This document discusses short circuit calculations according to ANSI and IEC standards. It provides information on different types of faults including three-phase bolted faults, line-to-line faults, and line-to-ground faults. It also discusses contributions to fault current from generators, motors, and transformers. Methods for calculating fault current include per unit calculations and considering the X/R ratio of circuits. Typical fault duties in the Philippine power system range from 15-2500 MVA depending on system voltage. Examples are provided to demonstrate short circuit calculations for three-phase faults at an infinite source and available fault duty.

1. SHORT CIRCUIT
CALCULATION:
Comparing the ANSI
and IEC 60909
PRESENTED BY:
Engr. Noel R. Angeles, PEE, ACPE
IIEE Senior Member

2. PURPOSE OF FAULT CALCULATIONS
• Proper selection of protective equipment ratings as circuit
breakers or fuses that suit to system requirements;
• Realistic arming up of protective relays to trigger operation
of circuit breakers once faults do occur;
• Proper coordination of operation of these protective devices
to effect selective interruptions of the only required
breakers;

3. CONTRIBUTING SOURCE
• Utility generation thru the industry substation.
• Local generation.
• Synchronous motors and synchronous condensers.
• Induction motors.

4. Generators, synchronous motors,
and induction motors all produce
short-circuit current.

5. Short-circuit current
produced by Generator

6. Short-circuit current
produced by Generator

7. Three reactances in a generator:
X”d - subtransient reactance
- apparent reactance of the stator winding at the instant short circuit occurs.
- it determines the current row during the first few cycles of a short circuit.
X’d - transient reactance
- apparent initial reactance of the stator winding, if the effect of all windings is
ignored and only the field winding considered.
- effective up to ½ sec or longer, depending upon the design of the machine.
Xd - synchronous reactance
- apparent reactance that determines the current flow when a steady-state condition
is reached.
- it is not effective until several seconds after the short circuit occurs.
- it has no value in short-circuit calculations for the applications of circuit breakers,
fuses, and contractors but is useful for relay-setting studies.

8. Why short-circuit
current asymmetrical?
Symmetrical short-circuit current and generated voltage
for zero-power-factor circuit.
Symmetrical short-circuit current:
In a circuit mainly containing reactance a short circuit occurs
at the peak of the voltage wave,
The short-circuit current would start at zero and trace a sine
wave which would be symmetrical
about the zero axis.

9. Why short-circuit
current asymmetrical?
Asymmetrical short-circuit current:
A circuit containing a small ratio of reactance to
resistance,
A short circuit occurs at the zero point of the voltage
wave, the current will start at zero.
Asymmetrical short-circuit current and generated
voltage in zero-power-factor circuit. Condition is
theoretical and is shown for illustration purposes only.

10. Why short-circuit
current asymmetrical?
If in a circuit containing only reactance, the short
circuit occurs at any point except at the peak of
the
Voltage wave, there will be some offset of the
current.
The amount of offset depends upon the point on
the voltage wave at which the short circuit occurs.
It may vary from zero to a maximum.

11. D-C component of
asymmetrical short-
circuit currents.
The asymmetrical alternating current
behaves exactly as if there were two
component currents flowing
simultaneously.
a. Symmetrical a-c component.
b. D-c component.
The initial magnitude of the d-c
component is equal to the value of
the a-c symmetrical component at
the instant of short circuit.

12. DECREMENT
If the circuit
had zero
resistance, the
direct current
would flow at a
constant value.

13. X/R RATIO OF A CIRCUIT
- Is the ratio of the reactance to the resistance of the circuit.
- The decrement or rate of decay of the d-c component is proportional to the x/r
ratio of the complete circuit from generator to short circuit.
- If x/r ratio is infinite (zero resistance), the d-c component never decays.
- If x/r ratio is zero (all resistance, no reactance), it decays instantly.

14. TYPES OF POWER SYSTEM
SHORT CIRCUITS
• Three-phase bolted shorted circuits.
• Line-to-line bolted short circuit.
• Line-to-ground bolted short circuit.
• Arcing short circuits.

15. TYPES OF POWER SYSTEM
SHORT CIRCUITS
Shunt faults
These types of short circuits are also
referred to as “ shunt faults”, since all
four are associated with fault currents
and MVA flows diverted to paths
different from the pre fault “series”
ones.

16. TYPES OF POWER SYSTEM
SHORT CIRCUITS
Three phase short circuits are often to be
the most severe of all. Thus, we perform
only three phase faults simulations when
searching for the maximum possible
magnitudes of fault currents.

17. TYPES OF POWER SYSTEM
SHORT CIRCUITS
Exception:
Single line-to-ground short circuit currents can be exceed three
phase short circuit current levels when they occur in the vicinity
of the following:
• The solidly grounded wye side of a delta-wye transformer of
the three-phase core (three-leg) design.
• The grounded wye side of a delta-wye autotransformerA
solidly grounded synchronous machine
• The grounded wye, grounded wye, delta-tertiary, three-
winding transformer

18. TYPES OF POWER SYSTEM
SHORT CIRCUITS
THREE-PHASE BOLTED SHORT CIRCUITS
- In this condition, the impedance between these conductors or
terminals is zero.
- Results in maximum thermal and mechanical stress in the system.
- Generally results in maximum short-circuit values.
LINE-TO-LINE BOLTED SHORT CIRCUIT
- Approximately 87% of three-phase bolted short circuit currents.
- Values are needed as basis in relay settings.

