This document discusses short circuit calculations for electrical systems. It explains that short circuits can be caused by insulation failures, flashovers, physical damage or human error. Symmetrical and asymmetrical faults are described. Short circuit calculations should be performed at protective devices to determine device ratings and settings, cable sizes, and motor starting capabilities. A 6-step process for short circuit calculations is outlined, involving drawing diagrams, applying a power base, calculating impedances, and determining fault currents. Equations for converting three-phase values to single-phase are provided. An example cable calculation and fault current determination is shown.

1. By: Irvan A Estrella,
Emil Andes &
Ranier de Asis

2.  Short circuits (S.C) are generally caused by
insulation failure, flashovers, short circuits,
broken conductors, physical damage or
human error. Short circuits involving all three
phases simultaneously are of symmetrical
nature, whilst those involving only one or two
phases are asymmetrical faults. The
balanced three phase faults are normally
analysed using equivalent single phase
circuits. Use of symmetrical components
helps to resolve the asymmetrical system
faults.

3. Short-circuit calculation should be
calculated at every bus and points where
relays or any short-circuit protective device
are installed in order to:

4. 1. Determine the duty rating protective
devices and busses.
2. Determine the proper size of the cables
3. Determine the setting of the relays
4. Properly coordinate the protective device
5. Determine whether the short-circuit MVA
is sufficient to start large motors w/o
excessive voltage dip

6. UTILITY SUPPLY
LOCAL GENERATORS
SYNCHRONOUS MOTORS
INDUCTION MOTORS
FRQUENCY CHANGERS

7. CABLE, BUSBAR, CIRCUIT BREAKERS,
REACTORS
TRANSFORMERS

8. 1. DRAW THE SYSTEM SINGLE LINE
DIAGRAM INDICATING VOLTAGE IN
ALL LEVELS, EQUIPMENT RATINGS
AND THEIR PARAMETERS AND
LENGTH AND TYPE OF CABLES.

9. 2. MARK POINTS WHERE FAULT
CURRENT IS DESIRED TO BE KNOWN
3. ASSUME A COMMON POWER BASE
4. CONVERT ALL PARAMETERS FROM
ORIGINAL POWER BASE INTO THE
NEW COMMON POWER BASE.
5. DRAW THE REACTANCE DIAGRAM
INDICATING PARAMETERS OF
EQUIPMENT DEVICES & CKT
ELEMENTS AS COMPUTED IN STEP 4.

10. 6. PERFORM THE OPERATION OF CKT
REDUCTION UNTIL SINGLE IMPEDANCE
IS LEFT FROM THE SOURCES OF FAULT
CURRENT UP TO THE POINT WHERE
FAULT CURRENT IS DESIRED TO BE
KNOWN. (IN THE PROCESS OF THE CKT
REDUCTION WHERE RESISTANCE IS
NEGLIGIBLE RELATIVE TO THE
INDUCTANCE, IMPEDANCE OF SERIES
CKT CAN BE RAGARDED AS INDUCTIVE
RACTANCE W/O APPLICABLE ERRORS.)

11. The voltage used in a short circuit
calculation is the line to line value.
The cable parameter to be used in a three
phase short ckt calculation is the line to
neutral while in a single phase must be line
to line value
The power base must be used throughout
the calculation at all voltage levels and is a
three-phase value.

12.  Utility power sc duty is given in three phase
value. For single phase sc application, the
computed per unit value based on a common
power base from three phase supply shall be
multiplied twice to get the per unit value.
 Impedance info. available from three phase
x’formers is the per unit value on the base
determined by its rating.
 Impedance in per unit for the 3 phase unit is
the same as that for each individual x’former.

13. In a 3-phase system;
and %Z voltage is
POWERBASEZIIVP LLL  2
3 33
1.% eq
V
ZI
Z
N
L


14. Multiplying eq.1 by VN/ VN gives,
 
2
3
3
3*% 22
L
LL
L
L
L
N
LN
N
N
N
L
V
ZIV
V
ZI
V
V
ZIV
V
V
V
ZI
Z 







  2.% 2
eq
V
ZPOWERBASE
Z
L


15. Multilying eq.1 by IL/IL gives,
Where IL=IF3ф
3
3
3*%
2
LL
L
L
POWERBASE
NN
L
N
N
N
L
IV
POWERBASE
I
VIV
ZI
I
I
V
ZI
Z 







3.
3%
POWERBASE
IF3 eq
ZVL


16. Given a cable connected from 480V,
3phase, 15MVA system.
L= 200ft, R/100ft=0.026ohms,
XL/100ft=0.0048ohms
solve for %Z based on 15MVA

17.     385.3
480
052.01015
% 2
6
2

x
V
RPOWERBASE
R
L
    625.0
480
0096.01015
% 2
6
2

x
V
XPOWERBASE
X
L
 46.1044.3625.0385.3%%% jXjRZ

18. Determine the 3phase short ckt current at
the load end of the cable in example 1.

19.  
A
ZVL
4805
34807548.3
15x10
3%
POWERBASE
I
6
F3 

20.  The short-circuit MVA of an equipment is
equal to its rated MVA divided by its %Z or
%X. Example:
3phase induction motor, 0.5MVA, 2.3kV,
%X=0.25.
MVAsc=0.5MVA/0.25=2
 In sc calculation;
KVA base= HP rating (for induction motor and
80%pf sync motor)
KVA base=0.8HP rating (for unity pf sync.
Motor)