237
©2014 Eaton
ShortCircuit Current Calculations
Basic PointtoPoint Calculation Procedure
Step 1.Determine the transformer full load amps (F.L.A.) from
Multiplier = 100*%Z
transformer
3Ø Faultsf =1.732 xL xI
3Ø
C x n x E
LL
1Ø LinetoLine (LL) Faults 2 xL xI
LL
See Note 5 & Table 3
f =C x nxE
LL
1Ø Lineto
N
eut
r
al (L
N
) Faults2 xL xI
L
N
†
See Note 5 & Table 3
f =C x n xE
L
N
Wh
e
r
e
:
L=
length (feet) of conductor to the fault.
C=
con
s
tant fro
m
T
a
b
le
4
of
“C”
v
alue
s
for conductor
s
and
T
a
b
le
5
of
“C”
v
alue
s
for
b
u
sw
a
y
.
n=
N
u
mb
er of conductor
s
p
er
p
ha
s
e (ad
j
u
s
t
s
C
v
alue for
p
arallel run
s
)
I=
A
v
a
i
la
b
le
s
hort

c
i
rcu
i
t current
i
n a
m
p
ere
s
at
b
eg
i
nn
i
ngof c
i
rcu
i
t.
E=
V
oltage of c
i
rcu
i
t.
MAINTRANSFORMERH.V. UTILITYCONNECTIONI
S.C. primary
I
S.C. secondary
I
S.C. secondary
I
S.C. primary
M =11 +f
I
S.C. sym. RMS
= I
S.C.
xM
3Ø Transformer(I
S.C. primary
andf =I
S.C. primary
x V
primary
x1.73 (%Z)
I
S.C. secondary
are100,000
xkVA
transformer
3Ø fault values)1Ø Transformer
(I
S.C. primary
and
I
S.C. secondary
aref =
I
S.C. primary
x V
primary
x (%Z)1Ø fault values:100,000
xkVA
transformer
I
S.C. secondary
is LL)
M = 11 +f
I
S.C. secondary
=V
primary
xM xI
S.C. primary
V
secondary
either the nameplate, the following formulas or Table 1:Step 2.Find the transformer multiplier. See Notes 1 and 2
* Note 1.
Get %Z from nameplate or Table 1. Transformer impedance (Z) helps to determine what the short circuit current will be at the transformer secondary.Transformer impedance is determined as follows: The transformer secondary is shortcircuited. Voltage is increased on the primary until full load current flows in the secondary. This applied voltage divided by the rated primary voltage (times 100) is theimpedance of the transformer.Example: For a 480 Volt rated primary, if 9.6 volts causes secondary full load current toflow through the shorted secondary, the transformer impedance is 9.6/480 = .02 = 2%Z.
* Note 2.
In addition, UL 1561 listed transformers 25kVA and larger have a ± 10%impedance tolerance. Short circuit amps can be affected by this tolerance. Therefore, for high end worst case, multiply %Z by .9. For low end of worst case, multiply %Z by 1.1.Transformers constructed to ANSI standards have a ±7.5% impedance tolerance (twowinding construction).
Step 3.Determine by formula or Table 1 the transformer letthrough shortcircuit current. See Notes 3 and 4.I
S.C.
= Transformer
F.L.A.
x Multiplier
Note 3.
Utility voltages may vary ±10% for power and ±5.8% for 120 Volt lighting services. Therefore, for highest short circuit conditions, multiply values as calculated instep 3 by 1.1 or 1.058 respectively. To find the lower end worst case, multiply results instep 3 by .9 or .942 respectively.
Note 4.
Motor short circuit contribution, if significant, may be added at all fault locationsthroughout the system. A practical estimate of motor short circuit contribution is to multiply the total motor current in amps by 4. Values of 4 to 6 are commonly accepted.
Step 4.Calculate the f factor.Step 6.Calculate the available short circuit symmetrical RMS current at the point of fault. Add motor contribution, ifapplicable.Step A.Calculate the f factor (IS.C. primary known)Step B.Calculate M (multiplier).Step C.Calculate the shortcircuit current at the secondary of thetransformer. (See Note under Step 3 of Basic PointtoPoint Calculation Procedure .)
† Note 5. The LN fault current is higher than the LL fault current at the secondary terminals of a singlephase centertapped transformer. The shortcircuit current available(I) for this case in Step 4 should be adjusted at the transformer terminals as follows: AtLN center tapped transformer terminals, ILN= 1.5 x ILLat Transformer Terminals.