19. TYPES OF POWER SYSTEM
SHORT CIRCUITS
LINE-TO-GROUND BOLTED SHORT CIRCUITS
- In solidly grounded systems, can almost equal to the three-
phase bolted short circuit current.
- Seldom necessary in solidly grounded low voltage systems.
- Usually needed in medium voltage systems for relay setting
purposes.
ARCING SHORT CIRCUITS
- Much lower level short circuit current than the bolted ones at
the same fault point.
- Fall in the range from 40% to 50% of the bolted values.
- Single-line-to-ground arcing faults are the most frequent faults
experienced in any power system.

20. TYPES OF POWER SYSTEM
SHORT CIRCUITS
Series faults
Refers to one of the following system
unbalances:
• Two lines open.
• One line open.
• Unequal impedances. Unbalanced
line impedance discontinuity.

21. INDUSTRY STANDARDS FOR SHORT
CIRCUIT CURRENT CALCULATIONS
International Standard IEC 60906
North American ANSI

22. INDUSTRY STANDARDS FOR SHORT
CIRCUIT CURRENT CALCULATIONS
IEEE standards covering short circuit
current calculations for low voltage
electrical systems (below 1000 V), are:
IEEE Standard 242-1986
IEEE Standard 241-1990
IEEE Standard C37.13-1990
IEEE Standard 141-1993
IEEE standards dealing with short circuit
current calculations for medium and high
voltage electrical networks are:
IEEE Standard 141-1993
IEEE Standard C37.5-1979
IEEE Standard 241-1990
IEEE Standard 242-1986
IEEE Standard C37.010-1979

23. INDUSTRY STANDARDS FOR SHORT
CIRCUIT CURRENT CALCULATIONS
Depending on the time frame of interest considered from the origin of the fault:
First-cycle fault currents also called momentary fault currents, are the currents at 1/2 cycle
after fault initiation.
These currents pertain to the duty circuit breakers face when “closing against” or
withstanding fault currents.
Bearing in mind that low voltage circuit breakers operate in the first cycle, their breaking
ratings are compared to these currents.

24. INDUSTRY STANDARDS FOR SHORT
CIRCUIT CURRENT CALCULATIONS
GE Publications:
“a circuit breaker can’t interrupt a circuit at the instant of inception of a short. Instead, due to
the relay time delay and breaker contact parting time, it will interrupt the current after a period
of five to eight cycles, by which time the DC component would have decayed to nearly zero
and the fault will be virtually symmetrical.”

25. INDUSTRY STANDARDS FOR SHORT
CIRCUIT CURRENT CALCULATIONS
“closing a breaker against an existing fault makes it possible to intercept the peak
asymmetrical short circuit current.”
“it is for this reason that modern circuit breakers have built-in momentary rating equal to 1.6
times the symmetrical current rating for medium voltage circuit breakers and typically 1.25
times for low voltage circuit breakers.”

26. INDUSTRY STANDARDS FOR SHORT
CIRCUIT CURRENT CALCULATIONS
IEEE Standard C37.5-1979
Older rating structure are assessed on the total asymmetrical short circuit current, or total
fault MVA.
Short-circuit current calculations are bounded by minimum parting time.
IEEE Standard C37.0101979
The newer circuit breaker structure was defined on their symmetrical basis.
The symmetrical short circuit currents calculated using this method can be sufficient since
certain degree of asymmetry is included in the rating structure of the breaker depending on
the actual operation conditions and overall system X/R ratio.

27. THREE-PHASE FAULTS AT AN INFINTHREE-
PHASE FAULTS AT AN INFINITE SOURCEITE
SOURCETHREE-PHASE FAULTS AT AN
INFINITE SOURCE
FLC of 500 KVA Trafo @ 230v
= 500kva/ 3x230
= 1255 A
SC current = (100/4.5) x 1255
= 27.9 kA
SC MVA = 3 x 230V x 27.9kA
= 11.11 MVA
Isc / IFL = 27.9/1.255
= 22.23
NON-MOTOR
LOADS
500 KVA,
230V, 3 PH
%IZ = 4.5%
3 PH SHORT
CIRCUIT
INFINITE
SOURCE
ISC = ?
THREE-PHASE FAULTS AT
AN INFINITE SOURCE

28. THREE-PHASE FAULTS AT AN INFINTHREE-
PHASE FAULTS AT AN INFINITE SOURCEITE
SOURCETHREE-PHASE FAULTS AT AN
INFINITE SOURCE
FLC of 1 MVA Trafo @ 230v
= 1 MVA/ 3x230
= 2510 A
SC current = (100/5.5) x 2510
= 45.6 kA
SC MVA = 3 x 230V x 45.6kA
= 18.2 MVA
Isc / IFL = 45.6/2.5
= 18.24
NON-MOTOR
LOADS
1000 KVA,
230V, 3 PH
%IZ = 5.5%
3 PH SHORT
CIRCUIT
INFINITE
SOURCE
ISC = ?
THREE-PHASE FAULTS AT
AN INFINITE SOURCE

29. THREE-PHASE FAULTS AT AN
INFINITE SOURCE
NOTE:
1. The larger the transformer, the larger is the fault current.
2. The larger the transformer, the impedance becomes
larger too, in small increments.

