Busway Circuits</p><p>Per cent reoctonce of 1000 circuit feet on D 1000-kva base</p><p>98.8 24.7 15.8</p><p>62 .4 15.6 10.0</p><p>45.2 11.4 7 . 3</p><p>17.2 4 .3 2.7</p><p>10.8 2.7 1.7</p><p>7.6 1.9 1.2</p><p>System voltage</p><p>Butuoy rating, amp</p><p>240 1 480 I 600</p><p>Although not gcnerally used in short-circuit calculations the resistance</p><p>of typical copper busway circuits is giveu in Table 1.25.</p><p>TABLE 1.25 Resistance of Typical Copper Busway Circuits</p><p>Current Capacity Resistance.</p><p>of Bvsroy, Amp Ohms pel 1000 Ft</p><p>250 0.114</p><p>400 0.033</p><p>600 0.023</p><p>800 0.016</p><p>1000 0.012</p><p>I350 0.0096</p><p>1600 0.0073</p><p>2000 0.0055</p><p>102 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>FIG. 1.58 Chart showing reactance VI. spacing of rectangular bus bars 160 cycler).</p><p>FIG. 1.59 Chart showing reactance VI. spacing of rectangular bur bars I60 cycler).</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>FIG. 1.60 Chart showing reactance YI. spacing of rectangular bus bars (60 cycled.</p><p>104 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>FIG. 1.62 Chart showing reactance VI. spacing of channel bus bars I60 cycler).</p><p>REACTANCE AND RESISTANCE OF OVERHEAD LINES</p><p>To assist in obtaining the conductor spacings, two typical crossarm</p><p>arrangements are shown in Fig. 1.63. The arrangements used in practice</p><p>will vary from system to system, but hecause of space limitations only</p><p>these two are shown.</p><p>For ordinary single-phase circuits, the equivalent spacing is the dis-</p><p>tance between conduct,ors. For ordinary t,hree-phase circuits, the equiva-</p><p>lent spacing is exprcssed by the formula + A X t( X C where A , B , and</p><p>C are the distances, center to center, of the conductors as follows:</p><p>- ~ I-A-I-B-~</p><p>The resistance of overhead lines may not always he neglected without</p><p>significant error. In general, long runs of overhead lines (several miles)</p><p>at 2.4 t o 13.8 kv with small conductors 250 MCM or less have significant</p><p>resistance compared to reactance; therefore resistance should be con-</p><p>sidered in short-circuit calculations for short circuits a t the ends of such</p><p>long overhead lines. Resistance should he considered in all low-voltage</p><p>(600 volts or less) overhead lines.</p><p>Reactances and resistances may be taken from Table 1.26 for small</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 105</p><p>spacings (up to 8 ft) and from Figs. 1.64 and 1.65 for spacings up t o</p><p>20 ft.</p><p>Under usual application conditions, transmissiodine reactance</p><p>varies over quite a narrow range. Table 1.27 includes the usual varia-</p><p>tions as well as “average ohms per mile” which are normally satis-</p><p>factory for quick estimating work. Very large conductors, used to</p><p>carry unusually large amounts of power for short distances, have abnor-</p><p>mally low reactance so that this tahlr is not applicable.</p><p>L--- 67“</p><p>4- P IN CROSSARM</p><p>AND S P O O L- T Y P E</p><p>SECONDARY RACK</p><p>6 - P I N CROSSA-M</p><p>FIG, 1.63</p><p>tance on 2400/4160-Y or 48OO-volt circuits.</p><p>Spocing of pins on four- and six-pin crossarms for vie in calculating line reoc-</p><p>106 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>FIG. 1.64</p><p>hard-drown stranded copper conducton (60 cycle).</p><p>Chart showing poritive-phaie-sequence reactonce of transmission lines using</p><p>SHORT-CIRCUIT-CURRENT CALCUUTING PROCEDURfS 1 07</p><p>POSITIVE SEQUENCE 60 CYCLE</p><p>EQUIVALENTA SPACING OF CONDUCTORS IN FEET</p><p>FIG. 1.65</p><p>ACSR conductors (60 cycler).</p><p>Chart ahowing poritive-pha**.requence reactance of trammission liner uting</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>SHORT-CIRCUIT-CURRENT CALCUUTING PROCEDURES I09</p><p>TABLE 1.27 Approximate Equivalent Delta Spacing and Average</p><p>Reactance r Mi le of Three-phase 60-cycle Transmission Lines</p><p>40. 410 Awg</p><p>No. 6 Awg</p><p>~-</p><p>250 MCM</p><p>250 MCM</p><p>400 MCM</p><p>500 MCM</p><p>No. 4 Awg</p><p>No. 4 Awg</p><p>No. 2 Awg</p><p>No. 1 Awg</p><p>411 UIUDI sizes</p><p>Line in.ul.tion</p><p>class, kv</p><p>0.61</p><p>0.74</p><p>0.64</p><p>0.75</p><p>0.65</p><p>0.77</p><p>0.64</p><p>0.75</p><p>0.65</p><p>0.77</p><p>__</p><p>0.70-0.80</p><p>both copper</p><p>and oluminum</p><p>5</p><p>Wire sire</p><p>N 0 . 4 l o N 0 . 1 A w g _ _ . . .</p><p>No. I i O A w g t a 2 5 0 M C M</p><p>300 lo 750 MCM.. . . . . .</p><p>15</p><p>23</p><p>34.5</p><p>46</p><p>Per cent reoctance of I000 ci r~ui l feet on D 1000-kvo bore</p><p>180208~22345 .052 .156 .028 .833 .335 .8 2.19 0.7620.286 0.073</p><p>155 i8520238.846.550.724.8 29.732.4 2.06 0.6880.258 0.067</p><p>134 1631180 33.6 40.8 45.0 21.5 26.1 28.8 1.87 0.625 0.235 0.061</p><p>I</p><p>69</p><p>115</p><p>138</p><p>161</p><p>220</p><p>287</p><p>Approximate</p><p>equi*o1ent</p><p>dell. spocing of</p><p>conductors. ft</p><p>2.5</p><p>3.5</p><p>4</p><p>4.5</p><p>5.5</p><p>14 8</p><p>16</p><p>20</p><p>20</p><p>40</p><p>Normal tronrmiirion</p><p>Reoctonce,</p><p>ohms per mile</p><p>lslrmded ~oppei l</p><p>size</p><p>0.65</p><p>0.70</p><p>0.75</p><p>TABLE 1.28 Reactance of Typical Three-phase Medium- and</p><p>Low-voltage Distribution Circuits*</p><p>System Yoiloge.. , . . . , . ~ 230 ~ 460 575</p><p>Equivalent delta . spacing, . . . . . . . ~ 6 ~ 1 2 ~ 1 8 ~ 6 / 1 2 1 1 8 1 d / I 2 / 1 8 1 301 301 361 42</p><p>I". . . . . . . . .</p><p>110</p><p>Volts Equivalent dslto spacing,</p><p>(line-to-lid in.</p><p>* 115 12</p><p>230 12</p><p>460 I8</p><p>575 I8</p><p>2,300 30</p><p>4,160 30</p><p>6,900 36</p><p>13.800 42</p><p>22,000 48</p><p>33,000 54</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>Kv line- Equirdent delta rpmcing,</p><p>to.li". ft</p><p>44 5.5</p><p>66 8.0</p><p>I10 14.0</p><p>112 16.5</p><p>I54 20.0</p><p>220 29.0</p><p>APPROXIMATE REACTANCE O F LOW-VOLTAGE ClRCUll</p><p>BREAKERS AND DISCONNECTING SWITCHES</p><p>In some low-voltage circuit calculations, the reactance of such switch-</p><p>ing equipment may be significant. The reactance of circuit breakers</p><p>varies greatly, depending upon the rating and design. For approxima-</p><p>tion, however, the reactance in ohms of a circuit breaker may be taken as</p><p>0.2</p><p>continuous rating of circuit breaker in amperes</p><p>The reactance of lever switches and disconnecting switches for low-vol-</p><p>t,age circuits (600 volts and below) is of the order of magnitude ranging</p><p>from 0.000050 to 0.000080ohm per pole at fiO rycles,for sizes ranging from</p><p>4000 to 400 amp, respectively, depending on the ampere rating, design,</p><p>and phase spacing of the switches.</p><p>APPROXIMATE REACTANCE O F CURRENT TRANSFORMERS</p><p>These data are useful o~ily for calculation of short-circuit currents ill</p><p>circuits rat,ed 600 volts and below.</p><p>The reactance of current transformers depends 011 their current rating</p><p>and design arid varies over a wide range. Therefore, a sirigle value of</p><p>reactance applicable to a variety of current transformers cannot be</p><p>given.</p><p>Current Transformers with Primary Circuits of t he Wound Type.</p><p>Approximate data on renctarice a t 60 cycles for current transformers of</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES I l l</p><p>type W, covering current ratings from 100 to 800 amp based on tests</p><p>at short-circuit currents, are given in Table 1.30. The values in Table</p><p>1.30 apply to t,ransformers with a serondary burden of I volt-amp or</p><p>less at 5 amp or a t normal i:urrent. For higher burdens, the impedance</p><p>referred to the primary side will be somewhat increased, but the increase</p><p>is far less than that occurring a t normal currents, berause of the reduced</p><p>mutual inductance between primary and secondary windings. The</p><p>reactance values based on low burden are conservative fur calculations</p><p>of maximum short-circuit current.</p><p>TABLE 1.30 Over-all Reactance of Type W Current Transformers,</p><p>Referred to Primary Winding</p><p>Approximate Values at Short-circuit Cvrrenh with D-C Component, Rms Symmetrical Component</p><p>Ronging from 15,000 to 55.000 Amp</p><p>Current Rating of Reactance (11</p><p>Primary Winding, 60 Cycles,</p><p>Amp Ohms</p><p>100 0.0035</p><p>I50 0.0017</p><p>200 0.0010</p><p>250 0.00066</p><p>300 0.00050</p><p>400 0.00032</p><p>500 0.00022</p><p>600 0.00019</p><p>800 0.00012</p><p>These values are also representative of t,he order of magnitude of the</p><p>reactance for current transformers of the following types, rated a t 5000</p><p>volts: JW1, JW4, JW6, JW14, WC12, WFI, WF6, and WF12.</p><p>Reactances for other designs of current transformers of the wound</p><p>primary type may be estimated by applying the folloming approximate</p><p>factors to the values of Table 1.30.</p><p>Type of Current</p><p>Transformer</p><p>KF85-7,500 volt 1.8</p><p>JSI-15,000 volt 0.4</p><p>Foctor to Be Applied to</p><p>Reactance Vduer in Table 1.30</p><p>Current Transformers Having a Bar-type Primary Conductor. For</p><p>bar-type current transformers with currerit rat,ings from 1000 to 4000</p><p>amp, such as t,ypes bS2-GO0 volts, WC15-5000 volts, KC60 7500 volts,</p><p>the react,arice has an approximate order of magnit,ude of 0.000070 ohm</p><p>a t currents within the range from 10,000 to over 80,000 rms symmct,rical</p><p>amperes, with or ait,hout d-c component,.</p><p>The reactauce depends on the spacing bctweeu phases, since a COIL-</p><p>siderable amount of air flux links the primary bar conductor. The</p><p>value given is t,hat for !&in. phase spacirig wit,h the t,ransformers side by</p><p>side, reprcserit,ing an average value for the three phascs for t,hree-phase</p><p>short circuits. Strictly speaking, the reactance in the three phascs will</p><p>I12 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>he unequal i n a side-by-side assembly of current transformers, but for</p><p>short-circuit-current calculations an average value can ordinarily be</p><p>used without serious error.</p><p>To say that the reactance for bar-type transformers is equal to the</p><p>air reactance of the primary conductor, considering its length, size, and</p><p>shape, and the spacing between phases, is a fair approximation.</p><p>I</p><p>No. of Circ. Kva of</p><p>core, "Oil, r.g"l.tol</p><p>~ _ _</p><p>.. 2400 I ?</p><p>.. to I0</p><p>4800 96 _____</p><p>Amp</p><p>rO~Y1.101</p><p>1 2400 Allrolings</p><p>I lo to160omp</p><p>3 13,800 Over 160arnp</p><p>APPROXIMATE REACTANCE O F A-C REACTORS AND FEEDER REGULATORS</p><p>The reactance is proportional t o the rating.</p><p>The voltage drop through the reactor at rated current and frequency</p><p>divided hy the line-to-neutral voltage of the circuit gives the per-unit</p><p>reactance on the current rating of the reactor. (This will also he the</p><p>per-nnit reactance on the kva rating of the circuit if the rated reactor</p><p>current is the same as the rated current of the circuit.)</p><p>The reactanre of a given step regulator is modified by the position</p><p>of the tap changer and becomes a maximum at maximum voltage boost.</p><p>It is minimum at neutral position, while at maximum buck, the impedance</p><p>is higher than at neutral.</p><p>TABLE 1.31 Short-circuit ImDedance of Feeder Reaulators of</p><p>Per cent ,O(lCt(l"L*</p><p>Ion base of C ~ C . k r d</p><p>Min Avg Max</p><p>-~~</p><p>0.65 0 .85 1.00</p><p>_ _ ~</p><p>~ _ _ _ _</p><p>O.O+ .... 0.6</p><p>O.O+ .... 0.7</p><p>0.15 .... 1.0</p><p>No. of</p><p>Type phoier</p><p>!-</p><p>Indue.. ......... I or 3</p><p>Indw.. ......... I or3</p><p>Step ............ 1</p><p>Step. ........... 3</p><p>Step. ........... 3</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 113</p><p>REFERENCES</p><p>1. A I E C Committee Rrport. Simplified Calculation of Fault Currents, Eke. E’ng.,</p><p>Octobrr, 1912. pp. 509-511.</p><p>2. AII*X Committee Ilrport. Simplifird Calmlation ai Fault Currmts. Trans.</p><p>AlEE’. 1942, Vol. G I . pp. 113:3-11:35.</p><p>3. Revision Made to AIICI: Report, Simplified Calculation of Fault Currents, Efec.</p><p>Emf,.. February, 194d. p. 65.</p><p>4. Darling. A. G., 4-C Short Cirrriit Caleiilating Procdure for Lon--roltage Systems,</p><p>‘I‘mns. AIEE. l!)41, Vol. GO, pp. 1121-1136.</p><p>5. Srhurig, 0. It.. Fault Voltngr Drop and ImpPdanre a t Short-circuit Ciirrmts in</p><p>Low Voltngr Circuits. Trans. A I E E , 1941, Vol. 60, pp. 479-486.</p><p>6. AIEE Committw Rrport. Simplifird Calculation of Fault Currents, Trans. A I E E ,</p><p>1948, Vol, 67, p. 1433.</p><p>Chapter 2 by R, H. Kaufmann</p><p>Symmetrical Components as Applied</p><p>to Short-circuit-current Calculation</p><p>on Three-phase Systems</p><p>The unhalanred circuit problems eucountered in short-circuit analysis</p><p>can be resolved by using symmetrical-component analysis. This</p><p>analysis technique is used extensively by power-system invest,igators</p><p>and authors. Developed in this chapter are concepts and procedures</p><p>for the application of symmetrical-romponent aualysis t,o the deter-</p><p>mination of short-circuit currents. While this procedure is built up</p><p>from base fundamentals, i t is aimed expressly a t the solutiori of electrical-</p><p>systcm short-circuit problems. For other possihlc applications of sym-</p><p>metrical-component analysis such as the determination of unbalanced</p><p>currents in certain circuits or machines, it is suggested that reference be</p><p>made to a more elaborate texthook* which explores the full field of appli-</p><p>cation more completely.</p><p>THE USE OF COMPONENTS</p><p>The separation of an electrical vector quantity into components to</p><p>simplify computation procedure is familiar to all. It has been common</p><p>practire to consider an alternating voltage or alternating current to he</p><p>composed of two components a t right angles t,o each other. It should he</p><p>evident that the process is not limited to two quantities, nor is it necessary</p><p>tha t the components be 90" apart.</p><p>*Edith Clark?. "Cirruit Analysis of A. C. POIVPI Systrrns." vol. 1, John Wiky &</p><p>Sons, Inc., NPW York, 1943.</p><p>I14</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>For example, take the expression</p><p>E = I Z</p><p>115</p><p>It is entirely valid to express this as</p><p>E = ( I , + I , )Z = IIZ + I,%</p><p>I , + I , = I</p><p>E = (I1 + 1 2 + I J Z = I1Z + I2Z + IaZ</p><p>I , + I2 + I3 = I</p><p>It is applicable</p><p>so long as the equations are linear (as they will be in electrical-cirruit</p><p>work).</p><p>E = I Z I = EY I , = Z J , etc.</p><p>provided that</p><p>or as</p><p>provided that</p><p>Thus there is 110 mystery about the use of components.</p><p>SYMMETRICAL COMPONENTS</p><p>If the Z per phase as illustrated in Fig. 2.1 could he represented as a</p><p>Since the con- firm fixed value, the circuit analysis would be-simple.</p><p>ductors of the three phases are magneti-</p><p>cally coupled, the voltage drop in the A</p><p>phase depends not only on the current in</p><p>the A phase but on the current in the</p><p>other two phases as well.</p><p>Consider the induction-motor imped-</p><p>ance diagram of Fig. 2.2. Assume the</p><p>rotor to be turning a t normal speed in</p><p>the direction produced by an impressed</p><p>voltage of sequence ARC. What I Z drop will be produced in the -1 phase</p><p>because of a current I , alone? That is a tough one; although there are</p><p>some relationships of which we are sure.</p><p>Under the conditions of balanced currents of sequence ARC there</p><p>will be balanred terminal voltages of sequence ABC. With normal</p><p>rated voltage and light load the current will he of the order of one-fourth</p><p>or one-third rated value. Under this condition all three phases appear</p><p>to have identical impedances of 1/0.25 or 1/0.333 or three or four per-unit</p><p>(300 or 400 per cent).</p><p>On the other hand, had the impressed voltage been applied with</p><p>opposite sequence (.4CB), i t is evident that this would he equivalent to</p><p>z I A -b</p><p>JWVL Nl</p><p>FIG. 2.1 A simple iymmetr i cd</p><p>ryrtem,</p><p>116 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>plugging. There mould he a balanced set of currents, but this time the</p><p>application of rated voltage would cause currents of about six times</p><p>rated value. In other words, the impedance appears to be the same in</p><p>all three phases, but its value is now $6 = 0.16 per-</p><p>unit, or 16 per cent. The effect of mutual winding</p><p>coupling alone may make the effective impedance</p><p>per phase as low as 16 per cent or as high as 300</p><p>or 400 per cent.</p><p>So long as</p><p>the three currents are equal and separated by the</p><p>same angular displacement, the effect of currents I s</p><p>and I , on the voltage drop in phase A will be iden-</p><p>tical with the effect of currents I , and I , on the</p><p>of currents I , and I , on the voltage drop in phase</p><p>C . Thus the effective impedance will appear to be</p><p>voltage drop in phase B and also with the effect</p><p>identical in all three phases; that is, the impedance voltage drop in the A</p><p>phase will bear the same relationship to the current in the A phase as the</p><p>impedance drop in phase B bears to the current in phase B and as the</p><p>impedance drop in phase C hears to the current in phase C . Or expressing</p><p>this symbolically,</p><p>There is one significant observation. k!J</p><p>'IG, '.' Induction-</p><p>motor impedance dio-</p><p>grcm.</p><p>InZn I B Z B - ICZC _ = _ _ _</p><p>I , IS I C</p><p>Thus Z , = Z B</p><p>= Z c .</p><p>I,Z,, and I c Z c are separated by the same angles as I, , I , , and Ic .</p><p>symmetrical components.</p><p>This also identities the fact that the impedance voltage drops I a Z A ,</p><p>These are two very important facts which emphasize the value of</p><p>POSSIBLE SYMMETRICAL COMBINATIONS</p><p>There are but three possible symmetrical combinations in a three-phase</p><p>system in which the three phase quantities are equal and separated by the</p><p>same angle. The displacement angle must be a multiple of 120" since</p><p>the three phases of a three-phase system are separated by 120". This</p><p>is shown in the following three cases using currents for illustration.</p><p>Case 1.</p><p>Case 2 .</p><p>Case 3.</p><p>I , is 120' behind I , and Ic is 120° behind I,.</p><p>I , is 240" behind I d and I , is 240' behind I, .</p><p>I , is 360" behind I , and Ic is 360" behind I , .</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 117</p><p>The vector relationships represented by these three cases of sym-</p><p>metrical displarement are shown in Fig. 2.3. Henceforth reference will</p><p>he made to case 1 as the positive-sequence component denoted by a suh-</p><p>script 1 characterized by three equal vectors 120' apart in the normal</p><p>sequenre A R C ; to rase 2 as the negative-sequence component denoted by</p><p>a subscript 2 charact,eriaed by three equal vectors 120" apart hut with a</p><p>sequence A C B opposite normal; and to rase 3 as the zero-sequence com-</p><p>ponent denoted by a subscript 0 chararterized by three equal vectors</p><p>with zero angular separation (in phase with each other).</p><p>Even a t the risk of unnecessary repetition, the two important properties</p><p>of these three symmetrical components are repeated.</p><p>The circulation of any one of the three symmktrical three-phase current</p><p>patterns in a symmetrical three-phase circuit, even though the phase</p><p>windings are mutually coupled, yields a balanced three-phase impedance</p><p>voltage drop whose sequence pattern is identical with that of the current</p><p>pattern. Likewise, the application of any one of t,he three symmetrical</p><p>three-phase voltage patterns on the circuit will give rise to a balanced</p><p>three-phase current whose sequence patterti is ideutical with that of the</p><p>voltage.</p><p>1. Current flow of one sequence pattern produces voltage drops of the</p><p>same sequence pattern only.</p><p>2. Applied voltage of one sequence pattern produces currents of the</p><p>same sequence pat,tern only.</p><p>3. For each sequence pattern, the impedance can he regarded as a</p><p>definite fixed quantity identical in all three phases.</p><p>This then is the significance and identity of the symmetrical compo-</p><p>nents (of which there are three types in t,hree-phase systems) and may be</p><p>applied to voltages as well as currents.</p><p>I c</p><p>CASE I CASE 2 CASE 3</p><p>lPOSlTlVE SEOUENCEl (NEGATIVE SEWENCE) (ZERO SEOUENCE)</p><p>FIG. 2.3 Symmetrical patterns of current.</p><p>118 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>THE OPERATOR 0</p><p>In the application of symmetrical-component analysis there will be</p><p>repeated need to shift a particular vector by multiples of 120". Particu-</p><p>larly in analytical studies i t will be advisable merely to indicate the</p><p>desired operation, leaving the actual resolution until the final solution is</p><p>approached. Invariably it will be found that combinations of operations</p><p>appear modifying a particular vector which can be directly reduced to</p><p>much simpler form, or often simply vanish.</p><p>The small letter a is used to indicate an angular advance of 120' in the</p><p>vector to which i t is appended. Its use parallels the use of j as a 90"</p><p>advance operator, i.e., aIb would mean a vector of the same magnitude as</p><p>la but advanced 120'; while azZb would mean a vector of the same magiii-</p><p>tude as Ib but advanced 240'.</p><p>0 - 1 2</p><p>O+j1.732</p><p>- 0 2</p><p>/71:a2 ,, 1.5tJO.866</p><p>0.5tj0.866</p><p>f</p><p>/</p><p>/</p><p>/ /</p><p>I /</p><p>I / //</p><p>0.5tj0.866</p><p>/71:a2 ,, 1.5tJO.866 f</p><p>/</p><p>/</p><p>/</p><p>/ /</p><p>/ I /</p><p>//' a 3</p><p>\</p><p>I: a</p><p>\ + 0.5 Ir -0. -50.866 1.5-~0.866</p><p>FIG, 2.4 Functions of the 120' operator 0.</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 119</p><p>The significance of commonly encountered combinations of a operators</p><p>is indicated in Fig. 2.4. For instance (az-a)Is would indicate a vector fl</p><p>times as large as la and advanced 270' in angle.</p><p>Comparing the operators j and a in more detail to explain Fig. 2.4, a</p><p>vector 1 to the right on the horizontal, Fig. 2.4, when multiplied hy .i</p><p>would be 0 + jl. That same vector multiplied by a becomes (in terms</p><p>of j ) - 0.5 + j0.866; multiplied by u2 it hecomes - 0.5 - j0.866. 1-a</p><p>then becomes 1 - (- 0.5 + jO.866) = 1.5 - j0.866 or an advance of</p><p>270" and 4 times as large.</p><p>RESOLUTION OF SEQUENCE COMPONENTS</p><p>It develops that any possible patt,erri of three-phase currents or three-</p><p>phase voltages can be resolved exactly into rombinatioris of the three</p><p>types of symmetrical components. Some properties of the three sym-</p><p>metrical-sequence components will he of interest in showing the nature of</p><p>their independence and the manner in mhirh they may he separated.</p><p>Referring to Fig. 2.5 it will he seen that, if the vertor sum of the three</p><p>vectors of each component is made, the answer will be zero for the posi-</p><p>tive-sequence and negative-scquence systems and 3 for the zero-sequence</p><p>system. If first the B-phase quantity is advanced 120' and the C phase</p><p>advanced 240" and the vector sum then evaluated, the answers will he 3</p><p>for the positive-sequence system and zero for the negative- and zero-</p><p>sequence systems. But if first the B-phase quantity is advanced 240" and</p><p>the C-phase quantity advanced 120", the vector sum will then be zero for</p><p>ADVANCE B 120" ADVANCE 0 240-</p><p>,I C240* c 1200</p><p>FIG. 2.5 Properties of rymmetriccll-component quantitiei.</p><p>I20 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>the positive-sequence system, three for the negative-sequence system,</p><p>and zero for the zero-sequence system.</p><p>This suggests that the three sequence components have independent</p><p>degrees of freedom.</p><p>Suppose that the three actual line currents I,, I,, and I , are to he</p><p>resolved into three balanced-sequence components of types positive,</p><p>negative, and zero. If I,, I,, and l c are added vectorially, it may be</p><p>expected that whatever positive-sequence and negative-sequence com-</p><p>ponent were contained therein would add up to zero, and the answer</p><p>should he three times the value of the zero-sequence component.</p><p>I A + I B + I c = 31.0</p><p>Pi + I s + I c</p><p>3 I.0 =</p><p>If the B-phase currerit is first advanced 120' and the C-phase current</p><p>2.40' and then added, it can be expected that whatever negative-sequence</p><p>and zero-sequence component were coutained therein would add up to</p><p>zero, and the sum should thus he three times the positive-sequence</p><p>component.</p><p>1, + a l , + aZIc = 3I.,</p><p>11 + a l a + a21c</p><p>3 I., =</p><p>In similar fashion hy first advancing the B-phase current 240' and the</p><p>C-phase rurrent 120' the sum should then he three times the negative-</p><p>sequeuce component in the A phase.</p><p>I , + azIB + aIc = 3I,,</p><p>11 + a2Ia + aIc</p><p>3 1 - 2 =</p><p>Sinre each of thc sequence systems is symmetrical, one can immediately</p><p>Refer to identify the corresponding comporierits in the other phases.</p><p>Fig. 2.3 to cherk the angular positiou of phase components.</p><p>Zero sequence:</p><p>I , + I , + I ,</p><p>3 Id = Ih0 = 1 . 0 =</p><p>Positive sequenre:</p><p>1, + a l e + a21c -</p><p>2 I , , =</p><p>-</p><p>a21A + I S + a l c</p><p>3 IS, = al l . , =</p><p>aIA + a21a + I c</p><p>3 IC1 = aI., =</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 121</p><p>Srgativc sequence:</p><p>I , + a2IB + a l e</p><p>3 I,, =</p><p>.\I1 three i.urrctits whii.11 romprise each of the three component systems</p><p>The sum of all t,hree compotrcnt currents of now have been dekiiicd.</p><p>each phase should equal the original actual phase current.</p><p>Phase il :</p><p>Phase 13 :</p><p>I B = I b , + l a ? f IbO</p><p>a?IA + I I I + a l c</p><p>3</p><p>aIr -t I , + azIc</p><p>3</p><p>I , + I , + I c</p><p>3</p><p>= >SIA(aP + a + 1) + I a ( l + 1 + 1) + Ic (a + a2 + 1)</p><p>= !5(0 + 31, + 0) = I B</p><p>+ + - -</p><p>Phase C:</p><p>I c = I , , + I,, + I d + a21e + IC +</p><p>a'IA + a l a + IC I., + 1, + I c</p><p>3</p><p>= 4$IA(a + az +, I ) + IB(a* + a + 1) + Ic (1 + 1 + 1)</p><p>1 hus a means now has been devised of separat,ing the three actual line</p><p>currents (or voltages) into t,hree systems of symmetrical components, and</p><p>further it, has bee11 shown that the sum of the three component quantities</p><p>of earh phase does exactly equal the original true line current (or voltage).</p><p>Several fuiidamental equatioiis and commonly used relationships are</p><p>listed i n Tahle 2.1.</p><p>+ - -</p><p>3 3</p><p>= >$(O + 0 + 3 I c ) = I c</p><p>? ></p><p>INDEPENDENCE OF SEQUENCE SYSTEMS</p><p>The fact has been developed that, in symmetrical circuits, currents of</p><p>one sequence produce voltages of the same sequence only and likewise</p><p>122 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>impressed voltages of one sequence produce currents of like sequence only.</p><p>In other words, there is no mutual coupling between scquence systems.</p><p>Thus the voltage drops in impedances can be separately evaluated for earh</p><p>sequence componerit of current and the resulting volt,agc drops added to</p><p>get the total voltage drop. Thus in Fig. 2.1 the t,otal impedance drop</p><p>across the impedance Z in the direction of current flow is</p><p>Phase A :</p><p>(IZ). = IdZI + I,,Z, + I,OZ,</p><p>Similar expressions could he written for the other two phases, hut a</p><p>simpler attack is possible from concepts already acquired. The positive-</p><p>sequence drops will all be of equal magnitude and of positive sequence,</p><p>the negative-sequence drops will all be of equal magnitude and of ncgative</p><p>sequence, and the zero-sequence drops will he of cqual magnitude and of</p><p>zero sequence. Therefore,</p><p>Phase B :</p><p>(IZ), = a21.,Z, + aIa2Zr + I,,Zn</p><p>Phase C:</p><p>(IZ), = a,I.,Z, + aZIa2Zz + I,oZO</p><p>Here for the first time the advantage of the symmetrical-component</p><p>approach can be appraised. For each symmetrical-compoiierit system,</p><p>impedances can he regarded as having a definite fixed value identical in</p><p>all three phases. The impedauce values in the three component systems</p><p>may he widely different, howcver. That is, Z, may he altogether differ-</p><p>ent from Z2 or Zo. Unt,il the actual currents were resolved into sym-</p><p>metrical componcnt,s, there seemed no alternate to thc use of self atid</p><p>mutual impedances in each phase.</p><p>At this point note that under balanced-load (wnditioiis the current is</p><p>entirely of positive sequence. Thus t,he usual solution of balanced</p><p>operation is really a special case iiivolving only the positive-sequence</p><p>system, i.e., positive-sequence voltages, poshive-sequence currents, and</p><p>positive-sequence impedances.</p><p>The application of these principles to the solution of unbalanced-load</p><p>problems now may he studied. It seems appropriate at this point to</p><p>review some physical concepts of the three compi~neut systems.</p><p>The</p><p>winding pattern in the A phase will he repeated in the B phasc 120 elec-</p><p>trical degrees later and i n the C phase 240 electrical degrees later. Thus</p><p>identical voltages will be generated in each phase minding except that the</p><p>B phase mill be 120' behind the A phase arid the C phase will he 120"</p><p>behind the B phase.</p><p>All source machines generate only positive-sequence vokage.</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 123</p><p>TABLE 2.1 Fundamental Equations</p><p>With line currents I A , I*, I c known, sequence currents are</p><p>I n + Is + Ic</p><p>3 = I, , = I d 1.0 =</p><p>NOTE: Voltages E., Eb, and E. generated vithin halaneed-winding rotating machines</p><p>are entirely positive sequence.</p><p>Commonly Used Relationships:</p><p>Negative- and zero-sequence voltages result from the impedance drop</p><p>produced by the flow of negative- and zero-sequence components of cur-</p><p>rent. Generally, positive-sequence voltages will he greatest at the source</p><p>machines and diminish as one moves toward the short circuit. On the</p><p>other hand, negative- and zero-sequence voltages will he greatest a t the</p><p>124 SYMMETRICAL COMPONENT5 FOR THREE-PHASE SYSTEMS</p><p>short-circuit point and diminish as one approaches the source machines,</p><p>Positive-sequence voltages and currents produce (and are associated</p><p>with) magnetic fields within rotating machines which rotate in the same</p><p>direction as normal rotational dirertion.</p><p>Negative-sequence voltages and currents produce (and are associated</p><p>with) magnetic fields in rotating machines which rotate in a direction</p><p>opposite to normal rotation. The latter thus produce machine torques</p><p>tending to slow down a motor rotor, and the positive-sequence electrical</p><p>quantities must produce a torque equal to the load torque plus that</p><p>resulting from the negative-sequence current if normal running speed is</p><p>to be maintained.</p><p>For such</p><p>currents to flow a t all it is evident that the electrical neutral must be con-</p><p>nected to a fourth conductor or ground. Being in phase, the currents add</p><p>up numerically at the neutral ronnection and become 31.0 in the neutral</p><p>circuit, Zero-sequence currents produce a stationary pulsating magnetic</p><p>field in the rotating machine stator winding which is predominantly of</p><p>stator-leakage character, very little of which crosses the air gap to enter</p><p>the rotor. Zero-sequence current will rarely be found in motors since the</p><p>motor neutral is almost never grounded.</p><p>Zero-sequence currents are in phase in all three conductors.</p><p>PER-UNIT SYSTEM'</p><p>While symmetrical-component analysis is valid regardless of the system</p><p>of units used, it will be found desirable to adopt the per-unit system.</p><p>In the per-unit system, potentials are expressed as a fraction of an</p><p>arbitrarily assigned line-to-neutral voltage (usually the normal operating</p><p>voltage). Currents are expressed as a fraction of an arbitrarily assigned</p><p>circuit current. This base current is usually selected to correspond with</p><p>a convenient round-number kva such as 1000, 10,000, etc.</p><p>Only two quantities can be arbitrarily assigned, i.e., base voltage and</p><p>base kva or base voltage and base current. Unit values of all other</p><p>quantities become fixed as soon as the first two are assigned.</p><p>Unit base voltage and current are arbitrarily assigned a t some one part</p><p>of the system. The values of unit voltage and current at other partsof the</p><p>system become those which would result from the turn ratio of intercon-</p><p>necting transformers.</p><p>The per-unit impedances define the fraction of base voltage which will</p><p>be produced by the flow of unit base current.</p><p>The value of the per-unit system is a t once apparent. The impedances</p><p>of generators, motors, and transformers when expressed in per-unit on</p><p>their own rating as a reference base assume almost a constant numerical</p><p>*See Chap. 1, p. 52.</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 125</p><p>value throughout a wide range of physical size and voltage rating. For</p><p>example, the impedance of a transformer mill be about 0.05 per-unit (6 per</p><p>cent) on its own rat,ing as a base quite independent of size or voltage</p><p>rating. If expressed in ohms, the numerical value of Z would vary</p><p>widely wit,h 110 sigu of any common denominator. Also, in the per-unit</p><p>system a particular per-unit value of current flowing into one side of a</p><p>transformer comes out the other side as the same per-unit value.</p><p>Refer to Chap. 1, page 54, for a complete list of equations relating</p><p>~e r -un i t values.</p><p>SYSTEM APPLICATION</p><p>The approarh to circuit problems consists of writing the relations</p><p>existing between geuerated voltages and impedance drops in the usual</p><p>conventional manner except that three sequelice systems may he involved.</p><p>In t,he simple cirruit arrangement shoivu in Fig. 2.0 it cau he seen that</p><p>oue can directly evaluate (in terms of the A phase)</p><p>Positive sequence:</p><p>E. = r. ,(zol + zLl + zr,) + v.,</p><p>Val = 8. - Z,I(ZC, + ZL, + Z T d</p><p>Segative sequence:</p><p>0 = I.dZm + Z L 2 + ZTJ + Tio*</p><p>0 = I.o(Zo0 + Z L O + Zro) + v.0</p><p>v., = -ra2(zGs + zL2 + zr2)</p><p>vm0 = -r,,(zoo t z t o + zTd</p><p>Zero sequence:</p><p>Combined :</p><p>Ti" = v.1 + ve/02 + v.,o</p><p>= E , - Iai(Zoi + ZLI + Z T ~ - Za,(Zci + Zr.2 + Z T ~</p><p>- I ~ O ( 2 0 0 + Z L O + ZTO)</p><p>It will be useful to draw the individual sequence circuits such as indi-</p><p>Xote that the circuit for the positive sequence is cated on Fig. 2.7.</p><p>Z G ZL 1 A - h</p><p>" W Y .".. vb</p><p>".,. ... " V0</p><p>I6 -w</p><p>Ec/ E+T</p><p>.,.A .... . . tc * vc</p><p>FIG. 2.6 Typical symmetrical three-phase circuit.</p><p>126 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>262 z L2 zT2 102-</p><p>I Vaz I- SEQUENCE1</p><p>N</p><p>E a = 101 ( Z G I + Z ~ I + Z T ~ I ~ V O I</p><p>0 = 102 ( Z G Z t Z L 2 1 Z T Z 1 t VO2</p><p>0 - 100 (ZGO+ Z L O t Z T O I t YO0</p><p>VA = volt v02+v00</p><p>= Eo- Io l ( Z G l t Z L l tZT11-102 (ZGZ+ZLZ+ZTZI</p><p>-1w ~ ~ t z , ~ + z , l</p><p>FIG. 2.7 Equivalenl sequence circuits of Fig. 2.6 (in terms of the A phase)</p><p>identically that which mould be used alone for balanced-load prohlems.</p><p>In the treatment of unbalanced loads, two additional circuits are involved</p><p>(negative and zero sequence) which appear about the same evcept that</p><p>there are no generated voltages therein and the respective sequence</p><p>impedances are used.</p><p>TYPE OF APPROACH</p><p>Through experience in the application of symmet,rical-component</p><p>analysis, partirular types of approach, appropriate selection of reference</p><p>phase, and useful equivalent circuits have been discovered vhich lead to</p><p>a solution in the simplest manner.</p><p>Generalized solutions of problems presented in short-circuit studies of</p><p>three-phase systems (circuit-breaker selection or relay appliration)</p><p>include the following forms of short circuits:</p><p>1. Three-phase</p><p>2. Line-to-line</p><p>3. Line-to-ground</p><p>4. Double line-to-ground</p><p>THREE-PHASE SHORT CIRCUITS</p><p>The three-phase short-circuit condition represents a balanced three-</p><p>phase short circuit on the system. Only positive-sequenre quantities are</p><p>involved; hence only the positive-sequence impedance system will be</p><p>needed. The solution thus simplifies to an analysis of a single-rircuit</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 127</p><p>network involving only positive-sequence impedances and is done in the</p><p>familiar conventional mariner as follows, using Figs. 2.8 and 2.9:</p><p>Balanced operation [ ( l ) (2) (3) tied together]</p><p>For balanced load Z, per phase, make Z, = Z,</p><p>For thrce-phase short circuit, make Zx = 0</p><p>Reference phase: L4</p><p>E.</p><p>z, + z x I , = I., =</p><p>I , = a21A</p><p>I c = a l l</p><p>The solution becomes simply</p><p>I , = I,, = ~ E.</p><p>total ZI</p><p>I s = a21A</p><p>I , = aIn</p><p>FIG. 2.8 Actual three-phore circuit pattern.</p><p>_ _ _ ~ _ _ _ - _ - - 1 I b = l o l EO</p><p>I 2 , + Z X</p><p>I IS= 0216</p><p>N+ v;, z x I I C = o h</p><p>POSITIVE</p><p>SEQUENCE</p><p>- -v/JI.