Short circuit test in brief

Presented by
S.M.G. MUTHAR HUSSAIN
Transformer Engineer

 WHAT IS SHORT CIRCUIT?
 WHAT IS SHORT CIRCUIT CURRENT?
 WHAT IS THE DIFFERENCE BETWEEN SHORT CIRCUIT
AND OVERLOAD?
 WHAT CAUSES A SHORT CIRCUIT?
 WHAT IS THE EFFECT OF SHORT CIRCUIT?
 WHAT IS ABILITY TO WITHSTAND SHORT CIRCUIT TEST?
 WHY IS SHORT CIRCUIT TEST DONE ON DISTRIBUTION.
AND POWER TRANSFORMERS?
 TRANSFORMER WITH TWO SEPARATE WINDINGS
 RECOGNIZED MINIMUM VALUES OF SHORT CIRCUIT
IMPEDANCE
 SHORT CIRCUIT APPARENT POWER OF THE SYSTEM
 WHAT IS SYMMETRICAL SHORT CIRCUIT CURRENT?

 SYMMETRICAL SHORT CIRCUIT CALCULATIONS
 WHAT IS ASYMMETRICAL SHORT CIRCUIT CURRENT?
 ASYMMETRICAL SHORT CIRCUIT CALCULATIONS
 SC TEST PROCEDURES FOR TRANSFORMERS WITH 2
WINDINGS
 SC TEST SETUP IN HI-POWER LAB
 DETECTION OF FAULTS
 EVALUATION OF COMPLIANCE TO STANDARD
 FOOT NOTES
 REFERENCES
 ANNEXURE A (Sample Calculation)
 ANNEXURE B (Final Calculation Sheet)

 Whenever a fault occurs on a network such
that a large current flows in one or more
phases, a short circuit is said to have
occurred.

When a short circuit occurs, a heavy current
flows through the circuit. This is explained in the
figure attached here. The figure shows a single
phase generator of Voltage “V” and internal
impedance “Zi” (source impedance) is supplying
to a load “Z” (transformer impedance).

•Under normal load conditions, the current in
the circuit is limited by load impedance “Z”.
•However if the load terminals get shorted
due to some reason, the circuit impedance
is reduced to a very low value being “Zi” in
this case. As “Zi” is very small, therefore a
large current flows through the circuit. This
is called the short circuit current.

 People usually get confused with over load and short
circuit as both of them cause trouble to the system in
similar way. It is worthwhile to make a distinction
between a short circuit and an overload. When a
“short circuit” occurs, the voltage at fault point is
reduced to Zero and current of abnormally high
magnitude flows through the network to the point of
fault.
 On the other hand, an “overload” means that load
loads greater than the designated values have been
imposed on the system. Under such condition the
voltage at the overload point may be low, but not
Zero. The under voltage condition may extend for
some distance beyond the over load point into the
remainder of the balance. The currents in the
overloaded equipment are high but are substantially
lower than that in the case of a short circuit.

 A short circuit in the power system is the result of
some kind of abnormal conditions in the system. It
may be caused due to “internal” and or “external”
effects.
 Internal effects are caused by breakdown of
equipment or transmission lines from deterioration of
insulation in a generator, transformer etc… such
troubles may be due to ageing of insulation,
inadequate design or improper installation.
 External effects causing short circuit include
insulation failure due to lightning surges, contact of
live wires due to high speed winds, earth quakes,
falling of a tree, bird or snake shorting, overloading
of equipment causing excessive heating, mechanical
damage by public etc…

 In case of a short circuit, the current in the system
increases to an abnormally high value while the
system voltage decreases to a low value.
 The heavy current due to short circuit causes
excessive heating which may result in fire or
explosion. Sometimes short circuit takes the form of
an arc and causes considerable damage to the
system. For example an arc on a transmission line not
cleared quickly will burn the conductor severely
causing it to break, resulting in a long time
interruption of line.
 The low voltage created by the fault has a very
harmful effect on the service rendered by the power
system. If the voltage remains low for even a few
seconds, the consumer’s motors may be shut down
and generators on the power system may become
unstable.

 Due to the above effects of short circuit it is
desirable and necessary to disconnect the faulty
section and restore normal voltage and current
conditions as quickly as possible. Short circuit can
produce very high temperatures due to the high
power dissipation in the circuit. This high
temperature can be utilized in the application.
 “Arc welding” is a common example of the practical
application of the heating due to a short circuit. The
power supply for an arc welder can supply very high
currents that flow through the welding rod and the
metal pieces being welded. The point of contact
between the rod and the metal surfaces gets heated
to the melting point, fusing a part of the rod and
both surfaces into a single piece.