30. CONVERTING CIRCUIT ELEMENTS
TO PER UNIT VALUES
1. If utility fault MVA is given:
Xu pu = MVA base used in study / Available utility fault MVA
2. For transformers:
XT pu = (%IZ/100) x (MVA base / Trafo MVA)
3. For generators and motors:
X” dG pu = (%X dG / 100) x (MVA base / generator MVA)
X” dM pu = (%X dM/100) x (MVA base / motor MVA)
4. For cables:
Xc pu = (ohms x MVA base) / (KV base)2

31. CONVERTING CIRCUIT ELEMENTS
TO PER UNIT VALUES
5. For neutral grounding resistor or reactor:
3XNGR pu = 3 x (ohms x MVA base )/ (KV base)2
6. For non-rotating circuit elements (trafo, cables, etc.):
X1 (pos seq) = X2 (neg seq) = X0 (zero seq)
7. For transmission lines:
X1 = X2 but X0 might be higher

32. CONVERTING CIRCUIT ELEMENTS
TO PER UNIT VALUES
8. For rotating machineries (generators):
Pos seq reactance, X1 = X”d (subtransient reactance)
Neg seq reactance, X2 = (X”d + Xq) / 2
Where Xq is the quadrature axis reactance of the generator
If Xd is not given, X2 = X1 = X”d
Zero seq reactance, X0 = a value significantly lesser than X1 or X2,
the value of X0 must be sought for.
XG pu = %XG / 100 x (MVA base/generator MVA) (true to X1, X2 and X0)

33. TYPICAL AVAILABLE SHORT CIRCUIT DUTIES IN PHILIPPINE SYSTEM
SYSTEM VOLTAGE
USUAL RANGE OF SC
MVA
SC MVA RECOMMENDED FOR
USE IN CALCULATIONS
SYSTEM
2.4 KV 15 -150 150
From primary unit substation internal
to the industry.
3.6 KV 20-200 200
4.16 KV 25-250 250
6.9 KV 50-500 350
13.2 KV 100-1000 500 Typically from Electric Cooperatives
13.8 KV 100-1000 500
MEPZ1, MEPZ2, Davao Light &
Power, etc.
23 KV 150-1500 750 Visayan Electric Company
34.5 KV 150-1500 1000 MERALCO, CEPALCO
69 KV 150-1500 1500 VECO, NGCP
115 KV 250-2500 2500 NGCP

34. TYPICAL TRANSFORMER DATA (UP TO 34.5 KV)
KVA %IZ KVA %IZ
100 4.0-4.5 630 4.0-4.5
150 4.0-4.5 750 5.0-5.5
160 4.0-4.5 800 5.0-5.5
200 4.0-4.5 1000 5.0-5.5
225 4.0-4.5 1250 5.0-5.5
250 4.0-4.5 1500 5.6-6.0
300 4.0-4.5 1600 5.6-6.0
315 4.0-4.5 2000 5.6-6.0
400 4.0-4.5 2500 5.6-6.0
500 4.0-4.5 3150 5.6-6.0

35. THREE-PHASE FAULTS AT AN INFINTHREE-
PHASE FAULTS AT AN INFINITE SOURCEITE
SOURCETHREE-PHASE FAULTS AT AN
INFINITE SOURCE
Let MVA base = 1 MVA
KV base = 0.23
I base = 2,510 A
UTILITY Impedance, Xu pu:
Xu pu = MVA base / utility SC MVA
Xu pu = 1 / 500
Xu pu = 0.002 pu
Transformer Impedance, XT pu:
XT pu = (%IZ/100) x MVAbase/Trafo MVA
XT pu = (5.5/100) x 1 /1
XT pu = 0.055 pu
THREE-PHASE FAULTS AT
AN AVAILABLE FAULT DUTY
NON-MOTOR
LOADS
1000 KVA,
230V, 3 PH
%IZ = 5.5% 3 PH SHORT
CIRCUIT
13.2 KV UTILITY
Available SC MVA: 500
MVA
Xu = 0.002 pu ISC = ?

36. Total Impedance, Xeq:
Xeq = 0.002 + 0.055
Xeq = 0.057 pu
Fault Current, Isc 3-ph:
Isc pu = 1.0 / 0.057
Isc pu = 17.544
Isc 3-ph = 17.544 x 2,510
Isc 3-ph = 44,035 A
Xu = 0.002 pu XT = 0.055 pu
System Driving
Voltage
Vn = 1.0 pu
I sc = 44,035 A
THREE-PHASE FAULTS AT
AN AVAILABLE FAULT DUTY