- - - 2</p><p>_c</p><p>1.3</p><p>FIG. 2.9 Equivalent circuit for three-phore short-circuit analysis.</p><p>LINE-TO-LINE SHORT CIRCUITS</p><p>The generalized solution works out in the simplest manner by consider-</p><p>ing the short circuit to exist between the B and C phases, using phase A</p><p>as the reference, as illustrated in Fig. 2.10.</p><p>I28 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>The boundary coiiditioris a t the short-circuit point are</p><p>I , = 0</p><p>I , = - I c</p><p>v, = v c</p><p>LINE-TO-LINE SHORT CIRCUIT (SOLID1 E2F;z:: vc</p><p>SHORT CIRCUIT</p><p>II' - 0</p><p>I g = - I c</p><p>vg = v c</p><p>EQUIVALENT SEOUENCE CIRCUITS IN TERMS OF THE A PHASE</p><p>( b ) LINE-TO-LINE SHORT CIRCUIT (SHORT CIRCUIT IMPEDANCE ZF)</p><p>EcE5g vb</p><p>+ (a-a2)ia2 + 0 = 0</p><p>(a2-a)I., = (a2-a)I,z</p><p>I., = 1 - 2</p><p>Substituting t,his result into I , gives</p><p>IB = a21a1 + al,, 4 I.u = 0</p><p>= a21,1 + aI,l + I,, - I., + I,, = 0</p><p>(a' + a + 1)1., - 1-1 + I,o = 0</p><p>I,, = I d</p><p>I., = id = I,,</p><p>Thus:</p><p>This fact might have been evident by the geometry of line currents at</p><p>the short circuit. The sum of the three component currents in the B and</p><p>i n the C phase must be zero. Only if the individual component currents</p><p>are equal and 120' apart in both the B and C phases could this be possible.</p><p>But this would mean that in the A phase the component currents would</p><p>be equal and in phase.</p><p>Substituting = I,, and IaO = I., into the V A equation gives</p><p>v, = E. - I,,,Z, - I*,Z% - I.,ZO = 0</p><p>Ea = I , I ( Z I + Z , + Zu)</p><p>E. I - - z , + z* + z,</p><p>This suggests that the solution can he made in terms of an equivalent</p><p>circuit in whirh the generated voltage E. is impressed on the three imped-</p><p>anre networks Z, , Z 2 , and Z o in series. It is more accurate to think of</p><p>this to he in the form</p><p>Ea = I ,IZI + I d 2 + 1,oZo</p><p>This still suggests the series ronnection of the three networks hut</p><p>recognizes that the current in Z , is Ia1, in Z 2 is lo2, and in Zo is Io0. Since</p><p>1-1 = i 0 2 = Ia0 there is no conflict with Kirchhoff's law at the junction</p><p>between the individual sequence networks.</p><p>The important result is the equivalent-circuit concept by which the</p><p>sequence networks ran he interconnected to yield an answer for the value</p><p>of I., = I,, = 1,". The equivalent-circuit concept is helpful even when</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 133</p><p>the solution is to be obtained by numerical computation, hut i t is of</p><p>partirular importance if use is to he made of a d-e or an a-e network</p><p>analyzer (calcuiating hoard).</p><p>Knowing the value of I.,, the value of current in the fault (I,,) is</p><p>10 = I A = 1.1 + 1.2 + I,, = 31.1</p><p>whirh is t,hree times the current found directly from t,he equivalent</p><p>circuit.</p><p>Where t,here is impedance in the short circuit or in the neutral path,</p><p>the procedure outliiied above is modified as shown in Fig. 2.13.</p><p>z</p><p>. .</p><p>I I :-</p><p>L-G CONNECTION</p><p>THROUGH IMPEDANCE ZF</p><p>M 4 G h E TnE SYSTEM S</p><p>EXTEhDED THRObGh Q</p><p>BALAhCED C.RCJlT OF ZF</p><p>PER PhASE LZFI’ZF~=ZFO=ZFI</p><p>A SOLID SHORT CIRCUIT TO GRD</p><p>NOTE: SINCE IB=Ic=O, THE BEYOND THIS IMPEDANCE</p><p>RESULTS IN ZFCONNECTED</p><p>LINE-TO-GROUND.</p><p>LUSETHESAME PROCEDURE</p><p>INCLUSION OF ZF IN THESE</p><p>PHASES PRODUCES NO ERROR,</p><p>THbT I+:</p><p>FIG. 2.1 3 External impedance in the line-to-ground connection.</p><p>I _ - - - _ _ - _ _ - -1 I</p><p>S E Q U E N C E zo ZF I</p><p>N</p><p>ZERO</p><p>VVAv “20 *,,. **,.+ - 1</p><p>Iao --t</p><p>FIG. 2.14 Equivalent circuit for line-to-ground short-circuit analysis.</p><p>134 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>Summarizing, the solution becomes (see Figs. 2.14 and 2.8)</p><p>Line-to-neutral connection (line A to ground)</p><p>(1) zonnected to ground; (2) open; (3) open</p><p>For a line-to-ground impedance Z,, make Zx = Z ,</p><p>For a line-to-ground short-circuit, make Z , = 0</p><p>Referenee phase: A</p><p>[Boundary conditions: I , = 0, Z C = 0, V ( , ) = 01</p><p>E.</p><p>ZI + z* + zo + 32" + 32 ,</p><p>I., = I., = I,o =</p><p>I A = I d f I d + I d =</p><p>3E.</p><p>z, + Z? + zo + 3z,2 + 32,</p><p>Other Cases. The equivalent circuits by which other common cir-</p><p>cuit conditions can be evaluated are worked out in a similar manner as,</p><p>for example, a double line-to-ground fault would he solved as follows</p><p>using Figs. 2.8 and 2.15:</p><p>Double line-to-ground solid fault (line B t o C to ground)</p><p>(2) and (3) connected to ground; (1 ) open; Zx = 0</p><p>Reference phase: A</p><p>[Boundary conditions: I , = 0, V(,) = V(,) = 01</p><p>ZZO v,, = v,, = Vao = E, - I,,Z,= I,, ~ z, + Z"</p><p>Rotating-machine Characteristics. Positive-sequence currents are</p><p>associated with mmf patt,erns which rotat,e at synchrouous speed in the</p><p>normal rotational dirertion. The effective pi)sitive-sr(ioetice reactance</p><p>is consequently influenced by time. For the first cyrle of short-circuit</p><p>current, the subtransient rcact,ance of synchronous machines and the</p><p>standstill reactance of induction machines apply. Within a few cycles</p><p>the subt,ransient effects have decreased t o negligible proportions and the</p><p>transient reactance of synrhronons marhines is i t i control while the effec-</p><p>tive impedance of induction marhines has inrreased to a value close to t,he</p><p>normal running impedanre ( i t1 the order of 100 per cent, 011 its o\vn base).</p><p>During t,he next, serond or t,wo, t,wo artions are taking place in the</p><p>synchronous machine. Induced field currents are decaying, and t,he</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS I35 -</p><p>FI -</p><p>I</p><p>-1 -- ' $+EO z& v:l</p><p>POSITIVE I</p><p>SEOUENCE</p><p>I I a i + I</p><p>I</p><p>I VOO</p><p>+ - - A</p><p>ZERO</p><p>SEOVENCE</p><p>100-</p><p>FIG. 2.1 5 Equivalent circuit for double line-to-ground short-circuit analysis.</p><p>effective machirie reactance is approaching the synohrooous reactance.</p><p>The effective voltage ahead of synchronous reactance is approaching the</p><p>value established by the steady-state field current and may he influenced</p><p>by the operat,ion of an automatic voltage regulator.</p><p>Rarely n i l1 it be neressary to evaluate short-circuit-current magnitudes</p><p>for prolonged time intervals, but it will he well to recognize that special</p><p>treatment will be needed to obtain correct results in such cases.</p><p>Negative- and zero-sequence impedances of rotating machines car1 he</p><p>considered as remaining constant regardless of the duration of short-</p><p>circuit-current flow.</p><p>TRANSFORMER CHARACTERISTICS</p><p>The zero-sequence circuit produced by various transformer connections</p><p>is often a source of trouble; so a considerable number of typical comhi-</p><p>nations are defined in Fig. 2.16.</p><p>The positive- and negative-sequence impedauces are equal as are those</p><p>of all stationary winding circuits.</p><p>There is one tricky aspect associated with Y-delta or delta-Y trans-</p><p>formers. There is an inevitahle phase displacement hetween the high-</p><p>and low-tension line circuits. Standard convention has agreed that the</p><p>terminals designated H , and XI shall be those which are only 30" apart.</p><p>Present st,andards also st,ate that when operated with electrical sequence</p><p>A B C ou t,erminals HI, H,, H 3 the high-tension system will lead the low-</p><p>tension system by 30'.</p><p>This displacement is the result of winding geometry and is not of the</p><p>136 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>CONNECTION</p><p>ZERO SEQUENCE CIRCUIl</p><p>ZT</p><p>4---- x x</p><p>131 ( - P H</p><p>-</p><p>-( F 3 2</p><p>---( -7:-</p><p>Y</p><p>SPECIAL CASE- 3- P H CORE TVPE</p><p>N</p><p>WYE-WYE WITH DELTA TERTIbRI</p><p>131 I-pn</p><p>P</p><p>N</p><p>* CLOSED IF THE CORRESPONDLNG TRANSFORMER NEUTRAL</p><p>IS GROUNDED. LT IS THE NORMAL TRANSFORMER Z (SAME</p><p>AS POSITIVE SEQUENCE 2 1</p><p>FIG. 2.16 Zero-sequence circuits clrrocioted with common three-phase transformer</p><p>connections.</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 137</p><p>nature of an impedanre voltage displacement angle. Thus if the standard</p><p>transformer is operated with reversed sequence, i.e., electrical sequence</p><p>.,IN' assuriated with terminals H,, H , , HI, the high-tension system will</p><p>lag t,he low-tension system by 30".</p><p>l3y reason of these fa&, in a Y-delta or delta-Y transformer with</p><p>standard ronnertions operating with normal sequence, the positive-</p><p>sequenre rurrerit and voltage iii the high-tension circuit will be advanced</p><p>30" with respect to that i n the low-tension circuit, while the negative-</p><p>sequenre mrrent i n the high-tension -cirruit will be retarded 30", as is</p><p>defined in Table 2.2.</p><p>The zero-sequence circuit pro-</p><p>duced by various transformer couiiectious is ofben a source of trouhle;</p><p>so a considerable number of typical combinations are defined in Fig. 2.1G.</p><p>By first examining the zero-sequence properties of simple winding</p><p>patterns, it ivill then be possible t o identify understandably the sero-</p><p>sequetirc circuits of more complicated practical transformer connections.</p><p>Zero-sequence current cannot flow in the</p><p>circuit to a deltt-connected 11-inding (see Fig. 2.17)</p><p>sinre there is no elee-</p><p>trical conuection to ground by which it could return, even though zero-</p><p>sequenre current can flow within the closed delta circuit. Thus the zero-</p><p>sequence circuit is always interrupted at a juiirt,ion with a delta-connected</p><p>minding.</p><p>Zero-sequence current cannot flow in a cir-</p><p>ruit ronnerted to a Y-connected winding if the neutral is not grounded</p><p>(see Fig. 2.18). Thus the zero-sequence circuit will be interrupted at the</p><p>jurirtion with a Y-connected winding if the neutral is.uugrounded.</p><p>Tronsformer Zero-sequence Circuits.</p><p>Delto Winding Connection.</p><p>Y Winding Connections.</p><p>Iao -</p><p>FIG. 2.17 A circuit connecting with (I delta-connected transformer winding.</p><p>FIG. 2.18 A circuit connecting with an ungrounded Y-connected transformer winding.</p><p>138 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>TABLE 2.2 Transformer Phase Shift</p><p>With standard d-lta-Y or Y-delta transformms, H I (high voltage) will hc 30" shcad</p><p>H I will hc 30" behind X , with oppo- of XI (law voltage) for normal phase sequence.</p><p>site phase sequence.</p><p>PHASE SHIFT I N A-> OR )-ATRANSFORHER</p><p>Standard, H , 30" ahead of X I</p><p>Many investigators pwfer to exprrss the relationship hetween high- and low-ten-</p><p>sion line currents in B slightly different manner so as to simplify the associated phase</p><p>shift opcration, for example,</p><p>Standard, H I 30" ahead of X I</p><p>I:, = -jr o, = +jZO</p><p>1;. = +jI-*</p><p>I. , = -jib, = + j 1 b x</p><p>I:, = i i I b , 1:, = -j1tIs</p><p>z;, = +I,, z: = +I</p><p>I:, = + j ~ * , Z' r2 = - j I a 9</p><p>ZC = - jZo1</p><p>H I 30" behind X ,</p><p>NOTE: If currents w e not in per-unit, the transformation ratio must also he fac-</p><p>tared in.</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS I39</p><p>Zero-sequence current, in a circuit connect,ed to a grounded-neutral</p><p>Y-connected winding can flow if zero-sequence rurrent, in t,he secondary</p><p>windings can he caused to flow in the direct,iori iiidicat,ed by the secoridary</p><p>arrows, (see~Fig. 2.19).</p><p>FIG. 2.19 A circuit connecting with 0 grounded Y-connected tronrformer winding.</p><p>If the secondary currents in Fig. 2.19 cannot flow, the primary zero-</p><p>sequence current is limited to the magnetizing current of the core (in t,he</p><p>order of 5 per cent of rated current for 100 per cent impressed voltage).</p><p>This represents a Z O of ahout, 2000 per cent on the transformer rating,</p><p>which for practical purposes may he regarded as infinite.</p><p>A n exception to this rulc is presented hy the thrce-phase core-type</p><p>design whose construction is as indicated in Fig. 2.20. The flow of zero-</p><p>sequence current, in the primary windirig produces magnetic flux whii,h</p><p>is in phase in the same direction in all three core legs. Since there are no</p><p>external core legs between upper and lower core yokes (as would exist in a</p><p>shell type of three-phase design),</p><p>the zero-sequenre flux must re-</p><p>turn largely through the air.</p><p>The steel tank walls provide a</p><p>fairly low reluctance path forpart</p><p>ofthereturn circuit, but thecross-</p><p>over to t,he core yoke at both the</p><p>topand bottom isdirectly through</p><p>air. The magnetizing reactance</p><p>c</p><p>FIG. 2.20 The three-phase c k - t y p e tronr-</p><p>represented by this flux path</p><p>usually he in the order of 30 to , -, . .</p><p>50 per cent on the t,ransformer</p><p>rating, which is low enough to have practical significance.</p><p>Zero-sequence current in a circuit connected to a grounded-neutral</p><p>Y-connected winding can flow if another set of transformer wiridiiigs is</p><p>connected in delta as in Fig. 2.21. The closed delta provides a circuit for</p><p>t h e flow of zero-sequence current. The impedance presented to the flow</p><p>of current is the interminding impedance Z, (the same as the normal</p><p>positive sequence ZT), Kote, however, that the zero-sequence currents</p><p>are not repeated in the outgoing line circuit but are short-circuited within</p><p>the delta winding.</p><p>140 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>111 a Y-Y-connected transformer& b&L neut,rals grounded, as in</p><p>Fig. 2.22, zero-sequence current can flow if the 'reflected zero-sequence</p><p>current in the other winding finds a closed circuit at some point along the</p><p>connected circuit. In this case the tramformer t,ransfers zero-sequence</p><p>current from primary circuit to a serondary circuit in the same manner</p><p>that it transfers positive- or negative-sequence current. The transformer</p><p>simply inbroduces a series impedance in t,he zero-sequence circuit which in</p><p>magnitude is identiral with the normal positive-sequence impedance Zr.</p><p>With this understanding of elemental behavior, the equivalent zero-</p><p>sequence circuits for the usual transformer connections can he directly</p><p>resolved.</p><p>When drawing zero-sequence circuits for extensive systems, it is a good</p><p>plan t,o designate transformers in the manner shown in Fig. 2.16, showing</p><p>an interruption of the zero-sequence circuit by an open gap. By this</p><p>method one ran be constantly aware that a break in the cirruit was inten-</p><p>tional and not the result of an oversight.</p><p>I n Fig. 2.23 is illustrated a particular</p><p>typical syst,em. The resuking composition of the positive-, negative-,</p><p>and zero-sequenre circuits is also portrayed. Suppose that the immediate</p><p>problem concerns the evaluation of various performance qualities on the</p><p>2.4-kv system radiating from bus L1.</p><p>The first step involves a resolution of equivalent impedances by which</p><p>the entire hulk system to the left of hus L4 is expressed as a single equiva-</p><p>lent impedanre. This would be accomplished by successively paralleling,</p><p>etc., until firially a single equivalent impedance value connecting with</p><p>bus L, is obtained which would then look like Fig. 2.24. In many cases</p><p>Some of the more rommon ones are identified in Fig. 2.16.</p><p>Circuit Resolution Example.</p><p>FIG. 2.21</p><p>delta winding on the same core structure.</p><p>A circuit connecting with a grounded Y-connected transformer winding with a</p><p>FIG. 2.22</p><p>another grounded Y winding on t he same core structure.</p><p>A circuit connecting with a grounded Y-connected transformer winding with</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 141</p><p>i t vill he at onre apparent that the impedanre of transformer Ts will he</p><p>the major rontrolling impedance in the circuit from Ai',. I n this case it</p><p>may be entirely reasonable to consider that rated voltage is sustained on</p><p>the high-tension side; or consider the short-circuit rapacity at the high-</p><p>tension terminals to be about equal to the interrupting rating of the</p><p>MOT MOT</p><p>. S. 2.23 Typical system example.</p><p>142 SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS</p><p>switching equipment on the bus M,. Obviously, some such approxi-</p><p>mation will be required in practically every problem since the actual line</p><p>interconnections will otherwise extend over the entire electrical distribu-</p><p>tion system of several states.</p><p>It will be of interest to note that the zero-sequence system is quite</p><p>discontinuous, which is typical of practical systems. In the present</p><p>illustrative problem the zero-sequence system associated with bus L4 is</p><p>independent of that on bus M2.</p><p>For comprehensive studies of extensive system networks, the equiva-</p><p>lent sequence circuits shown in Fig. 2.23 might he set up on the d-a or a-c</p><p>network analyzer. To examine an operating characteristic at the point</p><p>P I , each individual sequence circuit would be tapped at the point PI.</p><p>For each sequence system the correct impedance network is that</p><p>obtained from the tap lead P I and it,s own neutral bus N . The intercon-</p><p>nections between sequence networks will he governed by the type of</p><p>unbalance (see Figs. 2.8, 2.9, 2.11, 2.14, 2.15). Provision is made in the</p><p>network analyzer directly to measure current in or voltage across indi-</p><p>vidual branches of all three networks.</p><p>Useful measurement con-</p><p>nections by which a particular sequence quantity may be independently</p><p>resolved, or a particular sequence quantity excluded, are identified 011</p><p>Fig. 2.25.</p><p>(In</p><p>applying potential transformers for measuring EO on an urigrounded</p><p>neut,ral system,</p><p>line-to-line rated transformers and secondary loading</p><p>resistors should he used t o avoid overvoltage hazards.)</p><p>The delt,a-connected current transformer circuit (which excludes l o</p><p>in the output) is useful in providing internal-short-circuit protection for</p><p>grounding transformers.</p><p>The circuits for individually segregating the sequence quantities I , and</p><p>E', are only rarely used. Possible applications would be (1) making a</p><p>single-roil voltage regulator responsive to positive-sequence voltage of</p><p>Measurement of Individual Components.</p><p>The circuits for obtaining 10 or Eo alone are frequent,ly used.</p><p>FIG. 2.24 Simplification of Fig. 2.23 for study of performance on bur 14.</p><p>SYMMETRICAL COMPONENTS FOR THREE-PHASE SYSTEMS 143</p><p>the three-phase system, ( 2 ) providiug a protective relay which will trip</p><p>if the sustained negative-sequence voltage exceeds a preassigned value.</p><p>It is of interest to note that the usual open-delta line-to-line connected</p><p>potential transformer application excludes go in the secondary.</p><p>CURRENT IK=CTR.VIOI</p><p>ZERO SEQUENCE NEG4TIVE SEQUENCE</p><p>VOLTAGE l i = m R A T I O I</p><p>ZERO SEO'lENCE POS!TIVE SEQUENCE</p><p>JO 8661</p><p>NOTE - 8" INTERCWNGING LINES B B C</p><p>METER WILL READ Vo2</p><p>-</p><p>FIG. 2.25 Measuring circuits for segregating specific components.</p><p>Chapter 3 by Donald Beernan and R. H. Kaufrnann</p><p>Selection of A-C Short-circuit</p><p>Protective Devices and Circuit</p><p>Equipment</p><p>HOW TO BE SURE OF ADEQUATE</p><p>SHORT-CIRCUIT PROTECTION</p><p>To design an industrial power distribution system adequate from a</p><p>short-cirruit st,andpoiiit, the maximum short-circuit current at any point</p><p>should he less than the short-circuit rating of the equipment applied at</p><p>that piiint. When systems are so designed, it is common to speak of them</p><p>as having adequate short-circuit protert,ion. In other cases, they are</p><p>said to have adequate int,errupting capacity (IC). Horn can one be sure</p><p>that, a plant, dist,ribution system is adequate for all short-circuit eondi-</p><p>tions?</p><p>1. First accurately determine the available short-circuit currents a t all</p><p>sigriificant poir1t.s in t,he system, using the methods outlined in Chaps. 1</p><p>atid 2. These rdrulat,ing procedures have been verified by many tests on</p><p>actual systems and in short-cirruit testing lahoratories. Nariy former</p><p>fallarious ideas w1iir.h led to the installation of inadequate short-circuit</p><p>prot,ectire devires arid circuit elemerit,s hare beerr dispellcd. For exam-</p><p>ple, the idea that, ouly about 20,000 amp maximum short-circuit rurreiit</p><p>could he obtairied at -180 rolt,s has heen dispelled by actual measurements</p><p>of short-circuit currents of over 100,000 amp a t this voltage.</p><p>L'iitil t,he magnitude of short-cirruit currcnt is known, one cannot be</p><p>sure thitt adequate short-rircuit protcrtion is provided.</p><p>2 . Iiistall only short-circuit protect,ive devires such as circuit breakers,</p><p>s\ritvhes aiid fuses, and r~ombinat,iou motor starters of known adequate</p><p>Mere are the steps t,o follow:</p><p>I44</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 145</p><p>interrupting sating. Use circuit elements of known adequate short-</p><p>circuit-current rating. Equipment of adequate short-circuit rating can</p><p>be obtained to meet the requirements of all industries when proper con-</p><p>sideration is given to system design from a short-circuit-protectiorr</p><p>standpoint.</p><p>If the system is installed with circuit</p><p>breakers that are large enough only for present requirements, the cirruit</p><p>breakers will become too small from ail interrupting standpoint when</p><p>capacity is added. The system should be designed and the circuit</p><p>breakers should be selected on a hasis that will enable expaiisioii without</p><p>exceeding the circuit-breaker interrupting ratings. Short-circuit stresses</p><p>must be checked too, as the stresses increase as the square of the short-</p><p>circuit-magnitudes. Future expansion can be accomplished at prac-</p><p>tically no added expense in the initial installation by employing a modern</p><p>poiver-distribution-system layout (see Chaps. 11 to 15).</p><p>Main and auxiliary switchboards in hundreds of plants in operation</p><p>today were installed years ago when the plants were small. The power</p><p>demand was limited then, and only small transformers were required to</p><p>supply the plant. At tha t time the switchboards may possibly have been</p><p>adeqnat,e. Xew</p><p>feeders were added t o carry the new load, arid new transformers were</p><p>added to the bus to supply the added load. 111 many cases no thought</p><p>was given to t,he circuit, breakers because t,hey carried their load currents</p><p>satisfactorily. Hovever, when new transformers were added, the capac-</p><p>it,y of the power supply inrreased. Consequently, the available short-</p><p>circuit current also iiirreased. This higher short-circuit current imposed</p><p>added interrupting duty on the old circuit breakers when they were</p><p>required to clear a faulty feeder cable. Often this added short-circuit</p><p>current was sufficient t o bring the total short-circuit current beyond the</p><p>rating of those existing circuit breakers. However, through oversight</p><p>nothing was done about it, leaving the plant vulnerable to a major shut-</p><p>down if a fault occurred which one of these old circuit breakers failed to</p><p>clear.</p><p>Failure t,o consider the effect of increased short-circuit currents has</p><p>heeti one of the most common causes of many of the older installations</p><p>being unsafe.</p><p>Many systems which have been operating</p><p>for years have never had a major short circuit. Operators of these sys-</p><p>tems have come to believe t,hat short circuits never occur; so they do not</p><p>bother about interrupting rating of rircuit breakers. This belief com-</p><p>pares with the assumption that fire insurance is not necessary because the</p><p>factory has never burned down. The older the system grows, the weaker</p><p>the insulat,ion becomes and the greater the possibility of major short cir-</p><p>3. Prepare for load growth.</p><p>However, as the plants grew, more power was needed.</p><p>4. Do not he complacent,</p><p>146 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>cuits. The circuit breakers too are often inadequate in these old syst,ems.</p><p>Thus, when a short circuit does occur it is almost cert,ain to cause a major</p><p>shutdown with possible damage to other propert,y as well as loss of</p><p>production.</p><p>If the short-circuit,-protection proh-</p><p>lem is approached on an engineering basis instead of depending on good</p><p>luck, the plant investment can be more adequately protected and undue</p><p>risks eliminated. Good luck over a period of years may give a false</p><p>assurance that failures are never going t,o occur, but, good luck eventually</p><p>runs out as it has in so many cases. The cost of a loss due to a failure</p><p>then is far more than i t would have been to modernize the switchgear oil</p><p>a planned step-by-step basis.</p><p>In the engineering approach a study is made to determine t,he weak</p><p>spots in t,he electric system and remedy them hefore a major shutdown</p><p>occurs, with attendant financial and production loss. The engineering</p><p>approach is of a prevent,ive nature, i.e., finding the weak spots and correct-</p><p>ing them before a failure does occur.</p><p>No one would t,hink of running a boiler indefinitely just hecause “ i t</p><p>had never failed.” Preventive maint,enance involves continually repair-</p><p>ing and replacing weak parts hefore they fail. The results of the failure</p><p>of an inadequate circuit breaker can he as serious as a boiler failure; so the</p><p>same intelligent engineering approach should be used in providing safe,</p><p>adeyuat,e circuit breakers as is used with other machinery even thongh</p><p>one has heen lucky enough over a period of years to avoid the failure of an</p><p>inadequate circuit breaker. Luck might change for the worse tomorrow;</p><p>so it may pay real dividends not to be complacent ahout short-circuit</p><p>conditions.</p><p>To have a safe power system with low maintenance cost and high</p><p>service continuity, adequate circuit prot,ertive equipment is necessary</p><p>throughout the ent,ire system from the place where the</p><p>BY SYNCHRONOUS</p><p>OF TOTAL OSCILLOGRAM</p><p>ONLY TWO ENDS SHOWN</p><p>HERE. THIS REPRESENTS</p><p>THE BREAK BETWEEN</p><p>THE TWO PARTS.</p><p>OCCURS AT</p><p>THIS TIME.</p><p>FIG. 1.7 Trace of orcillograrn of hart-circuit current produced by a generator.</p><p>10 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>MAX, SUBTRANSIENT CURRENT- USE SUBTRANSIENT REACTANCE X"d</p><p>/-</p><p>TIME-</p><p>( 8 )</p><p>FIG 1.8 Variation of generotor short-circuit current wilh time.</p><p>current delivered by a generator.</p><p>given by the machine designer is the lowest value obtainable.</p><p>use will show maximum short-circuit current.</p><p>before a system analysis can he made.</p><p>The value of X i or X y generally</p><p>Hence, its</p><p>Certain characteristics of short-circuit currents must he understood</p><p>SYMMETRICAL AND ASYMMETRICAL SHORT-CIRCUIT CURRENTS</p><p>These terms are used to describe the symmetry of the a-c waves about</p><p>the zero axis. If the envelopes of the peaks of the current waves are</p><p>symmetrical about the zero axis, the current is called symmetrical current</p><p>(Figs. 1.9 and 1.10). If the envelopes of the peaks of the current waves</p><p>are not symmetrical about the zero axis, the current is called asymmetrical</p><p>ENVEWPES OF PEAKS</p><p>OF SINE WAVE ARE</p><p>SYMMETRIGAL ABOUT</p><p>THE ZERO AXIS.</p><p>ZERO</p><p>A X I S</p><p>FIG. 1.9 Symmelrical a-c wove.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>THE ENVELOPES OF PEAKS</p><p>ARE SVHHETRICAL ABOUT</p><p>ZERO AXIS</p><p>FIG, 1.10 Symmetrical d t e r n a t i n g current f rom a short-circuited generotor.</p><p>1 1</p><p>ENVELOPES OF PEAKS ARE NOT</p><p>SYMMETRICAL ABOUT ZERO AXIS</p><p>AX1 S</p><p>TOTALLY 0 F F SET</p><p>PARTIALLY OFFSEl</p><p>FIG. 1.11</p><p>for the purpose of illustration only.</p><p>circuits.</p><p>Asymmetrical (I-c waver. The conditions shown here ore theoreticol a n d ore</p><p>D-C component will r a p i d l y d e c a y to zero i n actual</p><p>FIG. 1.12 Trace of orc i l logram of a typ ica l short-circuit current</p><p>12 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>current (Fig. 1.11). The envelope is a line drawn through the peaks of</p><p>the waves, as shown in Figs. 1.9 to 1.12.</p><p>For the sake of explanation, many of the illustrations, such as Figs.</p><p>1.11, 1.15 to 1.19, show sine waves of current uniformly offset for several</p><p>cycles. It should be noted that in practical circuits the amount of asym-</p><p>metry decreases rapidly after the occurrence of the short circuit in the</p><p>system. This decrease of asymmetry is shown qualitatively in illustra-</p><p>tions such as Figs. 1.12, 1.20, 1.23, and 1.24.</p><p>Oscillograms show that short-circuit currents are nearly always asym-</p><p>metrical during the first few cycles after the short circuit occurs. They</p><p>also show that the asymmetry is maximum at the instant the short circuit</p><p>occurs and that the current gradually becomes symmetrical a few cycles</p><p>after the occurrence of the short circuit. The trace of an oscillogram of a</p><p>typical short-circuit current is shown in Fig. 1.12.</p><p>WHY SHORT-CIRCUIT CURRENTS ARE ASYMMETRICAL</p><p>In the usual industrial power systems the applied or generated voltages</p><p>are of sine-wave form. When a short circuit occurs, substantially s ine</p><p>wave short-circuit currents result. For simplicity, the following discus-</p><p>sion assumes sine-wave voltages and currents.</p><p>In ordinary power circuits the resistance of the circuit is negligible com-</p><p>pared with the reactance of the circuit. The short-circuit-current power</p><p>factor is determined by the ratio of resistance and reactance of the circuit</p><p>only (not of the load). Therefore the short-circuit current in most power</p><p>circuits lags the internal generator voltage by approximately 90" (see</p><p>Fig. 1.13). The internal generator voltage is the voltage generated in</p><p>the stator coils by the field flux.</p><p>If in a circuit mainly containing reactance a short circuit occurs at the</p><p>peak of the voltage wave, the short-circuit current would start at zero</p><p>and trace a sine wave which would be symmetrical ahout the zero axis</p><p>(Fig. 1.14).</p><p>If in the same circuit (i.e., one containing a large ratio of reactance to</p><p>resistance) a short circuit occurs at the zero point of the voltage wave, the</p><p>current will start a t zero but cannot follow a sine wave symmetrically</p><p>about the zero axis because such a current would be in phase with the</p><p>voltage. The wave shape must be the same as that of voltage hut 90'</p><p>behind. That can occur only if the current is displaced from the zero</p><p>axis, as shown in Fig. 1.15. In this illustration the current is a sine wave</p><p>and is displaced 90' from the voltage wave and also is displaced from the</p><p>zero axis. The two cases shown in Figs. 1.14 and 1.15 are extremes.</p><p>One shows a symmetrical current and the other a completely asym-</p><p>metricd current.</p><p>This is known as a symmetrical short-circuit current.</p><p>WORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 13</p><p>GENERATOR TRANSFORMER</p><p>INTERNAL VOLTAGE OF GENERATOR APPLIED HERE</p><p>ONE LINE IMPEDANCE</p><p>ioxazx 7x 0.m</p><p>REACTANCE, X = 19%</p><p>RESISTANCE. R = 1.4%</p><p>I</p><p>RESISTANCE I S LESS THAN OF THE REACTANCE HENCE MAY</p><p>BE NEGLECTED WITHOUT AN APPRECIABLE ERROR</p><p>INTERNAL VOLTAGE OF GENERATOR</p><p>NEARLY 90' - SHORT CIRCUIT CURRENT</p><p>DIAGRAM</p><p>SHOWING</p><p>SINE WAVES</p><p>CORRESPONDING</p><p>TO VECTOR</p><p>DIAGRAM</p><p>FOR ABOVE</p><p>CIRCUIT</p><p>FIG. 1.13 Diagrams Illustrating the phase relations of voltage and short-circuit current.</p><p>14 SHORT-CIRCUll-CURRENT CALCULATING PROCEDURES</p><p>GENERATED VOLTAGE</p><p>SHORT CIRCUIT CURRENT</p><p>ZERO</p><p>A X I S</p><p>SHORT CIRCUIT OCCURRED AT THIS POINT</p><p>FIG. 1.14</p><p>cirwit.</p><p>Symmetric01 short-circuit current and generoted voltage for zero-power-factor</p><p>-SHORT</p><p>CIRCUIT</p><p>CURRENT</p><p>FIG. 1.15 Asymmetrical short-circuit current and generated voltage in zero-power-factor</p><p>circuit. Condition i s theoretical and is shown for illustration purposes only.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES I S</p><p>If, in a circuit containing only reactance, the short circuit occurs a t any</p><p>point except a t the peak of the voltage wave, there will be some offset of</p><p>the current (Fig. 1.16). The amount of offset depends upon the point on</p><p>the voltage wave at which the short circuit occurs. It may vary from</p><p>zero (shown in Fig. 1.14) to a maximum (shown in Fig. 1.15).</p><p>I n circuits containing both</p><p>reactance and resistance, the s~,?&&,R&!~~</p><p>amount of offset of the short-</p><p>circuit current may vary be-</p><p>tween the same limits as for</p><p>circuits containing only react-</p><p>ance. However, the point on</p><p>the voltage wave a t which the</p><p>short circuit must occur to pro-</p><p>duce maximum asymmetry</p><p>dependsupon the ratioof react-</p><p>ance to resistance of the cir-</p><p>cuit. Maximum asymmetry</p><p>is obtained when the short cir-</p><p>cuit occurs a t a time angle</p><p>equal to 90" + 0 (measured</p><p>forward in degrees from the</p><p>zero point of the voltage wave)</p><p>CURRENT</p><p>where tangent 0 equals there- ASYMMETRICAL</p><p>actance-to-resistance ratio of</p><p>FIG. 1.16 Short-circuit current and generated</p><p>the circuit' The short-circuit voltage in zero-Dower-factor circuit. Short cir-</p><p>current will be symmetrical cuit occurred between the</p><p>when the fault occurs 90" from</p><p>that point onthe voltage wave.</p><p>point and peak of the generated voltctge wove.</p><p>This condition i s theoretical and for illustration</p><p>an example, assumeacir- purporer only. The short-circuit current will</p><p>gradually become symmetrical in practical cuit that has equal resistance</p><p>and reactance, i.e., the react-</p><p>ance-to-resistance ratio is 1. The tangent of 45" is I ; hence, maximum</p><p>offset is obtained when the short circuit occurs a t 135' from the zero</p><p>point of the voltage wave (Fig. 1.17).</p><p>CiTCUit.,</p><p>D-C COMPONENT OF ASYMMETRICAL SHORT-CIRCUIT CURRENTS</p><p>Asymmetrical alternating currents when treatedas a single current wave</p><p>are difficult to interpret for circuit-breaker application and relay-setting</p><p>purposes. Complicated formulas are also required to calculate their</p><p>magnitude unless resolved into components. The asymmetrical alter-</p><p>nating currents are, for circuit-breaker applications and relay-setting</p><p>16 SHORT-CIRCUIT-CURRENT CALCUUTING PROCEDURES</p><p>MAXIMUM OFFSET</p><p>Short-circuit current and generated voltage in circuit</p><p>power system</p><p>enters the plant down to t,he smallest motor or light.</p><p>An Example of What Can Happen When Available Short-circuit Cur-</p><p>rents Exceed the Interrupting Rating of Short-circuit Protective Devices.</p><p>An inadequate circuit breaker mas mounted in a svit,ch riiiim which was</p><p>part of the distribution system. A short circuit occurred in the outgoing</p><p>rable. The short-circuit duty was well above the interrupt,ing rating of</p><p>the circuit breaker i n the switch house. As a result, the circuit breaker</p><p>attempted to open the circuit hut did not havetheability todoso. There-</p><p>fore, the circuit breaker failed, blew up, and when it did two things hap-</p><p>pened. First, the circuit breaker at the source had to clear the fault in</p><p>t,he failed circuit hreaker and thus drop all the load instead of just the one</p><p>load on the fauky hranch. This meant unnecessary loss of prodwt‘ * ion.</p><p>Second, a fire resulted and completely destroyed the switch house.</p><p>5. Use an engineering approach.</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 147</p><p>FIG. 3.1 Rerull of foilure of inodequote oil circuit breaker on heovy short circuit.</p><p>Fortunately the switrh house was isolated from nl.her I~uildit~gs, and orily</p><p>the switrh house burned dn\vn. llad this fsilurc ocwrrcd i n a fiict,ory</p><p>tiuilding, the damage could have been much more cxLensive.</p><p>Ihe picture, Fig. 3.1, tells the st,nry of what happened hotter thaii ii</p><p>book of mords could.</p><p>The irony of this fiiilurc was that, the plant, erigineer had ri~cogriizetl the</p><p>inadequacy of the circuit hrcakers in this swit,ch house aiid was replacing</p><p>t,hem with adequat,e ones. The ot,her circuit breakers in this switi.h</p><p>house had already hcen rcplaced mit,li adequat,e unes, and t.liey \wre</p><p>destroyed too.</p><p>One can never tell how long hia luck will last wii.h inat1t.quat.e circuit</p><p>breakers or fuses.</p><p>r .</p><p>It, may rim out sooner tliaii one thinks.