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affected by frequency, while reactance varies directly with frequency. Single-phase Circuits. UhIN SOURCE BUS 48O"OLTS ,.P"**E 6 0 C I C L E S SHORT CIRCUIT C W l R E N T i 4CCOOAYP CABLES IN PARALLEL 250 YCY 3IC INTERLOCKED ARMOR FIG. 1.55 System diagram used as on example to illustrate the determination of short-circuit currenk a t the end of feeder circuits. SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 85 Short-circuit current at bus A? Example of Application-Fig. 1.55. Source short-cirruit current = 40,000 amp Equivalent single cable feeder length = 1595 = 75 ft From curve Fig. 1.52 (4GO-volt short-circuit current scale; 250-MCM feeder Irngt,h scale) : Contribut,ion via feeder cable = 23,000 amp Motor contribution, bus A = 5 X 310 = 1,550 24,550 amp Motor contribution, bus R = 5 X 03 = 315 Short-circuit current bus A = 24,8G5 amp ~ ~ Short-circuit current a t bus B? Source short-circuit current for section 2 = 24,550 amp (say 25,000) Feeder lengt,h = 75 ft From curve (4GO-volt short-circuit current scale) interpolate between the 7 5 f t point on ;To. 210 and KO. 4 feeder length scales-Ko. 2 about one- third of the way from Xo. 4 to No. 2/0. Contribution via feeder cable Motor ront,ribution, bus R = 5 X G3 = Short-circuit current bus R = 11,000 amp 315 = 11,315 amp ~ CIRCUIT ANALYSIS-GENERAL CASE The circuit, problem involved in resolving short-circuit-current magni- tudes in low-voltage feeder systems is outlined in Fig. 1.56. I n general, low-voltage short-circuit current,s are expressed in terms of three-phase average asymmetrical rms amperes during the first cycle of currcnt flwv. Since main low-voltage source systems exhibit an X / R rat,io of about, 10, it, is standard convention t o multiply the symmet,rical short,-rirruit, current, by 1.25 to obtain the short-circuit current a t the main buses (this corresponds with an X / R ratio of 12) (see Table 1.2). Therefore, at the main bus 1.25 E Short-circuit current = 1.25 X I symm = - X - v5 z* 1.25 E & short-circuit current z , = - x Considering the source system X / R ratio = 12 1.25 E z . = - x 4 short-circuit current (A + jl) = R. + j X . 86 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 2, (obtained from reference tables) = R, + j X , 2, (impedance to end of feeder) = R, + R, + j ( X . + X I ) X, /R, ratio a t end of feeder = x. + XI - x, _ - R. + Izi Rt M is the factor to account for d-c offset and is a direct function of the X , / R , ratio XdRt ratio.. ......... 2 K ................... I 1;;s 1 I:* I 1:s I l!l I I i 6 1 1.02 I, is the local motor contribution, and the three-phase average assym- metrical rms value may be taken as five times the motor full-load rated amperes. Available short-circuit current at X = I, (three-phase avg assym- metrical rms) + I, 61 2 -SOURCE SVSTEY IMPEDANCE '1 Rg+ j X s OHMS/PHASE I MAIN LOW-VOLTaGE BUS FEE0ER:Zf:Rf t j x f OHMSIPHASE IFROH TABLES) VAIL4ELE SHORT ClRCUlT CURRENT DESIRED HERE FIG. 1.56 One-line diagram for rhort- circuit-current calculation ot the end of IS'CURRENT CONTRIBUTION FROM ly*CURRENT CONTRlBUTlON FROM SOURCE *"STEM LCCAL YOTORS feeder circuits-genernl core. : \J 4 'I LOAD SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 07 TABLES SHOWING EFFECl OF CABLE LENGTH Another useful way of showing the effect of length of cable in reducing short-circuit currents is given in the Tables 1.7 to 1.10. These show how much cable length is required to reduce the short-circuit current from one protective-device rating level to another for circuits GOO volts and less. Standard protective-device rating levels are: 100,000 amp rms