37. THREE-PHASE FAULTS AT AN INFINTHREE-
PHASE FAULTS AT AN INFINITE SOURCEITE
SOURCETHREE-PHASE FAULTS AT AN
INFINITE SOURCE
Let MVA base = 1 MVA
KV base = 0.23
I base = 2,510 A
UTILITY Impedance, Xu pu:
Xu pu = MVA base / utility SC MVA
Xu pu = 1 / 500
Xu pu = 0.002 pu
Transformer Impedance, XT pu:
XT pu = (%IZ/100) x MVAbase/Trafo MVA
XT pu = (5.5/100) x 1 /1
XT pu = 0.055 pu
X”dM pu = (%X”dM / 100) x (MVAbase / MVAM)
= (25 /100) x (1.0 / 0.3)
= 0.8333 pu
THREE-PHASE FAULTS AT
AN AVAILABLE FAULT DUTY
NON-MOTOR
LOADS
1000 KVA,
230V, 3 PH
%IZ = 5.5%
3 PH SHORT
CIRCUIT
13.2 KV UTILITY
Available SC MVA: 500
MVA
Xu = 0.002 pu
ISC = ?
MOTOR LOADS
TOTAL HP = 300
X”d = 25%

38. Total Impedance, Xeq:
Xu/T1 pu = Xu pu + XT1 pu
= 0.002 pu + 0.055 pu
= 0.057 pu
Xeq pu = 1 / (1/Xu/T1 pu + 1/X”dM pu)
= 1 / (1/0.057 pu +
1/0.8333 pu)
= 1 / (17.54 pu + 1.2 pu)
= 1 / (18.74 pu)
= 0.0534 pu
THREE-PHASE FAULTS AT
AN AVAILABLE FAULT DUTY
Xu = 0.002 pu XT = 0.055 pu
System Driving
Voltage
Vn = 1.0 pu
I sc = ?
X”dM = 0.8333 pu

39. Fault Current, Isc 3-ph:
Isc pu = 1.0 / 0.0534
Isc pu = 18.73
Isc 3-ph = 18.73 x 2,510
Isc 3-ph = 47 kA
THREE-PHASE FAULTS AT
AN AVAILABLE FAULT DUTY
Xu = 0.002 pu XT = 0.055 pu
System Driving
Voltage
Vn = 1.0 pu
I sc = 47 kA
X”dM = 0.8333 pu

40. THREE-PHASE FAULTS AT
AN AVAILABLE FAULT DUTY
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41. NOCECO
@ 13.2 KV,
3PH
52.6 MVA
Available SC
MVA
3PH FAULT
1.25/1.75 MVA,
AA/FA,CAST RESIN
TRANSFORMER,
3PH, 13.2-0.48 KV,
60HZ, 5.5% IZ
FAULT
POINT
1
Available SC Fault MVA = 52.6 MVA
(data from UTILITY)
I3p SC = 52.6 MVA / (1.732 x 13.2 KV)
I3p SC = 2.3 KA (fault pt. 1) @ 13.2 KV
SINGLE LINE DIAGRAM SHOWING FAULT AT 13.2 kV SIDE

42. 3PH FAULT
FAULT
POINT
2
NOCECO
@ 13.2 KV,
3PH
52.6 MVA
Available SC
MVA
T1:
1.25/1.75 MVA,
AA/FA,CAST RESIN
TRANSFORMER,
3PH, 13.2-0.48 KV,
60HZ, 5.5% IZ
MOTOR GROUP
Total HP = 444 HP
X”d = 25%
Fault current
from motor
contributions
NON-MOTOR
LOADS FROM
DP 3.0
NON-MOTOR
LOADS FROM
DP 4.0
NON-MOTOR
LOADS FROM
DP 5.0
T2 T3 T4
TOTAL FAULT
CURRENT
Fault current
contribution
from NOCECO
480V BUS, 60 HZ
MVA, KV, I BASES
Let,
MVAbase = 1.25 MVA
KVbase = 0.48 KV
Ibase = 1.25 MVA / (1.732 x 0.48 KV) = 1.5 KA
Xu pu = MVAbase / MVAu
= 1.25 MVA / 52.6 MVA
= 0.02376 pu
XT1 pu = (%IZ / 100) x (MVAbase / MVAT1)
= (5.5 / 100) x (1.25 / 1.25)
= 0.055 pu
SINGLE LINE DIAGRAM SHOWING FAULT AT 480V BUS

43. 3PH FAULT
FAULT
POINT
2
NOCECO
@ 13.2 KV,
3PH
52.6 MVA
Available SC
MVA
T1:
1.25/1.75 MVA,
AA/FA,CAST RESIN
TRANSFORMER,
3PH, 13.2-0.48 KV,
60HZ, 5.5% IZ
MOTOR GROUP
Total HP = 444 HP
X”d = 25%
Fault current
from motor
contributions
NON-MOTOR
LOADS FROM
DP 3.0
NON-MOTOR
LOADS FROM
DP 4.0
NON-MOTOR
LOADS FROM
DP 5.0
T2 T3 T4
TOTAL FAULT
CURRENT
Fault current
contribution
from NOCECO
480V BUS, 60 HZ
MCC
CONNECTED MOTOR
KVA
MCC 1 150
MCC 2 72
MCC 3 150
MCC 4 72
TOTAL 444
X”dM pu = (%X”dM / 100) x (MVAbase / MVAM)
= (25 /100) x (1.25 / 0.444)
= 0.7038 pu
TOTAL CONNECTED MOTORS
SINGLE LINE DIAGRAM SHOWING FAULT AT 480V BUS