</p><p>SELECTION OF THE TYPE OF</p><p>SHORT-CIRCUIT PROTECTIVE DEVICE</p><p>r l 1 here are marly features to he cnnsitlered in t,he sr?lcvtioii of short-</p><p>circuit protective devices t,n provide adequate short-circuit prntertinn for</p><p>an industrial power syst,em. One of the most import,ant, is that t h e short-</p><p>eircuit protective device be adequate for the service. The adrqiuicy of</p><p>circuit breakers, fuses, or motor st.art,ers can be determined from the pro-</p><p>cedures outlined in Chaps. I and 2. Ariother important function of mnst</p><p>short-circuit, prot ive devices is that t,hey also provide a means of</p><p>switehi!ig circuits nder normal operat,irig conditions. To m(</p><p>requirements fully and eomplctcly for both circuit switching and short-</p><p>148 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>circuit protection, a protective and switching device should fulfill the</p><p>following basic specifications:</p><p>1. It should be capable of being safely closed in on any load current or</p><p>short-circuit current within the momentary rating of the device.</p><p>2. It should safely open any current that may flow through i t up to the</p><p>interrupting rating of the device.</p><p>3. It should automatically interrupt the flow of abnormal currents up</p><p>to the interrupting rating of the device.</p><p>There are two fundamental devices that are commonly used for or have</p><p>as one of their functions short-circuit protection.</p><p>1. Circuit breakers</p><p>2. Fuses</p><p>Some motor starters are used for short-circuit protection, hut in general</p><p>these have either circuit hreakers or fuses as the short-circuit protective</p><p>element.</p><p>A basic comparison of fuses and circuit breakers will be made and their</p><p>area of application outlined. More detailed comparisons are made on</p><p>the basis of syst,ems voltage classes, i.e., (1) 600 volts and below and (2)</p><p>above 600 volts.</p><p>These are:</p><p>CIRCUIT BREAKERS-GENERAL</p><p>Meets All Requirements. A modern circuit breaker meets all the basic</p><p>requirements listed above. It is designed and rated to be capable of</p><p>heirig safely closed in on any current within its momentary rating (some</p><p>oil circuit breakers do not fully meet this requirement). It can safely</p><p>open any current within its interrupting rating. When proper relays or</p><p>tripping devices are applied, i t is capable of automatically opening any</p><p>current which is above the pick-up setting of the tripping device and</p><p>below its interrupting rating. It combines in one unit a device for safely</p><p>switching the circuit under normal as well as abnormal load conditions</p><p>and to automatically open abnormal rurrents up to its interrupting rating.</p><p>Circuit breakers, in all except a few special</p><p>cases where single-phase switching is used in transmission-line circuits,</p><p>open all ungrounded conductors of a circuit. Therefore, the probability</p><p>of single phasing of three-phase circuits is eliminated from a practical</p><p>standpoint in so far as the circuit protective equipment is concerned.</p><p>The total time to operate</p><p>under various overcurrent conditions is adjustable for practically all cir-</p><p>cuit breakers. The adjustment is either in the built-in tripping devices or</p><p>in the relays associated with the circuit breakers. The adjustability of</p><p>time of operation makes the circuit breaker ideally suited to selective</p><p>operation as is required for circuit protective service in a system.</p><p>Circuit breakers in general are suitable for elec-</p><p>Eliminates Single Phasing.</p><p>Adjustable Tripping Time and Pickup.</p><p>Electrical Operation.</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT I49</p><p>trical operation, which means they can be used for automatic control,</p><p>remote operation, etc. Furthermore, auxiliary circuits are available on</p><p>practically all electrically operated circuit breakers for the control of</p><p>external auxiliary or process circuits.</p><p>Various types of</p><p>relays with special characteristics to meet particular service requirements</p><p>can be used with circuit breakers to broaden their scope of application.</p><p>For example, time-delay overcurrent relays which match motor-heating</p><p>curves can be used to enable the circuit breaker to be used for motor start-</p><p>ing and running and short-circuit protection. Or the relays may he</p><p>specially designed to protect transformers or any ot,her piece of equipment</p><p>or circuit. This makes the circuit breaker and its associated relays</p><p>almost universally applicable as a short-circuit protective and switching</p><p>device .</p><p>Circuit breakers can repeatedly open abnor-</p><p>mal currents without destroying t,he interrupting element, Of course,</p><p>inspection and some maintenance may be required after each duty cycle</p><p>at or near their interrupting rating. When the circuit, openings are</p><p>repeated a few cycles or seconds apart, derating factors must be applied.</p><p>But fundamentally the circuit breaker does permit repeated operations</p><p>without destroying itself or affecting the accuracy of operatirlg time.</p><p>Since when a circuit</p><p>breaker operates i t does not destroy itself, there is little likelihood of</p><p>affecting calibration of time and pickup settings; hence the same protec-</p><p>tion is afforded all the time.</p><p>Most circuit-breaker time overcurrent</p><p>tripping devices and relays are not appreciably affected by temperature.</p><p>Hence, greater accuracy as a function of ambient t,emperature can be</p><p>maintained than by devices that depend upon t,hermal conditions to</p><p>activate them.</p><p>Circuit breakers in general are not so</p><p>fast in operation a t high overcurrents as are most fuses.</p><p>Circuit breakers are available up to</p><p>4000 amp cont,inuous current rating at GOO volts and less and up to 1200</p><p>to 5000 amp a t higher voltages. Trip-coil ratings run from 15 amp up.</p><p>Interrupting levels are available from 5000 to 100,000 amp a t GOO volts or</p><p>less and from 15 mva to 25,000 mva at higher voltages.</p><p>Circuit breakers are made under rigid</p><p>industry standards which prescribe complete interrupting ratings for</p><p>them and methods of test for establishing interrupting ratings. These</p><p>permit the application engineer to apply them on a sound safe basis and</p><p>within their rating. Proper derating factors</p><p>must be applied for repeti-</p><p>tive-duty cycles and high-altitude applications. *</p><p>Wide Selection of Time-current Characteristics.</p><p>Repeated Operations.</p><p>Same Degree of Protection after Operation.</p><p>Minimum Temperature Effect.</p><p>Moderate Operating Speed.</p><p>Wide Choice of Current Ratings.</p><p>Rigid Industry Standards.</p><p>* Refer to applicable NEMA standards.</p><p>150 A-C SHORT.CIRCUIT PROTECTIVE DEVICES AND ClRClllT EQUIPMENT</p><p>FUSES-GENERAL</p><p>Fuses are often considered for circuit protection because of their low</p><p>first cost. Before selecting fuses in place of circuit breakers, there are</p><p>certain general characteristics and limitations which must be recognized</p><p>and considered as well as cost.</p><p>While fuses have their proper applications, one must look rarefully a t</p><p>the fuse picture in general and then more closely a t specific fuses to see</p><p>how many of the hasic requirements are met.</p><p>Generally Do Not M e e t All Requiremsnts. One of the first and fore-</p><p>most considerations is that fuses in themselves do not meet the basic</p><p>requirements for a complete short-circuit protective device. Fuses alone</p><p>(except t,he oil-fuse cutouts) do not incorporate any switching means to</p><p>permit closing in on high currents or to switch load currents. T o meet</p><p>the basic requirements i t is necessary that a fuse other than an oil-fuse</p><p>cutout be used in conjunct,ion with a properly rated interrupter or safety</p><p>switch. In this combination the fuse provides the ability to open ahnor-</p><p>ma1 currents automatically. The switch should provide the ability to</p><p>open load currents and moderate overcurrents which are below the blow-</p><p>ing point of the fuse and should provide the ability t,o safely close in on</p><p>short-circuit currents up to the interrupt,ing rating of the fuse. When</p><p>the switch is in the closed position, i t should be able to carry safely what-</p><p>ever current the fuse will pass.</p><p>The operation of fuses in combination with interrupter switches at</p><p>moderate overcurrents imposes problems not easily overcome. The</p><p>fundamentals of the problem can be seen by referring to Fig. 3.2. To</p><p>illustrat,e one phase of the problem, let us assume that i t takes $6 see only</p><p>to close and open a switch manually. Should there be a moderate over-</p><p>load when the switch is opened and closed rapidly, as there may well be</p><p>because of connected motors, etc., the switch would have to open perhaps</p><p>several times its rating because the operation took place so quickly that</p><p>the fuse did not have time to melt. This area is represented by the</p><p>crosshatched section of Fig. 3.2. For example, an interrupter switch</p><p>might he rated to make 20,000 amp, carry 20,000 amp momentarily, and</p><p>to open 100 amp. This switch, when used with a 100-amp E-rated fuse*</p><p>or even a much smaller rated fuse, may not be adequate on moderate</p><p>values of current,. At 1000 amp, for example, the blowing time of the</p><p>fuse may be 3 see. An operator may close the switch and open i t within</p><p>36 see. The fuse would not have had time to melt, and the switch vould</p><p>be required to open 1000 amp, or ten times its rating. Whenever</p><p>the circuit interruption takes place in two separate devices which are</p><p>* E-rated fuses will carry their rated eurrmt eontinuouslv and blow in 5 to 10 min</p><p>at 200 to 264 per cent of rated current.</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 151</p><p>interdependent on each other for complete functioning over a wide range</p><p>of currents, there is always this problem of operation on moderate over-</p><p>loads which is much more difficult to overcome than operation a t very</p><p>high short circuits where the time for the fuse to clear is very short.</p><p>As a precautionary measure for increased safety, the switch element of</p><p>fused switches should be closed with a fast positive action and not opened</p><p>immediately. This will give the fuse a chance to melt on moderate over-</p><p>currents before the switch is again opened.</p><p>BASIC CHARACTERISTICS OF FUSES</p><p>Possible Single Phasing. Fuses are single-phase devices; therefore,</p><p>one fuse may blow, leaving a multiphase circuit supplied with only single-</p><p>phase power.</p><p>When fuses are used, their</p><p>pickup setting and time-current setting are changed by changing the size</p><p>or type of fuse.</p><p>It may not completely isolate a faulty circuit.</p><p>Nonadjustable Tripping Time or Pickup.</p><p>AMPERES</p><p>FIG. 3.2 Interrupter-switch rating and fuse time-current characteristics showing per-</p><p>formonce on moderate overcurrent..</p><p>152 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>Limited Choice of Characteristics. Because fuses are thermal devices,</p><p>the choice of shape of time-current characteristics is very limited for</p><p>coordination purposes.</p><p>Fuses, once they have operated, must be</p><p>replaced. Certain types of fuses have removable links which permit</p><p>salvaging part of the fuse after i t has blown. Replacement cost of fuses</p><p>varies widely, depending on type and size of fuse.</p><p>Because the</p><p>interrupter destroys itself, care must be taken to replace a blown fuse</p><p>with one of the same rating and characteristics. Otherwise, protection</p><p>may he lost. There is always the danger that if no fuses are available</p><p>short bars or wires may be inserted to keep power on. When this is done,</p><p>all protection may be lost.</p><p>Fuses are thermal devices; there-</p><p>fore, their operation is subject t o variation due t o ambient temperature</p><p>changes. This effect in fuses is much greater than in relays or cir-</p><p>cuit-breaker tripping elements. It is less important in high-voltage</p><p>circuits.</p><p>Fuses are generally divided into two classes:</p><p>(1) non-current-limiting and (2) current-limiting. The current-limiting</p><p>fuses possess two important advantages, particularly for branch-circuit</p><p>protection: (1 ) Berause of extremely fast operation, they limit the damage</p><p>due to the flow of short-circuit current. (2) They actually limit the short-</p><p>circuit-current magnitude to far less than the available short-circuit cur-</p><p>rent, thereby allowing the use of smaller conductors and equipment in</p><p>branch circuits.</p><p>This current-limiting ability is one of the most useful characteristics of</p><p>the fuses in branch-circuit protection.</p><p>Non-current-limiting fuses also operate faster than circuit breakers at</p><p>currents near their interrupting rating.</p><p>The fast operat,ion of most types of fuses, however, makes i t difficult</p><p>and often impossible to coordinate them with other short-circuit protec-</p><p>tive devices located beyond the fuse in the circuit. Therefore, fuses in</p><p>general are best suited for branch-circuit protection where they need not</p><p>operate selectively with other protective devices between the fuse and the</p><p>load.</p><p>Fuses are now available for low voltages</p><p>(600 volts and less) up to 4000 amp. For circuits above 600 volts, the</p><p>upper limit of fuse ratings is in t,he range of 100 to 400 amp.</p><p>Fuses are generally limited in size hecause of thermal considerations.</p><p>Large fuses may produce so much heat that ventilation and mounting</p><p>become severe problems. Also, as current-limiting fuses become larger</p><p>Nonrepetitive Operation.</p><p>Protection M a y Be Reduced or Lost after Operation.</p><p>Affected by Ambient Temperature.</p><p>Fast Operating Speed.</p><p>Choice of Current Ratings.</p><p>A-C SHORT-CIRCUIT PROTECTIVE DNICES AND CIRCUIT EQUIPMENT I53</p><p>and larger, they lose more and more of their current-limiting ability.</p><p>Sinre the current-limiting ability of fuses is most useful in branch-circuit</p><p>protection, the handirap of having to use small ratings to get effective</p><p>rurreut-limiting artion is not so pronounred, as most branch circuits are</p><p>of small rurrent rating anyway.</p><p>Fuses above 600 volts are made according to</p><p>indnstry st,andards esrept, that standardized levels of interrupting ratings</p><p>are not set up. Low-voltage fuses have no a-c interrupting standards,</p><p>although surh st,andards may be available in the future. See further</p><p>disrussion nuder voltage classification.</p><p>Fuses and their asso-</p><p>ciated switches for low-current circuits, i.e., about 200 amp or less, are</p><p>simpler mechanically than circuit breakers. For higher current circuits</p><p>t,he switrh, if built, t o have the necessary momentary and interrupting</p><p>abilit,y, loses its advantage of mechanical simplicity.</p><p>Industry Standards.</p><p>Mechanical Simplicity a t Low Current Ratings.</p><p>CIRCUIT BREAKERS VS. FUSES-GENERAL</p><p>I11 selecting circuit breakers YS. fuses, the techniral ronsideratious cer-</p><p>tainly favor the rirruit breakers in most rases. Because of this, circuit</p><p>breakers are generally considered the only acceptable protective devices</p><p>by most engineers for all'lorations in industrial plants where switching</p><p>and short-rirruit protectioii is required except for some hranch circuits</p><p>and control circuits and motor starters. Fuses and switches are pre-</p><p>ferred for some hranch rircuits because of the fast operation of the fuse.</p><p>While cost</p><p>is very important, i t is secondary to the technical considerations noted</p><p>above and secondary to select,ing the devire that has an adequate inter-</p><p>rupting rating for t,he servire. Berause there may he in some cases a</p><p>wide difference in rost between circuit hreakers and fuses, there is a tend-</p><p>ency to get so involved in economic issues in the selection of circuit</p><p>breakers vs. fuses that technical ronsiderations are lost sight of. AS a</p><p>result many hazardous syst,ems are installed to save a few dollars in first</p><p>cost, a saving that may soon be lost because of the poor performance and</p><p>higher maintenanre of inadequate equipment, particularly in low-voltage</p><p>circuits. It is for that reason and because the technical cansiderations</p><p>vary somewhat with voltage that the technical considerations are reviewed</p><p>in further detail as a function of voltage class.</p><p>There are other factors in the selection of fuses for overcurrent protec-</p><p>tion. These factors involve mainly coordination with relay time-current</p><p>characteristics or the time-current characteristics of built-in devices on</p><p>circuit breakers (see Chap. 9).</p><p>Besides the technical roiisiderations, economirs is a factor.</p><p>I54 A-C SHORT-CIRCUIT PROTECTIVE DEVICES A N D CIRCUIT EQUIPMENT</p><p>* Standai</p><p>SHORT-CIRCUIT PROTECTIVE EQUIPMENT</p><p>FOR SYSTEMS 600 VOLTS AND LESS</p><p>For low-voltage systems rated 600 volts or less, there are three com-</p><p>monly used types of short-circuit protective devices for protecting main</p><p>power circuits, secondary feeders, and branch circuits. These devices are</p><p>1. Large air circuit breakers (sometimes referred to as magnetic circuit</p><p>breakers) of which the one shown in Fig. 3.3 is typical.</p><p>2. Molded-case circuit breakers of which those shown in Fig. 3.9 are</p><p>typical.</p><p>3. Fused safety switches of which the one shown in Fig. 3.11 is an</p><p>example of a high-quality safety switch and fuse.</p><p>There are panel boards which are used for protection of small branch</p><p>circuits. These are used mainly in lighting and in small power systems</p><p>and employ either small molded-case-type circuit breakers or fuses as</p><p>their overcurrent protective means.</p><p>rating8 are 15, 20, 25,35, 50,70,90, 100, 125, 150, 175, 200, 225, 250,275,</p><p>LARGE AIR CIRCUIT BREAKERS</p><p>Description. The large air circuit breaker, Fig. 3.3, consists of an</p><p>operating mechanism, contacts, an arc interrupter, and usually a built-in</p><p>overrurrent tripping device. These circuit breakers are characterized by</p><p>their sturdy construction, ample electrical clearances, availability in high-</p><p>current-carrying and interrupting and momentary ratings. The tripping</p><p>devices are adjustable as to their pickup setting and operating time, and</p><p>various shapes of time-current characteristics are available. The ratings</p><p>available are listed in Table 3.1.</p><p>TABLE 3.1 Ratings of Low-voltage Large Air Circuit Breakers for A-C Service</p><p>1nterrvpting roting., rm, amp a.ymmetric.al</p><p>240 volts</p><p>and below</p><p>__. ~ ~</p><p>30,000</p><p>50,000</p><p>75,000</p><p>l00,000</p><p>150,000</p><p>241-480 volts</p><p>25,000</p><p>35,000</p><p>60,000</p><p>75,000</p><p>100,000</p><p>_____</p><p>600 volts</p><p>_____</p><p>15,000</p><p>25,000</p><p>50,000</p><p>75,000</p><p>I00,000</p><p>Range l r i p - d ratings,' omp</p><p>~</p><p>240 voitl</p><p>and below 241-480 volts</p><p>30-225 25-225</p><p>150-600 100-600</p><p>60+1,600 400-1.600</p><p>2,000-3,000 2,000-3,000</p><p>4,000 4,000</p><p>600 volts</p><p>15-225</p><p>35-600</p><p>200-1,600</p><p>2,000-3.000</p><p>4,000</p><p>.> - - ., .__.) -__., ....</p><p>Application. These circuit breakers are intended primarily for appli-</p><p>cation in main switchboards where pou'er may be generated a t low voltage</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 155</p><p>FIG. 3.3 Large air circuil</p><p>breakers mounted in drawout</p><p>metal-enclosed low-voltoge</p><p>rwifchgear.</p><p>or where it may he received from the utility at low voltage and for the</p><p>secondary svitchgear of load-center unit substations or in main subdis-</p><p>tribution centers, Fig. 3.4. They are also applicable for individual</p><p>branch-circuit prokction where t,he highest qualit,y device is required and</p><p>where special time-current characteristics are necessary for coordination.</p><p>They are particularly applicable for braneh-circuit protection for larger</p><p>loads over 200 amp or for smaller loads where, as stated above, highest</p><p>quality protection is desired or electrical operation is required. These</p><p>circuit breakers have longer life built into them than do other types of</p><p>low-voltage circuit breakers and are, therefore, suitable for many more</p><p>operations, particularly where there is moderately repetitive duty imposed.</p><p>Large air circuit breakers may</p><p>be used either in selective tripping systems or in cascade systems. Selec-</p><p>tive tripping systems, Fig. 3.5, are those in which the circuit breakers are</p><p>set to trip selectively so that the one nearest the fault operates first so</p><p>that only the faulty portion of the circuit is deenergized. In this case all</p><p>circuit breakers should have adequate interrupting ratings, that is, their</p><p>rating should be equal to or greater than the short-circuit duty a t the</p><p>Selective Tripping vs. Cascading.</p><p>156 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND ClRCUlT EQUIPMENT</p><p>point of application. There are additional problems of selecting the</p><p>time-current settings which are discussed more fully in Chap. 9.</p><p>In cascaded operation, Figs. 3.6 and 3.7, circuit breakers may he used</p><p>under certain circumstances beyond their interrupting rating. This</p><p>applies where the main circuit breaker (commonly referred to in applica-</p><p>tion tables as the A' circuit breaker) has adequate interrupting rating,</p><p>that is, its rating is equal to or greater than the short-circuit duty imposed</p><p>a t the point of application. The feeder circuit breakers (commonly</p><p>referred to in application tables as the B circuit breaker) in this case,</p><p>Fig. 3.6, may be used to twice their interrupting rating provided that the</p><p>following conditions are met. The total kva of connected synchronous</p><p>motors should not exceed 25 per cent of the supply transformer or</p><p>I</p><p>A / A LOAD CENTER</p><p>nln UNIT SUBSTATION</p><p>GENERATOR</p><p>- A</p><p>Y</p><p>MAIN SECONDARY 3 BREAKER</p><p>SU0 - DISTRI0UTlON</p><p>CENTER</p><p>FURNACE LOAD MOTOR WELDER</p><p>200 HP</p><p>FIG. 3.4 One-line diogrom showing typical applications of large air circuit breakers.</p><p>ELECTRICALLY</p><p>OPERATED</p><p>I 5 7 A-C SHORT-CIRCUIT PROTECnVE DEVICES AND CIRCUIT EQUIPMENT</p><p>generator rating. In addition to the main circuit breakers having ade-</p><p>quate interrupting ratings, their instantaneous tripping attachment must</p><p>be set t o operate when the current through the backed-up or B circuit</p><p>breakers is not more than 80 per cent of the interrupting rating of the</p><p>backed-up or B circuit breakers. This ensures that the main circuit</p><p>breakers will operate whenever the short-circuit duty exceeds the inter-</p><p>rupting rating of the B circuit breakers.</p><p>& 1500 KVA LOAD CENTER - UNIT SUESTDTION</p><p>A</p><p>MDIN CIRCUIT BREAKER</p><p>RATE0 DT LEAST 50.000</p><p>DMP INTERRUPTING</p><p>ERANCH FEEDER</p><p>CIRCUIT BREDK-</p><p>/</p><p>SHORT CIRCUIT DUTY</p><p>DT THIS POINT 32.000 ~ U E - B U ~</p><p>DMP RMS DSYMMETRICDL 7</p><p>FEEDER CIRCUIT EREDKERS</p><p>RDTED 50,000AMP INTERRUPTING</p><p>SHORT CIRCUIT DUTY DT THIS POINT</p><p>50.000 DMP RMS ASYMMETRICDL</p><p>WOOD DMP FROM THE TRANSFORMER</p><p>DND 9000</p><p>DMP FROM THE MOTORS</p><p>+FEEDER CABLE</p><p>A /I' ERDNCH FEEDER CIRCUIT</p><p>DMP INTERRUPTING</p><p>V</p><p>)BREDKERS RATED 25000</p><p>SHORT CIRCUIT DUTY</p><p>DT THIS POINT 22000</p><p>DMP RMS DSYMMETRICDL</p><p>*</p><p>NOTE: SHORT CIRCUIT LEVELS</p><p>DT SUB E u s s E s n am</p><p>REDUCED DUE FEEDER</p><p>CAELE IMPEDDNCE</p><p>FIG. 3.5</p><p>tripping system.</p><p>tain selectivity.</p><p>One-line diagram showing large air circuit breakers applied in selective</p><p>Time settings of overcurrent trip elements must be properly set to ob-</p><p>158 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>I</p><p>i</p><p>& I 5 0 0 KVA LOAD CENTER - UNIT SUBSTATION</p><p>MAGNETIC TYPE</p><p>SHORT CIRCUIT DUTY</p><p>AT THIS POINT 50.000AMP</p><p>RMS ASYMMETRICAL</p><p>I ' I</p><p>MOTOR</p><p>CONTRIBUTION</p><p>9000AMP</p><p>I</p><p>NOTE: INSTANTANEOUS TRIP ELEMENT ON MPIN</p><p>BREAKER A MUST BE SET TO TRIP AT</p><p>16400 AMP THIS IS 0.8 OF 2 0 5 0 0 AMP</p><p>20500 AMP IS THE CURRENT FLOWING</p><p>FROM THE MAIN TRANSFORMER THRU</p><p>BREAKER A WHEN CURRENT FLOWING</p><p>THRU FEEDER BREAKER 0 IS 25.000 AMP</p><p>THE RAyiNGOF BREAKEdB</p><p>FIG. 3.6 One-line diagram showing large oir circuit breakers applied in cascade with</p><p>only one source of low-voltage power.</p><p>Motor contribution must be considered. The duty including motor</p><p>contribution should not exceed twice the interrupting rating of the</p><p>backed-up circuit breaker. However, the motor contribution may not</p><p>come through the main circuit breaker. Therefore, the main A' circuit-</p><p>breaker instantaneous trip setting may be less than 80 per cent of the</p><p>backed-up circuit-breaker interrupting rating because the main A' circuit</p><p>breaker must trip instantaneously when the total rms asymmetrical short-</p><p>circuit current through the backed-up circuit breaker is 80 per cent or</p><p>more of the interrupting rating of the backed-up B circuit breaker. For</p><p>example, in Fig. 3.6 if the backed-up or B circuit breakers are rated</p><p>25,000 amp interrupting rating, the short-circuit duty a t the point of</p><p>application of the B circuit breaker should not exceed 50,000 amp rms</p><p>asymmetrical. This may he made up of 41,000 amp from the main</p><p>source and 9000 amp from the motors. The main-source circuit breaker</p><p>must trip instantaneously a t 0.8 X 20,500 or 16,400 amp rms asym-</p><p>metrical. It makes no difference whether the circuit breaker is applied</p><p>a t the bus or a t some point remote from the bus. When the backed-up</p><p>circuit breakers are applied a t points remote from the bus, such as circuit</p><p>breakers B' in Fig. 3.7, the interrupting duty a t the circuit breaker ahead,</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 159</p><p>B in Fig. 3.7, may be in excess of twice the interrupting rating of the</p><p>backed-up B' circuit breaker, but because of cable impedance the short-</p><p>circuit current a t the point of application of the backed-up circuit breaker</p><p>B' must be limited to twice its interrupting rating. Circuit breakers</p><p>operating a t beyond their interrupting rating in cascade mustbe inspected</p><p>after each operation and may require more than normal maintenance</p><p>after interrupting currents beyond their rating even though the main</p><p>circuit breaker does open.</p><p>Another qnalification is that the circuit breakers must be of the same</p><p>manufacture and of similar characteristics. Feeder circuit breakersshould</p><p>be electrically operated because the forces incident to closing against</p><p>short circuits in excess of the circuit-breaker rating may preclude success-</p><p>ful manual closing. Circuit breakers of two widely different interrupting</p><p>I</p><p>f</p><p>&I000 KVA LObD CENTER</p><p>-UNIT SUBSTATION</p><p>I I I -knunar</p><p>I</p><p>HERE 26000 AMP RMS</p><p>NOTE! INSTbNTANEWS snom CIRCUIT DUTY</p><p>TRIP ELEMENT ON</p><p>FEEDER BREIKER B ISYMUETRICbL</p><p>MUST BE SET bT</p><p>12000 bMP(OQ X I50001</p><p>FIG. 3.7</p><p>remote from the main source of power.</p><p>One-line diagram showing large air circuit breakers in cascade applied</p><p>160 A-C SHORT-CIRCUIT PROTEtTlVE DEVICES AND CIRCUIT EQUIPMENT</p><p>ratings cannot be cascaded. As a guide to this, refer to Table 3.2 which</p><p>shows the maximum interrupting rating circuit breaker which can he</p><p>used to back up any given interrupting rating feeder circuit breaker.</p><p>The ratio of the columns may be more than 2 : 1. The higher interrupt-</p><p>ing duty in the main circuit breaker often comes about because of having</p><p>to select it for continuous current-carrying rating rather than inter-</p><p>rupting rating. Regardless of the interrupting rating of the main cir-</p><p>cuit breaker, the duty cannot exceed twice the interrupting rating of the</p><p>cascaded B feeder circuit breaker.</p><p>TABLE 3.2 Range of Large Air Circuit Breakers Which Can Be Cascaded with</p><p>Main Circuit Breaker A'</p><p>Interrupting Rating,</p><p>Each Other</p><p>Minimum Interrupting Rating of</p><p>Coscaded Feeder Circuit Breaker B,</p><p>Amp Rmr Amp Rmr</p><p>25,000 15.000</p><p>30,000 15,000</p><p>50.000 15.000</p><p>60,000 25,000</p><p>75,000 25,000</p><p>100,000 50.000</p><p>120,000 75,000</p><p>150.000 100.000</p><p>Where there are two or more sources of current to a bus with cascaded</p><p>feeder rircuit breakers, the following rule applies, Fig. 3.8. All main A'</p><p>circuit breakers (i.e., A : , A: , A : ) must be tripped instantaneously when</p><p>the total short-circuit current through the hacked-up B circuit breaker</p><p>exceeds 80 per cent of its interrupting ratings.</p><p>The example in Fig. 3.8 shows what the various instantaneous over-</p><p>current trip settings of the main circuit breakers should be for a given</p><p>case. The rule is that the instantaneous setting must be proportioned to</p><p>the short-circuit current delivered through the main circuit breaker in</p><p>question. The interrupting rating of the B circuit breakers is 50,000</p><p>amp. When the total current reaches 40,000 amp, the current delivered</p><p>by these various sources is 6000 amp, 8000 amp, and 18,800 amp. The</p><p>motor contribution is 7200 amp. All currents are rms asymmetrical.</p><p>Cascaded operation is a means of lowering the cost of short-circuit pro-</p><p>tection in secondary systems. In the cascaded system, smaller feeder</p><p>circuit breakers are used than in the selective system; therefore this</p><p>differential favors the cascaded system from an economic standpoint.</p><p>I t must be recognized, however, that the service reliability of a cascaded</p><p>system is poorer than that of a selective system because in a cascaded</p><p>system, whenever a feeder short circuit draws a current in excess of 80</p><p>per cent of the interrupting rating of the feeder circuit breaker, the main</p><p>circuit breaker is tripped out and service on all feeders served by that</p><p>main circuit breaker or breakers is lost until the service is restored by</p><p>reclosing the main circuit breaker. This application has proved satis-</p><p>factory from a service-reliability standpoint for many industrial processes.</p><p>A-C SHORT-CIRCUIT PROTECTIVE OEVICES AND CIRCUIT EQUIPMENT 161</p><p>However, mhere critica1 processes are iiivolved, selective tripping is gen-</p><p>erally considered essential.</p><p>As a guide to selertion of</p><p>Iarge air circuit breakers for selective or cascade service, three-phase and</p><p>single-phase, see Tablc 3.3.</p><p>Selection of Large Air Circuit Breakers.</p><p>d:3J) 7 4 7 , 0 0 0 AMP</p><p>Y</p><p>TOTAL SHORT CIRCUIT</p><p>CURRENTAT THIS POINT</p><p>100,000 AMP RMS</p><p>ASYMMETRICAL 18,000 AMP</p><p>WHEN TOTAL CURRENT THRU BREAKER B IS 0.8</p><p>OF ITS RATING I.E. 40,000 AMP BREAKERS n'i,</p><p>"'2 8 d j MUST TRIP. THE CURRENT FLOWING</p><p>IN THIS CASE AND THE INSTANTANEOUS TRIP</p><p>ELEMENT SETTINGSX ON BREAKERS ~) .n; .e A; ARE:-</p><p>MOTOR</p><p>CONTR IBUTION</p><p>7200 AMP</p><p>THIS FAULT DRAWS</p><p>40.000 AMP RMS</p><p>ASYMMETRICAL</p><p>r</p><p>FIG. 3.8</p><p>ihan m e source of power to ihe main low-volloge bur.</p><p>One-line diogram showing lorge air circuil breakers in cascade wilh more</p><p>- m</p><p>b</p><p>J</p><p>TA</p><p>BL</p><p>E</p><p>3.</p><p>3</p><p>A</p><p>ir</p><p>-c</p><p>ir</p><p>cu</p><p>it-</p><p>br</p><p>ea</p><p>ke</p><p>r</p><p>A</p><p>pp</p><p>lic</p><p>at</p><p>io</p><p>n</p><p>T</p><p>ab</p><p>le</p><p>s-</p><p>C</p><p>as</p><p>ca</p><p>de</p><p>S</p><p>ys</p><p>te</p><p>m</p><p>a</p><p>nd</p><p>S</p><p>el</p><p>ec</p><p>tiv</p><p>e</p><p>S</p><p>ys</p><p>te</p><p>m</p><p>60</p><p>0</p><p>V</p><p>ol</p><p>ts</p><p>o</p><p>nd</p><p>L</p><p>es</p><p>s</p><p>R</p><p>at</p><p>in</p><p>gs</p><p>re</p><p>qu</p><p>ire</p><p>d</p><p>fo</p><p>r</p><p>eq</p><p>ui</p><p>pm</p><p>en</p><p>t f</p><p>or</p><p>I</p><p>ro</p><p>ns</p><p>fo</p><p>rm</p><p>er</p><p>a</p><p>nd</p><p>f</p><p>ee</p><p>d</p><p>er</p><p>c</p><p>ir</p><p>c~</p><p>it</p><p>l,</p><p>w</p><p>ith</p><p>s</p><p>el</p><p>ec</p><p>tio</p><p>n</p><p>o</p><p>f</p><p>ci</p><p>rc</p><p>ui</p><p>t</p><p>br</p><p>ea</p><p>ke</p><p>r</p><p>8</p><p>on</p><p>b</p><p>as</p><p>is</p><p>o</p><p>f</p><p>ca</p><p>sc</p><p>ad</p><p>e</p><p>sy</p><p>ds</p><p>m</p><p>a</p><p>nd</p><p>s</p><p>el</p><p>ec</p><p>tiv</p><p>e</p><p>tr</p><p>ip</p><p>s</p><p>ys</p><p>te</p><p>m</p><p>.</p><p>R</p><p>ef</p><p>er</p><p>t</p><p>o</p><p>m</p><p>on</p><p>vf</p><p>oc</p><p>tu</p><p>re</p><p>r f</p><p>or</p><p>o</p><p>th</p><p>er</p><p>l</p><p>irn</p><p>ito</p><p>tio</p><p>ns</p><p>.</p><p>O</p><p>th</p><p>e</p><p>r</p><p>fa</p><p>d</p><p>o</p><p>is</p><p>th</p><p>an</p><p>s</p><p>ho</p><p>rt</p><p>-c</p><p>irc</p><p>ui</p><p>t</p><p>du</p><p>ty</p><p>o</p><p>re</p><p>i</p><p>m</p><p>p</p><p>o</p><p>rt</p><p>a</p><p>n</p><p>t i</p><p>n</p><p>t</p><p>he</p><p>s</p><p>el</p><p>ec</p><p>tio</p><p>n</p><p>o</p><p>f</p><p>ci</p><p>rc</p><p>ui</p><p>t</p><p>br</p><p>ea</p><p>ke</p><p>rs</p><p>f</p><p>o</p><p>r</p><p>se</p><p>le</p><p>ct</p><p>iv</p><p>e</p><p>tr</p><p>io</p><p>n</p><p>in</p><p>e</p><p>.</p><p>S</p><p>ho</p><p>rt</p><p>-c</p><p>irc</p><p>ui</p><p>t</p><p>cu</p><p>rr</p><p>en</p><p>t,</p><p>rr</p><p>ni</p><p>a</p><p>m</p><p>p</p><p>.Z</p><p>m</p><p>pe</p><p>.&</p><p>(o</p><p>v</p><p>e</p><p>ra</p><p>g</p><p>e</p><p>t</p><p>hr</p><p>ee</p><p>-p</p><p>ha</p><p>se</p><p>T</p><p>m</p><p>ns</p><p>fo</p><p>rm</p><p>er</p><p>ra</p><p>tin</p><p>g,</p><p>th</p><p>re</p><p>e</p><p>-p</p><p>h</p><p>o</p><p>re</p><p>R</p><p>ec</p><p>om</p><p>m</p><p>en</p><p>de</p><p>d</p><p>in</p><p>te</p><p>rr</p><p>u</p><p>p</p><p>tin</p><p>g</p><p>r</p><p>at</p><p>in</p><p>g</p><p>o</p><p>f</p><p>o</p><p>ir</p><p>c</p><p>ir</p><p>w</p><p>it</p><p>b</p><p>re</p><p>ak</p><p>er</p><p>N</p><p>or</p><p>m</p><p>0</p><p>(s</p><p>ee</p><p>fl</p><p>gu</p><p>re</p><p>r</p><p>ab</p><p>ov</p><p>e)</p><p>lo</p><p>ad</p><p>K</p><p>vo</p><p>_</p><p>_</p><p>_</p><p>_</p><p>_</p><p>_</p><p>_</p><p>_</p><p>_</p><p>_</p><p>_</p><p>_</p><p>11</p><p>2.</p><p>5</p><p>4</p><p>15</p><p>0</p><p>1 4</p><p>22</p><p>5</p><p>5</p><p>Im</p><p>pe</p><p>d.</p><p>on</p><p>ce</p><p>,</p><p>p</p><p>e</p><p>r</p><p>C</p><p>eP</p><p>30</p><p>0</p><p>5</p><p>50</p><p>0</p><p>~</p><p>5</p><p>75</p><p>0</p><p>5</p><p>%</p><p>__</p><p>No</p><p>rm</p><p>.</p><p>lo</p><p>ad</p><p>ti"</p><p>"O</p><p>"l</p><p>:".,</p><p>O"</p><p>t</p><p>C</p><p>on</p><p>-</p><p>a</p><p>m</p><p>p</p><p>__</p><p>31</p><p>3</p><p>41</p><p>7</p><p>62</p><p>5</p><p>83</p><p>4</p><p>1,</p><p>38</p><p>8</p><p>2,</p><p>08</p><p>0</p><p>2,</p><p>78</p><p>0</p><p>4,</p><p>17</p><p>0</p><p>20</p><p>8Y</p><p>/1</p><p>20</p><p>V</p><p>ol</p><p>ts</p><p>.</p><p>T</p><p>hr</p><p>ee</p><p>P</p><p>h</p><p>o</p><p>ie</p><p>I</p><p>3 rn</p><p>24</p><p>0</p><p>V</p><p>ol</p><p>ts</p><p>,</p><p>T</p><p>hr</p><p>ee</p><p>P</p><p>ha</p><p>se</p><p>2</p><p>S</p><p>ho</p><p>rt</p><p>-c</p><p>irc</p><p>ui</p><p>t</p><p>cu</p><p>rr</p><p>en</p><p>t,</p><p>5 z</p><p>(a</p><p>v</p><p>e</p><p>ra</p><p>g</p><p>e</p><p>th</p><p>re</p><p>e-</p><p>ph</p><p>as</p><p>e</p><p>(s</p><p>ee</p><p>fl</p><p>gu</p><p>re</p><p>s</p><p>ab</p><p>ov</p><p>e)</p><p>5</p><p>om</p><p>po</p><p>re</p><p>.)</p><p>,</p><p>A</p><p>R</p><p>ec</p><p>om</p><p>m</p><p>en</p><p>de</p><p>d</p><p>in</p><p>te</p><p>rr</p><p>up</p><p>tin</p><p>g</p><p>ra</p><p>ti</p><p>n</p><p>g</p><p>o</p><p>f</p><p>a</p><p>ir</p><p>c</p><p>irc</p><p>ui</p><p>t</p><p>br</p><p>ea</p><p>ke</p><p>r</p><p>rm</p><p>r</p><p>am</p><p>p</p><p>></p><p>B</p><p>6</p><p>ti</p><p>*=</p><p>c?</p><p>,</p><p>co</p><p>de</p><p>;I</p><p>I</p><p>c</p><p>re</p><p>le</p><p>c-</p><p>tr</p><p>ip</p><p>Tr</p><p>an</p><p>s-</p><p>10</p><p>0%</p><p>al</p><p>o</p><p>n</p><p>e</p><p>lo</p><p>ad</p><p>A"</p><p>! A' I</p><p>ca</p><p>+</p><p>8.</p><p>40</p><p>0</p><p>1,</p><p>35</p><p>0'</p><p>9</p><p>,7</p><p>50</p><p>50</p><p>,0</p><p>00</p><p>11</p><p>,2</p><p>00</p><p>1.</p><p>80</p><p>0</p><p>13</p><p>.5</p><p>50</p><p>50</p><p>,0</p><p>00</p><p>13</p><p>,4</p><p>00</p><p>2,</p><p>70</p><p>0</p><p>16</p><p>,1</p><p>00</p><p>50</p><p>,0</p><p>00</p><p>17</p><p>.9</p><p>00</p><p>3.</p><p>60</p><p>0</p><p>21</p><p>5</p><p>00</p><p>7</p><p>5</p><p>0</p><p>0</p><p>0</p><p>95</p><p>,9</p><p>00</p><p>1</p><p>50</p><p>00</p><p>0</p><p>I0</p><p>00</p><p>00</p><p>5</p><p>00</p><p>00</p><p>1</p><p>00</p><p>.0</p><p>00</p><p>1'</p><p>1</p><p>'I,</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT I63</p><p>I64 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>Standards. The XEMA Standards that, apply to all large air circuit</p><p>breakers are KO. SG3-1951.