 Ability to withstand short circuit test is the
requirement of IEC 60076-5 standard for
distribution and power transformers to
demonstrate the thermal and dynamic ability to
sustain without damage the effects of over
currents originated by external short circuits.
 The general requirements with regard of this
test is that transformers together with all
equipment and accessories shall be designed and
constructed to withstand without damage the
thermal and dynamic effects of external short
circuits.
 External short circuits are Line-to-Line, double-
earth, Line-to-neutral etc…

 Short circuit test is a simulation of the
effects of repetitive short circuit events
likely to occur in service.
 To prove that identical transformers of
similar design can withstand external short
circuits without any damage to parts.
 To ensure safe and reliable operation at site
 To determine compliance with the
requirements of the standard.

 Three phase transformers with separate LV
and HV windings are categorized into three
 CATEGORY 1 – Up to 2500kVA
 CATEGORY 2 – 2501 to 10 000kVA
 CATEGORY 3 – above 10 000kVA

Rated power Minimum short circuit impedance (%)
Up to 630kVA 4.0
631 – 1250kVA 5.0
1251 – 2500kVA 6.0
2501 – 6300kVA 7.0
6301kVA – 25MVA 8.0
25 – 40MVA 10.0
40 – 63MVA 11.0
63 – 100MVA 12.5
Above 100MVA > 12.5
Table 1 - Short circuit impedance (%Zt) at rated current and at principal tapping

Highest voltage for equipment, Um (kV) Short circuit apparent power (MVA)
7.2, 12, 17.5, 24 500
36 1000
52 and 72.5 3000
100 and 123 6000
145 and 170 10000
245 20000
300 30000
362 35000
420 40000
525 60000
765 83500
The short circuit apparent power in MVA of the system at the
test lab should be specified to calculate the value of the
Symmetrical short circuit current to be used during SC testing.
If the short circuit apparent power of the system is not
specified, the values are taken from Table 2 of IEC 60076-5.
Table 2 - Short circuit apparent power of the system

 A fault current with No DC offset is a
symmetrical short circuit current.
The symmetrical short circuit current in the
three phases is displaced equally by 120° due
to the symmetry that exists when all the
three phases are shorted. It is the most
severe type of fault involving largest r.m.s
current, but it occurs rarely. For this reason
balanced short- circuit calculation is
performed to determine these large
currents.

 Symmetrical short circuit current (r.m.s value) is calculated using
the measured short circuit impedance “Zt” of the transformer
and the System impedance “Zs”.
For transformers falling under CATEGORY 1, the System
impedance “Zs” is neglected if is equal to or less than 5% of the
transformer impedance “Zt”.
 To be more precise, Symmetrical short circuit current value in
r.m.s is got by simply dividing the rated line current by %Z.
 We know the Power in a three phase circuit is P=√3 x VL x IL
 From this we can calculate line current IL
 IL (HV) = P/√3 x VL (HV) (Amperes)
 IL (LV) = P/√3 x VL (LV) (Amperes)
 The Symmetrical short circuit current is then calculated as follows
 ISC (HV) = (IL (HV) /%Z) x 100 / 1000 (Kilo Amperes)
 ISC (LV) = (IL (LV) /%Z) x 100 / 1000 (Kilo Amperes)
 NOTE: While calculating the ISC (Sym), we need to add the System
(Lab) impedance “Zs” with the transformer impedance “Zt”. i.e.
%Z becomes (Zt + Zs). If “Zs” is less than 5% of “Zt” then “Zs” is
neglected in the calculation.

 A fault current with DC offset is an Asymmetrical
short circuit current. Asymmetrical Short circuit
involves only one or two phases. Hence there exists
asymmetry and the three phases become unbalanced.
 Asymmetrical short circuit often occurs between line-
to-ground or between lines. In single line-to-ground
fault, one conductor comes in contact with the
ground or the neutral conductor. A line-to-line fault
occurs when two conductors are short circuited
 During the first half of a cycle, the fault current is at
its largest magnitude occurring at a moment when
the voltage wave is passing the reference axis i.e.
Zero. This asymmetry is brought on by the DC offset
current. At the half cycle mark, the peak value of the
asymmetrical current is about ~1.6 times the
symmetrical current.