44. 3PH FAULT
FAULT
POINT
2
NOCECO
@ 13.2 KV,
3PH
52.6 MVA
Available SC
MVA
T1:
1.25/1.75 MVA,
AA/FA,CAST RESIN
TRANSFORMER,
3PH, 13.2-0.48 KV,
60HZ, 5.5% IZ
MOTOR GROUP
Total HP = 444 HP
X”d = 25%
Fault current
from motor
contributions
NON-MOTOR
LOADS FROM
DP 3.0
NON-MOTOR
LOADS FROM
DP 4.0
NON-MOTOR
LOADS FROM
DP 5.0
T2 T3 T4
TOTAL FAULT
CURRENT
Fault current
contribution
from NOCECO
480V BUS, 60 HZ
EQUIVALENT REACTANCES IN pu
Xu/T1 pu = Xu pu + XT1 pu
= 0.02376 pu + 0.055 pu
= 0.07876 pu
Xeq pu = 1 / (1/Xu/T1 pu + 1/X”dM pu)
= 1 / (1/0.07876 pu + 1/0.7038 pu)
= 1 / (12.7 pu + 1.4209 pu)
= 1 / (14.12 pu)
= 0.0708 pu
SINGLE LINE DIAGRAM SHOWING FAULT AT 480V BUS

45. Van = 1.0 pu
+
N
Xu pu
= 0.02376
XT1 pu
= 0.055
X”dM pu
= 0.7038
3PH
FAULT
FAULT
POINT
2
Van = 1.0 pu
+
N
Xeq pu
= 0.0708
3PH
FAULT
FAULT
POINT
2
Van = 1.0 pu
+
N
Xu/T1 pu
= 0.07876
X”dM pu
= 0.7038
3PH
FAULT
FAULT
POINT
2
(a)
(b)
(c)
THREE PHASE SHORT CIRCUIT
Therefore,
I3P SC pu = Van / Xeq pu
= 1.0 / 0.0708 pu
= 14.12 pu
I3P SC @ 480 V bus = I3P SC pu x Ibase
= 14.12 pu x 1.5 kA
= 21.18 KA
MVA3P SC @ 480 V SWGR BUS
= 1.732 x 0.48 KV x 21.18 KA
= 17.61 MVA
SINGLE LINE DIAGRAM SHOWING FAULT AT 480V BUS

46. Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
3PH
FAULT
FAULT
POINT
3
Van = 1.0 pu
+
N
Xeq pu
= 0.05704
3PH
FAULT
FAULT
POINT
3
(a)
(b)
MVA, KV, I BASES
Let,
MVAbase = 0.3 MVA
KVbase= 0.24 KV
Ibase = 0.3 MVA / (1.732 x 0.24 KV)
Ibase = 721.7 A
Available fault MVA at 480V bus = 17.61 MVA
THREE PHASE FAULT AT THE 240V LV/LV SECONDARY SIDE
(fault point 3)

47. Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
3PH
FAULT
FAULT
POINT
3
Van = 1.0 pu
+
N
Xeq pu
= 0.05704
3PH
FAULT
FAULT
POINT
3
(a)
(b)
REACTANCES IN pu
XT2 pu = (%IZ / 100) x (MVAbase / MVAT1)
= (4.0 / 100) x (0.3 / 0.3)
= 0.04 pu
Using the 17.61 MVA available SC at the LV swgr,
XLV swgr pu = MVAbase / MVALV swgr
XLV swgr pu = 0.3 MVA / 17.61 MVA
XLV swgr pu = 0.01704 pu
THREE PHASE FAULT AT THE 240V LV/LV SECONDARY SIDE
(fault point 3)

48. Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
3PH
FAULT
FAULT
POINT
3
Van = 1.0 pu
+
N
Xeq pu
= 0.05704
3PH
FAULT
FAULT
POINT
3
(a)
(b)
EQUIVALENT REACTANCE IN pu
Xeq pu = XLV swgr pu + XT2 pu
Xeq pu = 0.01704+ 0.04
Xeq pu = 0.05704 pu
THREE PHASE FAULT AT THE 240V LV/LV SECONDARY
SIDE (fault point 3)

49. Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
3PH
FAULT
FAULT
POINT
3
Van = 1.0 pu
+
N
Xeq pu
= 0.05704
3PH
FAULT
FAULT
POINT
3
(a)
(b)
THREE PHASE SHORT CIRCUIT
Therefore,
I3P SC pu = Van / Xeq pu
= 1.0 / 0.05704 PU
= 17.53 pu
I3P SC @ 240 V = I3P SC pu x Ibase
= 17.53 pu x 721.7 A
= 12.65 KA
MVA3P SC @ 240V = 1.732 x 0.24 KV x 12.65 KA
= 5.26 MVA
THREE PHASE FAULT AT THE 240V LV/LV SECONDARY SIDE
(fault point 3)

50. Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
(a)
ZC1 pu
= 0.30328
3PH
FAULT
FAULT
POINT
4
Van = 1.0 pu
+
N
Xeq pu
= 0.36032
3PH
FAULT
FAULT
POINT
4
(b)
MVA, KV, I BASES
Let,
MVAbase = 0.3 MVA
Kvbase = 0.24 KV
Ibase = 0.3 MVA / (1.732 x 0.24 KV)
Ibase = 721.7 A
Available fault MVA at 480V bus = 17.61 MVA
FAULT DUTIES AT DOWNSTREAM OF LV/LV TRANSFORMER
(fault point 4)

51. Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
(a)
ZC1 pu
= 0.30328
3PH
FAULT
FAULT
POINT
4
Van = 1.0 pu
+
N
Xeq pu
= 0.36032
3PH
FAULT
FAULT
POINT
4
(b)
For C1:
From PEC 2017 Table 10.1.1.9 AC Resistance and Reactance for 600V Cables
R(3-ph) = 0.16 ohm/305m for 38 sq mm in steel conduit
X(3-ph) = 0.057 ohm/305m for 38 sq mm in steel conduit
Z(3-ph) = (0.162 + 0.0572) = 0.16985 ohm
Cable impedance = (0.16985 ohm/305 m) x 111 m
= 0.06181 ohms
Z C1 pu = ohms x MVAbase / KV2
base
Z C1 pu = 0.06181 x 0.3 MVA / (0.24 KV)2
Z C1 pu = 0.3219 pu
FAULT DUTIES AT DOWNSTREAM OF LV/LV TRANSFORMER
(fault point 4)

52. FAULT DUTIES AT DOWNSTREAM OF LV/LV TRANSFORMER
(FAULT PT. 4)
Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
(a)
ZC1 pu
= 0.30328
3PH
FAULT
FAULT
POINT
4
Van = 1.0 pu
+
N
Xeq pu
= 0.36032
3PH
FAULT
FAULT
POINT
4
(b)
XT2 pu = (%IZ / 100) x (MVAbase / MVAT1)
= (4.0 / 100) x (0.3 / 0.3)
= 0.04 pu
Using the 17.61 MVA available SC at the LV swgr,
XLV swgr pu = MVAbase / MVALV swgr
XLV swgr pu = 0.3 MVA / 17.61 MVA
XLV swgr pu = 0.01704 pu

53. FAULT DUTIES AT DOWNSTREAM OF LV/LV TRANSFORMER
(FAULT PT. 4)
Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
(a)
ZC1 pu
= 0.30328
3PH
FAULT
FAULT
POINT
4
Van = 1.0 pu
+
N
Xeq pu
= 0.36032
3PH
FAULT
FAULT
POINT
4
(b)
EQUIVALENT REACTANCE IN pu
Xeq pu = XLV swgr pu + XT2 pu + ZC1 pu
Xeq pu = 0.01704 + 0.04 + 0.3219
Xeq pu = 0.37894 pu

54. Van = 1.0 pu
+
N
XLV SWGR pu
= 0.01704
XT2 pu
= 0.04
(a)
ZC1 pu
= 0.30328
3PH
FAULT
FAULT
POINT
4
Van = 1.0 pu
+
N
Xeq pu
= 0.36032
3PH
FAULT
FAULT
POINT
4
(b)
THREE PHASE SHORT CIRCUIT
Therefore,
I3P SC pu = Van / Xeq pu
= 1.0 / 0.37894 PU
= 2.64 pu
I3P SC @ 240 V = I3P SC pu x Ibase
= 2.64 pu x 721.7 A
= 1905 A
MVA3P SC @ 240V = 1.732 x 0.24 KV x 1905 A
= 0.79 MVA
FAULT DUTIES AT DOWNSTREAM OF LV/LV TRANSFORMER
(fault point 4)

55. IEC 60906 - International Standard (published in 1988) distinguishes four types resulting in four different
calculated short circuit currents:
The initial short-circuit I”k
The peak short-circuit current Ip
The breaking short-circuit current Ib
The steady-state fault current Ik
NOTE:
The peak short circuit currents are the maximum fault currents reached during the first cycle from a beginning of a fault’s and are importantly different from the
first-cycle fault currents described in IEEE Standards, which are total asymmetrical RMS short circuit currents.
IEC 60906

56. IEC 60909 calculation methodology:
Maximum short circuit currents - used for sizing circuit breakers.
Minimum short circuit currents - are used for setting protective relays.
IEC 60906

57. I - rms value current
Ib - short circuit breaking current
Ik - steady state short circuit current
Ik” - initial symmetrical short circuit current
Ir - rated current of a generator
IS - design current
u - instantaneous voltage
U - network phase to phase voltage with no
load
Un - network nominal voltage with load
Sn - transformer kva rating
Scc - short circuit power
IEC 60906

58. IEC 60906

59. Different types of short-
circuits and their currents
as per IEC 60909

60. Different types of short-
circuits and their currents
as per IEC 60909

61. Consequences of short circuits
1. Damage to insulation.
2. Welding of conductors.
3. Fire and danger to life.
4. Deformation of the busbars.
5. Disconnection of cables.
6. On near-by networks, voltage dips, shutdown of a part of
the network, disturbances in control circuits, etc.