At some distance from the terminals, depending upon wire size, the LN fault current is lower than the LL fault current. The 1.5 multiplier is an approximationand will theoretically vary from 1.33 to 1.67. These figures are based on change inturns ratio between primary and secondary, infinite source available, zero feet fromterminals of transformer, and 1.2 x %X and 1.5 x %R for LN vs. LL resistance andreactance values. Begin LN calculations at transformer secondary terminals, thenproceed pointtopoint.
Step 5.Calculate M (multiplier) or take from Table 2.Step 6A.Motor short circuit contribution, if significant, may beadded at all fault locations throughout the system. A practical estimate of motor short circuit contribution is tomultiply the total motor current in amps by 4. Values of 4to 6 are commonly accepted.
Calculation of ShortCircuit Currents When Primary Available ShortCircuit Current is Known
Use the following procedure to calculate the level of fault current at the secondaryof a second, downstream transformer in a system when the level of fault current atthe transformer primary is known.
238
©2014 Eaton
ShortCircuit Current Calculations
ThreePhase Short Circuits
M
213
System A
Available UtilityInfinite Assumption1500 KVA Transformer480V, 3Ø, 3.5%Z,3.45% X, 0.56%RI
f.l.
=1804A25’  500kcml Cu3 Single Conductors6 Per PhaseMagnetic Conduit2000A SwitchKRPC 2000SP Fuse400A SwitchLPSRK400SP Fuse50’  500 kcmil Cu3 Single ConductorsMagnetic ConduitMotor Contribution*
System B
Available UtilityInfinite Assumption1000 KVA Transformer480V, 3Ø, 3.5%Z,I
f.I.
=1203A30’  500kcml Cu3 Single Conductors4 Per PhasePVC Conduit1600A SwitchKRPC 1500SP Fuse400A SwitchLPSRK350SP Fuse20’  2/0 Cu3 Single Conductors2 Per PhasePVC Conduit225 KVA Transformer208V, 3Ø1.2%Z
Fault X
1
Step 1.I
f.l.
=1500
X
1000= 1804.3A480
X
1.732Step 2.Multipler 100= 31.7463.5
X
0.9
†
Step 3. I
s.c.
= 1804.3
X
31.746 = 57,279AI
s.c. motor contribution**
= 4
X
1804.3 = 7217AI
total s.c. sym RMS
= 57,279 + 7217 = 64,496A
Fault X
1
Step 1.I
s.c.
=1000
X
1000= 1202.8A480
X
1.732Step 2.Multipler =100 = 31.7463.5
X
0.9
†
Step 3. I
s.c.
= 1202.8
X
31.746 = 38,184A
Fault X
2
Step 4.
f
=1.732
X
25
X
57,279= 0.038822,185
X
6
X
480Step 5.M =1 = 0.96261 + 0.0388Step 6. I
s.c. sym RMS
= 57,279
X
0.9626 = 55,137AI
s.c. motor contribution**
= 4
X
1804.3 = 7217AI
total s.c. sym RMS
= 55,137 + 7217 = 62,354A
Fault X
2
Step 4.
f
=1.732
X
30
X
38,184= 0.038726,706
X
4
X
480Step 5.M =1 = 0.96271 + 0.0387Step 6. I
s.c. sym RMS
= 38,184
X
0.9627 = 36,761A
Fault X
3
Step 4.
f
=1.732
X
50
X
55,137= 0.448422,185
X
1
X
480Step 5.M =1 = 0.69041 + 0.4483Step 6. I
s.c. sym RMS
= 55,137
X
0.6904 = 38,067AI
s.c. motor contribution**
= 4
X
1804.3 = 7217AI
total s.c. sym RMS (X3)
= 38,067 + 7217 = 45,284A
Fault X
3
Step 4.
f
=1.732
X
20
X
36,761= 0.11612
X 11,424X
480Step 5.M =1 = 0.89601 + 0.1161Step 6. I
s.c. sym RMS
= 36,761
X
0.8960 = 32,937A
Fault X
4
Step A.
f
=32,937
X
480
X
1.732
X
(1.2
X
0.9) = 1.3144100,000
X
225Step B.M =1 = 0.43211 + 1.3144Step C. I
s.c. sym RMS
= 480
X
0.4321
X
32,937 = 32,842A208
*See note 4 on page 240.**Assumes 100% motor load. If 50% of this load was from motors. I
s.c. motor contrib.
= 4
X
1804 X 0.5 = 3,608A
†
See note 2 on page 240
1234
OneLine DiagramOneLine Diagram
This example assumes no motor contribution.
239
©2014 Eaton
ShortCircuit Current Calculations
SinglePhase Short Circuits
Short circuit calculations on a singlephase center tapped transformer systemrequire a slightly different procedure than 3Ø faults on 3Ø systems.