X/R 1 1.5 2 3 4 5 6 8 10 14
k x √2 1.51 1.64 1.76 1.95 2.09 2.19 2.27 2.38 2.46 2.55
Asymmetrical Short circuit PEAK current is calculated by just multiplying
the Symmetrical Short circuit current r.m.s value with the PEAK FACTOR.
Peak factor is k x √2
To find the value of k x √2 we need to calculate first the X/R ratio of the
transformer and refer Table 4 of IEC 60076-5.
LV Peak current ISC-peak (LV) = ISC (LV) X (k x √2) Kilo Amperes
HV Peak current ISC-peak (HV) = ISC (HV) X (k x √2) Kilo Amperes
Table 4 – Values for k x √2
If not otherwise specified, in case X/R >14 the factor k x √2 is
assumed equal to 1.88 x √2 = 2.55 for transformers of CATEGORY 2
1.9 x √2 = 2.69 for transformers of CATEGORY 3

 S.C. test is carried on a new transformer with the protection accessories
like the Buchholz relay and Pressure relief device mounted on it.
 Before the S.C. test, the transformer shall be subjected to Routine tests
as per IEC-60076-1 and the test results made available.
 LI test is not required at the start and is performed after the completion
of SC test as a verification test.
 Measure the Resistance and Reactance of the winding with tapping (i.e.
HV winding) at Tap 1, Tap 3 and Tap 5 at which SC test will be carried
out.
 The temperature of the windings shall be between 10 to 40°C before the
start of SC test.
 During the tests, winding temperature may increase due to the
circulation of the short circuit current.
 For thermal ability to withstand short circuit only symmetrical short
circuit current Isc (sym) is applied. The test duration is 2s.
 Tolerance on the symmetrical r.m.s value of the short circuit test current
is 10% from the specified value.
 The maximum permissible value of the average temperature of each
winding after short circuit is 250°C for Copper and 200°C for Aluminum
for Class A insulation.

 For three phase transformers, a three phase supply is used.
 The winding closer to the core is to be short circuited in order to avoid
saturation of the magnetic core which could lead to an excessive
magnetizing current superimposed on the short circuit current during the
first few cycles.
 For dynamic ability to withstand short circuit Asymmetrical short circuit
current Isc (asym) is applied. The test duration is 0.5s for Category 1 and
0.25s for Category 2 with a tolerance of ±10%
 Tolerance on the Asymmetrical peak value of the short circuit test
current is 5% from the specified value
 To obtain the initial peak value of the Asymmetrical current in the phase
winding under test, the moment of switching ON shall be adjusted by
means of a synchronous switch.
 The values of Isc (asym) and Isc (sym) are checked using the Oscilloscope and
Oscillographic recordings are made.
 In order to obtain the maximum asymmetry of the current in one of the
phase windings, the switching ON must occur at the moment the voltage
applied to this winding passes through Zero.
 The frequency of the test supply shall be the same as the rated
frequency of the transformer and the voltage of the test supply is
suitably adjusted with respect to the rated voltage of the transformer.

 Hence LV windings which are close to the core are shorted through shunt
resistance
 For three phase transformers, a three phase supply is used.
 The winding closer to the core is to be short circuited in order to avoid
saturation of the magnetic core which could lead to an excessive
magnetizing current superimposed on the short circuit current during the
first few cycles.
 Hence LV windings which are close to the core are shorted through shunt
resistance
 Short circuit current is applied from the HV side in each phase by
adjusting the supply voltage such that rated LV Short circuit current flows
through it.
 Preliminary adjustment test (calibration at 70% of the specified current)
is done to check the proper functioning of the test setup with regard to
the moment of switch ON, the current setting, the damping and the
duration.
 For categories 1 and 2 three phase transformers, the total number of
tests shall be nine. I.e. three tests on each phase.
 The nine tests on a three phase transformer with tapping are made in
different tap positions i.e., three tests in the position corresponding to
the highest voltage ratio on one of the outer phases, three tests on the
principal tapping on the middle phase and three tests in the position
corresponding to the lowest voltage ration on the other outer phase.

 Test Sequence simplified:
 Following sequence of short circuit current
applications is specified for three phase
transformers classified under Category 1 and
2 as per IEC 60076-5
 3 shorts in φU at TAP 1
 3 shorts in φV at TAP 3
 3 shorts in φW at TAP 5

Short circuit testing laboratories are required in order to test the transformers at
different short circuit currents. The short circuit testing set up is given in Fig
11.15.
It consists of a short circuit testing generator in association with a master circuit
breaker, reactor and measuring devices. A make switch initiates the short circuit
and the master circuit breaker isolates the test device from the source at the end
of a pre-determined time set on a test sequence controller. Also the master circuit
breaker can be tripped if the test device fails to operate properly. Short circuit
generators with induction motors as prime movers are also available.