63. As per IEC 60909, calculate the following:
a. The initial short-circuit current I”k
b. The peak short-circuit current ip
during a three-phase fault at the fault point.
Available data:
a. The impedance of the connection between the
supply and transformer may be neglected.
b. Cable L is made up of two parallel cables with three
conductors each, where:
l = 4 m of 3x185 sq mm Al
ZL = (0.208 + j0.068) ohm/km
c. The short-circuit at Fault 1 is assumed to be far from
any generator.
SrT = 400 kVA
UrTHV = 20 kV
UrTLV = 410 V
Ukr = 4%
PkrT = 4.6 kW
R(0)T / RT = 1.0
X(0)T / XT = 0.95
Supply network
UnQ = 20 kv
I”kQ = 10 kA
T (Dyn5)
Cable L
l = 4 m
Un = 400 V
Fault 1
Example calculations of short-circuit using IEC 60909

64. Solution:
Impedance of the supply network (LV side)
𝑍𝑄𝑡 =
𝑐𝑄𝑈𝑛𝑄
3𝐼”𝑘𝑄
x
(𝑈𝑟𝑇𝐿𝑉)2
(𝑈𝑟𝑇𝐻𝑉)2 =
1.1 𝑥 20
3 𝑥10
x
(0.41)2
(20)2 = 0.534 mΩ
It is assumed that
𝑅𝑄
𝑋𝑄
= 0.1, hence:
XQt = 0.995ZQt = 0.531 mΩ
RQt = 0.1XQt = 0.053 mΩ
ZQt = (0.053+j0.531) mΩ
SrT = 400 kVA
UrTHV = 20 kV
UrTLV = 410 V
Ukr = 4%
PkrT = 4.6 kW
R(0)T / RT = 1.0
X(0)T / XT = 0.95
Supply network
UnQ = 20 kv
I”kQ = 10 kA
T (Dyn5)
Cable L
l = 4 m
Un = 400 V
Fault 1
Example calculations of short-circuit using IEC 60909

65. Impedance of the transformer
𝑍𝑇𝐿𝑉 =
𝑢𝑘𝑟
100
x
𝑈𝑟𝑇𝐿𝑉
2
𝑆𝑟𝑇
=
4
100
x
(410)2
400 𝑥 102 = 16.81 mΩ
𝑅𝑇𝐿𝑉 = 𝑃𝑘𝑟𝑇
𝑈𝑟𝑇𝐿𝑉
2
𝑆𝑟𝑇
2 = 4,600
(410)2
(400𝑥103 )2 = 4.83 mΩ
𝑋𝑇𝐿𝑉 = (𝑍𝑇𝐿𝑉
2 − 𝑅𝑇𝐿𝑉
2) = 16.10 mΩ
ZTLV = (4.83 + j16.10) mΩ
𝑋𝑇 = 𝑋𝑇
𝑆𝑟𝑇
𝑈𝑟𝑇𝐿𝑉
2 = 16.10x
400
4102 = 0.03831
SrT = 400 kVA
UrTHV = 20 kV
UrTLV = 410 V
Ukr = 4%
PkrT = 4.6 kW
R(0)T / RT = 1.0
X(0)T / XT = 0.95
Supply network
UnQ = 20 kv
I”kQ = 10 kA
T (Dyn5)
Cable L
l = 4 m
Un = 400 V
Fault 1
Example calculations of short-circuit using IEC 60909

66. Impedance correction factor can be calculated as:
𝐾𝑇 = 0.95
𝑐𝑚𝑎𝑥
1+06𝑥𝑇
= 0.95
1.05
1+0.6(0.03831)
= 0.975
𝑍𝑇𝐾 = 𝐾𝑇𝑍𝑇𝐿𝑉 = (4.71 + 𝑗15.70) 𝑚𝛺
Impedance of the cable
ZL = 0.5(0.208+j0.068)(4x10-3) = (0.416+j0.136) mΩ
Total impedance seen from the fault point
Zk = ZQt+ZTK+ZL = (5.18+j16.37) mΩ
SrT = 400 kVA
UrTHV = 20 kV
UrTLV = 410 V
Ukr = 4%
PkrT = 4.6 kW
R(0)T / RT = 1.0
X(0)T / XT = 0.95
Supply network
UnQ = 20 kv
I”kQ = 10 kA
T (Dyn5)
Cable L
l = 4 m
Un = 400 V
Fault 1
Example calculations of short-circuit using IEC 60909

67. Calculation of I”k and ip for a three-phase fault
𝐼”𝑘 =
𝑐𝑈𝑛
3𝑍𝑘
=
1.05(400)
3(17.17)
= 14.12 kA
𝑅
𝑋
=
𝑅𝑘
𝑋𝑘
=
5.18
16.37
= 0.316
ҟ = 1.02 +0.98𝑒−3
𝑅
𝑋 = 1.4
ip = ҟ 2 x I”k = 1.4 2 x 14.12 = 27.96 kA
SrT = 400 kVA
UrTHV = 20 kV
UrTLV = 410 V
Ukr = 4%
PkrT = 4.6 kW
R(0)T / RT = 1.0
X(0)T / XT = 0.95
Supply network
UnQ = 20 kv
I”kQ = 10 kA
T (Dyn5)
Cable L
l = 4 m
Un = 400 V
Fault 1
Example calculations of short-circuit using IEC 60909