</p><p>MOLDED-CASE CIRCUIT BREAKERS</p><p>Malded-case circuit breakers, Fig. 3.9, are smaller in dimension, less</p><p>sturdily constructed, and do not have t,he electrical &ararrces that large</p><p>air circuit breakers have. They are distinguished from large air cir-</p><p>cuit hrcakcrs primarily because of the fact that t,hey are mounted in a</p><p>molded plastic case. These circuit, breakers have built-in trip element,s,</p><p>and in some cases they are adjustable. Also marly functions cannot, be</p><p>huilt into these smaller molded-case circuit-breaker tripping elements that</p><p>can be huilt into the large air circuit-breaker tripping e1ement.s. It, is</p><p>not easy t o make t,hem electrically operated or to provide large numbers</p><p>of auxiliary swit,ches.</p><p>Ratings Available. Ratings are available as given in Table 3.4.</p><p>FIG. 3.9 Molded-care air circvit breakers mounted in a panel board.</p><p>165 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 165 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>MOLDED CASE CIRCUIT {h 1 1 )</p><p>BREAKERS I 1 I I</p><p>MAGNETIC</p><p>C I R C U I T</p><p>BREAKERS</p><p>LOAD CENTER</p><p>SUBSTATION</p><p>WITH</p><p>MAGNETIC</p><p>C I R C U I T</p><p>BREAKERS</p><p>INDIVIDUAL</p><p>MOLDED CASE</p><p>BREAKERS</p><p>I</p><p>&75D KVA MAX</p><p>A 1:</p><p>+MOLDED ) CASE</p><p>BREAKERS</p><p>IN PLUG-IN</p><p>DEVICE</p><p>DISTRIBUTION CENTER</p><p>MOLDED CASE</p><p>BREAKERS</p><p>FIG. 3.10</p><p>plied in a low-voltage power distribution system.</p><p>One-line diagram showing where molded-core air circuit breakers can be ap-</p><p>Application. Because of their small size and lower cost, the molded-</p><p>case circuit breakers find application for branch-circuit, protection where</p><p>the interrupting duty is within their interrupting rating, Fig. 3.10. They</p><p>also find applicabion on the secondaries of some small light-duty Ioad-</p><p>center unit substations.</p><p>These circuit breakers are not</p><p>suitable for cascade operation wit,h large air circuit breakers berause they</p><p>operate so fast that the large air circuit breakers are not able to protect</p><p>them (see iVEhlA Standards for Large Air Circuit Breakers, Section</p><p>SG3-3.43). Xeither are they suitable for cascading vi th ot.her molded-</p><p>case circuit breakers. This conclusion mas reached after exhaustive tests.</p><p>Not Suitable for Cascade Operation.</p><p>166 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>Selection of Interrupting Ratings. As a guide, the portion of Table 3.3</p><p>referring to circuit hreakers for selective operation may be used mhere the</p><p>continuous current is less than 600 amp and iiiterrupting duty is within</p><p>the available ratings of molded-case circuit breakers.</p><p>Table 3.4 gives the interrupting ratings as defined by applicable</p><p>NEMA standards.</p><p>TABLE 3.4 Interrupting Rotings of Molded-cose Circuit Breakers for</p><p>A-C Service</p><p>7,500</p><p>20,000</p><p>25,000</p><p>50,000</p><p>30,000</p><p>Interrupling iatingi, r m i amp orymmetrical</p><p>15,000 15.000</p><p>20,000 15.000</p><p>35,000 25.000</p><p>25,000 25,000</p><p>241-480 volts 600 ~011s i 240 ~o l ts</p><p>and below</p><p>I</p><p>Range of trip-coil</p><p>rrrtingr amp</p><p>~-</p><p>15-100</p><p>15-100</p><p>125-225</p><p>125-225</p><p>125-600</p><p>)O, 125, 150, 175, 200, 225, 250,</p><p>Standards. At preseiit there are no applicable NEMA standards for</p><p>molded-case air circuit breakers.</p><p>FUSED SWITCHES</p><p>Fused switches, Fig. 3.11, consist of an interrupter switch and a fuse</p><p>mouuted on a common base and usualiy in a metal enclosure. There are</p><p>many types and varieties available.</p><p>The most common variety</p><p>is the standard NEC (Kational Electrical Code) cartridge fuses. These</p><p>fuses practically a11 corisist, of a fusible link enclosed in a cylindrical cart-</p><p>ridge with connectors a t each end to slip into the fuse clips of the switch.</p><p>Xew and improved designs of fuses and switches for low-voltage service</p><p>have been developed recently. The fuses are mainly of the current-</p><p>limiting high interrupting capacity silver-sand type, typical of which is</p><p>the General Electric type EJ-6 fuse shomn in the smitch in Fig. 3.11.</p><p>Typical of the improved switches is t,he type HCI switc,h as manufac-</p><p>tured by the Trumhull Components Department of the General Electric</p><p>Company. To be specific in the follomiiig discussion of the improved</p><p>types of fuses and switches, the type HCI smitch and EJ-6fuseivill beused.</p><p>There is a wide variety of lon-voltage fuses and</p><p>switches available. These ratings run from as low as a few amperes up to</p><p>There are severa1 types of fuses available.</p><p>Ratings Availoble.</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 167</p><p>FIG. 3.1 1</p><p>current-limiting silver-rand fuses (EJ-6).</p><p>several hiindred amperes. Cnfortunately most, of them do not have</p><p>short-cirruit rat,ings assigned. Again, t,o be specifio, t,lie type IICI</p><p>switch and E.J-6 fuse will be used t,o illusirate a-c short-circuit abilities</p><p>which have hem established hy test,. The preserit availahle data are</p><p>listed in Table 3.5.</p><p>High-copocity interrupting (HCI) enclosed switch with high interrupting-rating</p><p>TABLE 3.5 Interrupt ing Ability of Type HCI Switches and EJ-6</p><p>Current- l imit ing Fuses (1954)</p><p>I</p><p>Type HCI switch I Type EJ-6 fuses</p><p>I</p><p>Volts Amperes Volts 1 Amperes ! -I--, I</p><p>15-20-30</p><p>200</p><p>lnterrvpling ability of</p><p>combination HCI rwilch</p><p>and EJ-6 fuse. amp</p><p>b y m l</p><p>100.000</p><p>100.000</p><p>100,000</p><p>l00,000</p><p>Application. All small hiRh-interrupting-ability loix~-voltage fuses are</p><p>current,-limit,ing in t,heir action, hence are very fast in their operation, and</p><p>from this staidpoint they are partieiilarly well suited to branch-circuit</p><p>168 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMEW</p><p>L LOAD CENTER - UNIT SUBSTATION</p><p>INDIVIDUAL WALL MOUNTED</p><p>HCI SWITCHES AND EJ- 6 FUSES</p><p>FIG. 3.12</p><p>the lost protective device in low-voltage distribution circuits.</p><p>One-line diagram showing whsre safety switches m d fuses may be applied as</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 169</p><p>AVAILABLE SHORT CIRCUIT CURRENT-</p><p>ASYMMETRICAL RMS AMPERES</p><p>1.25 I SYMMETRICAL (AVERAGE FOR THREE CONDUCTORS)</p><p>AVAILABLE SHORT CIRCUIT</p><p>FIG. 3.1 3</p><p>current-limiting fuses (60 cycler).</p><p>Curves showing the current-limiting choracterirtics of type EJ-6 silver-sand</p><p>I70 A-C SHORT-CIRCUIT PROTECTIVE DEVICES</p><p>AND CIRCUIT EQUIPMENT</p><p>will withstand 9000 amp rms for 0.2 cycle. So, the 30-amp fuse vill pro-</p><p>tect a wire which will be required to carry 30-amp load current.</p><p>This current-limiting feature, in addition to protecting small wires in</p><p>systems of high short-circuit-current capacity, can protect small switching</p><p>devices. It is for this reason that the type HCI switch can he used with</p><p>type EJ-6 fuses 011 circuits where the available short-cirruit-current duty</p><p>is as high as 100,000 amp.</p><p>The t,ype HCI switch and EJ-G fuse combination has high interrupting</p><p>rating arid is current-limit,ing in its operation which enables it to beusedin</p><p>many places where molded-case circuit breakers would not have adequate</p><p>interrupting rating and where large air circuit breakers would be too ex-</p><p>pensive, too large, or not applicable from an engineeriug standpoint. For</p><p>example, a circuit breaker for a 30-amp circuit fed from a certain low-</p><p>voltage bus may require a circuit breaker with 100,000 amp interrupting</p><p>rating. The wire or cable mould have to be of the order of 350 MCM to</p><p>withstand the short-circuit current. In the first place, a 100,000-amp</p><p>interrupting rating circuit breaker cannot be built with a 30-amp trip coil</p><p>that will withstand the short-circuit forces or heating. In the second</p><p>place, any 30-amp load devire mould not have terminals that would</p><p>accommodat,e 350-MCM cable, the size required to withstand 100,000</p><p>amp. The use of an EJ-G current-limiting fuse and the HCI switch rated</p><p>30 amp would provide adequate short-circuit protection, and the current-</p><p>limiting effect of the fuse mould enable a wire of smaller size to be used.</p><p>The switch and fuse comhinat,ion is not generally suitable for main feeder</p><p>circuit protection because of the fact that it is difficult to make the fast</p><p>current-limiting fuses operate selectively with other overcurrent protec-</p><p>tive devices that would be in the circuit between the fuse and the load.</p><p>Information for standards on fuses may be obtained from</p><p>the Underwriters Laboratories, Incorporated, bulletin, Standard for</p><p>Fuses.</p><p>Information on st,aridards for enclosed switches (safety switches) may</p><p>be obtained from IJnderwriters Laboratories, Incorporated, bulletin,</p><p>Standard for Enclosed Switches or NEMA Publication No. 42-78,</p><p>Enclosed Switch Standards.</p><p>Standards.</p><p>SHORT-CIRCUIT PROTECTIVE EQUIPMENT</p><p>FOR SYSTEMS ABOVE 600 VOLTS</p><p>There are in general two types of short-circuit protective equipment</p><p>1. Power circuit breakers</p><p>2. Power fuses</p><p>available for systems above GO0 volts. These are:</p><p>A-C SHORT.CIRCUlT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 171</p><p>POWER CIRCUIT BREAKERS</p><p>There are many types of power circuit breakers availahle, but basically</p><p>they are divided into the oil t,ypc and the nillcss type. In the field 2.4- to</p><p>13.8-kv t,he oilless-type cirruit breaker, Fig. 3.14, has largely superseded</p><p>the oil-t,ype circuit breaker. In indoor metal-enclosed switehgear of the</p><p>st,ation t,ypc for circuits 13.8 Lo 34.5 kv, the air-type circuit breakers are</p><p>in general superseding the oil-type vircuit breakers. I n the field above</p><p>11.A kv for outdoor switchgear, oil circuit breakers are most commonly</p><p>used, Fig. 3.15. For the sake of the discussion here relative to dec t ion</p><p>of equipment> fiom a short,-rircuit standpoint, it makes no difference</p><p>whcthcr the rircirit breakers are of t,he nil or oilless type.</p><p>High-voltage power circuit breakers are availahle</p><p>in ratings from 2.4 kv up to over 300 kv and in interruptirig ratings from</p><p>15 mva up to 25,000 mva. Complete listings of power circuit breakers</p><p>can he found iii the latest copy of S E R l A Standards SG&l954. The cir-</p><p>cuit, breakers most comtnonly used in industrial plants are the oilless or</p><p>air type, sho\rn i n Fig. 3.14. The available ratings of this type of cir-</p><p>cuit breaker are given in Table 1.1 (Chap. I).</p><p>Ratings Available.</p><p>FIG. 3.14</p><p>c i tw i ts rated 2.4 to 13.8 kv.</p><p>Typical ille err (air) power circuit breaker ar wed in metal-clad switchgear for</p><p>172 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>FIG. 3.15</p><p>This circuit breoker i s rated 34.5 kv.</p><p>Outdoor frome-type oil circuit breaker 01 used in circuits rated above 15 kv.</p><p>Application. Power circuit breakers are applicable anywhere in the</p><p>syst,cms rated 2.4 kv up t,o the highest a-c voltages in use today. They</p><p>combine all the essential characteristics for circuit switching and protec-</p><p>t,ion and therefore may be used at main buses supplied by generators or</p><p>transformers or in connection with unit substations. They are also</p><p>applirable at, loral switching points and for protection of primary branch</p><p>circuits (see Fig. 3.16).</p><p>Certain of the power circuit</p><p>breakers, particularly the oilless type, are suitable for motor-st,arting duty</p><p>within the limitations outlined by the manufacturer. It should be noted</p><p>that compared with contact,ors the principal limitation of power circuit</p><p>breakers for motor-starting duty is the degree of repetitive duty that can</p><p>be withstood. Contactors are designed for more operations and longer</p><p>life under severe operating duty cycles than are power circuit breakers.</p><p>Motor Starting or Other Repetitive Duty.</p><p>A-C SHORT-CIPCUIT PROTECTIVE DEVICES ANO CIRCUIT EQUIPMENT 173</p><p>Selection of Interrupting Ratings. The selection of interrupting</p><p>ratings of power circuit breakers for industrial applications is out,lined in</p><p>Chap. 1. A detailed description of the various faetors to consider in</p><p>applying oilless eircuit breakers as used in metal-clad switchgear is given</p><p>there.</p><p>I I 69 KV</p><p>OUTDOOR POWER</p><p>CIRCUIT BREAKERS Q P Q GENERATOR T TYWI</p><p>TRANSFORMER</p><p>SECONDARY</p><p>CIRCUIT BREAKER</p><p>GENERATOR</p><p>CIRCUIT BREAKER</p><p>' MAlN FEEDER ! CIRCUIT BREAKER</p><p>AHEAD O F L I N E</p><p>OF LIMITAMP</p><p>MOTOR STARTERS</p><p>A</p><p>LARGE OU</p><p>HIGH VOLTAGE</p><p>MOTORS</p><p>FIG. 3.16 One-line diogrorn rhowing where oilless power circuit breakerr in metal-clad</p><p>rwitchgeclr and outdoor power cirwit brecikerr may be applied in industrial power dir-</p><p>tribution ryrtemr.</p><p>174 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>Standards. Poiver circuit breakers are eovered by NEMA Standards</p><p>SG1-19%</p><p>POWER FUSES AND OIL-FUSE CUTOUTS</p><p>There are many types of power fuses available for circuits rated 2.4 kv</p><p>and above. These t,ypes of fuses, generally speaking, divide t,hemselves</p><p>into three categories. The first is the power fuse, typical examples of</p><p>which are shown in Fig. 3.17 which are for high-rapacity power circuits.</p><p>The second type that is slightly differeni, i n construct,ion i s the oil-fuse</p><p>cutout, which i s really a combination of a cntout and a fuse immersed</p><p>in a container of oil, Fig. 3.18. The third type of fuse is used mainly in</p><p>distribntion cutouts for overhead opcir-wire outdoor distriliutioii systems</p><p>of utilit.ics in urban and suburban areas, Fig. 3.119.</p><p>FIG. 3.17</p><p>enpulrion silver-rand type, (right] "on-current-limiting expulsion outdoor type.</p><p>Typical high-voltage (above 600 volts1 power furer: Ifeft) current-limiting non-</p><p>176 A-C SHORT-CIRCUIT PROTECTIVE DNICES AND CIRCUIT EQUIPMENT</p><p>The last type of fuse mentioned is applicable toindustrial power systems</p><p>for outdoor installations only where the interrupting rating is less than</p><p>the duty on the system. This fuse is not metal-enclosed and is not for</p><p>indoor installation.</p><p>In general, power fuses divide thcmselves into two classes, i.e., current-</p><p>limiting and non-current-limiting. Typical of the current-limiting</p><p>category are the silver-sand fuses, Fig. 3.17(left). Typical of the non-cur-</p><p>rent-limiting type are the oil-fuse cutout, Fig. 3.18, the expulsion fuses,</p><p>Fig. 3.17(right), as well as the “boric acid” fuses and “liquid” power fuses.</p><p>A further classification is that some are expulsion type, i.e., expel hot</p><p>gases when they operate. These are not suitable for indoor application</p><p>because of the hazard of the expelled hot gases. Such fuses are the</p><p>expulsion fuse, Fig. 3.17(right), and the “boric acid”</p><p>fuse without a con-</p><p>denser and the “liquid fuse”. Typical of the nonexpulsion type are the</p><p>silver-sand fuse, Fig. 3.17(left), and the “boric acid’’ fuse with condenser.</p><p>All types of power fuses operate faster than</p><p>power circuit breakers a t or near their interrupting ratings. Because of</p><p>the fast operating time of the fuses, they are generally employed as the</p><p>last circuit protective device in each voltage level in a primary power sys-</p><p>tem, as shown in Fig. 3.20. Typical applications are in motor starters</p><p>and ahead of primaries of transformers stepping down to a lower volt-</p><p>age. The silver-sand fuse, Fig. 3.17(left), is often the preferred type of</p><p>fuse for power circuits because of its fast operating time and current-</p><p>limiting ability. However, in some cases where coordination is required,</p><p>it may be necessary to use non-current-limiting types of fuses which have</p><p>longer time delay. However, when the longer time delay is obtained,</p><p>the benefits of reduction of damage to the circuit through which short-</p><p>circuit current passes is lost to a large degree.</p><p>Nonexpulsiori-type power fuses suit-</p><p>able for indoor use are often applied in a metal enclosure with an inter-</p><p>rupter switch to form a switch-and-fuse cornbination for high-voltage</p><p>circuits. Interrupter switches are desirable for this application because</p><p>they have interrupting ratings usually in the range of 100 to 400 amp.</p><p>Plain disconnecting switches are generally not satisfactory for this service</p><p>because they have no interrupting ability, and therefore the combination</p><p>of the plain switch and fuse cannot be used as a load-switching device.</p><p>The oil-fused cutouts combine in one unit the fuse and the interrupter</p><p>switching element. Interrupter slyitches and fuses and oil-fused cutouts</p><p>find wide application in industrial plants as the primary swit,ching and</p><p>protecting section of a load-center unit substation (see Chap. 11).</p><p>Open-structure</p><p>switches or disconnect,ing mountings without current-interrupting ability</p><p>are often used with power fuses. These can be considered for isolation</p><p>purposes only. Hazards in operations are materially increased in this</p><p>Application-General.</p><p>Interrupter Switches and Fuses.</p><p>Application of Fuses in Open Switching Structures.</p><p>A-C SHORT-CIRCUIT PROTECTIVE DNICES AND CIRCUIT EQUIPMENT 177</p><p>type of appliration. That is the reason that such applications should be</p><p>limited to outdoor structures Ivhere the operator is a considerable distance</p><p>from the disconnecting switch when operating it. The use of such</p><p>isolatiug switches i n series with fuses in indoor metal-enclosed structures</p><p>is not coilsidered safe practice bemuse of thc proximity of the operator</p><p>to t,he sivitrh and the possibility of the operator inadvertently operating</p><p>the switch under roiiditions i u which the switch will hare to interrupt or</p><p>close in on currents ronsiderably beyond its ability. Failure may result</p><p>eveti though there is a fuse in series with such switches.</p><p>33 KV</p><p>UTDOOR TYPE FUSE P</p><p>SMALL POWER</p><p>IyTy\ TRANSFORMER</p><p>I</p><p>LIMITING</p><p>AHEAD OF SMALL LOAD CENTER UNIT SUBSTATIONS-</p><p>USE INTERRUPTER SWITCH AND POWER FUSE OR</p><p>FUSED OIL CUTOUTS.</p><p>FIG. 3.20</p><p>may be applied in industrial power distribution systems.</p><p>One-line diagram rhowing where high-voltage (above 600 VOllS) Power</p><p>178 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>Selecting Fuse-interrupting Rating. Fuses are generally rated in</p><p>amperes interrupting ahility. CMculate the short-circuit duty in rms</p><p>amperes asymmetrical at the first half cycle as outliiied in Chap. 1, and</p><p>select a fuse whose interrupting rating is greater than the duty imposed.</p><p>Equivalent three-phase iiiterrupt,ing ratings may also be considered.</p><p>Since the ratings of fuses are not too well st,andardized, refer to the fuse</p><p>manufacturer for complete data before applying fuses.</p><p>Power fuses are covered by S E M A Standards, Cutouts,</p><p>I’orer Fuses, and Current-limit,ing Resistors, Publication S(2-1954,</p><p>and AIEE Standards S o . 25.</p><p>Standards.</p><p>MOTOR STARTERS</p><p>There are in general three kinds of motor starters:</p><p>1 . The contactor</p><p>2. The combination motor starter</p><p>3. The circuit breaker</p><p>Contactors are in general of two types, the most common variety being</p><p>t,hose which have an interrupting rating of only ten times normal rated</p><p>current. These are completely inadequate for short-circuit protection</p><p>and must have addit,ional short-circuit protection provided by either</p><p>fuses or circuit breakers.</p><p>When a short,-circuit protective device such as a fuse or circuit breaker</p><p>FIG. 3.21 Typical low-</p><p>voltage 1600 volts and</p><p>below) combination motor</p><p>starter with current-limit-</p><p>ing silver-rand furer for</p><p>short-circuit protection.</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 179</p><p>is used in comhinatiori with contactors, it forms what is commonly called</p><p>a combination motor starter.</p><p>In systems 600 volts and less</p><p>there are t w i types of cornhination motor starters, both employing the</p><p>same type of cont,act,or. The first employs a fuse disconnecting sm.itch</p><p>alirad of t,he conta,ctor, and the other a circuit, breaker, usually a molded-</p><p>case-type circuit breaker, ahead of t,he cout,actor. The select,ion between</p><p>t,hese two is based mainly on the fuudamerital differelice betveen fuses</p><p>and circuit, breakers as short-circuit protective devices. The fused</p><p>combinatirin motor starters have an over-all inter-</p><p>rupting ahility so that the combination motorst,arter</p><p>can successfully irit,errupt an available short-circuit</p><p>current equal t,o 50,000 amp rms asymmetrical wheri</p><p>equipped wit,h high-interrupt,ing-capacity current-</p><p>limiting fiises. This is for a short circuit outside ”</p><p>Circuits 600 Volts and Less (Fig. 3.21).</p><p>the case of t,he mot,or starter and using type E.14</p><p>fuses.</p><p>The molded-case circuit-breaker comhiiiat,ion</p><p>mot,or starters are limited to a maximum duty of</p><p>15,000 or 25,000 amp rms asymmet,rical.</p><p>Circuits above 600 Volts (Fig. 3.22). For circuits</p><p>of 2.4 kv aud up to 5 kv, the combination motor</p><p>starter commonly used consists of current-limiting</p><p>silver-sand fuses and contactors with the fuses</p><p>mount,ed in disconnecting-type supports and placed</p><p>in a metal enclosure s o interlocked that the fuses</p><p>cannot be disconneeted unless the coritactor is in the</p><p>open position. In this way the disconnecting fuse</p><p>mounting has no current to interrupt. Since the</p><p>fuses are for short-circuit protection only, suit,able (2,4 to 4,8 kv)</p><p>running overload relays should he provided in the b i n t i o motor</p><p>motor st.art,er. These motor starters have inter- starter using current-lim-</p><p>rupting ratings of 150 mva at 2.4 kv and 250 mva iting rilver-rand power</p><p>furel for short-circuit a t 4.16 kv. From a short-circuit standpoint they</p><p>protection.</p><p>may be appIied up to their momentary and int,er-</p><p>rupting rat,iiig. Since these devices contain fuses as the short-circuit</p><p>protective element, they are naturally best suited to application as the</p><p>last protective device in the circuit. When used as motor starters, they</p><p>are t,he last protective device, and therefore the fast operating time of</p><p>the fuse is a very dist,inct advantage in limiting damage in the motors</p><p>when a failure occurs. The fast operating time of the fuse also permits</p><p>low settings on other relays further back in t.he system.</p><p>FIG.3.22 Typicalhigh-</p><p>180 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>SELECTION OF CONDUCTORS AND OTHER CIRCUIT</p><p>COMPONENTS FROM A SHORT-CIRCUIT STANDPOINT</p><p>The floiv of short-circuit current in an electric system imposes mechani-</p><p>cal and thermal st,resses (heating) on all component,s of the system through</p><p>which such currents flow. This includes cables, bus bars, current trans-</p><p>formers, disconnecting switches, as \veil as circuit breakers, fuses, and</p><p>motor starters. The following is intended t,o aid in the selection of circuit</p><p>component,s, ot,her than circuit breakers, fuses, and motor starters, from</p><p>a short-circuit,-current</p><p>standpoint.</p><p>POWER-CABLE SELECTION FROM A SHORT-CIRCUIT STANDPOINT</p><p>Multiple-conductor power cables possess high mechanical strength</p><p>because of the compact conductor lay and the continuous concentric bind-</p><p>ing arsist,ed many times by armor or lead sheath. KO limit on mechanical</p><p>stresses in such cables has been assigned.</p><p>This is not true with regard to thermal effects. In common with ot,her</p><p>current-carrying parts of the electric system during short-circuit-rurrent</p><p>flow, the abrupt elevation in conductor temperature will be limited only</p><p>by the ability of the conductor metal to absorb the heat developed. The</p><p>magnitude of the temperature increase is greater (1) as the current magni-</p><p>tude becomes greater (as the square of the current), (2) as the conductor</p><p>cross section becomes smaller, and (3) as the duration of current flow</p><p>becomes greater.</p><p>Temperature limits. Power-system short-circuit-current magnitudes,</p><p>feeder-conductor cross section, and short-circuit protective device inter-</p><p>rupting time should be coordinated to avoid severe permanent damage to</p><p>cable insulation during an interval of short-circuit-current flow in the</p><p>system. The effect should be limited to a moderate reduction in useful</p><p>cable life (possibly 1 per cent of normal life).</p><p>Reasonable maximum-peak transient temperatures for various cable</p><p>insulations and operating potentials have been designated and in general</p><p>are approximately 150 C (see Table 3.6). At a slightly higher tempera-</p><p>ture (approximately 175 C), destructive disint,egration of organic mate-</p><p>rials may occur, accompanied by smoke and combustible vapors. At</p><p>somewhat higher temperatures large quantities of combustible vapors are</p><p>expelled which increases the risk of explosion and fire.</p><p>It is important to note that the abnormal temperature persists much</p><p>longer than t,he duration of short-circuit-current flow. For example, the</p><p>flow of 20,000 amp in a KO. 4/0-Awg copper conductor will elevate the</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>copper temperature from an initial temperature of 75 C to 150 C i i i ahout</p><p>34 see. With the current then redured to zero, about 1000 see d l be</p><p>required for the copper temperature to return to 75 C in a 30 C ambient.</p><p>181</p><p>TABLE 3.6 Conductor Rated Maximum Continuous Operating Temperature</p><p>and Peak Transient (Momentary) Temperature for Various Types</p><p>and Operating Voltages</p><p>Cable type</p><p>Vc type V or VL, single conductor or three conductor.. .......</p><p>Impregnated paper (slid), single conductor.. ..............</p><p>lmpregnalod paper (did), three mnductor. ...............</p><p>Type R*. ............................................</p><p>Tips RH ............................................</p><p>Coronol .............................................</p><p>-</p><p>lollogl</p><p>d.**,</p><p>kv</p><p>-</p><p>I</p><p>5</p><p>8</p><p>I5</p><p>1</p><p>5</p><p>8</p><p>I5</p><p>1</p><p>5</p><p>8</p><p>I5</p><p>I</p><p>5</p><p>8</p><p>15</p><p>1</p><p>5</p><p>8</p><p>1</p><p>5</p><p>8</p><p>15</p><p>-</p><p>85</p><p>85</p><p>84</p><p>77</p><p>85</p><p>85</p><p>85</p><p>81</p><p>85</p><p>85</p><p>85</p><p>81</p><p>60</p><p>60</p><p>60</p><p>60</p><p>75</p><p>75</p><p>75</p><p>80</p><p>80</p><p>80</p><p>80</p><p>-</p><p>M O X</p><p>lronrienl</p><p>copper</p><p>temp.</p><p>C</p><p>I50</p><p>145</p><p>135</p><p>120</p><p>I50</p><p>I45</p><p>140</p><p>135</p><p>I40</p><p>135</p><p>I30</p><p>125</p><p>I40</p><p>135</p><p>130</p><p>125</p><p>I50</p><p>145</p><p>140</p><p>I50</p><p>145</p><p>I40</p><p>I30</p><p>* Applies to new type R (1947 code specification).</p><p>t Actual operating temperature may be lompr because of consprvative application</p><p>or a favorable ambient temperature.</p><p>Conductor Heating. On the basis that all heat produced by short-</p><p>circuit-current flow is initially absorbed by the rondurtor metal (wbirh</p><p>I82 A-C SHORT-CIRCUIT PROTECTIVE DNICES AND CIRCUIT EQUIPMENT</p><p>TABLE 3.7 Quick Estimating Table of Minimum Conductor Sire'</p><p>A. Low-voltoge Air-circuit-breaker Protection</p><p>Interrupting kvo at</p><p>Short-circuit current.</p><p>11.25 X rymmetricall</p><p>omp</p><p>Duration of hort-sircuit current</p><p>5,000</p><p>I0,OOO</p><p>15,000</p><p>25,000</p><p>35,000</p><p>5 o . m</p><p>75,000</p><p>100.000</p><p>No. 2 Awg</p><p>No. 2 Awg</p><p>No. 2 Awg</p><p>No. 2 Awg</p><p>No. 1 Awg</p><p>No. 1 Awg</p><p>No. I /O Awg</p><p>No. 2/0 Awg</p><p>No. 2/0 Awg</p><p>Na 3/0 Awg</p><p>No. 4/0 Awg</p><p>300 MCM</p><p>350 MCM</p><p>400 MCM</p><p>500 MCM</p><p>600 MCM</p><p>Short-circuit</p><p>current, amp</p><p>(1.0 x</p><p>symmetricoll</p><p>No. 2 Awg</p><p>No. 1 Awg</p><p>No. 1 Awg</p><p>No. 1/0 Awg</p><p>No. 2/0 Awg</p><p>No. 2/0 Awg</p><p>No. 3 / 0 Awg</p><p>No. 3/0 Awg</p><p>No. 4/0 Awg</p><p>250 M C M</p><p>300 M C M</p><p>400 MCM</p><p>500 M C M</p><p>600 M C M</p><p>750 M C M</p><p>750 M C M</p><p>3,000-3.500</p><p>3,500-4.000</p><p>4.000-4.500</p><p>4.500-5.000</p><p>......</p><p>25 mva</p><p>50 mvo ......</p><p>00 m w</p><p>......</p><p>5 0 m r o</p><p>5,000-6.000</p><p>6.000-7.000</p><p>7,000-8,000</p><p>25mvo ....... 75 m w No.6Awg</p><p>........................... No. 4 Awg</p><p>............. 50 mva ....... No. 4 Awg ........................... No. 4 Awg</p><p>........................... No. 2 Awg</p><p>50 m w ....... I50 mva No. 2 Awg ........................... No. 2 Awg ............. 100 mva ....... No. 1 Awg</p><p>........................... No. 1 Awg</p><p>No. 2/0 Awr ........................... No. 3/0Awr</p><p>.............. 250 mvo No. 1/0 Awg</p><p>100 mva 150 m w .......</p><p>150 m w 250 m w 500 m w No. 4/0 AWI ........................... 250 M C M</p><p>250 mva ....... 750 mvo 300 M C M ....... 500 mvm ....... 3 5 0 M C M</p><p>8,000-9.000</p><p>9,000-1 0,000</p><p>10,000- 12.500</p><p>12.500-15,000</p><p>15.000-20.000</p><p>20,000-25.000</p><p>25,000-30.000</p><p>30.000-35.000</p><p>35.000-40.000</p><p>~</p><p>1.5 to 2 cycles</p><p>linrt. trip)</p><p>No. 8 Awg</p><p>No. 4 Awg</p><p>No. 2 Awg</p><p>No. I Awg</p><p>No. 1/0 Awg</p><p>No. 3 / 0 Awg</p><p>300 M C M</p><p>350 M C M</p><p>~~</p><p>>g s*c</p><p>No. 4 Awg</p><p>No. 1 Awg</p><p>No. 2/0 Awg</p><p>No. 4/0 AWQ</p><p>300 M C M</p><p>400 M C M</p><p>600 M C M</p><p>800 M C M</p><p>No. 2 Awg</p><p>No. 1/0 Awg</p><p>No. 3/0 Awg</p><p>300 M C M</p><p>400 M C M</p><p>600 MCM</p><p>800 M C M</p><p>I000 M C M</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQVIPMENT 183</p><p>has been proved to be valid for canductor sizes of No. 8 Awg or larger*),</p><p>the conductor heating is governed by the following:</p><p>For Copper:</p><p>t = duration of current flow, see</p><p>I = rms amperes during entire interval of current flow</p><p>em = conductor cross sect,ion, cir mils</p><p>TI = initial copper temperature, C</p><p>T2 = final copper temperature, C</p><p>To simplify a n application, these relationships are presented graphically</p><p>in the large chart in Fig. 3.23. The permissible time for various ternpera-</p><p>ture ranges can be quickly evaluated with the aid of the auxiliary curve B,</p><p>shown in Fig. 3.23. For quick estimating purposes, minimum safe con-</p><p>ductor size is given in Table 3.7, subject to application conditions as</p><p>shown.</p><p>The problem of joining and terminating aluminum</p><p>conductors without creating local “hot spots” deserves very careful</p><p>attention. There are available, however, materials and methods which</p><p>laboratory tests and experience have proved to be satisfactory.</p><p>In the absence of abnormal local heating, a rough approximation of per-</p><p>missible current duration may he made on the basis of the same limiting</p><p>temperatures as for copper. (For a particular current and conductor</p><p>cross section, the permissible duration of short-circuit-current flow will</p><p>he 45 per cent of that for copper.) It may be more convenient to make</p><p>an artificial correction in current. Consider the current to be 150 per</p><p>cent of the actual value, and proceed on the chart (Fig. 3.23) as if the</p><p>conductor were copper.</p><p>Rms current as used here is defined as the root-rnean-</p><p>square value for the total interval of short-circuit-current flow. The</p><p>temporary d-c component encountered in a-c circuits increases the rms</p><p>current, but to a lesser extent as the interval of current flow becomes</p><p>longer. The appropriate factor K , by which the symmetrical current</p><p>value shall be multiplied to determine the true rms current is given in</p><p>chart A , Fig. 3.23, for several typical ratios of circuit 60-cycle reactance</p><p>* B. W. Jones and J. A. Seott, Short-time Current Ratings for Aircraft Wire and</p><p>Cable. AIEE Technical I’sper, 1946.</p><p>For Aluminum.</p><p>Rms Current.</p><p>184 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT 18.5</p><p>"I. -8.</p><p>FIG. 3.23 Short-time bhort-circuit) heating limits of copper cables and correlation of</p><p>current and time to elevate the copper temperature from</p><p>75 to 150 C (dl heat is oirumed</p><p>to be stored in the copped.</p><p>to resistance (distribution circuits will generally fall in thc region of</p><p>X / R = 10 or less).</p><p>Circuit X / R ratio is generally not known and requires numerous circuit</p><p>constants for an evaluation. Conservative factors (K,) for the more</p><p>common application conditions are</p><p>Low-voltage circuit breakers tripped instantaneously</p><p>Power circuit breakers, eight cycle, instantaneously tripped</p><p>Any industrial power-distribution problem with current dura-</p><p>K , = 1.25</p><p>K I = 1 . 1</p><p>K , = 1.0 tion of 35 sec or more</p><p>Short-circuit Protective-device Interrupting Time. Circuit Breakers.</p><p>The minimum time duration of short-circuit-current flow in a rircuit</p><p>protected by a circuit breaker tripped by an instantaneous element will</p><p>vary with the type of circuit breaker used. Typical values are shown in</p><p>the lower left-hand corner of the large chart in Fig. 3.23.</p><p>When interruption is purposely delayed by time-delay relays or time-</p><p>delay trip coils, the duration of current flow will be governed by the time-</p><p>delay relay or trip coil plus the inherent delay in the circuit breaker.</p><p>Fuses (Current-limiting) , Current-limiting fuses (silver-sand and</p><p>National Electrical Code low voltage) tend progressively to limit the</p><p>time interval of current flow to lesser values as the magnitude of current</p><p>increases. As the current magnitude increases toward the maximum</p><p>interrupting ability of the fuse, the magnitude of Z't approaches a fixed</p><p>value (approximately) for a particular fuse ampere rating. This is</p><p>equivalent to a fixed temperature rise in a particular size of conductor.</p><p>Data accumulated indicate that a fuse (of the types mentioned in this</p><p>paragraph) whose ampere rating is not greater than 1.5 times the conduc-</p><p>tor continuous-current rating will protect against dangerous conductor</p><p>106 A-C SHORT-CIRCUIT PROTECTIVE DNICES AND CIRCUIT EQUIPMENT</p><p>temperatures for severe overcurrents up to the maximum interrupting</p><p>rating of the fuse.</p><p>Table 3.8 shows the wire sizes which will have less than 75 C conductor</p><p>TABLE 3.8 Silver-sand Fuse Protection at High Overcurrents Based</p><p>on Copper Conductor</p><p>Fuse</p><p>roting,</p><p>amp</p><p>S m a l l ~ t wire</p><p>normally vied,</p><p>RH insulation</p><p>30</p><p>60</p><p>I00</p><p>IS0</p><p>200</p><p>Sm.lle.t wire</p><p>protected</p><p>No. 10 Awg</p><p>No. 6 Awg</p><p>No. 3 Awg</p><p>No. 1/0 Awg</p><p>No. 3/0 Awg</p><p>No. 14 Awg</p><p>No. I2 Awg</p><p>No. 10 Awg</p><p>No. 0 Awg</p><p>No. 6 Awg</p><p>temperature rise because of the flow of short-circuit current when pro-</p><p>tected by silver-sand fuses.</p><p>Fuses (Nou-current-limiting). Non-current-limiting fuses accomplish</p><p>current interruption at a normal current zero, and thus the current con-</p><p>duction time cannot be reduced below that of the first current loop of</p><p>short-circuit-(.urrent flow which may be as much as about one cycle of</p><p>the power frequenry. Applications should thus recognize one cycle as</p><p>the minimum time of short-circuit-current flow.</p><p>Application Procedure. 1. Evaluate the symmetrical short-circuit</p><p>current or currents that may be critiral.</p><p>2. Define the short-circuit protertive device clearing time at this or</p><p>these current magnitudes.</p><p>3. Apply the rms correction factor to allom for the d-c component for</p><p>each time interval involved.</p><p>4. Make an initial check on the current-time chart for the smallest</p><p>conductor size being considered (permissible time should exceed short-</p><p>circuit protective-device interrupting time).</p><p>5 . If critical, i t is advisable to rorreet for the exart temperature range</p><p>(see Table 3.6 and temperature-range correction curve).</p><p>F. If an oversize ronduetor is considered, but the continuous-load rur-</p><p>sent is to remain fixed, advantage can be taken of the lower initial ron-</p><p>ductor temperature.</p><p>EXAMPLES</p><p>Example 1. A transformer feeder cable is being selected to accom-</p><p>The rated current of the t,rans-</p><p>The trans-</p><p>modate a 1000-kva 2.4-kv transformer.</p><p>former (240 amp) indirates a rahle conductor of 250 MCM.</p><p>A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT I87</p><p>former iri question is good for full short-circuit current (sixteen times</p><p>normal) for 5 sec. It is desired that the feeder cable have the same</p><p>ability.</p><p>Solution: Rms symmetrical amperes = rated current X 16 = 240 x</p><p>16 = 3900 amp. The duration of this current as defined by the condi-</p><p>tions of the problem is 5 sec.