1. It is necessary that the proper impedance be used to represent the primary system.For 3Ø fault calculations, a single primary conductor impedance is used from thesource to the transformer connection. This is compensated for in the 3Ø short circuitformula by multiplying the single conductor or singlephase impedance by 1.73.However, for singlephase faults, a primary conductor impedance is considered fromthe source to the transformer and back to the source. This is compensated in thecalculations by multiplying the 3Ø primary source impedance by two.2. The impedance of the centertapped transformer must be adjusted for the halfwinding (generally linetoneutral) fault condition.The diagram at the right illustrates that during linetoneutral faults, the full primarywinding is involved but, only the halfwinding on the secondary is involved.Therefore, the actual transformer reactance and resistance of the halfwinding condition is different than the actual transformer reactance and resistance of the fullwinding condition. Thus, adjustment to the %X and %R must be made when considering linetoneutral faults. The adjustment multipliers generally used for thiscondition are as follows:
ã 1.5 times full winding %R on full winding basis.ã 1.2 times full winding %X on full winding basis.
Note:
%R and %X multipliers given in “Impedance Data for Single PhaseTransformers” Table may be used, however, calculations must be adjusted toindicate transformer kVA/2.
3. The impedance of the cable and twopole switches on the system must be considered “bothways” since the current flows to the fault and then returns to thesource. For instance, if a linetoline fault occurs 50 feet from a transformer, then100 feet of cable impedance must be included in the calculation.The calculations on the following pages illustrate 1Ø fault calculations on a singlephase transformer system. Both linetoline and linetoneutral faults are considered.
Note in these examples:
a.The multiplier of 2 for some electrical components to account for the singlephasefault current flow,b.The halfwinding transformer %X and %R multipliers for the linetoneutral fault situation, and
ABCPrimarySecondaryShortCircuit
PrimarySecondaryShort CircuitL
2
N L
1
NL
1
L
2
50 Feet
Short Circuit
240
©2014 Eaton
ShortCircuit Current Calculations
SinglePhase Short Circuits
23
LinetoLine (LL) Fault
Step 1.I
f.l.
=75
X
1000= 312.5A240Step 2.Multipler =100 = 79.371.4
X
0.9
†
Step 3. I
s.c. (LL)
= 312.5
X
79.37 = 24,802A
LinetoNeutral (LN) Fault
Step 1.I
f.l.
=75
X
1000= 312.5A240Step 2.Multipler =100 = 79.371.4
X
0.9
†
Step 3*. I
s.c. (LN)
= 24,802
X
1.5 = 37,202AStep 4.
f
=2
X
25
X
24,802= 0.232922,185
X
1
X
240Step 5.M = 1 = 0.81111 + 0.2329Step 6. I
s.c. (LL) (X2)
= 24,802
X
0.8111 = 20,116
Fault X
3
Step 4.
f
=2
X
50
X
20,116= 1.75574774
X
1
X
240Step 5.M =1 = 0.36291 + 1.7557Step 6. I
s.c. (LL) (X3)
= 20,116
X
0.3629 = 7,300A
Fault X
2
Step 4.
f
=2
X
25
X
37,202= 0.698722,185
X
1
X
120Step 5.M =1 = 0.58871 + 0.6987Step 6*. I
s.c. (LN) (X2)
= 37,202
X
0.5887 = 21,900A
Fault X
3
Step 4.
f
=2
X
50
X
21,900** = 3.83234774
X
1
X
120Step 5.M =1 = 0.20731 + 3.823Step 6*. I
s.c. (LN) (X3)
= 21,900
X
0.2073 = 4,540A
Fault X
2
†In addition, UL 1561 listed transformers 25kVAand larger have a ± 10% impedance tolerance.Short circuit amps can be affected by this tolerance. Therefore, for high end worst case,multiply %Z by 0.9. For low end of worst case,multiply %Z by 1.1. Transformers constructed to ANSI standards have a ±7.5% impedance tolerance (twowinding construction).* Note 5. The LN fault current is higher than the LLfault current at the secondary terminals of a singlephase centertapped transformer. The shortcircuit current available (I) for this case in Step 4 should beadjusted at the transformer terminals as follows: At LNcenter tapped transformer terminals, ILN= 1.5 x ILLatTransformer Terminals.**Assumes the neutral conductor and the line conductor are the same size.
System A
Available UtilityInfinite Assumption75KVA, 1Ø Transformer.1.22%X, 0.68%R1.40%Z120/240V25’  500kcml CuMagnetic Conduit3 Single Conductors400A SwitchLPNRK400SP Fuse50’  3 AWG CuMagnetic Conduit3 Single ConductorsOneLine Diagram
Fault X
2
Fault X
1
Fault X
1
Fault X
1
Fault X
1