# OİL AND DRY TYPE TRANSFORMERS SHORT-CIRCUIT IMPEDANCE APPLİCATİONS

SHORT-CIRCUIT IMPEDANCE
Users have sometimes particular requirements regarding the short-circuit impedance. Such
requirements may be determined by:
• parallel operation with existing units,
• limitation of voltage drop,
• limitation of short-circuit currents.
The transformer designer can meet the requirements in different ways:
• The size of the core cross-section. A large cross-section gives a low impedance and vice
versa,
• A tall transformer gives a low impedance and vice versa.
For each transformer there is, however, a smaller range which gives the optimum transformer from
an economic point of view, that is the lowest sum of the manufacturing costs and the capitalised
value of the losses.
Regarding smaller distribution transformers a high degree of standardisation in the manufacturing
may in certain cases constrain the designer’s ability to meet the requirements.
Short-circuit impedance Z is often expressed as uz in p.u. or in % according to the following
formulas:
z r r u = I ⋅ Z /U p.u. (24)
z r r u = 100 ⋅ I ⋅ Z /U % (25)
Formulas (24) and (25) are valid for for single-phase transformers, where Ir and Ur are rated values
of current and voltage on either side of the transformer. For 3-phase transformers the nominator
must be multiplied with √3.
Based on measured short-circuit voltage the value of Zk expressed in ohm can be calculated from
the following formula:

Z u U = ⋅ ⋅
= ⋅ Ω (26)
for single phase transformers and
r
2
z% r
r
z% r
S
U
100
u
3 100 I
Z u U = ⋅
⋅ ⋅
= ⋅ Ω (27)
for 3-phase transformers. Sr is the rated power of the transformer.
From formulas (26) and (27) it appears that Z expressed in ohm is different depending on which
side of the transformer it is referred to, because the ratio Ur/Ir or Ur
2/Sr is different on each side of
the transformer, unless the turn ratio is equal to 1.
It follows that
2
2
1
2
r2
r1
2
1
N
N
U
U
Z
Z
 

 

=  

 

= (28)
where the indices 1 and 2 refers to the two sides of the transformer. The relation between the
short-circuit impedance expressed in ohm referred to the two sides of the transformer is equal to
the square of the turn ratio or square of the rated voltage ratio.
Z has the active and reactive components R and jX. To find these components the load losses (PL)
are registered by means of wattmeters during the short-circuit measurement. Then
R= PL/Ir
2 Ω (29)
X = Z2 − R2 Ω (30)
The real part of the short-circuit voltage is
r
r
r U
R I
u

= p.u. (31)
100
U
R I
u
r
r
r% ⋅

= % (32)
For three-phase transformers a factor √3 is inserted in the nominator. There is a simple relation
between ur, the load losses PL and the rated power Sr of the transformer:
r
L
r r
2
r
r
r
r S
P
U I
R I
U
R I
u =

# ⋅

= p.u. (33)
The relation ur = PL/Sr is also valid for 3-phase transformers. The imaginary part of uz can be
calculated as
2
r
2
x r u = u − u p.u.
or %
(34)
X = Z2 − R2 Ω (35)
While X0 is mainly linked to the magnetic field in the core, X is linked to magnetic leakage field,
which mainly runs through the windings and the ducts between windings.

When the transformer is energized but not loaded, the practically the whole magnetic field goes in
the core. When loading the transformer with a lagging current, a part of the magnetic field is drawn
out from the core to the core window where the windings are situated. See Figure 6-10. The
magnetic flux in the core becomes smaller. Typically for distribution transformers the high voltage is
applied to the outer winding. The secondary low voltage winding is the inner winding where the
voltage decreases.
If a capacitive load is added the resulting current decreases because the capacitive current partly
compensates the inductive current. The leakage field becomes smaller and the flux in the core
increases and makes the voltage drop on the secondary side smaller.
If the transformer is loaded with a leading current, that is a current with a resulting reactive current,
which is capacitive, this capacitive load supplies reactive power to the transformer. The magnetic
flux in the core and the secondary voltage increase.
The reactive power of the magnetic energy of the leakage field is

Q = ω⋅ LI2 = 2πfLI2 = XI2 VA(r) (36)
Here I is the load current flowing in either of the windings and X the reactance in ohm referred to
same side of the transformer as I.
X can be calculated based on the geometric dimensions of the windings and the number of turns.
For a simple two-winding arrangement the formula is:
k N 2 f
h
L L
)
3
t t
X 4 10 (t 2
R
w
1 2 1 2
12
7 ⋅ π

• +
= π ⋅ − +
Ω (37)
Dimensions in meter. N is the number of turns in the winding on the side to which X shall be
referred.

In the formula (37) L1 and L2 are the mean circumference for the inner and outer windings
respectively. The formula is based on the assumption that the magnetic field lines goes vertically
along the whole height hw of the windings. This is in reality not the case. It can be seen from Figure
10 that some of the field lines towards the ends of the windings have an increasing horizontal or
radial component. The German professor W. Rogowski made a careful study on how to
compensate for this deviation from the assumption in the formula. The result of this study
(published early in the last century) was a correction factor,

1 (38)
This factor has is known as the the Rogowski factor. In most practical cases its numerical value is
within the range 0,95 – 0,99. It makes the length of the leakage field duct a little longer, hw/kR
The formula for ux expressed as a percentage of the rated voltage Ur is
( ) 2
t0
R
w
1 2 1 2
x 12 10
50
f
e
IN
k
h
L L
)
3
t t
u 2(t − + +
= + % (39)
where et0 is the number of volts per turn in no load condition.
Corresponding hand formulas exist also for more complicated winding arrangements. The shortcircuit
reactance or short-circuit voltage calculated by means of these formulas comply very well
with the measured values. However, for high currents an addition for the reactance in the lead