 Any difference between the results of measurements
made before and after the test may be used as a
criterion for determining possible defects.
 The Oscillographic recordings of applied voltage,
currents at 70% and 100% and combustible gasses
collected in the Buchholz relay all serves as
reference for detection of faults. Further the visual
inspection of the tank from outside helps detection
of faults.
 The variation in reactance measured (2% for helical
winding and 4% for foil winding) before and after SC
test forms the basis for detection of faults.
 The Active part shall be removed from the tank for
inspection of core and windings. The possible
defects, possible cause and verification test are
summarized as below.

Defect Possible Cause Check
Change in Oscillographic
recordings of applied voltages
and currents
Displacement of windings from
position resulting asymmetry
SFRA test
Increase in Short circuit
reactance Bulging of coils
All the reactance measurements
made per phase shall be to a
repeatability of better than ±0.2%
Noise and Vibration Loose core and clamping, loose
assembly
Noise level test
Abnormal current between tank
and earth
Grounded core / Shorted core No load loss test and excitation
current measurement
Tank Bulging Excess pressure developed
during gas formation and burning
of windings
Pressure deflection test
Combustible gas collected in
Buchholz relay and tripping
Gasses formed due to burning of
insulation and oil
DGA test
Change in magnetizing current Core disturbed SFRA test
Inter turn to turn fault Weak /damaged turn to turn
insulation, loose winding and gap
between turns
Measure short circuit reactance
from HV as well as LV side
Movement / Shifting of leads and
position
Leads not held properly during
manufacturing / improper /
inadequate support and
tightening
LI test

 In order to consider the transformer as having passed the short
circuit test the following condition shall be fulfilled.
 No variation in the Oscillographic recordings when compared with
those at the start of the SC test and at the end of the test.
 The short circuit reactance values in ohms, calculated on per
phase basis or each phase at the end of the tests do not differ
from the original values by more than 2% for helical winding and
4% for LV foil winding.
 Pass in 100% Routine test including di-electric test.
 Pass in LI test performed after the completion of SC test.
 No displacement of core, deformation of windings, movement of
leads and supports in the out-of-tank inspection.
 No signs of combustible gases in the Buchholz relay and tripping
during SC test.
 No traces of any internal electrical discharge.
 If any of the above conditions are not met, the unit shall be
dismantled as necessary to establish the cause of the deviation.

 SEC distribution transformers fall under CATEGORY 1.
 PRIVATE distribution transformers fall either in CATEGORY 1 or CATEGORY
2.
 Both SEC and PRIVATE power transformers fall under CATEGORY 3.
 For transformers falling under CATEGORY 1, the System impedance “Zs” is
neglected if is equal to or less than 5% of the transformer impedance
“Zt”.
 Thumb rule: Short circuit current = Rated current x how many times
greater short circuit the transformer can withstand.
 Z=4% means, the transformer can withstand 25 times greater short
circuits.
circuits.
circuits.
 The currents resulting from a short circuit in the windings are called over
currents.
 In a three 1φ transformers forming a 3φ bank, the value of power applies
to three phase bank rating.
 The Short circuit apparent power and ratio of Z0/Z of the system (test
lab) should be specified by the purchaser during the enquiry stage itself
in order to calculate the Isc(sym) to be used in design and testing. If not
specified the values can be taken from Table 2 of IEC-60076-5: 2000
edition.

 For Um up to 24kV, the short circuit apparent power is 500MVA.
 For Um = 36kV, the short circuit apparent power is 1000MVA.
 Z0/Z should be considered between 1 and 3.
 For transformers with 2 separate windings, only the three phase are short
circuit is taken into account as the consideration of this case is
substantially adequate to cover also the other possible types of fault.
 This paper deals with the short circuit test and calculations for
distribution transformers up to 5MVA, 36kV class and does not cover
regulating transformer, booster transformer, 3 winding transformer, Unit
generator transformer, Arc furnace transformer, traction transformer
etc…..
 Tap changer (both Off-load and ON-load) shall be capable of carrying the
same over currents due to short circuit as the windings.
 The neutral terminal of the Star connected windings shall be designed for
the highest over current to flow through it.
 The mounting of accessories does not have any influence on the behavior
during SC test.
 Where K-factor is unknown, it is determined by switching on during
preliminary adjustment tests (i.e. calibration at 70%) at a maximum of
the line-to line voltage.
 K- Factor can be found from Oscillograph of the line currents.