</p><p>Assume X / R ratio = 10 or less</p><p>From chart A of Fig. 3.23, K 1 = 1; ( X / R ratio of 10 and time of 5 sec)</p><p>Henre, the total rms amperes affecting cable heating = K , X 3900</p><p>= 1.0 X 3900 = 3900 amp</p><p>On the large rhart of Fig. 3.23, locate the intersection of the horizontal</p><p>3900-amp line and the 250-MCM conductor diagonal line. The per-</p><p>missible time (read on the bottom scale) is indicated to be 12 sec (75 to</p><p>150 C hasis).</p><p>The 250-MCM cable will adequately meet the 5-sec requirement.</p><p>Example 2. Feeder circuits are t o be run from a 480-volt 60-cycle</p><p>load-center unit substation at which point the short-circuit duty is</p><p>25,000 amp (20,000 symmetrical rms amperes). What is the smallest</p><p>reasonable feeder conductor size based on the use of a 25,000-amp inter-</p><p>rupting rating air circuit breaker which trips instantarieously (1.5 cycles)</p><p>a t currents in excess of fifteen times the normal rating?</p><p>solulion:</p><p>Symmetrical current = 20,000 amp</p><p>Time duration = 1.5 cycles</p><p>Rms amperes = 20,000 X 1.25 = 25,000</p><p>See preceding text for explanation of 1.25 factor K ,</p><p>On the large rhart of Fig. 3.23, locate the intersection of the horizontal</p><p>25,000-amp line and the vertical 1.5-cycle line. The minimum size con-</p><p>ductor (75 to 150 C basis) whose curve is above the intersection is a</p><p>KO. 1 Awg.</p><p>A 4-kv feeder is t o be run from a substation at which the</p><p>symmetrical short-circuit current is 25,000 amp. A continuous load</p><p>caparit,y of 1000 kva is desired (113 amp), and a KO. 2/0-Awg coronol</p><p>cable run is being considered. Line relaying is to consist of standard</p><p>time-overcurrent relays on the &amp tap and S o . 5 time-lever setting</p><p>v i th 250/5-amp rurrent transformers. Instantaneous attachments are</p><p>not planned, but could be used if set at 3000-amp line current.</p><p>Example 3.</p><p>Solution:</p><p>Symmetriral short-circuit current = 25,000 amp</p><p>Case 1. No instantaneous attachment on relay</p><p>Rms symmetrical short-circuit current = 25,000 amp</p><p>Relay operating time = 50 cycles; (From published time-current</p><p>Circuit-breaker operating time = 8 cycles</p><p>curves)</p><p>188 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>Total time = 50 + 8 = 58 cycles</p><p>Assume X / R ratio = 15 or less</p><p>From chart A of Fig. 3.23, K 1 = 1</p><p>Hence, total rms amperes affecting cable heating = KI X 25,000</p><p>= 1.0 X 25,000 = 25,000 amp</p><p>On the large chart of Fig. 3.23, locate the intersection of the 25,000-amp</p><p>horizontal line and t,he 58-cycle vertical line. The smallest conductor</p><p>whose curve lies above this intersection is a 500 MCM. Therefore, a</p><p>Xo. 2/0-Amg conductor is inadequate.</p><p>Case 2. Instantaneous attachment, on relay set to operate at and above</p><p>3000 line amperes.</p><p>Two point,s must, he checked: (1) a current of 3000 amp and time delay</p><p>of overcurrent relay (just below the operating current of the instantaneous</p><p>element) and (2) the maximum current with instantaneous relay operation.</p><p>1. From published relay data, the relay time a t 3000 line amperes is 66</p><p>cycles, circuit-breaker time is 8 cycles, making a total time of 66 + 8 =</p><p>74 cycles.</p><p>From chart A of Fig. 3.23 for X / R ratio of 10 and time of 71 cycles,</p><p>KI = 1.</p><p>Total rms amperes affecting cable heating = KI X 3000 = 1.0 X 3000</p><p>= 3000 amp.</p><p>The intersection of 3000 amp and 71 cycles on the large chart of Fig.</p><p>3.23 shom that a Xo, 2/0-Awg conductor is amply large to carry the</p><p>3000 amp for 74 cycles.</p><p>2. At the maximum current, instantaneous relay operation will be</p><p>obtained.</p><p>The total current duration will be the relay time ?,g cycle plus circuit-</p><p>breaker time 8 cycles, or 836 cycles.</p><p>For 8>i-cycle time interval, K , = 1.1.</p><p>Total</p><p>rms current affecting cable heating = K , X 25,000 = 1.1 X</p><p>25,000 = 27,500 amp.</p><p>The intersection of the 27,500-amp horizontal line and the 84g-cycle</p><p>vertical line on the large chart of Fig. 3.23 indicates a No. 4/0-Awg con-</p><p>ductor (75 to 150 C basis) and shows that point 2 cont,rols the cable size.</p><p>However, a No. 4/0-Awg conduct,or mould operate at less than rated</p><p>temperature. A specific check may show that a KO. 3/0-Awg conductor</p><p>is adequate.</p><p>Rated conductor temperature coronol cable = 80 C (see Table 3.6),</p><p>ambient temperature = 40 C. Xormal temperature rise produced by</p><p>rated current = 80 - 40 = 40 C.</p><p>Rated continuous current for No. 3/0-Awg coronol cable = 185 amp.</p><p>The temperature rise will be roughly proportional to the square of the</p><p>current.</p><p>189 A-C SHORTT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT r Time- seconds</p><p>I</p><p>190 A-C SHORT-CIRCUIT PROTECTIVE DEVICES AND CIRCUIT EQUIPMENT</p><p>Hence, the normal conductor temperature of a No. 3/0-Awg conductor</p><p>operating a t 143 amp would he expected to be</p><p>(g)’ (full-load rise) + ambient =</p><p>= 63.8 C, or 64 C</p><p>The maximum momentary temperature for coronol at 5 kv is 145 C</p><p>From detail chart B , Fig. 3.23, the correction factor K for an initial</p><p>From the large chart of Fig. 3.23, the permissible time for 27,500 amp</p><p>The permissible time corrected t o a 64 to 145 C basis is K X 6.7</p><p>Therefore, a No. 4/O-Awg conductor is the correct selection since a</p><p>(see Table 3.6).</p><p>conductor temperature of 64 C and final of 145 C is K = 1.13.</p><p>in No. 3/0-Awg conductor (75 to 150 C basis) is 6.7 cycles.</p><p>= 1.13 X 6.7 = 7.6 cycles.</p><p>No. 3/O-Awg conductor would fail t o meet the 8.5-cycle requirement.</p><p>FUSING CURRENT TIME FOR COPPER CONDUCTORS</p><p>The fusing current time curves for copper conductors are shown in</p><p>The curves are based on the folloiving assumptions:</p><p>1. Radiatiou may be neglected because of the short time involved.</p><p>2. Resistance of 1 cu cm of copper at 0 C is 1.589 microhms.</p><p>3. Temperature-resistance coefficient of copper a t 0 C is 1/234.</p><p>4. Melting point of copper is 1083 C.</p><p>5. Ambient temperature is 40 C.</p><p>Data are an adaptation from the eight,h edition of “Standard Handbook</p><p>* A . E. Knowlton (editor-in-chief), “Standard Handhook for Electrical Engineers,”</p><p>Fig. 3.21.</p><p>for Elect,rical Engineers.”*</p><p>8th ed., Chap. 4, McGraw-Hill Book Company, h e . , S e w York, 1949.</p><p>Chapter 4 by W. R. Crites and Maynord N. Halberg*</p><p>Voltage-Standard Ratings, A llowa ble</p><p>Variations, Reduction of Variations,</p><p>Calculation of Drops</p><p>The purpose of any industrial power system is to maintain voltage a t</p><p>the terminals of power-using equipment. This voltage should be-</p><p>within acceptable limits-equal to the rated voltage of this equipment.</p><p>The standard voltage ratings for utilization equipment are discussed in</p><p>this chapter, along with the standard voltage ratings for power generation</p><p>and distribution equipment.</p><p>KO practical power system can maintain voltage a t rated value a t the</p><p>utilization equipment a t all times. The voltage variations allowable and</p><p>the methods which can be used in the design of a power system to keep</p><p>the variations within acceptable limits are discussed. I t is necessary to</p><p>calculate the voltage drop in the power system for steady-state conditions</p><p>and during the starting of the larger motors to determine whether or not</p><p>the voltage mill remain within acceptable limits. Methods of calculating</p><p>these voltage drops are presented in this chapter.</p><p>* The following men, formerly in Industrial Pawcr Enginwring. General Electric</p><p>Company, made substantial contributions to the material in this chapter: W. K.</p><p>Boice, General Electric Company, l e w Haven. Conn.; D. F. Capehart. General</p><p>Electric Company, Cincinnati, Ohio; J. R. Eliason, General Electric Company,</p><p>Sehenectady, N.Y.</p><p>191</p><p>192 V O L T A G F S T A N D A R D RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>VOLTAGE DESIGNATIONS *</p><p>It is necessary t o have an understanding of the voltage names of sys-</p><p>tems and t,he voltage rat,ings of various pieces of apparatus used in the</p><p>system before start,ing a discussion on system-voltage problems so tha t</p><p>the proper voltage identification can be used throughout. It is also</p><p>necessary t,o know v h y the voltage designat,ions are applied to help in</p><p>understairding the system-voltage disussion in the following sections.</p><p>The volt,age-identification structure is summarized iu Table 4.1. For</p><p>each of the nominal syst,em voltages listed, t,he table gives voltage ratings</p><p>of generators, transformers, motors, and (in some cases) lamps. To</p><p>illustrate the use of Table 4.1, consider a 13,800-volt system. The</p><p>generators would be rated 13,800 volts. Transformers stepping power</p><p>down from transmission voltage would have secondary windings (I?,</p><p>Fig. 4.1) rated 13,800 volts. Transformers steppiug power down to</p><p>utilization vokage in load-center substations would have primary mind-</p><p>ings (C, Fig. 4.1) rated 13,800 volts. Motors connerted directly to the</p><p>13,800-volt bus would lie rated 13,200 volts.</p><p>From the foregoing summary and Table 4.1 i t is evident that care must</p><p>tie exercised in using the proper voltage ident,ifiration for each piece of</p><p>equipmelit as well as for the system. Some fundamental rules are as</p><p>follo\vs :</p><p>1. When speaking of equipment, the rated voltage is used, aud it is the</p><p>voltage to which the operating characteristics are referred.</p><p>2. When speaking of systems, rat.ed voltage is not an applicable term</p><p>because various piwes of equipment in a given system often have different</p><p>voltage ratings. Therefore, t,he term n o m i n a l sys t em vollage is used for</p><p>convenient designation of systems and circuits to define the voltage class.</p><p>The problem of proper identification would be easier if all apparatus</p><p>of a given voltage class had the same vokage rating. Then, of course,</p><p>tem voltage could have that same value. Possibly if the</p><p>industry were starting over again, vokage ident,iticatioii mould be made</p><p>that simple. But, as syst,ems grew, voltages were ini,hed up to compen-</p><p>sate for t,he voltage drop between source arid load.</p><p>As a result, of t,hese changes that have taken pla(.e over a period of years,</p><p>transformer arid generator voltage rat,ings are generally higher than</p><p>utilization-eiiuipment vnltagc rat,ings. There is logic in this in that the</p><p>voltage rating of transformers, for example, is t,heir no-load rating.</p><p>Since most plants are supplied by transformers, the concept has beeri</p><p>acceptcd that, supply equipment will have a higher voltage rating than</p><p>utilization equipment,. This means that in a 480-volt system, for cxam-</p><p>* For a iiirthrr rrpansion of t h i s srihjpet FW l</p><p>with equal reactance and</p><p>The</p><p>FIG. 1.17</p><p>resistance.</p><p>short-circuit current will gradually become symmetrical in practical circuits.</p><p>purposes, arbitrarily divided into simple components, which makes it</p><p>easy to calculate the short-circuit magnitude a t certain significant times</p><p>after the short circuit occurs.</p><p>The asymmetrical alternating current behaves exactly as if there were</p><p>two component currents flowing simultaneously. One is a symmetrical</p><p>a-c component and the other a d-c component. The sum of those two</p><p>components a t any instant is equal to the magnitude of the total asym-</p><p>metrical a-c wave a t the same instant.</p><p>The d-c component referred to here is generated within the a-c system</p><p>with no external source of direct current being considered. In some cases,</p><p>particularly in the neighborhood of the d-c railways, direct current from</p><p>the railways flows through neighboring a-c systems. This type of d-c</p><p>current is not considered in this discussion or in the calculating procedures</p><p>which follow.</p><p>As an example of the resolution of asymmetrical alternating currents</p><p>into components, refer to Fig. 1.15 which shows an asymmetrical short-</p><p>circuit current which is resolved into a symmetrical a-c and a d-c compo-</p><p>nent in Fig. 1.18. If the instantaneous values of the two components</p><p>(dashed lines) are added a t any instant, the resultant will be that of the</p><p>asymmetrical current wave.</p><p>This condition i s theoretical and is shown for illustration purposes only.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>F I N S T A N T AT WHICH SHORT CIRCUIT OCCURS</p><p>17</p><p>ASYMMETRICAL</p><p>AC COMPONENT</p><p>FIG. 1.18</p><p>current.</p><p>Theoretical Ihort-circuit-cvrrent wove illustrating components of asymmetrical</p><p>In practical circuits, d-c component would decay to zero in o few cycler.</p><p>INSTANT OF SHORT CIRCUIT</p><p>T O T A L C U R R E N T</p><p>D C C O M P O N E N T</p><p>A C C O M P O N E N T</p><p>ZERO A X I S</p><p>a = b = D C C O M P O N E N T</p><p>FIG. 1.19</p><p>at some point between the zero point and peak of the generated voltage wave.</p><p>lhsoretical condition similar to that shown in Fig. 1.18.</p><p>Components of asymmetrical short-circuit current in which short circuit occurred</p><p>This is a</p><p>I8 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>As mentioned previously, the examples shown in Figs. 1.13 and 1.18 are</p><p>In practical circuits the d-c component for purposes of illustration only.</p><p>decays very rapidly, as shown in Fig. 1.20.</p><p>INITIAL MAGNITUDE OF D-C C O M P O N E N T</p><p>The magnitude of the d-c component depends upon the iustant, the</p><p>short circuit occurs and may vary from zero, as in Fig. 1.14, to a maximum</p><p>initial value equal to the peak of the a-c symmetrical compoiieiit, as i n</p><p>Figs. 1.15 and 1.18. When the short circuit occurs at any other point,</p><p>such as shown in Fig. 1.19, the initial magnitude of the d-c componciit is</p><p>equal to the value of the a-c symmct,riral component a t thc instant of</p><p>short circuit. The above limit,s hold true for the initial magiiitudc of d-c</p><p>eomporient in a system regardless of the reactance and resistance. Ilow-</p><p>ever, the d-c componeut does not continue to flo~v a t a constant value, as</p><p>shown in Figs. 1.18 and 1.19, unless there is zero resistauce i i i the circuit.</p><p>DECREMENT</p><p>There is uo d-c voltage in the system to sustaiu the flax of direct</p><p>current; therefore the energy represeuted by the dirert. component of</p><p>current will be dissipated as ZZR loss from the direct current flowiug</p><p>through the resistance of the circuit. If the circuit had zero resistance,</p><p>the direct current would flow at a constant value (Figs. 1.18 and 1.19)</p><p>TOTAL ASYMMETRICAL CURRENT</p><p>C COMPONENT</p><p>AC COMPONENT</p><p>FIG. 1.20</p><p>short-circuit currenl gradually becomes symmetrical when d-c component diroppearr.</p><p>Trace of orcillogrom showing decay of d-c component and how orymmetricd</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 19</p><p>until the circuit was interrupted. However, all practical circuits have</p><p>some resistance; so the d-c romponent decays as shown in Fig. 1.20. The</p><p>combination of the decaying of d-c and symmetriral a-(* components gives</p><p>an asymmetrical wave that changes to a symmetriral wave whcti the</p><p>d-c component has disappeared. The rate of decay of the currents is</p><p>called the decrement.</p><p>X / R RATIO</p><p>The X / R ratio is the ratio of the reactance to the resistance of the cir-</p><p>cuit. The decrement or rate of decay of the d-c component is propor-</p><p>tional to the ratio of reactance to resistance of the complete circuit from</p><p>generator to short circuit. The theory is the same as opening the circuit</p><p>of a battery and an inductive coil.</p><p>If the ratio of reactance to resistance is infinite (i.e., zero resistance),</p><p>the d-c component never decays, as shown in Figs. 1.18 and 1.19. On the</p><p>other hand, if the ratio is zero (all resistance, no reartance), it decays</p><p>instantly. FOF any ratio of reactarice to resistance in between these</p><p>limits, the d-c component takes a definite time to decrease to substantially</p><p>zero, as shown in Fig. 1.20.</p><p>! I n generators the ratio of subtransient reactance to resistance may be as</p><p>?much as 70: l ; so i t takes several cycles for the d-c component to dis-</p><p>appear. In circuits remote from generators, the ratio of reactance to</p><p>resistance is lower, and the d-c component decays more rapidly. The</p><p>higher the resistance in proportion to the reactance, the more IaR loss</p><p>from the d-c c.omponent, and the energy of the direct current is dis-</p><p>sipated sooner.</p><p>D-C TIME CONSTANT</p><p>Often i t is said that generators, motors, or circuits have a certain d-c</p><p>This refers again to the rate of decay of the d-c compo- time constant.</p><p>O C COMPONENT</p><p>a = 37Y. OF b (APPROX )</p><p>C- TIME</p><p>CONSTANT IN OF D C COMPONENT</p><p>SECONDS</p><p>FIG. 1.21 Graphic illustration of time constant.</p><p>20 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>nent. The d-c time constant is the time, in seconds, required by the d-c</p><p>component to reduce to about 37 per cent of its original value a t the</p><p>instant of short circuit. I t is the ratio of the inductance in henrys to the</p><p>resistance in ohms of the machine or circuit. This is merely a guide to</p><p>how fast the d-c component decays.</p><p>Stated in other terms, it is the time in seconds for the d-c component to</p><p>reach zero if it continued to decay a t the same rate it does initially</p><p>(Fig. 1.21).</p><p>RMS VALUE INCLUDING D-C COMPONENT</p><p>The rms values of a-c waves are significant since circuit breakers, fuses,</p><p>and motor starters are rated in terms of rrns current or equivalent kva.</p><p>The maximum rrns value of short-circuit current occurs at a time of about</p><p>one cycle after short circuit, as shown in Fig. 1.20. If there were no</p><p>decay in the d-c component, as in Fig. 1.18, the rrns value of the first</p><p>cycle of current would be j .732 times the rrns value of the a-c component.</p><p>I n practical circuits there is always some d-c decay during the first cycle.</p><p>An approximate rrns value of one cycle of an offset wave whether it is</p><p>partially or totally offset is expressed by the equation</p><p>where C = rrns value of offset or asymmetrical current wave over one</p><p>cycle</p><p>a = rrns value of a-c component</p><p>b = value of d-c component at one-half cycle</p><p>MULTIPLYING FACTOR</p><p>Calculation of the precise rrns value of an asymmetrical current a t any</p><p>time after the inception of a short circuit may be very involved. Accu-</p><p>rate decrement factors to account for the d-c component a t any time are</p><p>required, as well as accurate factors for the rate of change of the apparent</p><p>reactance of the generators. This precise method may he used if desired,</p><p>but simplified methods have been evolved whereby the d-c component is</p><p>accounted for by simple multiplying factors. The multiplying factor</p><p>converts the rrns value of the symmetrical a-c wave into rms amperes of</p><p>the asymmetrical wave including a d-c component.</p><p>The magnitude of the d-c component depends upon the point on the</p><p>voltage wave a t which the short circuit occurs. For protective-device</p><p>application,</p><p>is, as</p><p>INCOMING</p><p>MASTER UNIT</p><p>SUBSTATION 4 (PRIMARY SUBSTATIONI</p><p>\</p><p>1 (A IPRIMARI WINDING u</p><p>ml SECONDARY WINDING</p><p>(IF USEDI</p><p>X</p><p>P L A N T PRIMARY DISTRIBUTIDN VDLTAGE</p><p>LOAD CENTER UNIT</p><p>SUBSTATION</p><p>(SECONDARY SUBSTATION IN FACTORYI</p><p>PRIMARY WINDING</p><p>WINDING</p><p>FIG. 4.1 Typicol industrial plont power ryrtern</p><p>194 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>evident from Table 4.1, to use the same voltage rating for all traiis-</p><p>former windings connected to a given system voltage. This is true</p><p>whether the transformers are stepping down to this system or steppiug</p><p>down from this system.</p><p>TABLE 4.1 Boric Pattern of Voltage Identification</p><p>No m i n 0 I</p><p>system</p><p>*olt.ge</p><p>Genordor Transformer</p><p>rated secondory</p><p>voltage rated voltage</p><p>Transformer</p><p>primary</p><p>rated voltage</p><p>Motor and L.mp</p><p>rottoget YoltDge</p><p>control rated rated</p><p>Three-phase Systems</p><p>120 or 120/240 I20 or 120/240 I20 or 120/240</p><p>208Y/120 208Y/120 208Y/120</p><p>240 or 120/240 240 or 120/240 240 or 120/240 230</p><p>115 118or120</p><p>120</p><p>240</p><p>I20</p><p>208Y/120'</p><p>240</p><p>480*</p><p>600</p><p>2,400'</p><p>4.160'</p><p>4,800</p><p>6,900*</p><p>12,000</p><p>13,200</p><p>13.800'</p><p>23,000</p><p>34,500</p><p>46,000</p><p>69,000</p><p>1 1 5,000</p><p>20sY/l20</p><p>240</p><p>480</p><p>600</p><p>2.400</p><p>4,160</p><p>4,800</p><p>6,900</p><p>12.500</p><p>13.800</p><p>13.800</p><p>. . . . . . . .</p><p>. . . . . . . .</p><p>. . . . . . . .</p><p>208Y/120</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4,160</p><p>4.800</p><p>6,900</p><p>12,000</p><p>13,200</p><p>13,800 . . . . . . . .</p><p>. . . . . . . . . . . . . . . .</p><p>. . . . . . .</p><p>208 or 120</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4,160</p><p>4.800</p><p>6,900</p><p>13.200</p><p>13.800</p><p>22,900</p><p>34,400</p><p>43,800</p><p>67,000</p><p>I I0,OOO</p><p>12,000</p><p>220 or 208 208.118. or 120</p><p>220 I 236</p><p>440 165</p><p>2,300</p><p>4,000</p><p>4,600</p><p>6,600</p><p>11,000</p><p>13,200</p><p>13.200</p><p>* In ~ P I V installations, or W ~ P ~ P Y P ~ a srlwtion oi voltngr can l i p ~nadr. thrsr i ~ r c prc-</p><p>ferrrd s y s t m valtagrs.</p><p>t Specifying t h e w valiirs for motor voltsgcs is itnportarrt: For instnnw. motors to</p><p>opprste on -IltiO-. GWC-, or 18,800-volt systrins should Iw rntcil 4000. (i(iO0. or 1:1,200</p><p>volts, resp2ctively.</p><p>The one-line diagram (Fig. 4.1) shows a typ i id method of distributing</p><p>power in industrial plants and will be used as referenre to identify some</p><p>portions of the systems and equipment referred to.</p><p>RATED VOLTAGES OF TRANSFORMERS</p><p>Transformer voltage ratings are hased on the no-load values, and the</p><p>The transformers have a voltage rating for each xindiiig. These are</p><p>ratio of primary to secondary rated wltages is equal to the turn ratio.</p><p>VOLTAGSSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 195</p><p>the voltages a t which characteristics are measured. What then are</p><p>standard transformer voltage ratings for industrial plants?</p><p>First, consider primary or master unit substations and transformers</p><p>which step down from some voltage above 15 kv to plant primary dis-</p><p>tribution voltage, which is generally below 15 kv (see Fig. 4.1, top sub-</p><p>station). The standard primary-winding ( A , Fig. 4.1) voltage ratings of</p><p>this class of substation and transformers are 110, 67, 43.8, 34.4, and 22.9</p><p>kv.</p><p>They are derived</p><p>from the old rating structure based on secondary ratings in multiples of</p><p>115 volts. When secondary ratings were boosted to multiples of 120</p><p>volts, the high side rating was raised to maintain the same turn ratio.</p><p>For instance, 33,000-2300 volts was once a standard rating. Thc corre-</p><p>sponding present-day transformer would be rated 34,400-2400. The</p><p>familiar designations 115, 69, 46, 34.5, and 23 kv refer to the classes of</p><p>insulation used with these transformers.</p><p>Secondary-winding (B , Fig. 4.1) voltage ratings of this class of industrial</p><p>substat,ions and transformers are 13.8, 13.2, 12, 6.9, 4.8, 4.16, and 2.4 kv .</p><p>S e x t consider transformers in load-center unit substations (see Fig. 4.1,</p><p>bottom substation) used in t,he industrial plants for stepping down from</p><p>plant primary distribution voltage to utilization voltage. As stat,ed</p><p>above, the plant, primary voltage is usually less than 15 kv. Therefore,</p><p>the list belox includes only voltages below 15 kv.</p><p>The primary-winding (C, Fig. 4.1) voltage ratings of load-center unit</p><p>substations are 13.8, 13.2, 12, 6.9, 4.8, 4.16, and 2.4 kv.</p><p>Note that the primary voltage rating of this class of transformers (bot-</p><p>tom, Fig. 4.1) i s the same as the secondary voltage rat,ing of the primary</p><p>substation transformers (top, Fig. 4.1).</p><p>The voltage ratings of secondary substations in the plant which supply</p><p>motors and other utilization equipment are divided into two classes-</p><p>those for serving utilization equipment above 600 volts and those for</p><p>serving utilization equipment below BOO volts. Standard rat,ings are</p><p>listed in Table 4.2.</p><p>These are the actual transformer-minding ratings.</p><p>.</p><p>TABLE 4.2 Transformer Secondary Voltage Ratings ( I ) , Fig. 4.1)</p><p>Supplying Utilizoti0n Supplying Utilization</p><p>Equipment Roted Equipment Roled 600</p><p>Above 600 Volts, Kv Volt, or 0e1ox. Volt.</p><p>6 .9 600 IY or delta1</p><p>4.8 400 IY or delta1</p><p>4.16 240</p><p>2.4 208Y/l20</p><p>All standard unit substation transformers have taps in the primery</p><p>winding to allow compensation for voltages that vary from the trans-</p><p>former rating. The most common are four 255 per cent taps, two above</p><p>196 VOLTAGkSTANDARD RATINGS, VARIATIONS. CALCULATION OF DROPS</p><p>aiid two below normal, giving a total adjustment of plus or minus 5 per</p><p>cent,. With these t,aps in the primary winding, a transformer actually</p><p>has five different ratios. I t vould he very cumbersome to refer to all five</p><p>of these ratios in all discussions; therefore, when in the following dis-</p><p>cussion a transformer is referred to as having, for example, a rating of</p><p>2400-480 volt,s, the discussion will apply equally well whether the trans-</p><p>former is operated 00 the cenher tap or other taps. Regardless of the tap</p><p>used, the t,raiisformer will still be referred to as a 2100-480-volt transformer,</p><p>Comhined light arid power systems are frequently used where motors</p><p>are supplied a t 180 volts, for example, and lights are supplied at 120 volts</p><p>from the same 480-volt system, using dry-type transformers. The</p><p>standard primary volt,age ratings for t,hese light,ing transformers are 600</p><p>volts, 480 volts, arid 240 volts, aiid the standard secondary vohage ratings</p><p>are 208Y/120 volts and 120/240 volts. Two rated kva 5 per cent below</p><p>normal t,aps are provided in these transformers t,o allow for operation of</p><p>120-volt lamps near t,heir rated voltage when the voltage on the 480-volt</p><p>system is below 480 volts as it normally vill be.</p><p>TRANSFORMER VOLTAGE REPRESENTATIONS</p><p>Transformer voltage designations become rather complex. For</p><p>illstance, windings may have series-parallel connections. Or they may</p><p>be designed for connection line-to-neutral on higher rated volt,age sys-</p><p>tems, such as 3400-volt transformers which are suitable for line-to-</p><p>neutral operation OIL 4160-volt systems. These and other complex</p><p>arrangements make exact identification desirable.</p><p>These variables in t,ratisformer voltage ratings have long been expressed</p><p>by various symbolic met,hods. Such methods are essential because to</p><p>fully describe the \\-indings of transformers often would require a fairly</p><p>lengthy paragraph. However, to bc of any value a transformer rating so</p><p>expressed should meao the same to everyone. To further a consistent</p><p>use of symbols, hot,h KERIA and ASh standards have been established</p><p>t,o rci~ommend a standard transformer “shorthand.”</p><p>Four symbols are used: the dash (-), t,he slant (/), the X, and the Y.</p><p>In general terms, their uses are as follows:</p><p>Dash (-). Used to separate the voltage ratings of separate windings in</p><p>a specific transformer.</p><p>Slant (I). Used to separate voltages to be applied to or obtained from</p><p>the same windiug.</p><p>X. Used to designate separate vokagcs obtainable by reconnection of</p><p>the coils of a winding in series or multiple combinations.</p><p>Y. The absence of t!sed to designat,e a winding t,hat is Y-connected.</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS I97</p><p>this symbol</p><p>in a three-phase transformer rating indicates that the winding</p><p>is delta-connected.</p><p>The use of the dash, slant, and Y can he easily illustrated by the voltage</p><p>rating of the transformer for a typical load-center suhstation.</p><p>4160-480Y/277: Note that this meaus the 4160-volt high-voltage wind-</p><p>ing is delta-corinected while the 480-volt winding is Y-couiiected with t,he</p><p>neutral brought out. A three-winding t,ransformer might have this</p><p>voltage rating: 13,800-2400-480Y/277.</p><p>In three-phase transformers the slaut is ofteri used to indicate wiiidiiigs</p><p>connectable either in delta or Y. For iiistauce, a 2400/4lCiOY windiug</p><p>can be couiiected either for 2400 volts deka or 4160 volts Y. Xote that</p><p>the delta voltage is expressed first. When a Y-connected winding has the</p><p>neutral brought out it is siguified like this: 2OSY/lZO. Here the line</p><p>voltage is expressed first, fol1oir.d by the line-to-neutral voltage. If the</p><p>neutral is brought out with reduced insulation, that fact is shoivu by 208</p><p>Grd Y/120.</p><p>Another use of the slant is to indicate taps, especially 011 single-phase</p><p>transformers. For instance, a 240-volt wiuditig with a midtap is expressed</p><p>240/120. When a single-phase t,ransformer with a series-multiple wind-</p><p>ing is vound to be suitable for three-wire service on the series conoectioii,</p><p>i t is designat,ed 120/240. When a winding has several taps close to the</p><p>rated volt,age, it is cust,omary to specify them as illustrated in t,his specific</p><p>case: four 255 per ceut rated kva taps, t x o above and tI5-o below rated</p><p>voltage.</p><p>The X symbol is used to separate t,he volt,ages obtainable in a series-</p><p>mukiple minding not, suitable for three-wire operation. For example, a</p><p>minding rated 120 X 240 can be connected with t,he coils in parallel to</p><p>obt,ain 120 volts or Tr.it,h the coils iu series for 240 rolt,s.</p><p>RATED VOLTAGES OF GENERATORS</p><p>Siiice the generator is a source of elect,ric poir-er aud is ofteu iu parallel</p><p>wit,h primary substation transformers (see Fig. 4,1), its voltage aud ('oii-</p><p>scquently its rat,itig is in practically all cases the same as the transformer</p><p>in a giveu voltage class. Listed in Table 4.3 are the three-phase generator</p><p>ratings that, are recommended by the latest EEI-SE5I.i report.</p><p>TABLE 4.3 Generator Voltage Ratings*</p><p>208Y/120 "Olt. 2,400 volts</p><p>240 volts 4,160 volts</p><p>600 volts 6,900 volts 14,400 volts</p><p>480 volts 4.800 volts 13.800 "011.</p><p>* Ratings of 11,500 and 12,500 volts are n s ~ d for genrrators on smnr rstablislird</p><p>Thc corrcsponrling trnnsfornii,r systems hut are, not rrrommmdrd for nmv systim~s.</p><p>rating is 12,000 w i t s and transformcr taps sllon for paralirl oprration.</p><p>198 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>The 14,400-volt rating has been adopted largely in large generating</p><p>stations where the input is transformed up to higher voltage in a unit</p><p>transformer generator arrangement (see Fig. 4.2).</p><p>2</p><p>mTwI</p><p>4 FIG. 4.2 Unit transformer generator arronge- I HIGH VOLTAGE BUS merit,</p><p>RATED VOLTAGES OF MOTORS</p><p>At the other end of the system are the motors, and their rat,ings reflect</p><p>the fact that voltage at utilizatioii equipment is somewhat loirer t,haii a t</p><p>the sources of power because of voltage drop.</p><p>Single-phase motors are usually rated at 115 or 230 volts.</p><p>The standard voltage rat,ings of polyphase motors are given in Table</p><p>4.1.</p><p>TABLE 4.4 M o t o r Voltage Ratings</p><p>1 10 "0111 550 "011. 6,600 Volt.</p><p>208 volt. 2,300 ~011s I1.000 volt,</p><p>220 wit. 4,000 ~ o l t i 13,200 volts</p><p>440 rolls 4,600 volts</p><p>hlot,or-cotit,rol equipment has the same voltage rating as the associated</p><p>motor.</p><p>RATED VOLTAGES OF LAMPS</p><p>Inrandescent lamps are standardized at 120 volts. Higher voltages</p><p>have not in general heeo found sat,isfactory.</p><p>Fluorescent lamps offer a wider range of operation and are commotily</p><p>rat,rd a t 118, 208, 230, and 265 volt,s (for line-t,o-neut,ral on 480-volt</p><p>systems).</p><p>OTHER APPARATUS</p><p>Some other types of equipment such as capacitors and industrial heat-</p><p>ing equipment have compromised between the extremes of generator</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS I99</p><p>rating and motor rating in a given voltage class.</p><p>heating devices are rated at 115,230, 4G0, and 575 volts.</p><p>rated at 230, 460, 575, 2400, 4800, 7200, 12,470, and 13,800 volts.</p><p>For instance, industrial</p><p>Capacitors are</p><p>NOMNAL SYSTEM VOLTAGES</p><p>The choice of the numerical value t o represent nominal system voltage</p><p>is purely arhitrary and does not attempt to indicate an average system</p><p>voltage. However, it is very desirable that a con-</p><p>sistent practire in designating nominal voltages be followed. When used</p><p>properly, the nominal voltage should give a good picture of the voltage</p><p>struct,ure of a system with a minimum of misunderstandings. The</p><p>standard values for nominal system voltage correspond t,o the ratings of</p><p>source equipment.</p><p>It is merely a name.</p><p>TABLE 4.5 Standard Nominal System Voltages</p><p>Singlo Phase</p><p>120</p><p>120/240</p><p>240</p><p>Three Phore</p><p>208Y/l20 4,800 34,500</p><p>240 6,900 46,COO</p><p>480 12,000 69.000</p><p>600 13,200 115,000</p><p>2,400 13,800</p><p>4,160 23,000</p><p>Table 4.5 is not complete but is representative of industrial practice.</p><p>To repeat, it is extremely important to identify properly the voltage</p><p>rating of each piece of apparatus in a system as well as to identify the</p><p>nominal system voltage. The voltage ratings of the various pieces of</p><p>apparatus, as ran he seen from the foregoing, may be different even</p><p>though the apparatus is for use on the same given voltage class system.</p><p>Therefore, correct identification of each piece is of paramount importance.</p><p>For example, if one is buying equipment to supply a 180-volt system, the</p><p>secondaries of t,he transformers should he specified as -180-volt rating.</p><p>The motors and control should be specified as 440-volt rating. The sys-</p><p>tem nominal voltage is referred to as 480 volts. Other apparatus on this</p><p>system may have different voltage ratings. For example, capacitors</p><p>would be rated 460 volts; heating equipment would be rated 460 volts.</p><p>It is also important to remember that transformer and generator voltage</p><p>200 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>ratings are always higher than utilization-device ratings. This is logical</p><p>because the transformer voltage ratings are the no-load voltage ratings,</p><p>and as load is applied to the system the voltage drops to near the name-</p><p>plate rating of the lower rated utilization apparatus.</p><p>VOLTAGE SPREAD AND FLICKER REQUIREMENTS*</p><p>STEADY-STATE VOLTAGE REQUIREMENTS</p><p>An ideal electric power system is one which will supply constant fre-</p><p>quency and volt,age at rated name-plate value to every piece of apparatus</p><p>in the system. In modern power systems, frequency is a minor problem.</p><p>It is impractical, however, to design a power system which will deliver</p><p>absolutely constant rated name-plate voltage to every piece of apparatus.</p><p>Since this cannot he attained, what are the proper limits of voltage varia-</p><p>tion in a n industrial plant? These should be determined by the charac-</p><p>teristics of the utilization apparatus. First, certain definitions are</p><p>essential to underst,arid clearly the discussion of this problem.</p><p>Voltage spread is the difference between the maxi-</p><p>mum and minimum voltages which appear at any location in a system</p><p>under riormal operating conditions. Voltage spread is not intended to</p><p>cover momentary voltage changes uf a transitory nat,ure such as those</p><p>due to switching surges, motor starting, welders, etc. The first part</p><p>of this discussion is primarily concerned with voltage spread a t utiliaatiori</p><p>equipment. This is the diKercnce between the maximum and minimum</p><p>voltages a t the terminals of the utilization equipment under normal</p><p>system operating conditions (Fig. 4.3). Maximum values usually appear</p><p>during light load and minimum values a t full load on the electric system.</p><p>Another important type of voltage spread is primary or supply voltage</p><p>spread which is the difference</p><p>between the maximum and the niinimum</p><p>voltage a t the service entrance or plant primary bus of a particular plant</p><p>under normal operating conditions.</p><p>Voltage Zone. Voltage zone is the envelope of all voltage spreads for</p><p>a particular voltage class of system.