68. Factors that affect the accuracy of short-circuit studies results:
1. Depends on the modelling accuracy,
2. System configuration.
3. Equipment impedances.
4. Modelling of electrical machines, generators, grounding point, other system components and different
operating conditions.
IEC 60906

69. 1. Different system modelling.
2. Different computational techniques.
3. IEC 60909 generally provide higher short circuit current values.
4. Short circuit DC current decrement described in IEC 60909 does not always rely on a single X/R ratio.
5. Short circuit AC current decrement considered by IEC 60909 depends on the fault location and the
standard quantifies rotating machinery’s proximity to the fault.
6. IEEE standard recommends system-wide modelling of the AC decrement.
7. Steady-state short circuit current calculation in IEC 60909 considers excitation settings of the synchronous
machines.
8. In order to be in line with IEC standards, ANSI C37.06-1987 introduced peak fault current to the preferred
ratings as an alternative to total asymmetrical currents.
Differences between the IEEE C37
and IEC 60909

72. VOLTAGE DROP CALCULATIONS

73. VOLTAGE DROP CALCULATIONS
PEC 2017 2.10.2.2(A) FPN No. 4:
Conductors for branch circuits as defined in Article 1.1, sized to prevent a voltage
drop exceeding 3% at the farthest outlet of power, heating and lighting loads, or
combination of such loads, and where the maximum total voltage drop on both
feeders and branch circuits to the farthest outlet does not exceed 5%, provide
reasonable efficiency of operation.
PEC 2017 2.15.1.2(A)(1)(b) FPN No. 2:
Conductors for feeders as defined in Article 1.1, sized to prevent a voltage drop
exceeding 3% at the farthest outlet of power, heating and lighting loads, or
combination of such loads, and where the maximum total voltage drop on both
feeders and branch circuits to the farthest outlet does not exceed 5%, will provide
reasonable efficiency of operation.

74. VD must not exceed 3%
VD must not exceed 5%
feeders
branch circuit
VD must not exceed 3%
feeders
FPN No. 4 PEC 2.10.2.2(A)
FPN No. 2 PEC 2.15.1.2(A)

75. Using Basic Formulas:
VD = (1.732) x D x I x Z (3-phase)
VD = 2 x D x I x Z (1-phase)
%VD = (VD/Vs) x 100
Where:
“VD” = voltage drop
“I” = line current, amperes
“Z” = AC impedance for 600 V cable 3-ph 60 Hz 75 deg C
(PEC 2017 Table 10.1.1.9)
“Vs” = voltage supply
“D” = distance the load is located from the power supply

77. EXAMPLE:
For 38 mm2 (1 AWG) THW, 111.72 A load, at 240 V,
located 57 m (187 ft) from DP:
From PEC 2017 Table 10.1.1.9 AC Resistance and
Reactance for 600V Cables
R(3-ph) = 0.16 ohm/305m for 38 sq mm in steel conduit
X(3-ph) = 0.057 ohm/305m for 38 sq mm in steel conduit
Z(3-ph) = (0.162 + 0.0572) = 0.16985 ohm
Cable impedance = (0.16985 ohm/305 m)
= 0.0005569 ohms/m
VD = 1.732 x I x Z x D
VD = 1.732 x 111.72 x 0.0005569 x 57
VD = 6.14 volts
Voltage at load = 240 – 6.14 = 233.86 volts
%VD = [6.14/240] x 100
%VD = 2.56%

78. From PEC 2017 Table 10.1.1.9 AC Resistance and
Reactance for 600V Cables
R = 0.49 ohm/305m for 14 sq mm in steel conduit
X = 0.064 ohm/305m for 14 sq mm in steel conduit
Z(3-ph) = (0.492 + 0.0642) = 0.4942 ohm
Cable impedance = (0.4942 ohm/305 m)
= 0.00162 ohms/m
VD = 2 x I x Z x D
VD = 2 x 65 x 0.00162 x 55
VD = 11.6 volts
Voltage at load = 230 – 11.6 = 218.4 volts
%VD = [11.6/230] x 100
%VD = 5.0 %
Ex.
A 230 volt, 65 amp
heater is located 55 m
from a panel fed with
two 14 sq mm THW
conductors. Let’s find
the voltage drop of
the circuit.

79. From PEC 2017 Table 10.1.1.9 AC Resistance and
Reactance for 600V Cables
R = 0.31 ohm/305m for 22 sq mm in steel conduit
X = 0.06 ohm/305m for 22 sq mm in steel conduit
Z(3-ph) = (0.312 + 0.062) = 0.31575 ohm
Cable impedance = (0.31575 ohm/305 m)
= 0.001035 ohms/m
VD = 2 x I x Z x D
VD = 2 x 65 x 0.001035 x 55
VD = 7.4 volts
Voltage at load = 230 – 7.4 = 222.6 volts
%VD = [7.4/230] x 100
%VD = 3.2 %
Ex.
A 230 volt, 65 amp
heater is located 55 m
from a panel fed with
two 22 sq mm THW
conductors. Let’s find
the voltage drop of
the circuit.

80. THANK YOU…