asymmetrical 75,000 amp rms asymmetrical 50,000 amp rms asymmetrical 25,000 amp rms asymmetrical 15,000 amp rms asymmetrical 5000 amp rms asymmetrical The tables show how long a cable with a given con<w%or size is required to reduce the short-circuit current from 25,000, 50,000, or 100,000 amp to 5000, 15,000, 25,000, and 50,000 amp. The tables give the length L of cable a t various voltages which would change the available short-circuit current from I , to I , where I . is the available short circuit a t the source end of the cable and I , the short- circuit current a t the end of the cable of length L. These calculations were based on the assumptions that the impedance hack of the beginning of the cable is primarily reactive and that the fault i s symmetrical for all three phases. 4 R X CABLE I I I FIG. 1.57 Equivalent circuit for determining cable lengths given in Tables 1.6 lo 1.9. From the equivalent circuit per phase shown in Fig. 1.57 and using the nomenclature of Fig. 1.57, a general expression for the length of cable to limit the short-circuit current can be derived. The equation is L ~ Z 2 1 , 2 / I , 2 - R2 - X E = 221, Where I J I , is large or R is small, the equation reduces to 88 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES In these equations R is the resistance, X is the reactance, and Z is the impedance of the cable per unit length. For any voltage not given, the length a t the new voltage is t o the length a t a given voltage as the new voltage is to the given voltage, i.e., the length is directly proportional to the voltage 4.7 7.4 11.7 18.3 28.0 42.6 63.0 75.2 87.2 100.8 114.2 127.8 135.5 where L, = length a t voltage E, Lo = length a t voltage Eo The lengths L for all conductor sizes from No. 1 Awg to 250 MCM were put in the table for comparative purposes. There are certain minimum sizes of conductors and hence certain minimum lengths of cable necessary a t various values of I, to keep the cable from being damaged before the protective circuit breaker operates. Referring to Chap. 3, i t will be noted, for example, that a t 50,000 amp (I,, Tables 1.7 to 1.10) the mini- mum size cable which a 50,000-amp interrupting rating low-voltage air circuit breaker will protect is No. 4/0 Awg. Hence, the only values in the right-hand column of Tables 1.7 to 1.10 that have any practical signifi- cance are the two at the bottom of the column. The values above that are of academic interest only. TABLE 1.7 Lirnitina Effect of Cable on Short-circuit Currents at 400 Volts. 27.5 43.6 69.5 110.0 171.5 265.5 407.0 497.0 606.0 723.0 852.0 990.0 1072.0 Conductor size 109.4 170.3 263.0 402.0 No. 14 A x g . . No. 12 Awg . . No. 10 Awg. . No. 8 Awg . . . No. 6 Awg . . . N e . 4 A w g ... No. 2 Awg. . . No. I Awg , , . No. 110 Awg . No. 2 '0 A x g . No. 3/0 Awg . No. 4/0 Awg . 250MCM.. . . 34.2 53.0 81.0 122.1 I , = avail 488.0 592.0 706.0 827.0 960.0 038.0 Coble length 1, ft 146.8 175.0 206.0 237.5 271.0 290.5 - 26.9 42.6 67.5 106.5 165.0 254.0 384.5 468.0 564.0 664.0 775.0 890.0 962.0 - 7.3 11.4 17.9 28.0 42.6 63.7 91 .O 111.0 126.8 144.8 162.0 180.0 190.5 le short-circu: I f = short-circuit current :::: I 1::; 69.0 21.8 9.1 14.4 22.8 36.0 56.1 86.3 131 .O 159.1 192.8 228.5 267.0 308.0 333.0 - 5.3 8.5 13.4 21.1 32.7 43.8 75.8 91 .4 110.7 128.8 149.1 171.2 184.1 - __ 2.4 3.8 5.9 9.3 14.2 21.4 31.8 38. I 44.0 50.9 55.4 64.7 69.0 - urrent in kiloamperes a t source end of cable kiloamperes ior short circuit a t end of cable of length L SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 89 TABLE 1.8 limiting Effect of Cable on Short-circuit Currents at 480 Volts. Three Phase Three Single-conductor