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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