77
9.5 Calculation Procedures
Table 9.5 Necessary Equations
9.5.1 Computation Methods
Regardless of method, the aim is to obtain the total
impedance to the short-circuit point. One of two com-
mon methods is used, depending upon whether a
percentage or ohmic value is required.
1. Percentage Impedance Method
This method is convenient in that the total can be
derived by simply adding the individual impedances,
without the necessity of conversion when a voltage
transformer is used.
Since impedance is not an absolute value, being
based on reference capacity, the reference value must
first be determined. The reference capacity is normally
taken as 1000kVA; thus, the percentage impedance
at the transformer capacity, the percentage imped-
ance derived from the power supply short-circuit ca-
pacity, and also the motor impedance must be con-
verted into values based on 1000kVA (Eqs. 13 and
14). Also, the wiring and bus-duct impedances that
are given in ohmic values must be converted into per-
centage impedances (Eq. 12).
2. Ohmic Method
In calculating short-circuit currents for a number of
points in a system, since the wire and bus-duct im-
pedances will be different in each case, it is conve-
nient to use Ohm’s law, in that if, for example, the
total supply impedance (Z
s
) is derived as an ohmic
value, the total impedance up to the short-circuit point
can be obtained by simply adding this value to the
wire and bus-duct impedances, which are in series
with the supply. For total 3-phase supply impedance
(Z
s
), refer to Table 9.4 (which shows calculations of
Z
s
based on standard transformers) to eliminate
troublesome calculations attendant to the motor im-
pedance being in parallel with Z
s
.
9.5.2 Calculation Examples
1. 3-phase Circuit
For the short circuit at point S in Fig. 9.5, the equiva-
lent circuit will be as shown in Fig. 9.6. The 3-phase
short-circuit current can be obtained by either the %-
impedance method or Ohm’s law, as given in Table
9.6.
3-phase
Impedance
Ohmic method
% impedance method
Remarks
I
as
= K
3
·
I
s
.......................Eq. 4
Key
I
s
V
Z
I
as
P
%Z
I
B
K
3
3-phase short-circuit current (A, sym)
Line-line voltage (V)
Circuit impedance (1-phase component)
3-phase short-circuit current (A, asym.)
Reference capacity (3-phase component, VA)
% impedance of circuit (single-phase component, %)
Reference current (A)
3-phase asymmetrical coefficient
:
:
:
:
:
:
:
:
Á
Conversion from percentage value to
ohmic value
Where P is the capacity at which %Z
was derived.
Á
Power supply impedance seen from
primary side
Á
Supply impedance seen from second-
ary side
Á
Conversion from ohmic value to per-
centage value
Á
Eq. 2 is derived from Eqs. 1, 1' and 2'.
Á
Eq. 3 is derived from Eqs. 1 and 1'.
Á
Because Eq. 1 can be obtained from
Eqs. 2 and 12, it can be seen that
I
s
of
the % impedance method is not affect-
ed by the selection of the reference ca-
pacity.
Á
The single-phase short-circuit current
in a 3-phase circuit is
M
3/2 times the 3-
phase short-circuit current. Conse-
quently, a 3-phase circuit can be exam-
ined via the 3-phase short-circuit
current.
Á
Eqs. 9 and 12 are derived from Eqs. 1'
and 2', and Eqs. 3' and 4'.
Á
As the supply impedance is defined as
100% at short circuit capacity, for Eq.
13 conversion to reference capacity is
made.
Á
When the supply short-circuit capacity
is unknown, the impedance is taken as
0.0040+j0.0999 (%) for 3-phase sup-
ply, and 0.0080+j0.1998 (%) for a 1-
phase supply (see Table 9.6).
Á
The motor and transformer impedanc-
es are converted from %Z at their
equipment capacities into %Z at refer-
ence capacity, using Eq. 14.
Á
Eq. 14 for motor impedance becomes
Transformer impedance, motor im-
pedance:
Á
Conversion to %Z at reference capac-
ity
Power-supply impedance:
I
S
=
..................................Eq. 1
M
3 · Z
V
Z =
· %Z x 10
–2
Ω
........................Eq. 9
P
V
2
%Z =
· Z x 100%......................Eq. 12
V
2
P
%Z =
x 100......Eq. 13
short-circuit capacity
reference capacity
%Z =
x
............................................Eq. 14
equipment capacity
reference capacity
%Z at equip-
ment capacity
(4.11 + j24.66) x
(For details see Table 9.6.)
equipment capacity
reference capacity
Z =
.............Eq. 10
short-circuit capacity
(primary voltages)
2
Z =
primary-side
power supply x
impedance
(
)
2
..........................................Eq. 11
primary voltage
secondary voltages
I
S
=
x 100 .................Eq. 2
M
3 · V · %Z
P
%Z =
x 100 ........................Eq. 1'
V/
M
3
I
B
· Z
P =
M
3 · V ·
I
B
...............................Eq. 2'
=
x 100 .............................Eq. 3
%Z
I
B
K
3
=
{
1 + 2e
e
+ 2
1 +
}
x
2
π
R
–
x
2
π
R
–
3
1
2
1