</p><p>For any specific voltage class designated by a nominal system voltage</p><p>there inherently exists an appreciable range of operat,ing voltages between</p><p>the systems having the highest and lowest voltages for this class. Coun-</p><p>trywide, this zoue is larger thaii the voltage spread at, ariy one location</p><p>because of recognized differences in practices of different companies.</p><p>* The data in this sretion arc l a r~c ly adapted from an AIEE Industrial Power Sys-</p><p>tem Coinmittre 1Lpurt. Industrid Voltag- Ilrquirpmeats, Elec. Eng., vol. 6 i , 1948,</p><p>pp. 358-374.</p><p>Voltage Spread.</p><p>::</p><p>s z</p><p>r</p><p>2 4 0 0 -</p><p>Y 0</p><p>></p><p>E</p><p>Y v)</p><p>-</p><p>k</p><p>T.</p><p>5</p><p>> 2 2 0 0 -</p><p>I</p><p>9</p><p>2 2300-</p><p>(L</p><p>Y</p><p>0</p><p>I *</p><p>E</p><p>L</p><p>CURVE A -TRANSFORMER OPERATING ON HIGHEST TAP-</p><p>RATIO 2520-480 VOLTS AT NO LOAD.</p><p>CURVE 8 TRANSFORMER OPERATING ON LOWEST TAP-</p><p>RATIO 2260-480 VOLTS AT NO LOAD.</p><p>FIG. 4.3 Examples of voltage zone, spread, and drop.</p><p>3.3 7. z</p><p>PRIMARY 5 , LONGEST SECONDARY FEEDER</p><p>SYSTEM</p><p>NO LOAD VOLTAGE _________________ ~_-------------- 480</p><p>1 2500--</p><p>PRIMARY VOLTAGE SPREAD. NO LOAD TO FULL LOAO AT</p><p>PLANT SERVICE ENTRANCE</p><p>SPREAD</p><p>IN 0 SECONDARY</p><p>Y</p><p>0 N</p><p>> E</p><p>Q</p><p>2 SYSTEM</p><p>TRANSFORMER VOLTAGE DROP</p><p>FEEDER VOLTAGE DROP P</p><p>NO LOAD VOLTAGE</p><p>PRIMARY VOLTAGE SPREIO, NO LOAD TO FULL LOAD AT</p><p>PLANT SERVICE LNTRANCE</p><p>VOLTS</p><p>I</p><p>0</p><p>MINIMUM FULL TRANSFORMER VOLTAEE DROP</p><p>LOAD VOLTAGE</p><p>202 VOLTAGFSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>difference in voltage in various parts of the power system.</p><p>cause is primary voltage spread a t the service entrance of the plant.</p><p>The other</p><p>480</p><p>2</p><p>2 470-</p><p>4 6 0 ~</p><p>EFFECT OF VOLTAGE DROP</p><p>To show the effect of voltage drop in a plant it will be assumed that the</p><p>primary voltage is maintained a t a constant value regardless of plant load.</p><p>The simple circuit shown in Fig. 4.4 will be used as an illustration. The</p><p>primary voltage is assumed to be of such magnitude that the secondary</p><p>voltage on the transformer is 480 volts a t no load. Referring to Fig. 4.5,</p><p>at extremely light load there is essentially no voltage drop through the</p><p>transformer or in any of the secondary circuits connected to the trans-</p><p>former. Consequently, the voltage is substantially the same throughout</p><p>the plant, and any lights or other incidental load connected a t this time</p><p>is subject to practically the no-load voltage. It is particularly significant</p><p>a t this point to recognize that transformer voltage ratings are the no-load</p><p>400 VOLTS ZERO VOLTAGE DROP .-</p><p>A</p><p>SECONDARY BUS</p><p>TRANSFORMER</p><p>CIRCUIT</p><p>FIG. 4.4 Typical industrial plant power circuit,</p><p>_ _ _ _ _ ~. TRANS FA NO LOAD VOLTAGE-480 VOLTS</p><p>VOLTAGE DROP VOLTAGE DROP IN</p><p>y) ::rp ] THRU 15 VOLTS TRANSFORMER-</p><p>9 460</p><p>FIG. 4.6</p><p>spread.</p><p>SECONDARV FEEDER-IOVOLTS l,z; IN BRANCH DROP</p><p>A</p><p>No primary voltage</p><p>---- ----____________________ 3</p><p>CIRCUIT-</p><p>5 VOLTS</p><p>TOTAL VOLTAGE --- -___ _</p><p>450 sE~!~48oro~?p~"2~Ts G?!!E? ________ _________-__</p><p>Full-load voltage conditions for circuit shown in Fig. 4.4.</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 203</p><p>ratios. For example, a transformer rated 4160-450 volts will produce</p><p>480 volts a t no load with 4160 volts applied to the primary.</p><p>When load is connected to the transformer, current flows, and this</p><p>causes a voltage drop in the secondary circuits as shown in Fig. 4.6. At</p><p>t,he secondary bus the voltage drop caused by the current flowing through</p><p>the transformer is assumed to be 15 volts. With constant primary volt-</p><p>age the secondary bus voltage varies from 450 volts a t no load to 465</p><p>voks at full load--the voltage spread a t this point is 15 volts. There are</p><p>assumed additional drops of 10 volts in the secondary feeder and 5 volts</p><p>in the branch circuit, making a total drop to load A of 30 volts. If the</p><p>lowest voltage in the plant exists a t load A , then the maximum voltage</p><p>spread is 30 volts (450 a t no load to 450 volts a t full load, or 30 volts).</p><p>In designing an industrial power system the voltage spread should be</p><p>kept to a minimum consistent with reasonable first cost. If the spread is</p><p>too great,, the voltage may be too high a t light load, causing equipment</p><p>operating during that period to burn out, or voltage may he too low a t full</p><p>load a t much of the utilization apparatus, impairing the performance and</p><p>reducing the production obtained from the equipment,</p><p>The second cause of voltage spread is the primary voltage spread a t the</p><p>plant service connection. This may be caused by voltage drop in the</p><p>primary system, or it may be due to regulation of the primary system by</p><p>voltage regulators. To show the effect of primary voltage variation,</p><p>assume that the primary voltage drops as load comes on in the plant.</p><p>The transformer taps have been selected so that the no-load voltage is</p><p>450 volts as in Fig. 4.5. When load comes on the power syst,em, the same</p><p>voltage drop occurs as in Fig. 4.6, but in addition, the primary system</p><p>voltage is assumed t,o drop sufficiently to cause an additional 10-volt drop</p><p>in the vokage at the secondary of the transformer. This primary voltage</p><p>spread adds to the total voltage spread in the plant, making the spread</p><p>480 to 440 volts or a total of 40 volts as is shown in Fig. 4.7 instead of only</p><p>30 volts as shown in Fig. 4.8 where there was no primary voltage variation.</p><p>The primary voltage spread may not always be in the direction shown</p><p>in Fig. 4.7. The primary voltage may rise when the load comes on</p><p>because of voltage regulators in the primary feeder circuit or because of</p><p>other voltage regulators in the primary power system. This voltage rise</p><p>of the primary reduces the voltage spread in the plant, as shown in Fig. 4.5.</p><p>Very weak primary systems with a high drop or regulated primary sys-</p><p>tems whose load cycle does not coincide with the load cycle of the plant</p><p>may cause excessive voltage spread in the plant-beyond the limits shown</p><p>in Table 4.9. Automatic voltage regula-</p><p>tion is required in such cases to bring the voltage spread within the limits</p><p>shown in Table 4.9. Changing transformer taps to increase the vo1t:ige</p><p>a t full load will not solve the problem because that will increase the</p><p>no-load voltage beyond 450 volts.</p><p>This is illustrated in Fig. 4.9.</p><p>204 VOLTAGLSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>-</p><p>480-</p><p>470-</p><p>----</p><p>VOLTAGE DROP THRU VOLTAGE DROP IN</p><p>TRANSFORMER</p><p>15 VOLTS</p><p>SEOWDARI FEEDER</p><p>VOLTAGE DROP IN</p><p>y1</p><p>9</p><p>450 -</p><p>TOTAL VOLTAGE SPREAD</p><p>480 TO 440 VOLTS 140 VOLTS1</p><p>440 I _________________________________ --- - _ _ _ _ _ _</p><p>470 ~</p><p>460;</p><p>4SO</p><p>G 440</p><p>430</p><p>420</p><p>410</p><p>J</p><p>FIG. 4.7</p><p>volt baris) primary voltage spread.</p><p>minimum a t full load.</p><p>Full-load voltage conditions for circuit shown in Fig, 4.4 with 10 volts (on 480-</p><p>Primary voltage varies from maximum at no load to</p><p>_____ _________NO LOAD VOLTAGE - 480 VOLTS</p><p>PRIMARY VOLTAGE SPREAD - 40 VOLTS</p><p>___--- VOLTAGE DROP IN</p><p>VOLTAGE DROP THRU SECONDARY FEEDER</p><p>TRANSFORMER lo VOLTAGE DROP ;1 2s__vw3</p><p>TOTAL VOLTAGE SPREAD</p><p>480 TO 410 VOLTS 170 VOLTS) ..</p><p>VOLTAGE DROP IN</p><p>SECONDARY FEEDER-</p><p>10 VOLTS</p><p>FIG. 4.8</p><p>volt basis) primary voltage spread.</p><p>maximum at full load.</p><p>Full-load voltoge condition3 for circuit shown in Fig. 4.4 with 10 volt. (on 480-</p><p>Primory voltage varier from minimum at no load to</p><p>VOLTAGbSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 205</p><p>EFFECT OF VOLTAGE SPREAD O N UTlLlZATlON EQUIPMEN?</p><p>General Effects. Whenever the voltage a t the terminals of a utiliza-</p><p>tion device varies from name-plate rating of the de\.ice, something is</p><p>sacrificed either in life or performanre of t,he equipment. The effert, may</p><p>be minor or serious, depending upon the chararteristirs of the device,</p><p>how the device</p><p>is applied, and the amount the voltage deviates from the</p><p>device rating. KESIA Standards provide for rert,ain tolerances whirh</p><p>may he taken advantage of without seriously affertiiig the performanre of</p><p>the apparatus. However, with usbge of electrir pover for precise opera-</p><p>tions, there is often a major sacrifire in produrtion for volt,age variations</p><p>of considerably less than given in t,he NERlA Standards.</p><p>So that the plant engineer can better judge the effect 11f vokage varia-</p><p>tion on t,he electric equipment in his plant, the rharacteristics of many</p><p>commonly used derires are given here. I t is these rhararteristirs rvhirh</p><p>have been used as a st,arting point for establishing the desired voltage</p><p>spread of Tables 4.8 and 4.9.</p><p>Induction motors are the most rommoir</p><p>utilization derires in industrial plants. Thr variatioii i n rharactrristiw</p><p>as a function of voltage for the widely used inductiotr motors is shoivn i n</p><p>Table 4.G. The material in this section deals only n-ith the cffert 011</p><p>motor chararterist,ies of rhaiiges in voltage magnitude. The effect, of</p><p>unbalanced voltages is also very importatit and shonld he rotrsiderrd.</p><p>The rurrent may hecomc esressive for only a small voltage iuihalanre.</p><p>The XEBIA St,andards should be consulted for detailed information on</p><p>this subject,.</p><p>The most sig-</p><p>nificant effects of too lox voltage are reduction in starting torque a t i d</p><p>increased full-load t,emperature rise. The redurtion of st,arting torque</p><p>may be significant i n mot,or applications driving high-inertia rqnipmeirt.</p><p>The lower torqne i d 1 result, in longer armleration periods. Torque</p><p>mot,ors are also very materially affected hy redured voltage as thi. torque</p><p>decreases as the square of the voltage; thus a t 10 per reut helow normal</p><p>voltage, the torque is redured 19 per cent.</p><p>The increased heating at low voltage aiid full load rediirrs thr lifr of</p><p>the insulat,ion.</p><p>Principal Effects of High Voltage on Induction Motors. The most,</p><p>significant efferts of too high voltage are inrreased tnr(lue, inr,rrasrd</p><p>starting rurrent, and decreased porer factor.</p><p>The increased torque may muse rouplings to shear off or damage t o</p><p>driven equipment. Increased starting curretit raiiscs greater voltage</p><p>drop in the power system, henre increases light, flirker. Uecreased po~vzr</p><p>factor is particularly disadvantageous where power-fartor peualty rlanses</p><p>Effect on Induction Motors.</p><p>Principal Effects of l ow Voltage on Induction Motors.</p><p>206 VOLTAGE-STANDARD RATiNGS, VARIATIONS, CALCULATION OF DROPS</p><p>TABLE 4.6 General Effect of Voltage Variation on</p><p>Induction-motor Characteristics</p><p>Starting and maximum running</p><p>torque... .................</p><p>Synchronous speed.. ..........</p><p>Per cent dip. . ...............</p><p>Full-load speed. ..............</p><p>Efficiency:</p><p>Full load.. ................</p><p>9% load. ..................</p><p>)i load . . .................</p><p>Full land.. ................</p><p>I( load.. .................</p><p>36 load. ..................</p><p>Full-load ~urrent . .............</p><p>Starting wrrenl . . .............</p><p>Temperature rise, full load. .....</p><p>Maximum torque capocity.. ....</p><p>Magnetic n0ire.m load in parlicu-</p><p>lor.. .....................</p><p>Power faclor;</p><p>I Voltage Variotion</p><p>Decrease 19%</p><p>No change</p><p>Increase 23%</p><p>Decreore 136%</p><p>Decrease 2 points</p><p>Proclicolly no change</p><p>Increase 1 to 2 point$</p><p>Increase 1 point</p><p>Increase 2 lo 3 point!</p><p>Incrcoie 4 lo 5 points</p><p>Increase I1 Yo</p><p>Decrease 10 to 12%</p><p>Increose 6 to 7 C</p><p>Decrease 19%</p><p>Decrease slightly</p><p>90% voltage Functionof voltage 110% voltage</p><p>(Voltage)’</p><p>Cons1.nt</p><p>1 (voltagel~</p><p>ISyn. speod--.llpl</p><p>InCreOle 21</p><p>No change</p><p>Decrease 17%</p><p>Increase 1 %</p><p>.............. ..............</p><p>..............</p><p>.............. ..............</p><p>..............</p><p>..............</p><p>Voitoge</p><p>IV0l togeJ~</p><p>..............</p><p>..............</p><p>Small increo*e</p><p>Decrease 1 to 2 points</p><p>Procticdiy no change</p><p>Decrease 3 points</p><p>Decrease 4 points</p><p>Decrease 5 lo 6 points</p><p>Decrease7%</p><p>Inc,eo.e 10 to 12%</p><p>Decrease I lo 2 C</p><p>Increa3e 21 %</p><p>Increase slightly</p><p>This table s h o w gencral effcets, which will vary somewhat for specific ratings.</p><p>are applied by the utilities. The higher the motor voltage rises, the lower</p><p>the power fartor mill become. This may result in a greater penalty and</p><p>hence a higher power bill.</p><p>While the temperature rise at full load on standard motors decreases</p><p>slightly for moderate overvoltages, the temperature rise may increase on</p><p>certain types of sperial motors a t even very small overvoltages. Over-</p><p>voltages of 10 to 1.5 per cent have caused numerous burnouts on special</p><p>four-speed grinder motors. Motors rated for intermittent load are also</p><p>materially affected by overvoltagcs.</p><p>While marry drive applications are not seriously affected by voltage</p><p>deviations as much as plus or minus 10 per cent from rated voltage, there</p><p>are import,ant applications that are.</p><p>Effect on Synchronous Motors. The effect of voltage variation on the</p><p>performance of synchronous motors is similar to that on induction motors.</p><p>However, while t,he starting torque varies as the square of the voltage,</p><p>the maximum or pull-out torque varies directly with the voltage.</p><p>From the above discussions it will be noted that, in general, voltages</p><p>slightly in excess of motor name-plate rating have less detrimental effect</p><p>VOLTAGkSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 207</p><p>on motor performance than voltage helow the name-plate rating. This</p><p>is one of the bases on which the voltage spreads in Table 4.9 mere deter-</p><p>mined. A s an example, the figures show a recommended spread of 420</p><p>to 180 volts for the 480-volt nominal system voltage, which is approxi-</p><p>mately 4 per cent below and 9 per cent above the 440-volt motor rating.</p><p>The light output and life of incan-</p><p>descent filament lamps are critically affected by the impressed voltage.</p><p>In Table 4.7 is shown the relationship of lamp life arid output to voltage</p><p>for a vokage range from 80 to 120 per cent of rated voltage.</p><p>In general i t may be said that for incandescent filament lamps a 1 per</p><p>cent deviation from rated voltage causes a change of 3 to 335 per cent in</p><p>light output. It can be seen from Table 4.7 that a 10 per cent reduction</p><p>in lamp voltage results in a 30 per cent reduction in light output. In</p><p>other words, when the voltage is 10 per cent low, the investment in the</p><p>lighting system is working at only 70 per cent efficiency-thus, 30 per</p><p>Effect on Incandescent lamps.</p><p>!i!</p><p>2</p><p>3</p><p>a</p><p>0</p><p>c PER CENT NORMAL VOLTS 3</p><p>9 a</p><p>FIG. 4.10</p><p>average of many lampr.</p><p>Characteristics of large gar-filled incandescent type C lampr. There are the</p><p>208 VOLTAGbSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>cent of the investment is lost. With an overvoltage of 10 per cent the</p><p>lamp-life is reduced to less than oue-third~-t,hus lamp-replacement costs</p><p>are three times as great as a t normal voltage. Other dat,a arc shown in</p><p>Fig. 4.10, from which it, should be noted that the lumens per watt., or lamp</p><p>efficieilcy, rises sharply at voltages above 100 per cent. In some cases,</p><p>operating eronnmies result from hurriing lamps at higher efficiency and</p><p>short life, or vice versa.</p><p>TABLE 4.7 Effect of Voltage Variations on Gar-filled</p><p>Incandescent-lamp Choracteristics</p><p>80</p><p>85</p><p>90</p><p>92</p><p>94</p><p>96</p><p>98</p><p>100</p><p>102</p><p>104</p><p>106</p><p>108</p><p>110</p><p>115</p><p>I20</p><p>Socket</p><p>voltage</p><p>~</p><p>96.0</p><p>102.0</p><p>108.0</p><p>110.4</p><p>112.8</p><p>115.2</p><p>117.6</p><p>120.0</p><p>122.4</p><p>124.8</p><p>127.2</p><p>129.6</p><p>132.0</p><p>138.0</p><p>144.0</p><p>~-</p><p>47</p><p>58</p><p>70</p><p>75</p><p>81</p><p>87</p><p>93</p><p>I00</p><p>105</p><p>115</p><p>I20</p><p>I30</p><p>I40</p><p>I60</p><p>185</p><p>Per cent Per cent</p><p>rated rated</p><p>voltage lighl output</p><p>Per</p><p>cent variatiim i n line voltage n-ill changc t,he lumeir oudput only about</p><p>1 per cent. Toltage is a factor in starting reliahility, and voltages l o w r</p><p>than recommeiided may result in unsatisfactory starting. It will be</p><p>noted that the ores-all efficiency (if the fluoresrerrt, lamp decreases if the</p><p>line volt,age is raised above normal. The increased line volt,age causes</p><p>the choke t,o pass more current to the lamp. This loivers the resistance</p><p>of the arc. column, rcsulting in a lower voltage drop i n the lamp itself.</p><p>The input, Ti-atts t o the lamp are slightly increased, and t,herefore the</p><p>lumen output increases over a cert,aiii range. In this condition, however,</p><p>the higher currcnt density priiduces the short ultraviolet radiation less</p><p>effirieutly; wilserpently t,he luminous efficiency of the lamp decreases.</p><p>Effect on Fluorescent lamps.</p><p>VOLTAGGSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 209</p><p>Fluorcsreiit lamps are fa r less af'ectrd hy circuit voltage variatioli tllan</p><p>filament lamps, from t,he standpoint of life.</p><p>The life of preheat-type lamps should he quite satisfartmy throrlghollt,</p><p>the range of published voltage fur the various I ial lasts; tlicsr volt:tg(.</p><p>ranges, iii general, are 110 ti] 125, 100 t o 2l(i, 220 to 250. :ind 240 t o 280.</p><p>There may be some derrease i n life performalire \\-3ir11 opwitcd a t maxi-</p><p>mum vokage as compared with that, a t miiiimiim vdt:igr. I I ~ i ~ e v t ~ r ,</p><p>there are a numher of other fartors, SWIM: of whidi arc i ~ i r p r ~ ~ d i r ~ t a l ~ l ~ ~ , tlr;lt,</p><p>affect life.</p><p>There is ?onsideral)le differenre i l l this rrsprct het ~ C ~ I I slimlillp a t i l l</p><p>regular preheat-type lamps. The iiistairt-start rathodr whivh is ~lsrd i l l</p><p>all slimline and instalit-start lamps van Iw o p e m t d ovrr ii ividr ~'angr ,if</p><p>current, from 120 to 430 ma, with rrlatiwly littlr d fc r t 1111 life.</p><p>Even though they mtsrt sprvitivatioll r~y~i~i rc , .</p><p>meiits, they have maiiufactnriiig toleranre and t h i w a r c drsigit r l i f fwvnws</p><p>be twen types.</p><p>Ballasts also affert life.</p><p>DECREASE0 LIGHT OUTPUT ANP INFERIOR LAMP PERFORYANCE</p><p>UNCERTAIN STARTING AND AN0 DANGER OF OVERHEATING</p><p>OPERATION MAY RESULT AT AUXILIARY MAY RESULT AT</p><p>EXCESSIVE UNDER VOLTAGE. EXCESSIVE OVER VOLTAGE.</p><p>/ \</p><p>i RECOMMENDED OPERATING RANGE x BEST PERFORMANCE I I I</p><p>LINE VOLTAGES</p><p>FIG. 4.1 1 Characteristics of fluorescent lornpr OI function of voltage applied to bollort.</p><p>210 VOLTAGE-STANDARD RATINGS, VARIATIONS. CALCULATION OF DROPS</p><p>Fluorescent lamps also differ from filament lamps in that the frequency</p><p>of start,irig is a factor iii the life obt,ained. Rated life is usually based on</p><p>3 hr of operation per start,. For 10 hr operation per start, the lamp's life</p><p>is increased approximat,ely 35 per cent.</p><p>Therefore, ally data 011 life vs. circuit voltage for the normal range in</p><p>operat,iiig voltage ivould have little significaiice. At voltages below the</p><p>lower limit, insufficient preheat current for proper cathode emission prior</p><p>to starting may result in short life. At voltages heyoiid the upper limit,</p><p>the overcurrent operat,ioii may rcsult in unsatisfartory lamp life.</p><p>Effect on Mercury Lamps. The effect of voltage variation on mercury</p><p>lamps is shown in Fig. 4.12.</p><p>Effect on Resistance Heating Devices. The energy input and there-</p><p>fore the heat output of resistaiice heaters varies in general with the square</p><p>of the impressed voltage. Thus a 10 per cent drop in voltage will cause</p><p>a drop of 19 per cent in heat, output. This, however, holds true only for</p><p>an operating range over which the resistance remains constant.</p><p>Many healing devices are conservat,ively designed arid if thermostati-</p><p>cally controlled may operate satisfactorily even if the voltage varies 10</p><p>per cent or more.</p><p>However, in many rases the designer must confine his heating units into</p><p>a miiiimum of space and must, therefore, operate them near maximum</p><p>rating. Also the temperature requirements for many heating applica-</p><p>tioiis IiecessiMe the operation of the heating units a t maximum tempera-</p><p>ture. h drop i n voltage meaiis a drop in heat input, varying with the</p><p>square of the voltage, and a loss in production. On the other hand,</p><p>excessive voltage will increase the temperature of the heating units and</p><p>therefore will reduce their life. This condition applies especially to fur-</p><p>naces operating at high temperatures near the maximum permissible for</p><p>u OC</p><p>60</p><p>40</p><p>I I I I / I / I I I</p><p>I I</p><p>I I I I I I</p><p>U I I I I 1 I I I I I</p><p>0 10 60 70 rn 90 wo 1 1 0 iao 130 140</p><p>P R I M V0LTIT.F - C C Y I 0 s TIIANSFORMER TAP SETTING</p><p>FIG. 4.12 Choracterirticr of mercury type H 400-watt lamps.</p><p>VOLTAG&STANDARD RATINGS, VARIATIONS, CALCULATtON OF DROPS</p><p>the type of heating unit used. To assure uniform high production and the</p><p>best operating conditions, the voltage should be maintained mithiu a</p><p>spread of plus or minus 5 per cent of rated voltage.</p><p>Although the filaments of the</p><p>lamps used in these installations are of the resistance type, the energy</p><p>output does not vary with the square of the voltage because the resistance</p><p>varies a t the same time. The radiated energy vs. voltage is shown in</p><p>Fig. 4.13 for the rating of 115 volts used on industrial infrared lamps.</p><p>The wattage input is nearly proportional to the energy output for a volt-</p><p>age range of 50 t o 150 per cent of rated voltage. The change in wattage</p><p>and radiated energy is only 7 per cent for a 5 per cent change in voltage.</p><p>However, this might he more harmful thau a larger change in typical</p><p>resistance heaters employing thermostatic controls, if the product dryiiig</p><p>is very sensitive to temperature differences. For the usual paint-drying</p><p>applications, no voltage coutrols are required with infrarcd lamps.</p><p>Uniformity of product speed in the oven is the usual objective for coii-</p><p>veyerized operations. Differences in heating requirements are rea,dily</p><p>accomplished by connecting the infrared lamps to a number of circuits,</p><p>so that some of the lamps can be switched on and off in accord with t,he</p><p>exact, heat,ing needs. In t,he cases vhere lamp sivitching cannot rom-</p><p>pensat,e for the volt,age variat,ions, i t may be necessary to use a voltage</p><p>regulator to maintain conveyer speed and product quality.</p><p>The current-carrying ability or emis-</p><p>siou of all elect,ronic tubes is affect,ed seriously by voltage deviation from</p><p>rating. Figure 4.14 shows typical emission curves plotted agairist</p><p>cathode heater voltage. Curve 1, entitled Oxide Coated, applies to most</p><p>of the thyratrons, pliotrons, and rereiving tubes. Curve 2 for thoriated</p><p>tungsten applies to the small transmitter tubes and some of the hattery-</p><p>211</p><p>Effect on Infrared Heating Processes.</p><p>Effect on Electronic Equipment.</p><p>FIG. 4.13</p><p>QI a function of impressed voltoge.</p><p>Radiant-energy output of General Electric Company industrial infrared lamps</p><p>212 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>heated tubes. Curve 3, Pure Tungsten, applies t o the oscillator tube</p><p>such as used in high-frequeiicy induction and dielectric heaters.</p><p>The rathode-life curve f o r pure tungsten indicates that, the life is</p><p>redured by half for esrh 5 per rent iiiwease iii cathode volt,age. This</p><p>redured life is due t o the higher rate of evaporation of the rathode mate-</p><p>rial. At voltages below rating, the loss d emission has very serious sec-</p><p>2 0 40LL 0</p><p>30 40</p><p>FIG. 4.14 Calculated values of electronic-tube emission and life</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 213</p><p>ondary effects. In a vacuum tube surh as the pliotron and kenet,ron a</p><p>small loss of eniission below that needed means rednced ont,put and some-</p><p>times excessive tube heating which is reflerted in a shorter life. However,</p><p>for gas-filled tubes such as thyratrons and phanotrons i n ivhirh t,he rurrent,</p><p>is not limited by the tube spare rharge, if insuffirient emission is available</p><p>to carry the load current, the gas molecules</p><p>bombard the rathode surfare</p><p>and may destroy t,he tube in a matter of minnt,es. Therefore, it is</p><p>extremely important that the rathode voltage be kept up near rating on</p><p>these tubes for sat,isfactory service.</p><p>In addition to the above factors there are ot,her important things to be</p><p>taken into ronsiderat,ion. If the volt,sge is too high, the evaporated</p><p>material from the cathode may contaminate the grid or anode and cause</p><p>grid current and arc-back, making the tuhe iuoperativc.</p><p>If the rathode voltage is too low in the gas-filled tuhe, the snrfare call-</p><p>not be activated properly and loses its emitting effiiknry very quirkly.</p><p>This permits bombardment, as explained above, and destruct,ion of the</p><p>cathode.</p><p>To permit the voltage t o fall helow rather than to rise slightly above</p><p>rating i s serious. Standard industrial t,uhes are desigued to operate \vith</p><p>a voltage tolerance of plus or minus 5 per cent. Iloivever, if a closer</p><p>tolerance than this can be maiutained, thc user will he amply repaid i n</p><p>increased tube life and reliable operatioil. If voltage sij-ings must he</p><p>tolerated, it is more desirable t,hat, t,he minimum s\ving be to not less thau</p><p>95 per cent, of rating even hhuugh the average voltage may he slightly</p><p>above rating. While this prartice \\-ill, of course, give somewhat redured</p><p>tube life, it is preferablc to low xwltage rr-hich rauees rapid tube drterio-</p><p>ration. While t,he effect of voltage change is most, important on the tube</p><p>cathode, it is also undesirable ill ot,her parts of the ririwit. Electrotiic</p><p>circuits, as all other electric cirruits, lost power mparity rapidly if the</p><p>voltage is decreased from rating. Although critiml circuits normally</p><p>contain voltage-regulator tubes and other mealis to hold a constant</p><p>reference vokage in spit,e of line-voltage variat,ions, economic reasons pre-</p><p>vent voltage regulation on t,he majority of rirruits, and henre thcir funr-</p><p>tion will naturally be impaired by excessive voltage variation. This is</p><p>especially true when magnetic sat,uration is part of the roiitrol function.</p><p>I n this group fall solenoids,</p><p>brakes, valves, and rlutrhes. The pull of the a-c solenoid varies approxi-</p><p>mately as t,he square of the voltage. There is some deviation from this</p><p>law, depending upon which part of the brake-horsepower cnrve the sole-</p><p>noid is working. The temperature rise, too, varies approximately a s th r</p><p>square of the vokage.</p><p>In general, solenoids are liberally designed and standard rommerrial</p><p>solenoids are designed to operate satisfartorily on 10 per cent overvoltage</p><p>Effect on Solenoid-operated Devices.</p><p>214 V O L T A G F S T A N D A R D RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>and 15 per cent undervoltage. Since an a-c solenoid has an inrush current</p><p>of approximately ten times the sustained value when sealed, the branch</p><p>circuit sJpplying it should be of ample capacity to prevent an excessive</p><p>voltage drop.</p><p>The corrective capacity of capacitors varies</p><p>with the square of the impressed voltage. A drop of 10 per cent in the</p><p>supply voltage, therefore, reduces the corrective capacity by almost 20</p><p>per cent, and where the user has made a sizable investment in capacitors</p><p>for power-factor correction, he loses the benefit of 20 per cent of this</p><p>investment.</p><p>Effect on Capacitors.</p><p>Nominal Commonly "red</p><p>Iyllem ulilizolion-device</p><p>volt.ge "Oltage rating.</p><p>RECOMMENDED VOLTAGE SPREAD AT UTILIZATION EQUIPMEN1</p><p>Rased on the foregoing effects of voltage variation on utilization equip-</p><p>ment and an extensive poll of industrial plant operating engineers, the</p><p>AIEE Committee on Industrial Power Applications established the</p><p>recommended voltage spreads at the terminals of devices in industrial</p><p>plants. These are shown in Tables 4.8 and 4.9.*</p><p>TABLE 4.8 Recommended Voltage Spread at the Terminals of Utilization</p><p>Devices in Industrial Distribution Systems 600 Volts and Below</p><p>Recommended limib</p><p>of volloge at terminals</p><p>of ulilizolion devices</p><p>480 440,* 460 420-480</p><p>A00 ! 550,* 575 525-600 !</p><p>Drsigriations for nominal system voltages are those commonly used in industrial</p><p>* ThPse are standard polyphase-motor voltage ratings.</p><p>t Polyphase power loads may not operate satisfactorily a t this l o m ~ r limit</p><p>In designing industrial power distribution systems, the system design</p><p>engineer should design for voltage spreads not in excess of those mentioned</p><p>in Tables 4.8 and 4.9. If anything, it would be desirable to design for</p><p>closer limits to allow for critical utilization apparatus that may be devel-</p><p>oped and widely used in the future. The history of electricity in indus-</p><p>trial plants has been to extend its use to more and more functions. As</p><p>plants.</p><p>* Thcse rwommcndstions are in iuhstantial agreement with thP recommmdations</p><p>of the joint EM-SEMA Committce whirh puhlishrd their findings in a report, Prc-</p><p>ferrpd Voltage Ratings of AC Systems and Equipmcnt.</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 21s</p><p>TABLE 4.9 Recommended Voltoge Spreod at the Terminols of Motors</p><p>Served ot Primory Voltoge</p><p>Nominal syitem</p><p>d t a g e</p><p>Motor-nome-plote</p><p>*oltoge rating</p><p>2400</p><p>2400 ~ 2300" 1</p><p>4160</p><p>2160</p><p>2250</p><p>3920</p><p>4500</p><p>6470</p><p>4800 4600 I</p><p>6900 1 6600 I</p><p>2380</p><p>2480</p><p>4320</p><p>5000</p><p>71 30</p><p>Recommended limits af</p><p>voltage at terminalr of</p><p>high-voltage moiors</p><p>* I'rmrnt standard rnot,or voltagc rating.</p><p>well as driving the utilization equipment, it is alço used for a11 types of</p><p>rritical proccss control systems; therefore, its role is hecorniiig exceedingly</p><p>important, and to fulFiI1 this role effectively, good voltage must he rnain-</p><p>taiiied iii industrial plants.</p><p>L I G H T FLICKER V O L T A G E REQUIREMENTS</p><p>Relatively slom chaiiges in voltage are associated mith voltage spreads</p><p>as discussrd iii tlie foregoiiig. There are, however, maiiy types of voltage</p><p>changes 1rhii.h are of a traiisient nature aiid last only a feiv cycles. Thcse</p><p>are commiiiily referred to as voltage flicker, aiid its primary effect is to</p><p>cause flicker iii th r light ciiitput of lamps. The arnount of voltage varia-</p><p>tioii as a fiiiirtioii of frequency of variation which can be xvithstood on</p><p>iiicaiidesrent larnps aiid not cause ohjei:tionahle psychological effects is</p><p>shown iii Fig. 4.15. These curves were preseiited in the General Electric</p><p>Review, hugust, 1925.</p><p>Fluoresceiit lamps are less suhject to flicker over a range of voltage that</p><p>is beloiv that whirh mil1 piit them out. Iii industrial plants, voltage</p><p>flicker i s caiised primarily hy the followiiig types of load: repetitive motor</p><p>starting, large rei,iprocatiiig cornpressors, punch presses, etc., which dram</p><p>a fluctiiating load; resistarice wcldcrs; aiid arc furnaces.</p><p>To elimiiiatc objcctionable light flicker, the design of the systcm should</p><p>be siich that the lirnits of Fig. 4.15 are adhered to. Wider lirnits may be</p><p>iiscd uiider certaiii coiiditioiis without cornplaiiit from the personnel</p><p>orrupyiiig tlie affei,tcd arca. Ho!rcv&, this subject is so cornplicated</p><p>aiid involved that general guides other than Fig. 4.15 would probably not</p><p>be of much use.</p><p>216 VOLTAGE-STANDARD RATINGS. VARIATIONS. CALCULATION OF DROPS</p><p>FLICKER OF INCANDESCENT LAMPS</p><p>CAUSED 81 RECURRENT VOLTAGE DIPS</p><p>I</p><p>5</p><p>0</p><p>Y</p><p>w 3</p><p>0</p><p>5</p><p>' t</p><p>t- z</p><p>Y</p><p>0</p><p>w</p><p>,'</p><p>a</p><p>0</p><p>D l l O PL" "0"I DlPI PLI1 SECOND</p><p>I F C O Y D L</p><p>10 82 6 J 2 I 30 12 L</p><p>Y l U U l L I</p><p>T IME BETWEEN DIPS</p><p>FIG, 4.15 Relation of magnitude of voltage dips to frequency of dips for incandescent</p><p>IWlPS.</p><p>METHODS OF REDUCING VOLTAGE SPREAD AND FLICKER</p><p>REDUCING VOLTAGE SPREAD (See Fig. 4.271</p><p>\Vitlr recommended values of voltage spread established by the N E E</p><p>Industrial Power Systems Committee a i d EEI-SEA\Z.%, it is possible to</p><p>study specitiv syst,ems to see hon. they romparc with these rcquiremerits.</p><p>Where voltage spreads arc found t,o be heyorid t,hose limits, there are four</p><p>11-ays of reducing the voltage spread.</p><p>1 . Carry the power further a t a higher voltage and a t a lesser dist,aim</p><p>at 1o\vcr voltage, i.e., use the load-center power system.</p><p>2. 1tediii.p the impedance of the systrm.</p><p>3. Use regiilat,iiig equipment to rompelisate for volt,age drop.</p><p>4. Use s \~i t i~I ied capacitors.</p><p>llaintaiiiiiig the volt,age at an average desirable I e i d also requires the</p><p>judicious use of traiisformer ratios and taps. Traiisformer taps (for</p><p>changing a t no load oilly) do trot, reduce the spread but affect only t,he</p><p>general voltage level arid particularly the light load voltage in the plallt.</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 217</p><p>Load-center Distribution Systems. The load-ceiiter distribiitioii sys-</p><p>tem is 11011- almost uiriversally used i n industry for, among othcr reasons,</p><p>it provides Ion- voltage drop, henre small voltage spread, berausc the</p><p>power is carried right to the load i.etiter at high Iwltage. Refer to Chap.</p><p>11 for a one-liiie diagram of a typical load-retiter system.</p><p>I t</p><p>is obvious from this table that the tiig gaiii is made by going from voltages</p><p>iii the (i00-volt class to voltages il l the 2.4- to 13.8-kv class for rarryitig</p><p>poll-er from the source to the load ceuter. To illust,rate furthcr, supposc</p><p>that the voltage drop i n a 480-volt system Tx-ith long serondary feeders is</p><p>20 per relit total in the secondary feeders oiily. Should this power he</p><p>carried a t 4160 iiistead of 480 volts, the percentage voltage drop x~ould</p><p>have been only slightly ovcr one-quarter of 1 prr rrnt. Siirrc the load-</p><p>center system minimizes the length of low-\.oltage feeders, it minimizes</p><p>one of the chief causes of voltage drop atid herive redures voltage spread.</p><p>Table 4.10 illustrates the advantage of higher voltage distribution.</p><p>TABLE 4.10 Per Cent Voltoge Drop as a Function of Circuit Voltage for a</p><p>Feeder of a Given Cross Section</p><p>circcuit Relative Per Cent</p><p>V0ltoge Voltage Drop</p><p>240 400</p><p>480 100</p><p>2,400 4</p><p>4,160 I .33</p><p>13.800 0.12</p><p>Some examples will serve t,o illustrate the better voltage conditions in</p><p>the load-cetit,er system. The average 480-volt load-renter substatioii is</p><p>rated 750 h a . With ail average load density of 10 va per sq ft, this sub-</p><p>statioir will servc ai l arca of 73,000 sq ft,. Ideally, the load area would be</p><p>a square, with the substatioii esartly i i i the renter; then the longest feeder</p><p>length ivould tie about l(i5.ft. Rut it i d 1 he assumed t,hat t,he area is</p><p>somewhat rcctaiigular atid that the suhstatioii rannot he lorated exactly</p><p>at the center. The artual length of the longest feeder might then he</p><p>ahout 200 ft.</p><p>Figure 4.16 rontains charts showing the voltage profiles for this 480-volt</p><p>suhstatioii. The trairsformcr taps should lie set for 480 secondary volts</p><p>when the primary voltage is at its maximum atid with no load on the sub-</p><p>station. The highest, voltage that is eticoi~titered by ally equipment</p><p>served hy this substatioii is 480 volts. At maximum load, voltage drop</p><p>has its maximum effect. A 4 pcr rcnt voltage reduction iu the primary</p><p>system is assumed, to illustrate the Ion--voltage rondition. This could</p><p>he due to a dcrreasc i n the power-vompauy supply voltage with inrreased</p><p>load on its system. h drop of 15 volts due t,o traiisformer react,ancc can</p><p>he experted. Assuming the 200-ft feeder to ronsist of a 250-MCM cahle</p><p>per phase and to he fully loaded a t 80 per rent power factor, i t mill~iutro-</p><p>218 VOLTAGbSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>duce about another 7 volts drop. h final 5 volts may be lost in the branch</p><p>circuit. The result is a minimum voltage at the end of the branch circuit</p><p>of 433 volts. In this system, voltage varies between the limit,s of 433 volts</p><p>and 480 volts---a voltage spread that should, in general, be satisfactory.</p><p>The old-type system often uses a suhstatioii as large as 3000 kva at 480</p><p>volts. And not heing of unit substation construction, it has to be located</p><p>at one edge of the load area-probably with the t,rausformers outdoors.</p><p>With the same load density as before, 10 va per sq f t , the 3000-kva snb-</p><p>station must supply an area'of 300,000 sq ft,. I n this substation the</p><p>longest feeder will probably be ahout 900 ft . The corresponding feeder</p><p>voltage drop will be 29 volts. Here the voltage spread is from 411 to 480</p><p>volts. Such a spread is well heyond the recommeuded limits. A full</p><p>load voltage of 411 i s too low to be coiisidered good practice; 420 i s the</p><p>recommended minimum voltage for 440-volt motors.</p><p>750 KVA</p><p>SU0STATlDN</p><p>2 440-</p><p>></p><p>420 -</p><p>-</p><p>-</p><p>460</p><p>2 440</p><p>420</p><p>400</p><p>NO LOAD VOLTAGE CONDITIONS WITH PRIMARY</p><p>VOLTAGE a T MAXIMUM</p><p>--</p><p>SECONDARY FEEDER/</p><p>VOLTAGE DROP-7 VOLTS</p><p>VOLTAGE DROP-</p><p>VOLTAGE-STANDARD RATINGS. VARIATIONS. CALCULATION OF DROPS 219</p><p>60</p><p>48</p><p>w</p><p>0 a</p><p>3</p><p>0 ></p><p>k</p><p>n</p><p>3</p><p>U</p><p>U</p><p>w z 7 24l</p><p>201</p><p>W</p><p>J</p><p>z</p><p>(</p><p>ABLE FULL LOAL</p><p>51ZE AMP</p><p>U0.4 90</p><p>vo. I I40</p><p>000 210</p><p>500 MCM 40 -</p><p>-</p><p>-</p><p>5</p><p>SECONOARY FEEDER LENGTH(FEET1</p><p>Chart showing length of three-conductor 600-volt cable in iron conduit to FIG. 4.17</p><p>produce 2 3 per cent voltage drop at the most unfavorable power factor and full load</p><p>on the cable.</p><p>Tolerable Secondary-feeder Voltage Drop. Figure 4.17 offers a</p><p>guide as to about how far fully loaded cables for circuits 600 volts and less</p><p>can be run and not encounter voltage-drop troubles in a n average indus-</p><p>trial plant. Thcrc are many variables which can alter the maximum</p><p>feeder length materially, such as power factor of load, primary voltage</p><p>drop, load per feeder, etc. The chart of Fig. 4.17 is based on representa-</p><p>tive conditions, i.e., primary voltage drop 5 per cent, transformer drop</p><p>355 per cent, branch-circuit feeder drop 156 per cent. The remainder is</p><p>the secondary-feeder drop of 2>5 per cent, the basis of Fig. 4.17. The</p><p>allowable spread a t 480 volts is 480 t,o $20, or 60 volts or 1255 per cen-</p><p>the sum of the percentages just ment,ioned. Secondary-feeder drops</p><p>greater than 235 per cent should be cherked under conditions expected at</p><p>the plant, t,o see if t,hey can be tolerated without c-using undesirably wide</p><p>voltage spreads.</p><p>Looking at this another way, 480-volt secondary feeders longer than</p><p>250 f t for small cable sizes and 400 f t for larger cable sizes should be</p><p>220 V O L T A G G S T A N D A R D RATINGS. VARIATIONS, CALCULATION OF DROPS</p><p>avoided from a voltage-drop st,andpoint. If longer feeders must he used,</p><p>check the voltage drop. The tolerable secondary feeder lengths are some-</p><p>what longer for 600-volt cirruits, i.e., abont 300 and 500 ft, respectively,</p><p>aud considerably shorter a t 240 and 208 volts, i.e., about 125 arid 200</p><p>f t , respectively, a t 240 volts and 100 and 175 ft,, respectively, a t 208</p><p>volts.</p><p>Reducing Impedance. Sirirc volt,age drop is a product of current</p><p>t,imes impedance, anything t,hat, is done t,o reduce the impedance of a</p><p>circuit will reduce its voltage drop. The following are some suggestions:</p><p>1. t-se closely spaced cotidurtors, i.e., use cable instead (if open wire</p><p>with widely spaced (.onductors.</p><p>2. I'se interleaved huses, that is, bnses wit,h several cotiduvtors per</p><p>phase arratiged 8 , B , C; -4. B, C ; -4, R, C , etr., instead of liuses with all</p><p>conductors of one phase widely separated from the other phases.</p><p>3 . Use t x o smaller vahles in parallel instead of one larger cahle.</p><p>4. Use standard-rea(.tattce instead of high-rea(.tanre transformers.</p><p>€Iigh-reart,ance transformers reduce short-circuit rurrciits but increase</p><p>voltage drop, particularly lrith poor power-factor loads. A compromise</p><p>is necessary hecause a lover t,hari standard-reactanre transformer, n-hilc</p><p>reducing voltage drop, may invreasc short-circuit, rurrents so high as t o</p><p>require unreasonable switchgear for protection of the circuits fed by the</p><p>transformer.</p><p>5 . Keep feeders-particularly low-voltage feeders- as short as possible.</p><p>6. Use series caparitors t o neutralize the i n d u h v e reactance of a rir-</p><p>wit. There are few if any general applira.tions of the series capacitors</p><p>for this purpose</p><p>in indust,rial plants except in coiiriect,ion wit,h resistanre</p><p>welders as nokd later in this chapter. In a few rases they have heen used</p><p>in connection with motors t o maintain sufficient vokage a t the motor</p><p>terminals when starting a large motor oii a soft, system or t o neutralize</p><p>system impedance t,o maintain good voltage on lights, etc.</p><p>U s e Regulating Equipment. Even where the plaiit power system</p><p>uses a load-center system to rarry t.he power the great,est practical dis-</p><p>tance at high voltage, where impedances have been kept, to a minimum</p><p>and lorn-voltage feeder lengths as short as possible, it, may not be able to</p><p>meet the required voltage spreads because of too much voltage variation</p><p>in the primary supply system or because of a process requiring unusually</p><p>close voltage spread. In some old plants, low-voltage feeders may he</p><p>large and long, giving excessive voleage drop i n the secoiidary system.</p><p>Marry plants operate at 240 or 208Y/120 volts and have excessive voltage</p><p>spread that would be reduced to tolerable limits if 480 volts were used</p><p>instead. However, the change may not he practical or ecoriomical a t t,he</p><p>moment. In such cases voltage-regulating equipment provides the</p><p>answer to the problem.</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>Where a transmission or</p><p>disl,ribution hie supplies a plant with power whioh has a voltage spryad</p><p>greater than a’.iout 5 per rent,, it may tie difficult to mairrtairi the desired</p><p>voltage spread even wiLh the best designed plant power system. In these</p><p>cases sonic form of voltage regulatioii is often required. I f the supply is</p><p>at, high vdtage arid must, he stepped down t,o below 15 kv (commonly</p><p>-Ll(i0 or 13,800 volts) for distritiut,ion, regulation rain tie built int,o the</p><p>transformer. This regulation is accomplished by automatic tap changing</p><p>which xi11 operat,e under load (load-rat,io (:oiitrol). Usually t,here are 32</p><p>(76 per cent) steps to enable close volt,age control over a range of plus or</p><p>minus 10 per cent..</p><p>1,oad-ratio control for plus or niiriirs 10 per cent range is a very low cost</p><p>iii the over-a11 plant costs, arid yet because the load-ratio corit,rol provides</p><p>good voltage, it will improve prodii(tiiin and quality of maliufartured</p><p>goods. Iienc,e, the dividend from t,his small investment \rill often repay</p><p>the investmerit many t,imes over earh year. It is strongly rei~ommcrided</p><p>tha.t load-ratio control tie cmsidered itt every transformix stepping down</p><p>from voltages ahorc 15 k r to plant primary voltage iii the raiige of 2:I to</p><p>13.8 ku. The systems aliove 1.5 kv arc not dn-ays regulated to suit, t,he</p><p>industrial plant but for most, r f i r i i~nt oxr-al l opcratioli of t,lie poiver sys-</p><p>221</p><p>Voltage Regulation of M a i n Power Source.</p><p>FIG. 4.18</p><p>incorporated.</p><p>A typical outdoor packoged substation in which bod-ratio control con be</p><p>222 VOLTAGLSTANDARD RATINGS, 'VARIATIONS, CALCULATION OF DROPS</p><p>tem. When load-ratio control is installed, both the utility and industrial</p><p>plant can operate their systems independently and to their own best</p><p>advautage without interference voltage wise.</p><p>Figure 4.18 illustrates an outdoor substation, typical of those whose</p><p>transformers can include load-ratio control.</p><p>Voltage Regulators. If power is supplied by the utility at below</p><p>15 kv, the only transformation required is at the individual load-center</p><p>substations. Load-ratio control in each industrial load-center unit sub-</p><p>station is uneconomical and even may he impractical. Hence, where the</p><p>primary-voltage spread is wide enough to require voltage regulation,</p><p>separate voltage regulators should be installed in the primary supply,</p><p>Fig. 4.19. For this service either three-phase step voltage regulators</p><p>(Fig. 4.20) or induction voltage regulators (Fig. 4.21) can be used. Their</p><p>standard range of voltage regulation is plus or minus 10 per cent. The</p><p>question is sometimes raised as to whether two induction regulators</p><p>should be connected in open delta. This is slightly less expensive than</p><p>three regulators to regulate three-phase circuits. However, the open-</p><p>delta connection creates an unbalanced voltage condition that should be</p><p>avoided. The voltage unbalance is small but may be enough to increase</p><p>STEP OR</p><p>INDUCTION</p><p>VOLTAGE</p><p>REGULATOR</p><p>REGULATOR HOLDS</p><p>CONSTANT VOLTAGE</p><p>v v HERE</p><p>\</p><p>t Y</p><p>NOTE : THE BY-PASS PERMITS MAINTENANCE OF SERVICE</p><p>TO PLANT WHERE REGULATOR IS BEING MAINTAINED</p><p>FIG. 4.19 One-line dioorom rhowino the adicot ion of steD or induction voltom r e d o - - - , .</p><p>tori for holding constant voltage on the plant primary bur for plants served at primary</p><p>voltage.</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 223</p><p>FIG. 4.20</p><p>or minus 10 per cent voltage regulotion.</p><p>Typicol three-phore step voltage regulator roted 13,200 volts, 208 kvo, plus</p><p>FIG. 4.21</p><p>per cent voltoge regulation.</p><p>Typicol induction voltage regulator rated 225 kvo, 4330 volts, plus or minus 10</p><p>224 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>the heating appreciahly in fully loaded polyphase motors. For this rea-</p><p>son, best prartice avoids the open-delta conneition in favor of three-phase</p><p>regulation.</p><p>It is rerommended that serious consideration be given the addition of</p><p>t,hese regulat,ors in the plant supply lines whenever the expected voltage</p><p>spread in the primary supply lines exceeds ahout 5 per cent.</p><p>Regulators may he hypassed for maintenance and a t the same time</p><p>maint,ain unregulat,ed service to the plant. Itegulat,ors, like any other</p><p>piece of apparatus, must be given consideration from a short-circuit,</p><p>standpoint.</p><p>trt,ilit,ies often regulate individual feeders</p><p>at distribution voltage (2100 or 416F volts, for example) to compensate</p><p>Feeder Voltage Regulation.</p><p>480 VOLTS</p><p>SECONDARY FEEDER</p><p>INDUCTION VOLTAGE</p><p>REGULATOR</p><p>FEEDERS TO</p><p>MOTORS, ETC</p><p>LIGHTING FEEDER</p><p>LIGHTING LOAD</p><p>120 VOLTS</p><p>FIG. 4.22</p><p>regulators for secondary feeder regulation.</p><p>One-line diagram showing the opplication of air-cooled induction voltage</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 225</p><p>for the voltage drop i n that, feeder. The itidustrial plaut does iiot often</p><p>usually small, less illan 1 or 2 per cent. Thus, regulation of voltage a t</p><p>the main h s is more rwmmotily used.</p><p>While there seems to he little, jirstifiratioii for irrdividuul primary-feeder</p><p>regulation, there may be many appliratioris for individual seroiidary-</p><p>feeder reeulation. For examuk, tlie voltage spread may I ) satisfartory</p><p>ry, since \.oltage drops in individual primary ferde</p><p>._ . ,</p><p>for t,he majorit,y of utilization e q u i p</p><p>ment, such asmotors, welders, etr., but</p><p>not, considered good cliough for lights.</p><p>In such cases, t,he lighting feeder may</p><p>be regirlat,ed and the rest irnregulated,</p><p>Fig. .4.2%. For such applications, air-</p><p>~ooledregirlalorslike that shown in Fig.</p><p>4.23 may be used.</p><p>I n other cases, individual loads a t</p><p>GOO volts or less may require voltage</p><p>regulation t,o obtain the desired per-</p><p>formance from the equipment,. Rirh</p><p>loads might he heating unit,s, process</p><p>cont,rol, infrared ovens, hluepririt ma-</p><p>chines, lights, radio arid television</p><p>transmitt,ers, brooders, etc. Where</p><p>these loads are served at, utilizat,ion</p><p>voltage, aii iridrictimi regulator like</p><p>that, of Fig. 4.23 may be used.</p><p>lnductrol P o w e r Pock . A iie\\- dc-</p><p>velopment is a regulating loi\~-voltage</p><p>subst,atioII known as the Inductrol</p><p>Power Pack. It, is a itiiit made up</p><p>primarilyof an indurtioii voltage regu-</p><p>lator arid a dry-t,ype transformer.</p><p>The transformer is rat,ed 480 or 600</p><p>volts on t,he primary aiid %08Y/lZO or</p><p>120/240 volts on the secondary. A</p><p>primary switching arid protective de-</p><p>FIG, 4.23 A modern induction voltage</p><p>regulator for circuitr 600 volts and leis.</p><p>Typical of either single 01 three phase. -</p><p>vice arid secondary terminals complete the package. This unit may</p><p>be used for supplying regulat,ed lighting power from general-purpose</p><p>480-</p><p>or 600-volt feeders or for supplying any other loads with regulated 120-</p><p>volt power from 480- or 600-volt power systems.</p><p>Refer t o Chap. 8 for a comp1et.e discussion of the</p><p>application of shunt capacitors to improve voltage conditions.</p><p>In some cases where the general voltage level is</p><p>Shunt C a p a c i t o r s .</p><p>Auto trans formers .</p><p>226 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>lorn and transformt?r taps cannot he used t,o correct for it, autotrans-</p><p>formers may he used to provide a permanent boost i n voltage. The</p><p>autotrausformer does riot reduce t,he spread.</p><p>A t,ypiral appliration xvouid lie in the case of a, 208Y/120-voll system</p><p>supplying 220-nolt, mot,ors. The volt,ege may be proper lor the lights</p><p>hut not high eiiimgh for t,he 220-volt motors. An autolrmisformer could</p><p>he used l o step 208 volt,s up to 220 volts for the motors only.</p><p>Where power is generated by the</p><p>plant,’s oxvti geiir.ralors, the voltage on t,he powerhouse bns can be held</p><p>constant or exwi varied with load to compensate for voltage drop as load</p><p>comPs on. Problems of voltage rcgulat,ion where industrial generators</p><p>are operalnd i r i parallel with utility systems are referred to in Chap. 15.</p><p>Generator Voltage Regulators.</p><p>REDUCING FLICKER (See Fig. 4.27)</p><p>Reduction of flicker is often a much more difficult prohlem than the</p><p>reduction of voltage spread previously referred to.</p><p>Flicker due to reciprocating motor-driven loads such as compressors,</p><p>purich presses, et,c., can often be reduced by increasing the inertia of the</p><p>met:lranical system to smooth out the pulsations. Where this is not</p><p>One is to separate flicker-producing</p><p>load from the lights or critical load, i.e., use separate supply circuits. The</p><p>nther is to use a voltage stabilizer, Fig. 4.24, to feed the critical load.</p><p>Sometimes the critical load is fed through a motor-generator set to pro-</p><p>vide good voltage for tliet load. This, lioxvevrr, is more expensive over</p><p>all thaii voltage stabilizers and in gcmral offers no advantage in this</p><p>ive, t,wo t,liings may be done.</p><p>FIG. 4.24 Typicol voltage stabilizer.</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 227</p><p>application. Voltage regulators previously discussed are not fast enough</p><p>to correct flicker, but for single-phase circuits, and in small sizes, auto-</p><p>matic voltage stabilizers are available to hold voltage mit,hin very close</p><p>limits. A typical model is designed to maintain an output voltage of</p><p>115 volts with maximum variation of plus or minus 1 per cent, even</p><p>though the supply volt,age may vary between 95 and 130 volts. The</p><p>volt,age stabilizer has no moving parts and no electronic tubes; its opera-</p><p>tion is obt,ained from the properly coordinated characteristics of reactors</p><p>and caparitors.</p><p>Series capacitors can be of value in reducing volt-</p><p>age flicker. C:hapt,er 8 contains a complete discussion of the application</p><p>of series capacitors.</p><p>Series Capacitors.</p><p>SELECTION OF TRANSFORMER TAPS</p><p>All modern transformers in ratings above 100 kva and most or those</p><p>helow that kva rating have taps in the windings to change the turn ratio.</p><p>The taps do not materially affect the voltage drop through the trans-</p><p>former; they merely change the turn ratio, hence the no-load voltage</p><p>ratio. For example, a standard transformer rated 2400-480 volts may</p><p>have four 2>5 per cent taps in the 2400-volt winding. The standard for</p><p>these taps in transformers used in industrial systems is to have two 256</p><p>pcr cent, taps above 2400 volts and two 24i per cent taps below 2400volts.</p><p>The no-load ratios of such a transformer would be as given in Table 4.11.</p><p>TABLE 4.11 No-load Voltoge Ratios of Standard Transformer Rated</p><p>2400-480 Volts</p><p>2520-480 “0th 5% obove tap</p><p>2460-480 volts 236% obove top</p><p>2400-480 volts Norrnol rating top</p><p>2340-480 volts 2>P% below top</p><p>2280-480 volts 5% below tap</p><p>These taps do not improve voltage regulation but are only for changing</p><p>the general vokage level iq the plant. If a 2400-480-volt transformer is</p><p>connected to a system whose maximum voltage is 2520 volts, then the</p><p>2520-480-volt tap could be used which would provide a maximum of 480</p><p>volts no load on the system, as shown by curve A , Fig. 4.25. If, for</p><p>example, another system had a maximum no-load voltage of 2400 volts,</p><p>then the 240&480-volt tap could be used to provide 480 volts no load in</p><p>the plant. Similarly if a</p><p>plant had a maximum voltage of 2280 vo!ts, then the 2280-480-volt tap</p><p>could be used to provide a maximum of 480 volts no load in the plant, as</p><p>shown in curve C , Fig. 4.25. It will be noted that in all cases the second-</p><p>ary no-load voltage is 480 volts; so the secondary system does not know</p><p>This would be as shown in curve B, Fig. 4.25.</p><p>2600- - - -</p><p>- 480 VOLTS MAX</p><p>- U c</p><p>> -</p><p>></p><p>U -</p><p>(L</p><p>a</p><p>2 2 4 0 0</p><p>-</p><p>(r</p><p>I -</p><p>2300-</p><p>-</p><p>440 V</p><p>4 8 0 VOLTS MAX MIN ---- ---- - _ _ _ _ _ _ _ - -</p><p>-?</p><p>40</p><p>VOLTS</p><p>SPREAD</p><p>4 4 0 V</p><p>B</p><p>480 VOLTS MAX MIN --------- _ _ _ _ _ _ _ _</p><p>VOLTAGkSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>no-load voltage. By using the next tap up on the transformer, that is,</p><p>the one rated 2460-480 volts, the turn ratio of the transformer has now</p><p>been changed so that the no-load voltage is 472 volts, as shown in curve</p><p>B , Fig. 4.26. The voltage spread will be substant,ially the same, i.e.,</p><p>40 volts, so that the minimum voltage is now 432 volts, which is well</p><p>above the recommended minimum for plant distribution systems.</p><p>By judicious selection of the transformer tap t,he voltage within the</p><p>plant can he kept Tyithin acceptable 1imit.s provided that the primary</p><p>voltage does not vary more than about 5 per cent and that the plant dis-</p><p>tribution system is designed along modern lines with the load-center sys-</p><p>tem using short secondary feeders and transformers not larger than about</p><p>1500 kva a t 480 volts or proportional sizes a t other secondary volt,ages.</p><p>Changing taps cannot, correct conditions where voltage spread is t,oo</p><p>great. For example, suppose a plant suffered from low voltage at remote</p><p>points and had a large volt,age spread. To be specific, suppose the spread</p><p>was 80 volts and the minimum voltage at the remote end was 400 volts,</p><p>then the maximum voltage would be 480 volts. If taps are changed to</p><p>raise the general voltaga level, the spread will not change but the 400-volt</p><p>229</p><p>4 8 5 VOLTS MAX - -- -- - - - - - - - - - -_-</p><p>40</p><p>VOLTS</p><p>SPREAD I 440 V MIN</p><p>g 2400</p><p>a</p><p>I-</p><p>J</p><p>0</p><p>5</p><p>></p><p>a</p><p>a</p><p>I -</p><p>LL</p><p>P</p><p>FIG. 4.26</p><p>excessive no-load voltage by proper election of taps on transformer.</p><p>Voltage profile showing that rotisfactory voltages con be obtained without</p><p>230 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>Volloqe Correction For</p><p>TypicoI Feeder Circuits</p><p>FIG. 4.27 Summary of methods of improving</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 231</p><p>LOW LOAD VOLTAGE</p><p>"No b o d "</p><p>Except shunt</p><p>POI or seom.</p><p>2. Flmh. rssislmnre</p><p>welders.</p><p>3. Motor loads. such</p><p>01: sow mill^, Rubber</p><p>milli. Grinders.</p><p>4. Arc furnorer.</p><p>Correcl by m e o n r of</p><p>Series rapodor with welder to reduce dernond</p><p>by power.farlor correction</p><p>.eO</p><p>supply c i r w i t</p><p>Voltoge Ifobiliier for lighting circuil</p><p>re.artance</p><p>Voltage slmbilirer for lighting circuit</p><p>Series c"Po</p><p>only the maximum d-c component is considered, since the</p><p>circuit breaker must be applied to handle the maximum short-circuit</p><p>current that can occur in a system.</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 21</p><p>In the general case for circuits rated above 600 volts, the multiplying</p><p>factor to account for d-c component is 1.6 times the rms value of the a-c</p><p>symmetrical component at the first half cycle.</p><p>For circuits rated 5000 volts or less where there is no local generation,</p><p>that is, where the supply t,o the bus is through transformers or long lines,</p><p>the multiplying factor to ralculate the total current at the first half cycle</p><p>may be reduced to 1.5. For circuits 600 volts and less, t,he multiplying</p><p>factor to calculate the total current at the first half cycle is 1.25 when the</p><p>circuit breaker is applied on the average current in three phases. Where</p><p>single-phase conditions must be considered in circuits GOO volts and less,</p><p>then to account for the d-c component in one phase of a three-phase cir-</p><p>cuit a multiplying factor to calculate the total current at the first half</p><p>cycle of 1.5 is used.</p><p>For some calculations, rms current evaluations a t longer time intervals</p><p>than the first half cycle, such as three to eight cycles corresponding to the</p><p>interrupting time of circuit breakers, are required. Multiplying factors</p><p>for this purpose may be taken from the curve in Fig. 1.22.</p><p>Table 1.2 gives the multiplying factors commonly used for applying</p><p>e</p><p>FIG. 1.22</p><p>various X / R ratio of circuits.</p><p>Charts showing multiplying factors to account for decoy of d-c component for</p><p>22 SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES</p><p>short-circuit protective devices. These factors range from 1 to 1.6,</p><p>depending upon whether the short-circuit calculation is being made to</p><p>determine the interrupting or momentary duty on the short-circuit pro-</p><p>tective device.</p><p>SHORT-CIRCUIT RATIO OF GENERATORS</p><p>This term is referred to frequently in short-circuit discussions. With</p><p>present AIEE procedures of short-rircuit ralrulations, i t has become an</p><p>accessory with no practical significance from this standpoint. For the</p><p>sake of completeness, a definition is given here.</p><p>Short-circuit ratio</p><p>field current to produce rated voltage a t no load</p><p>field current to produce rated current at sustained short circuit</p><p>-~ -</p><p>No further mention will he made of short-circuit ratio.</p><p>TOTAL SHORT-CIRCUIT CURRENT</p><p>The total symmetrical short-rirruit current is made up of currents from</p><p>several sourves, Fig. 1.23. At the top of the figure is shown the short-</p><p>circuit current from the utility. This act,ually comes from ut,ility gener-</p><p>ators, but generally the industrial system is small and remote electrically</p><p>from the utility generators so that the Symmetrical short-rircuit current</p><p>is substant,ially constant,. If there are generators in the indust,rial plant,</p><p>then they cont,ribute a symmet,rical short-circuit rurreiit which for all</p><p>practical purposes is constant over the first few cycles. There is, how-</p><p>ever, a slight decrement, as indicated in Fig. 1.23.</p><p>The other sources are synchronous motors which act something like</p><p>plant generators, except that t,hey have a higher rate of decay of the sym-</p><p>metriral component, and induction motors whirh have a very rapid rate</p><p>of dccay of the symmetrical component of current. When all these cur-</p><p>rents are added, the total symmetrical short-circuit rurrent is typical of</p><p>that shown a t the bottom of Fig. 1.23.</p><p>The magnitude of the first few cycles of the t,otal symmetrical short-</p><p>circuit, current is further increased by the presence of a d-c compouent,</p><p>Fig. 1.24. The d-c component, offsets the a-c ware and, therefore, makes</p><p>it asymmetrical. The d-c component decays to zero within a few cycles</p><p>in most indust,rial power systems.</p><p>It is this total rms asymmetrical short-circuit current, as shown in Fig.</p><p>1.24, that must he determilied for short-circuit protective-derice applira-</p><p>tion. The problem of doing this has been simplified by standardized</p><p>procedures to a poiut xhere t o determine the rms asymmetriral current</p><p>one need only divide t,he line-to-neutral roltage by the proper reactance</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 23</p><p>RG. 1.23 Tracer of orcillogramr of rym- FIG. 1.24 Arymmelrical short-circuit current</p><p>metrical short-circuit currents from utility, from dl sources illustrated in Fig. 1.23 plus</p><p>panerator, synchronous motors, and induc- d-c component.</p><p>lion motors. The shape of the total com-</p><p>bined currents is illurtmted by the bottom</p><p>hace.</p><p>24 SHORT.CIRCUIT-CURRENT U L C U U l l N G PROCEDURES</p><p>or impedance and then multiply by the proper multiplying factor from</p><p>Table 1.2.</p><p>BASIS OF RATING A-C SHORT-CIRCUIT PROTECTIVE DEVICES</p><p>The background of the circuit-breaker rating structure as well as the</p><p>basic characteristics of short-circuit currents must be understood to</p><p>enable the engineer to select the proper rotating-machine reactances and</p><p>multiplying factors for the d-c component to determine the sbort-circuit-</p><p>current magnitude for checking the duty on a particular circuit breaker,</p><p>such as momentary duty or interrupting duty.</p><p>The rating structure of circuit breakers, fuses, and motor starters is</p><p>designed to tell the application engineer how circuit breakers, fuses, or</p><p>motor starters will perform under conditions where the short-circuit cur-</p><p>rent varies with time. In discussing these rating bases, and for the sake</p><p>of clarity, they will be arbitrarily divided into two sections, i.e., the rating</p><p>basis of high-voltage short-circuit protective devices above 600 volts and</p><p>the rating basis of low-voltage Short-circuit protective devices 600 volts</p><p>and below.</p><p>HIGH-VOLTAGE SHORT-CIRCUIT PROTECTIVE DEVICES (ABOVE 600 VOLTS)</p><p>Power-circuit-breaker Rating Basis. The standard indoor oilless power</p><p>circuit breakers as used in metal-clad switchgear will be used here t o</p><p>explain power circuit-breaker ratings. The same fundamental principles</p><p>apply to all other high-voltage power circuit breakers.</p><p>The circuit-breaker rating structure is complicated because of the time</p><p>of operation of the circuit breakers after a short circuit occurs.</p><p>The few cycles needed for the power circuit breaker to open the circuit</p><p>and stop the flow of short-circuit current consist of the time required for</p><p>(1) the protective relays to close their contacts, (2) the circuit-breaker</p><p>trip coil to move its plunger to release the breaker operating mechanism,</p><p>(3) the circuit-breaker contacts to part, and (4) the circuit breaker to</p><p>interrupt the short-circuit current in its arc chamber. During this time,</p><p>the short-circuit current produces high mechanical stresses in the circuit</p><p>breaker and in other parts of the circuit. These stresses are produced</p><p>almost instantaneously in phase with the current and vary as the square</p><p>of the current. Therefore, they are greatest when maximum current is</p><p>flowing. The foregoing discussion showed that t,he short-circuit current</p><p>is maximum during the first cycle or loop, because of the presence of the</p><p>d-c component and because the motors contribute the most short-circuit</p><p>current a t that time. Thus, the short-circuit stresses on the circuit</p><p>breakers and other parts of the circuit are maximum during the first loop</p><p>of short-circuit current.</p><p>During the time from the inception of the short circuit until the circuit-</p><p>breaker contacts part, the current decreases in magnitude because of the</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 25</p><p>decay of the d-c component and the change in motor reactance, as</p><p>explained previously. Consequently, the current that the circuit breaker</p><p>must interrupt, four or five cycles after the inception of t.he short circuit,</p><p>is generally of less magnitude than the maximum value of the first loop.</p><p>The fact that the current changes in magnitude with time has led to the</p><p>establishment of two bases of short-circuit-current ratings on power cir-</p><p>cuit breakers: (1) the momentary rating</p><p>Actual voltage drop = 2.75 per cent X 480 = 13.2 volts</p><p>Solution: Enter the chart on the horizontal scale a t 2000 kva.</p><p>Figure 4.30 applies to the 34.5-kv insulation class transformers in</p><p>ratings from 1500 to 10,000 kva. These curves can be used t o determine</p><p>the voltage drop for transformers in the 46- and 69-kv insulation classes</p><p>by using appropriate multipliers a t all power factors except unity. To</p><p>correct for 46 kv, multiply the per cent vokage drop obtained from the</p><p>chart by 1.065, and for 69 kv multiply by 1.15.</p><p>Find the per cent voltage drop in a 5000-kva 69,000-</p><p>13,800-volt three-phase 60-cycle liquid-filled transformer carrying 3500</p><p>kw a t 0.8 power factor.</p><p>Enter chart Fig. 4.30 a t 5000 kva and read per cent voltage</p><p>drop where this transformer size intersects the 0.8-power factor curve.</p><p>Example.</p><p>Solution:</p><p>Per cent voltage = 4.25 for 5000 kva</p><p>6</p><p>NOTE: CURVES ARE BASED ON 6 PERCENT</p><p>Q IMPEDANCE FOR 34.5 KV CLASS</p><p>5 I I I I I I I</p><p>05</p><p>54</p><p>w</p><p>I- 4</p><p>&</p><p>a</p><p>0</p><p>U</p><p>0</p><p>0</p><p>4</p><p>u 3</p><p>5</p><p>8 ,</p><p>I- z</p><p>Y</p><p>Y</p><p>U</p><p>Y I a</p><p>0</p><p>TRANSFORMER RATING-THREE PHDSE KVA</p><p>FIG. 4.30</p><p>age class.</p><p>Tronrformer voltage-drop curves for three-phase transformers, 34%-kv volt-</p><p>VOLTAGbSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 237</p><p>Transformer load = 3500 - - - 4380 kva</p><p>Multiplier for 69-kv insulation class = 1.15</p><p>Actual per cent voltage drop = 4.25 X 1.15 X - = 4.28 per cent</p><p>Voltage Drop in Cable.</p><p>0.8</p><p>4380</p><p>5000</p><p>Voltage-drop curves, Figs. 4.31 to 4.34, may</p><p>be applied with reasonable accuracy to all types of paper-insulated,</p><p>rubber-insulated, and varnished-cambric-irisulated cable insulated for 600</p><p>or for 5000 volts. Two charts were prepared for each of these two voltage</p><p>classes of cable to cover the different t.ypes of installat,ions for cable sizes</p><p>No. 14 to Xo. 4/0 Amg and 250 to 750 MCM. Voltage drop for loads</p><p>between 0.7 power factor lagging and unity is shown for t,his range of</p><p>cable sizes for three-conductor and three single-conductor cables in mag-</p><p>netic conduit.</p><p>The resistance and reactance used in preparing these charts are taken</p><p>from Chap. I. They are calculated values based on 75 C copper tempera-</p><p>ture and scattered tests. In determining reactances, it was assumed that</p><p>for three conductors in conduit the cables d l lie a t random in the hot-</p><p>tom of the conduit. If the cables are twisted together so that they oper-</p><p>ate in contact with each other, they should be regarded as a three-conduc-</p><p>tor cable.</p><p>For single-phase</p><p>circuits consisting of a two-conductor or two single cables in a conduit,</p><p>the. voltage drop measured line-to-line will be 16 per cent higher than</p><p>indicated in the charts.</p><p>First, select the chart apply-</p><p>ing to t,he problem with regard to voltage and type of installation.</p><p>Enter the chart a t the abscissa with the power factor of the load. Extend</p><p>a line vertically from this point to the correct size cable. On the ordinate</p><p>read the volts drop per 100 amp per 100 ft or per 10,000 amp-ft. Multi-</p><p>ply this value by the multiple of 10,000 amp-ft, for the problem under</p><p>consideration to get line-to-line voltage drop in a t,hree-phase system.</p><p>For a single-phase system multiply the three-phase drop by 1.IG.</p><p>Assume that a 500-ft three-conductor rubber-insulated size</p><p>KO. l/O-hwg cable in magnetic conduit is the feeder for a three-phase</p><p>440-volt 60-cycle 150-amp 0.8-power factor inductive load. Find the</p><p>voltage drop.</p><p>Solution: Enter chart, Fig. 4.31, at 0.8 power factor and move upward</p><p>to the KO. l/O-Awg cable curve. From the point of intersection move</p><p>to the left and read thc voltage drop as 2.08 volts per 10,000 amp-ft.</p><p>Ampere-feet in cable = 500 X I50 = 75,000</p><p>Actual voltage drop = E0 X 2.08 = 15.6 volts</p><p>The chart,s are prepared for three-phase voltages.</p><p>Use of Voltage-drop Charts for Cable.</p><p>Example.</p><p>10,000</p><p>238 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>W</p><p>LL</p><p>w</p><p>n</p><p>I</p><p>4</p><p>0</p><p>8</p><p>0</p><p>LL</p><p>w</p><p>0</p><p>u)</p><p>0 ></p><p>z</p><p>&</p><p>3</p><p>-</p><p>z</p><p>5</p><p>n</p><p>W</p><p>c3</p><p>4</p><p>0 ></p><p>VOLTAGGSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 239</p><p>FIG. 4.32</p><p>conduit.</p><p>Voltage-drop curves for three single-conductor 600-volt a b l e r in magnettc</p><p>240 VOLTAGbSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>FIG. 4.33</p><p>or interlocked-ormor cable.</p><p>Voltage-drop curves for three-conductor 5000-volt cable in magnetic conduit</p><p>242 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>Voltoge Drop in Busway. Figures 4.35 and 4.36 may be used to deter-</p><p>mine the approximate voltage drop in a busway. Figure 4.35 applies to</p><p>a busway that is designed specifically for low-voltage drop. Figure 4.3F</p><p>applies t o a typical feeder busway of the type used with plug-in switches.</p><p>Figure 4.35 gives the line-to-line voltage drop in volts for GOO-, 800-,</p><p>1000-, and 1350-amp low-voltage-drop busway. These curves apply</p><p>only for balanced loading of the busway at an operating temperature</p><p>of 70 C.</p><p>The voltage drops for other than rated load may be obtained by multi-</p><p>plying the voltage drop for rated load by the ratio of actual load to rated</p><p>load, Similarly, the voltage drop for lengths other than 100 ft may he</p><p>M 40 60 BO W O 20 40 60 80 100</p><p>LOAD POWER FACTOR LOAD POWER FACTOR</p><p>WAD POWER FACTOR</p><p>FIG. 4.35</p><p>operating temperature assumed.</p><p>Voltage-drop curves for low-voltage-drop burwoy ot rated load. 70 c</p><p>VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 243</p><p>obtained hy multiplying the voltage drop for 100 f t by the ratio of actual</p><p>length to 100 ft .</p><p>Actual line-to-:ine voltage drop</p><p>These corrections are expressed in the following formula:</p><p>actual load actual length</p><p>rated load 100 ft = voltage drop for 100 feet at rated load X</p><p>Example. Find the voltage drop on a 200-ft run of 800-amp husway</p><p>Solution: Enter Fig. 4.35 for au 800-amp husway at 90 per cent power</p><p>Follow a vertical line to its intersection</p><p>carrying a 600-amp load a t a 90 per cent power factor.</p><p>factor on the horizontal scale.</p><p>4 . 5 X 3 = 13.5 VOLTS</p><p>FIG. 4.36 Voltoge-drop curves for typical plug-in bvrwcly carrying rated load.</p><p>244 VOLTAGE-STANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS</p><p>00 the curve and proceed horizontally to the left. The intersection of</p><p>this line with the vertical scale is the voltage drop per 100 f t for an</p><p>800-amp busmay, 2.4 volts.</p><p>600 200</p><p>800 100 Line-to-line voltage drop = 2.4 X ~ X - = 3.6 volts</p><p>Single-phase voltage drops may he obtained by multiplying the three-</p><p>phase voltage drop times 1.16.</p><p>Figure 4.36 gives the line-to-line voltage drop in volts for a plug-in</p><p>type busmay. An example is given with the curves to illustrate their use.</p><p>Example of System Voltage-drop Calculation. The power system</p><p>shown in Fig. 4.37 is used t o illustrate the use of the foregoing charts and</p><p>formulas. Using the most critical feeders from the standpoint of voltage</p><p>33.5 KV TRANSMISSION L I N E</p><p>60 CYCLES</p><p>OVERHEAD LINE</p><p>10,000 KVA PERCENT 2 =</p><p>I . O P E R C E N T t i 6 . 0 P E R C E N T</p><p>34.400 -4160 VOLTS</p><p>BUS A LOAD</p><p>4160 VOLTS</p><p>9000 KVA 0.8 PF LAGGING</p><p>3- CONDUCTOR 250 MCM</p><p>1500 KVA, 4160-480</p><p>PERCENT t = I . O PERCENT</p><p>T t j 5 . 5 P E R C E N T</p><p>*-480 VOLTS</p><p>I N CONDUIT</p><p>1300 aus KVA 0.8 a LOAD PF LAGGING</p><p>480 VOLTS</p><p>I I I 3 - k S o O M C M V C L</p><p>OHMS ?=0.0072tj0.009 200 FEE1 A</p><p>W</p><p>250 KVA</p><p>440 VOLTS</p><p>0.7 PF LAGGING</p><p>FIG. 4.37</p><p>Calcdatio".</p><p>System one-line diagrom used 01 a baris for examples of system voltage-drop</p><p>VOLTAGkSTANDARD RATINGS, VARIATIONS, CALCULATION OF DROPS 245</p><p>drop, four solutions involving varying degrees of accuracy were made to</p><p>determine the operating voltage at the 4160- and 480-volt utilization</p><p>buses and at the load end of a 480-volt feeder.</p><p>In each solution except 4, it is assumed that the indicated load kva,</p><p>power factors, and efficiency remain constant for voltage variatious due</p><p>to regulation. I n other words, the load current varies with applied</p><p>voltage to keep the kva constant.</p><p>Table 5.12 lists the operating</p><p>or its ability to withstand</p><p>mechanical stresses due to high short-circuit current and (2) the inter-</p><p>rupting rating or its ability t,o interrupt the flow of short-circuit current</p><p>within its interrupting element.</p><p>What Comprises the Circuit-breaker-rating Structure. Circuit-</p><p>breaker-rating structures are revised and changed from time to time. It</p><p>is suggested that where specific problems require the latest information on</p><p>circuit-breaker ratings the applicahlc American Standards Association</p><p>(ASA), National Electrical Manufacturers Association (XEMA), or</p><p>American Instituteof Elect,rical Engineers (AIEE) standards he referred to.</p><p>To illustrate the various factors that comprise the circuit-breaker-</p><p>rating structure, an oilless power circuit breaker for metal-clad switchgear</p><p>rated 4.16 kv 250 mva* has been chosen. The complete rating is shown</p><p>on line 5, Table 1.1. The following will explain the meaning of the several</p><p>columns of Table 1.1, starting at the left. The rircuit-breaker-type</p><p>designation, column 1, varies among manufacturers. For the sake of com-</p><p>pleteness the General Electric Company nomenclature is used in this col-</p><p>umn. The remainder of the items are uniform throughout the industry.</p><p>1. Type of Circuit Breaker (AM-4.16-250)</p><p>AM = magne-blast circuit breaker</p><p>4.16 = for 4.16-kv class of circuits (not applicable to 4800- and 4800-</p><p>volt circuits)</p><p>250 = interrupting rating in mva a t 4.16 kv</p><p>2-4. Voltage Rating</p><p>2. Rated kv (4.16): the nominal voltage class or classes in which the</p><p>circuit breaker is rated.</p><p>3. Maximum design kv (4.76): the maximum voltage a t which the cir-</p><p>cuit breaker is designed to operate. The 4.16-kv circuit breakers,</p><p>for example, are suitable for a 1330-volt system plus 10 per cent for</p><p>voltage regulation or 4.76 kv.</p><p>(Note: 4330 is 4% X 2500.) Some utility syst.ems operate a t 1330</p><p>volts near the substation.</p><p>4. Minimum operating kv a t rated mva (3.85) : the minimum voltage a t</p><p>which the circuit breaker will interrupt its rated mva or in this case</p><p>it is 3.85 kv. At any voltages below this value, the circuit breaker</p><p>* blegavalt-amperes (see Appendix).</p><p>t</p><p>i.</p><p>16 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>I !</p><p>I (</p><p>t a</p><p>/</p><p>I</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 27</p><p>is not designed to interrupt the rated mva but will interrupt some</p><p>value less than rated mva.</p><p>This is very significant in the rating of power circuit breakers for,</p><p>as poiuted out later, the circuit hreaker will interrupt a maximum of</p><p>only so many amperes regardless of voltage. At any voltage less</p><p>than the minimum operating voltage the product of the maximum</p><p>kiloampere interrupting rating times the kv times the square root of</p><p>3 is less than the mva interrupting rating of the circuit breaker.</p><p>5-6. Insulation Level (Withstand Test)</p><p>5 . Low-frequency rrns kv (19): the 60-cycle high-potential test.</p><p>6. Impulse crest kv (60) : a measure of its ability to withstand lightning</p><p>This is applied with an impulse generator as a and other surges.</p><p>design test.</p><p>7-9. Current Ratings in Amperes</p><p>7. Continuous 60 cycles (1200 or 2000): the amount of load current</p><p>which the circuit breaker will carry continuously without exceeding</p><p>the allowable temperature rise.</p><p>8-9. Short-time Rating</p><p>8. Momentary amperes (60,000) : the maximum rms asymmetrical cur-</p><p>rent that a circuit breaker will withstand including short-circuit cnr-</p><p>rents from all sources and motors (induction and synchronous) and</p><p>the d-c component. This rating is independent of operating voltage</p><p>for a given circuit breaker.</p><p>This is just as significant a limitation as mva interrupting rating.</p><p>It defines the ability of the circuit breaker to withstand the mechani-</p><p>cal stresses produced by the very large offset first cycle of the short-</p><p>circuit current. This rating is nnusually significant because the</p><p>mechanical stresses in the circuit hreaker vary as the square of the</p><p>current. It is the only rating that is affected by the square law, and</p><p>therefore is one of the most critical in the application of the circuit</p><p>breakers. The rating schedules of power circuit breakers are so pro-</p><p>portioned that the momentary rating is about 1.6 times the maximum</p><p>interrupting rating amperes.</p><p>9. Four-second (37,500): the maximum current that the circuit breaker</p><p>will withstand in the closed position for a period of 4 sec to allow for</p><p>relaying operating time. This value is the same as the maximum</p><p>interrupting rating amperes.</p><p>10-13. Interrupting Ratings</p><p>10. Three-phase rated mva (250): the three-phase mva which the circuit</p><p>breaker will interrupt over a range of voltages from the maximum</p><p>design kv down to the minimum operating kv. In this case the</p><p>28 SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES</p><p>interrupting rating is 250 rnva between 4.76 and 3.85 kv. The mva</p><p>to be interrupted is obtained by multiplying the kv a t which the cir-</p><p>cuit breaker operates times the symmetrical current in kiloamperes to</p><p>be interrupted times the square root of 3. The product of these must</p><p>not exceed the rnva interrupting rating a t any operating voltage.</p><p>11. Amperes a t rated voltage (35,000): the maximum total rms amperes</p><p>which the circuit breaker will interrupt a t rated voltage, i.e., in the</p><p>case of the example used above 35,000 at 4.16 kv (4.16 X 35.000 x</p><p>fi = 250 mva). These figures are rounded. This figure is given</p><p>for information only and does not have a limiting significance of</p><p>particular interest to the application engineer.</p><p>12. Maximum amperes interrupting rating (37,500) : the maximum total</p><p>rms amperes that the circuit breaker will interrupt regardless of how</p><p>low the voltage is. At</p><p>minimum operating voltage, 3.85 kv, this corresponds to 250 mva,</p><p>and, for example, a t a voltage of 2.3 kv this corresponds to 150mva.</p><p>The circuit breaker will not interrupt this much current a t all volt-</p><p>ages, i.e., i t will not interrupt this much current if the product of</p><p>current, voltage, and the square root of 3 is greater than the mva</p><p>interrupting rating. This current limit determines the minimum kv</p><p>a t which the circuit breaker will interrupt rated mva (column 4). At</p><p>any voltage lower than that given in column 4, this maximum rms</p><p>total interrupting current determines how much the circuit breaker</p><p>will interrupt in mva. Therefore, when the voltage goes below the</p><p>limit of column 4, the mva which the circuit breaker will interrupt is</p><p>lower than the rnva rating given in column 10 by an amount propor-</p><p>tional to the reduction in operating voltage below the value of column 4.</p><p>13. Rated interrupting time (8 cycles on 60-cycle basis): the maximum</p><p>total time of operation from the instant the trip coil is energized until</p><p>the circuit breaker has cleared the short circuit.</p><p>What limits the Application of Power Circuit Breakers an on inter-</p><p>rupting-and Momentary-duty Basis? In so far as applying power cir-</p><p>cuit breakers on an interrupting-duty basis is concerned i t can be seen</p><p>from the foregoing that there are four limits, none of which should be</p><p>exceeded. These must all be checked for any application.</p><p>1. Operating voltage should never at any time exceed the limit of</p><p>column 3, Table 1.1, i.e., the maximum design kv.</p><p>2. Interrupting rnva should never be exceeded a t any voltage. This</p><p>limit is sig’nificant only when the operating voltage is between the limits</p><p>of columns 3 and 4, Table 1.1. It is not significant when the operating</p><p>voltage is below the limit of column 4, Table 1.1, because maximum inter-</p><p>rupting amperes limit the mva to values less than the rnva rating.</p><p>3. Maximum interrupting rating amperes should never be exceeded</p><p>In this example, this current is 37,500 amp.</p><p>SHORT-CIRCUIT.CURRENT CALCUUTING PROCEDURES 29</p><p>even though the product of this current times the voltages times the</p><p>square root of 3 is less than the interrupting rating in mva. This figure</p><p>is the controlling one in so far as interrupting duty is involved when the</p><p>voltage is below that of</p><p>column 4, Table 1.1 (minimum operating voltage</p><p>a t rated mva).</p><p>4. Momentary current should never be exceeded a t any operating</p><p>voltage. Modern power circuit breakers generally have a momeutary</p><p>rating in rms amperes of 1.6 times the maximum interrupting rating in</p><p>rms amperes. As a result, where there is no short-circuit-current contri-</p><p>bution from motors, a check of the interrupting duty only is necessary.</p><p>If this is within the circuit-breaker interrupting rating then the maximum</p><p>Short-circuit current, including the d-c component, mill be within the</p><p>momentary rating of the circuit breaker.</p><p>Where there is short-circuit contribution from motors, the momentary</p><p>rating of the circuit breaker may be exceeded, before the interrupting</p><p>rating is exceeded in a given cirruit. Whenever there are motors to be</p><p>considered in the short-circuit calculations, the momentary duty and the</p><p>interrupting duty should both be checked.</p><p>Siuce the</p><p>short-circuit current is maximum at the first half cycle, the short-circuit</p><p>current must be determined a t the first half cycle to determine the maxi-</p><p>mum momentary duty on a circuit breaker.</p><p>To determine the short-circuit current a t the first half cycle, it is neces-</p><p>sary to consider all sources of short-circuit current, that is, the generators,</p><p>synchronous motors, induction motors, and utility connections. The</p><p>subtransient reactances of generators, synchronous motors, and inductiou</p><p>motors are employed in the reactance diagram. Since the d-r component</p><p>is present a t this time, it is necessary to account for it by the use of a</p><p>multiplying factor. This multiplying factor is either 1.5 or l.G, as out-</p><p>lined in Table 1.2. Typical circuits where the 1.5 multiplying factor can</p><p>be used are shown in Fig. 1.25. The procedure is the same, regardless of</p><p>the type of power circuit breaker involved.</p><p>To check</p><p>the interrupting duty on a power circuit breaker, the short-circuit current</p><p>should be determined a t the time that the circuit-breaker contacts part.</p><p>The time required for the circuit-breaker contacts to part will vary over a</p><p>considerable range, because of variation in relay time and in circuit-</p><p>breaker operating speed. The fewer cycles required for the circuit-</p><p>breaker contacts to part, the greater will be the curreut to interrupt.</p><p>Therefore, the maximum interrupting duty is imposed upon the circuit</p><p>breaker when the tripping relays operate instantaneously. In all short-</p><p>circuit calculations, for the purpose of determining interrupting duties,</p><p>the relays are assumed to operate instantaneously. To account for</p><p>How to Check Momentary Duty on Power Circuit Breakers.</p><p>How to Check Interrupting Duty on Power Circuit Breakers.</p><p>SEPES-DIVEN</p><p>SEN-RIO-EIELI', tCA 1</p><p>30 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>2400</p><p>4160</p><p>4800 VOLT</p><p>INCOMING</p><p>LINE FROM</p><p>U T I L I T Y</p><p>TO PLANT LOAD</p><p>( 0 ) NO GENERATION</p><p>IN THE PLANT</p><p>u.-L NO GENERATION</p><p>ON THIS BUS</p><p>2400, 4160 OR</p><p>( C ) TO LOAD</p><p>13.6 KV</p><p>HIGH VOLTAGE</p><p>INCOMING LINE</p><p>$ :,oA,,60 4600 V BUS</p><p>TO PLANT LOAD</p><p>NO GENERATION (b)</p><p>IN THE PLANT</p><p>U U</p><p>USE 1.6</p><p>MULTIPLYING</p><p>FACTOR</p><p>NO GENERATION</p><p>FIG. 1.25</p><p>circuits rated less than 5 h.</p><p>One-line diogrom of carer where the multiplying factor 1.5 may be used on</p><p>. . , c .. .: .</p><p>, , ,.. . .</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 31</p><p>variation in the circuit-breaker operating speed, power circuit breakers</p><p>have been grouped into classes, such as eight-cycle, five-cycle, three-cycle</p><p>circuit breakers, etc. It is assumed for short-circuit-calculation purposes</p><p>that circuit breakers of all manufacturers, in any one speed grouping,</p><p>operate substantially the same with regard to contact parting time.</p><p>Instead of specifying a time a t which the short-circuit current is to he</p><p>calculated, it is determined by the simpler approach of specifying the</p><p>generator and motor reactances and using multiplying factors. These</p><p>factors are listed in Table 1.2.</p><p>In industrial plants, eight-cycle circuit breakers are generally used.</p><p>Normally, the induction-motor contribution has disappeared, and that of</p><p>the synchronous motors has changed from the subtransient to the transient</p><p>condition before the contacts of these circuit breakers part. Therefore,</p><p>in calculating the interrupting duty on commonly used power circuit</p><p>breakers, generator subtransient reactance and synchronous-motor</p><p>transient reactance are used and induction motors are neglected. The</p><p>elapsed time is so long that usually all the d-c component has disappeared.</p><p>What d-c component is left is more than offset by the reduction in a-c</p><p>component due to the increase in reactance of the generators. Hence, a</p><p>multiplying factor of one (1) is used.</p><p>In very large power systems, when symmetrical short-circuit interrupt-</p><p>ing duty is 500 mva or greater, there is an exception to this rule. In such</p><p>large power systems, the ratio of reactance to resistance is usually so high</p><p>that there may be considerable d-c component left when the contacts of</p><p>the standard eight-cycle circuit breaker part. To account for this, the</p><p>multiplying factor of 1.1 is used in determining the total rms short-circuit</p><p>mva that a circuit breaker may have to interrupt in these large systems.</p><p>The multiplying factor of 1.1 is not applied until the symmetrical short-</p><p>circuit value reaches 500 mva.</p><p>High-voltage fuses are either of the current-</p><p>limiting type, Fig. 1.26, which open the circuit before the first current</p><p>peak, or of the non-current-limiting type, which open the circuit within</p><p>one or two cycles after the inception of the short circuit. For the sake of</p><p>standardization, all fuse-interrupting ratings are on the basis of maximum</p><p>rms current that will flow in the first cycle after the short circuit occurs.</p><p>This is the current that will flow if the fuse did not open the circuit</p><p>previously, i.e., fuses are rated in terms of “available short-circuit</p><p>current.”</p><p>To determine the available short-circuit current a t the first cycle for</p><p>the application of high-voltage fuses, use the subtransient reactances of</p><p>all generators, induction motors, synchronous motors, and utility sources</p><p>and allow for the maximum d-c component. The multiplying factor for</p><p>allowing for d-c component is 1.6, the same as for allowing for d-c compo-</p><p>High-voltage Fuses.</p><p>TABLE 1.2 Condensed Table of Multiplying Factors and Rotating-machine Reactances</p><p>To Be Used for CaLdatina Swt-dreuit Cunanh for Circuit-breaker, Fuse, and Motor.rtartor Applicdons</p><p>Eight cycle or slower (general case). .......... Above 600 wlh</p><p>Rva cycle.. .............................. Above 600 volt,</p><p>Any ploee where symmetricmi I .O</p><p>short-circuit kva i s loss than 1.1</p><p>500 mva</p><p>........................... 1.6</p><p>Lar than 5 k.. .......................... 601 to 5000 volh Remote from generating do- 1.5</p><p>Generol GOSO.. Above 600 volt) Near generoting station</p><p>lion (X/R rotio l e u thon I01</p><p>High-voltaqe Fuses</p><p>All typos, including dl wrront-limiting fuses. .... Above 600 wih Anywhere in system I .O</p><p>w u</p><p>Interrupting duty 2</p><p>Subtransient</p><p>Subtransient</p><p>ii s s z Momentary duty</p><p>Subtransient</p><p>Subtransient</p><p>5</p><p>Three-phose Ino interrupting duly</p><p>Subhqndent 1 Transient 1 Neglect</p><p>Maximum rms ampere interrupting duty</p><p>I I</p><p>1 Generators. 1 I</p><p>0</p><p>a</p><p>w</p><p>C</p><p>frequency 1 changers I</p><p>All types, including dl current-limiting fuses.. ... Above 600 volt'</p><p>Non-current-limiting lypes only.. ............. 601 to 15,000 wlh Remote from generoting %to.</p><p>I Anywhere in system</p><p>1 tion ( X / R mtio leu lhm 41</p><p>1.6 Subtronsient Svbtronrient Svbwmrient</p><p>1 .? Subwoniiont Subhmrient Subtransient i i i</p><p>All h e p o w e r ratings.. .................... Anywhere in system 2400 and 4i60Y</p><p>Wlh</p><p>All horsepower rotingr.. .................... 2400 and 4160Y Anywhere in system</p><p>Yolh</p><p>1.0</p><p>I .6</p><p>CIrmit breaker w conladm lype. . ...........</p><p>Cirwit b r w b r or contocto~ lype. ............</p><p>Clrcvit b r e e b r or contartor type.. ...........</p><p>601 10 5000</p><p>volts</p><p>601 to MOO volts</p><p>601 lo 5000 volts</p><p>bywhere in system</p><p>lion lX/R ratio leis than 101</p><p>temote from gener.ting 11.-</p><p>m z Apparatus. 600 Volts and Below</p><p>1.6</p><p>1.5</p><p>Air circuit breakers or breaker-contactor combino.</p><p>lion motor stoners.. ....................</p><p>Low-voltacp furas or fused combination motor</p><p>Slarte" ...............................</p><p>Subtransient</p><p>Subtransient</p><p>600 volts and below</p><p>600 volt* and below</p><p>0</p><p>Subtrmdent Subtransient</p><p>Subtrmdent Subtransient 8</p><p>0</p><p>R</p><p>Interrupting or momentary duty</p><p>Anywhere in system</p><p>Anywhere in system</p><p>I .25 Subtransient Subtianrient Svbtronrienl</p><p>1 .25 Subtransient Subtransient Svbtraniient</p><p>34 SHORT-CIRCUIT.CURRENT CALCULATING PROCEDURES</p><p>nent when determining the momentary duty on a power circuit breaker</p><p>(see Table 1.2).</p><p>The interrupting rating of fuses in amperes is exactly parallel, in so far</p><p>as short-circuit+urent calculations are concerned, to the momentary</p><p>rating of power circuit breakers.</p><p>The ampere interrupting rating of high-voltage fuses is the only rating</p><p>that has any physical significance. For the sake of simplicity of applica-</p><p>tion in systems with power circuit breakers, some fuses are given inter-</p><p>rupting ratings in three-phase mva. The three-phase mva interrupting</p><p>rating has no physical significance, because fuses are single-phase devices,</p><p>each fuse functioning only on the current which passes through it.</p><p>WAVE OF AVAILABLE</p><p>THE FUSE ELEMENTS MELT</p><p>BEFORE PEAK VALUE OF</p><p>AVAILABLE SHORT CIRCUIT</p><p>CURRENT IS REACHED</p><p>1</p><p>FIG. 1.26</p><p>See Fig. 1.27 for method of determining available short-circuit current.</p><p>Grophic sxplonotion of the current-limiting action of current-limiting fuses.</p><p>SHORT-CIRCUIT-CURRENT CAKULATING PROCEDURES 35</p><p>These three-phase mva ratings have been selected so they will line up</p><p>with power-circuit-breaker ratings. For example, a high-voltage fuse</p><p>rated 150 mva and a power circuit breaker rated 150 mva can he applied</p><p>on the basis of the same short-circuit-current calculations. Of course, the</p><p>application voltage must he factored in each case.</p><p>High-voltage motor starters generally</p><p>employ for short-circuit protection either current-limiting fuses or power</p><p>circuit breakers. The short-circuit-current calculations for applying</p><p>these motor starters are the same as those for high-voltage fuses and</p><p>power circuit breakers, respectively.</p><p>High-voltoge Motor Starters.</p><p>LOW-VOLTAGE CIRCUIT PROTECTIVE EQUIPMENT (600 VOLTS AND BELOW)</p><p>low-voltage Air Circuit Breokers. The present designs of low-voltage</p><p>air circuit breakers differ from those of high-voltage power circuit break-</p><p>ers because they are substantially instantaneous in operation a t currents</p><p>near their interrupting rating. The contacts often begin to part during</p><p>the first cycle of current. Therefore, low-voltage air circuit breakers are</p><p>subject to interrupting the current a t the first cycle after short circuit and</p><p>withstanding the mechanical forces of that rurrent. It is necessary to</p><p>calculate the current a t only one time for the application of low-voltage air</p><p>circuit breakers. The current determined should be that of the first halt</p><p>cycle and should be determined on exactly the same hasis as for checking</p><p>the momentary duty of high-voltage power circuit breakers, except for a</p><p>change in the multiplying factor as discussed in the next paragraph. The</p><p>suhtransient reactances of generators, induction motors, and synrhronous</p><p>motors are used, and the d-c component is considered (see Table 1.2).</p><p>The multiplying factor for the d-c component is not so high in low-</p><p>voltage circuits as in some high-voltage circuits. This is due to the gener-</p><p>ally lower level of reactance-to-resistance ( X I R ) ratio in low-voltage</p><p>circLits, which causes the d-c component to decay faster than in some</p><p>high-voltage circuits.</p><p>In rating low-voltage air circuit breakers, the average d-c component of</p><p>the three phases is used, which is somewhat lower than that for the maxi-</p><p>mum phase.</p><p>The generally lower ( X / R ) ratio and the use of an average d-c compo-</p><p>nent for the three phases result in a considerably lower multiplying factor</p><p>in low-voltage circuits. The multiplying factor has been standardized</p><p>at 1.25 for the average for the three phases. This is equivalent to a</p><p>multiplier of about 1.5 to account for the d-c component in the maximum</p><p>phase.</p><p>Application of High-voltage Oil Circuit Breokers to 600-volt Systems.</p><p>In the 192Os, 5-kv oil circuit breakers were used extensively on 600-volt</p><p>36 SHORT-CIRCUIT-CURRENT CALCULAnNG PROCEDURES</p><p>systems. The procedure for determining short-circuit currents in sys-</p><p>tems of 600 volts and below is slightly modified for checking duty on oil</p><p>breakers of the 5-kv class as compared with low-voltage air circuit</p><p>breakers.</p><p>Both the momentary duty and interrupting duty must be checked for</p><p>the oil-circuit-breaker application. To check the momentary duty, use</p><p>the same procedure as for low-voltage air circuit breakers, i.e., generators,</p><p>utility sources, induction motors, and synchronous motors (subtransient</p><p>reactance). However, a multiplying factor of 1.5 is used instead of 1.25</p><p>as for low-voltage air circuit breakers. Oil-circuit-breaker momentary</p><p>ratings are based on the maximum current through any one pole, not on</p><p>the average current in the three phases which is employed in the rating of</p><p>low-voltage circuit breakers.</p><p>To determine the interrupting duty, use the generator subtransient</p><p>reactance and utility-source reactance plus the synchronous-motor</p><p>transient reactance and a multiplying factor of 1.0.</p><p>Low-voltage Fuses. Several low-voltage fuses with published a-c</p><p>interrupting ratings are appearing on the market. There are no industry</p><p>standards to follow, but most of these seem to be following air-circuit-</p><p>breaker standards, i.e., using the same rating base and same method of</p><p>determining short-circuit duty as is used for low-voltage air circuit</p><p>breakers. Hence, the procedure will not be repeated here except to</p><p>point out that the 1.25 multiplying factor is used (see Table 1.2).</p><p>So-called National Electrical Code (NEC) plug and cartridge fuses</p><p>have no established a-c interrupting ratings. Many tests have been made</p><p>to determine their a-c interrupting ability, but to date the industry has</p><p>not applied a-c interrupting ratings.</p><p>Low-voltage motor starters are of two</p><p>types: those using fuses and those using air circuit breakers for short-</p><p>circuit protection. Those using air circuit breakers for short-circuit</p><p>protection are applied 04 exactly the same basis as low-voltage air circuit</p><p>breakers in so far as short-circuit currents are concerned.</p><p>Motor starters using fuses for short-circuit protection are applied on</p><p>exactly the same basisas fuses in so far as short-circuit current is concerned.</p><p>Low-voltage Motor Starters.</p><p>AVAILABLE SHORT-CIRCUIT CURRENT</p><p>In determining the short-circuit current, the impedance of the circuit</p><p>protective device connected in the faulty feeder is neglected. The short-</p><p>circuit current is determined by’ assuming that the protective device is</p><p>shorted out by a bar of zero impedance (Fig. 1.27). The short-circuit</p><p>/</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 37</p><p>current which flows in such a circuit is commonly called available short-</p><p>circuit cumat. The procedure for determining the available short-circuit</p><p>current is based on setting up impedance or reactance diagrams. The</p><p>impedance of the short-circuit protective device that is nearest the short</p><p>circuit (electrically) is omitted from the impedance diagram.</p><p>Practically all protective devices are so rated and tested for short-</p><p>circuit interrupting ability; hence this procedure may be followed in</p><p>short-circuit calculations. This greatly simplifies the calculations and</p><p>removes the effect of impedance variations between different types and</p><p>makes of devices having the same interrupting rating. I t means that</p><p>one set of short-circuit-current calculations for a given</p><p>set of conditions</p><p>is all that is needed for applying any type of protective device, regardless</p><p>of the impedance of the devices themselves.</p><p>GENERATOR 0</p><p>TRANSFORMER</p><p>MOTORS</p><p>CABLE</p><p>C A B L E</p><p>SHORT</p><p>CIRCUIT</p><p>SHORT ClRCUlTED 81</p><p>JUMPER OF ZERO</p><p>IMPEDANCE</p><p>FIG. 1.27</p><p>circuit protective devices.</p><p>Connections for determining available short-circuit current for testing rhort-</p><p>38 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>HOW TO MAKE A SHORT-CIRCUIT STUDY</p><p>FOR DETERMINING SHORT-CIRCUIT CURRENT</p><p>FORMULAS FOR SHORT-CIRCUIT STUDY'</p><p>1. Changing ohms to per cent ohms, etc.:</p><p>(ohms reactance) (kva.base)</p><p>(kvt)*(lO)</p><p>(ohms reactance)(kva base)</p><p>(1.1) Per cent (%) ohms reactance =</p><p>Per-unit (90 ohms reactance = (kv)*( 1000) (1.2)</p><p>[see Eq. (1.34)]</p><p>(1.3)</p><p>(1.4)</p><p>(% reactance)(kv)2(10)</p><p>kva base Ohms reactance =</p><p>Per-unit ohms reactance =</p><p>per cent ohms reactance</p><p>100</p><p>2. Changing per cent or per-unit ohms reactance from one kva base to</p><p>another:</p><p>% ohms reactance on kva base 2</p><p>kva base 2</p><p>kva base 1 - - X (% ohms reactance on base 1) (1.5)</p><p>9f reactance on kva base 2</p><p>kva base 2</p><p>kva base 1</p><p>- - X (% ohms reactance on kva base 1) (1.36)</p><p>3. Converting utility-system reactance to per cent or per-unit ohms</p><p>a. If given in per cent ohms reactance on a kva base different than that</p><p>b. If given in short-circuit kva, convert to per-unit ohms thus:</p><p>reactance on kva base being used in study:</p><p>used in the study, convert according to Eq. (1.5).</p><p>(1.6)</p><p>kva base used in reactance diagram</p><p>short-circuit kva of utility system 9 i reactance =</p><p>c. If given in short-circuit amperes (rms symmetrical), convert t o per-</p><p>unit ohms thus:</p><p>Yi reactance =</p><p>d. If only the kva interrupting rating of the incoming line breaker is</p><p>* See pp. 54 to 57 for more prr-unit formulas</p><p>1 kv = line-to-line kilovolts.</p><p>(1.7)</p><p>kva base used in reactance diagram</p><p>(short-circuit current) (d$) (kv rating of system)</p><p>known,</p><p>SHORTT-CIRCUIT.CURRENT CALCULATING PROCEDURES 39</p><p>9f ohms reactance</p><p>(1.8) - kva base used in reactance diagram</p><p>kva interrupting rating of incoming line breaker</p><p>-</p><p>4. Determining kva base of motors:</p><p>The exact kva base of a motor = EI 43 (1.9)</p><p>where E = name-plate voltage rating</p><p>When motor full-load currents are not known, use the following kva bases:</p><p>Induction motors:</p><p>0.8-power factor synchronous motor:</p><p>1.0-power factor synchronous motors:</p><p>I = name-plate full-load current rating</p><p>kva base = horsepower rating</p><p>kva base = 1.0 (horsepower rating)</p><p>kva base = 0.80 (horsepower rating)</p><p>(1.10)</p><p>(1.11)</p><p>(1.12)</p><p>5. Changing voltage base when ohms are used:</p><p>Ohms on basis of voltage 1</p><p>- - ')* X (ohms on basis of voltage 2) (1.13)</p><p>In Eqs. (1.1) to (1.4), ohms impedance or ohms resistance may be sub-</p><p>The final product is then per-unit or per</p><p>(voltage 2)2</p><p>stituted for ohms reactance.</p><p>cent ohms impedance or resistance, respectively.</p><p>6 . Determining the symmetrical short-circuit kva:</p><p>Symmetrical short-circuit kva = ~ (kva base) (1.14)</p><p>% X *</p><p>-~ - y? '& (kva base) (1.15)</p><p>(1.16)</p><p>(1.16a)</p><p>(line-to-neutral voltage)2</p><p>ohms reactance X 1000</p><p>kv2 X lo00</p><p>ohms reactance</p><p>= 3</p><p>- - .</p><p>7. Determining the symmetrical short-circuit current:</p><p>(1.17) (100) (kva base)</p><p>(% X*)(v%(kvt)</p><p>(% X*)(&)(kvt)</p><p>Symmetrical short-circuit current =</p><p>(1.18)</p><p>(1.19)</p><p>- kva base</p><p>- - kvt X lo00</p><p>-</p><p>( d ) ( o h m s reactance)</p><p>* X = reactance or impedanoe.</p><p>t kv = line-&line kilovolts.</p><p>L 0 TABLE 1.3 Factor ( K ) to Convert Ohms to Per Cent or Per-unit Ohms for Three-phase Circuits*</p><p>216Y/125</p><p>240</p><p>480</p><p>600</p><p>2,400</p><p>4.1 60</p><p>4,800</p><p>6.900</p><p>7,200</p><p>l1,OOO</p><p>11.500</p><p>12,000</p><p>12,500</p><p>13.200</p><p>13,800</p><p>23,000</p><p>37.4M)</p><p>46,000</p><p>69,OCU</p><p>loot</p><p>P.r c*nt</p><p>'14</p><p>73</p><p>43.4</p><p>27.7</p><p>1.73</p><p>0.56</p><p>0.435</p><p>0.210</p><p>0.193</p><p>0.0825</p><p>0.0755</p><p>0.0695</p><p>0.064</p><p>0.0574</p><p>0.0525</p><p>0.0187</p><p>0.00711</p><p>0.00471</p><p>0.0021 2 -</p><p>Per-""it</p><p>2.14</p><p>1.73</p><p>0.434</p><p>0.277</p><p>0.0173</p><p>0.00576</p><p>0.00435</p><p>0.0021</p><p>0.001 93</p><p>0.000825</p><p>0.000755</p><p>0.000695</p><p>0.00064</p><p>0.000574</p><p>0.000525</p><p>0,0001 87</p><p>0.000071 I</p><p>0.0000471</p><p>0.0000212</p><p>Per cent</p><p>321.5</p><p>260.4</p><p>65.21</p><p>4.166</p><p>2.604</p><p>0.808</p><p>0.651</p><p>0.315</p><p>0.289</p><p>0.123</p><p>0.113</p><p>0.104</p><p>0.096</p><p>0.086</p><p>0.0787</p><p>0.0283</p><p>0.0107</p><p>0.00708</p><p>0.0031 5</p><p>-</p><p>1 50</p><p>Per-""it</p><p>3.215</p><p>2.604</p><p>0.6521</p><p>0.4166</p><p>0.02604</p><p>0.00808</p><p>0.00651</p><p>0.0031 5</p><p>0.00289</p><p>0.00123</p><p>0.00113</p><p>0.00104</p><p>0.00096</p><p>0.00086</p><p>0.000787</p><p>0.000283</p><p>0.000106</p><p>0.0000708</p><p>0.000031 5</p><p>Base kvo</p><p>200</p><p>Per cant</p><p>128</p><p>147</p><p>86.8</p><p>55.5</p><p>3.47</p><p>1.15</p><p>0.868</p><p>0.42</p><p>0.386</p><p>0.165</p><p>0.151</p><p>0.138</p><p>0.127</p><p>0.114</p><p>0.105</p><p>0.0378</p><p>0.0142</p><p>0.00945</p><p>0.0042</p><p>Por-un1t</p><p>4.28</p><p>3.47</p><p>0.868</p><p>0.555</p><p>0.0347</p><p>0.0115</p><p>0.00868</p><p>0.0042</p><p>0.00386</p><p>0.00165</p><p>0.00151</p><p>0.00138</p><p>0.00127</p><p>0.00114</p><p>0.00105</p><p>0.000378</p><p>0.000142</p><p>0.0000945</p><p>0.000042</p><p>300</p><p>Per cent</p><p>t43</p><p>a1</p><p>30.2</p><p>83.3</p><p>5.21</p><p>1.72</p><p>1.302</p><p>0.63</p><p>0.579</p><p>0.247</p><p>0.226</p><p>0.208</p><p>0.192</p><p>0.172</p><p>0.157</p><p>0.0567</p><p>0.0213</p><p>0.0141</p><p>0.0063</p><p>kva base , kva base</p><p>kv' X 10</p><p>* For per-unit, K = For per cent, K = kv = line-to-line kilovolts</p><p>kv' X 1wO</p><p>-</p><p>Per-""it</p><p>6.43</p><p>5.21</p><p>1.302</p><p>0.833</p><p>0.0511</p><p>0.0172</p><p>0.01302</p><p>0.0063</p><p>0.00579</p><p>0.00247</p><p>0.00226</p><p>0.00208</p><p>0.001 92</p><p>0,00172</p><p>0.001 57</p><p>0.000547</p><p>0.00021 3</p><p>0.0001 41</p><p>0.000063</p><p>500</p><p>__</p><p>Per cent</p><p>07 1</p><p>868</p><p>217</p><p>I38</p><p>8.68</p><p>2.88</p><p>2.17</p><p>1.05</p><p>0.965</p><p>0.413</p><p>0.377</p><p>0.347</p><p>0.32</p><p>0.286</p><p>0.262</p><p>0.045</p><p>0.0355</p><p>0.0236</p><p>0.0105</p><p>Per-""it v)</p><p>I</p><p>- 2</p><p>2</p><p>2.17 K</p><p>0.71</p><p>8.68</p><p>E</p><p>1.38 2</p><p>0.0868</p><p>0.0288</p><p>0.0105 n</p><p>0.00965</p><p>0.00413</p><p>0.00377 5</p><p>0.00347 f</p><p>0</p><p>0.0032</p><p>0.00286</p><p>0.00262</p><p>0.00045 c</p><p>0.000355 R</p><p>0.000236 v,</p><p>0.000105</p><p>0.0217 B -I</p><p>2</p><p>6</p><p>new base in kva,</p><p>100</p><p>t To determine multiplying factors far any other base use figures under 100-kvs base columns multiplied by</p><p>SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES</p><p>8. Determining the asymmetrical short-circuit current:</p><p>INFINITE</p><p>H BUSES</p><p>-SHORT CIRCUIT CURRENT GOES THROUGH HERE</p><p>41</p><p>(1.20)</p><p>Asymmetrical short-circuit current</p><p>Asymmetrical short-circuit kva</p><p>= (symmetrical current) (multiplying factor)</p><p>= (symmetrical kva) (multiplying factor)</p><p>DIAGRAMS</p><p>One-line Diagram. The first step in making a short-circuit study is to</p><p>prepare a one-line diagram showing all sources of short-circuit current,</p><p>i.e., utility ties, generators, synchronous motors, induction motors, syn-</p><p>chronous condensers, rotary converters, etc., and all significant circuit</p><p>elements, such as transformers, cables, circuit breakers, etc. (Fig. 1.28).</p><p>The second step is to</p><p>make an impedance or reactance diagram showing all significant react-</p><p>ances and resistances (Pig. 1.29). In the following pages this will be</p><p>Make an Impedance or Reactance Diagram.</p><p>C UTILITY SYSTEM I TRANS D GENERATOR</p><p>GENERATOR</p><p>CABLE E</p><p>SHORT</p><p>CIRCUIT</p><p>LARGE CABLE J</p><p>MOTOR</p><p>FIG. 1.28 e diagram c</p><p>480 VOLT</p><p>MOTORS</p><p>, typical large industrial power system.</p><p>FIG. 1.29 Reactonce diagram of system shown in Fig. 1.28.</p><p>42 SHORT-ClRCUIT.CURRENT CALCULAltNG PROCEDURES</p><p>referred to as an impedance diagram, recognizing of course that only</p><p>reactances will be used in many diagrams. The circuit element,s and</p><p>machines considered in the impedance diagram depend upon many</p><p>factors, i.e., circuit voltage, whether momentary or interrupting duty are</p><p>to be checked, etc.</p><p>The foregoing discussion and Table 1.2 explain when motors are to be</p><p>considered and what motor reactances are to he used for checking</p><p>the dut,y on a given circuit breaker or fuses of a given voltage class.</p><p>There are other problems, i.e., (1) selecting the type and location of the</p><p>short circuit in the system, (2) determining the specific reactance of a</p><p>given circuit element or machine, and (3) deciding whether or not circuit</p><p>resistance should be convidered.</p><p>SELECTION OF TYPE AND LOCATION OF SHORT CIRCUIT</p><p>Three-phase Short Circuits Generally Considered. I n most indus-</p><p>trial systems, the maximum short-circuit current is obtained when a</p><p>three-phase short circuit occurs. Short-rircuit-current magnitudes are</p><p>generally less for line-to-neutral</p>