 International Standard IEC 60076-5 Second
edition 2000-07
 Fundamentals of High Voltage Engineering
by Abdulrhman Al-Arainy, Mohammad Iqbal
Qureshi, Nazar Malik

A typical example of a 1000kVA, 13.8kV/0.4kV
distribution transformer is taken for short circuit
calculations.
 Rating: 1000kVA, 13.8kV (± 2 x 2.5%)/0.4kV
Frequency: 60Hz
 %Z = 6% @ 75°C at normal tap position (TAP 3)
 %R = 0.9% @ 75°C %X = 5.932%
 Symmetrical short circuit current value in r.m.s is
calculated by dividing the rated line currents by %Z.
 Power in a 3φ circuit is given by P=√3 x VL x IL.
From this we calculate line currents IL.
 IL (HV) = P/√3 x VL (HV) (Amperes)
 IL (HV) = 1000 /1.732 x 13.8 = 41.84 A
 IL (LV) = P/√3 x VL (LV) (Amperes)
 IL (LV) = 1000 /1.732 x 0.4 = 1443.42 A

Symmetrical short circuit current (r.m.s value)
calculation.
 While calculating the ISC(Sym), we need to add the
System (Lab) impedance “Zs” with the transformer
impedance “Zt”. i.e. %Z becomes (Zt + Zs). If “Zs” is
less than 5% of “Zt” then “Zs” is neglected in the
calculation.
 Here Zt = 6% and Zs given by the Hi-Power lab is
0.375%.
Therefore %Z = (Zt + Zs) = (6% + 0.375%) = 6.375%
 ISC (HV) = (IL (HV) /%Z) x 100 / 1000 (Kilo Amperes)
 ISC (HV) = (41.84 /6.375) x 100 / 1000 = 0.656 kA
 ISC (LV) = (IL (LV) /%Z) x100 / 1000 (Kilo Amperes)
 ISC (LV) = (1443.42 /6.375) x 100 / 1000 = 22.642 kA

Asymmetrical short circuit current (peak value)
calculation.
 Asymmetrical Short circuit peak current is calculated by
just multiplying the Symmetrical Short circuit current r.m.s
value with the peak factor.
 PEAK FACTOR is k x √2
 To find the value of k x √2 we need to calculate first the
X/R ratio of the transformer
 X/R = 5.932/0.9 = 6.591
 Now refer Table 4 of IEC 60076-5 to find the value of k x√2
corresponding to the X/R ratio calculated. It lies between
2.27 (6) and 2.38 (8).Difference is (2.38 – 2.27) = 0.11
Find the difference between the X/R and % Z.
Difference is (6.591- 6.375) = 0.216
Now multiply the differences, we get 0.11x 0.216= 0.02376
 From the Table 4 of IEC 60076-5, the k x √2 value for X/R
ratio 6 is 2.27. So add 0.0236 with 2.27 to get the
corresponding peak factor for X/R ratio 6.591.

 Final k x √2 value for X/R ratio 6.591 = 2.27 + 0.0236
= 2.29376
 LV Peak current ISC-peak (LV) = ISC (LV) X (k x √2) Kilo
Amperes
ISC-peak (LV) = 22.642 X 2.29376 = 51.935 kA
 HV Peak current ISC-peak (HV) = ISC (HV) X (k x √2)
Kilo Amperes
ISC-peak (HV) = 0.656 X 2.29376 = 1.504 kA
 In the same way calculate the Symmetrical short
circuit current ISC(sym) r.m.s values with %Z at TAP 1
and TAP 5.
 Likewise calculate the X/R ratios with the %X and %R
values at TAP 1 and TAP 5 to find k x √2 value (peak
factor) to be used in the Asymmetrical short circuit
current (peak value) calculations.

System Power 500MVA System Voltage 13.8kV
Tap Position 1 Max. 3 Nom. 5 Min.
Tap Voltage 14.49 kV 13.8 kV 13.11kV
% Impedance 6.15 6.00 5.85
% R at 75°C 0.88 0.9 0.92
% Reactance 6.087 5.932 5.777
TAP POSITION 1 Max.
Supply Voltage 14.49 kV Terminal Voltage 14.49 kV
Sym.SC Current Min. Value (90%) Rated (100%) Max. Value (110%)
HV Current 549.51 A 610.57 A 671.63 A
LV Current 19.90 KA 22.12 KA 24.33
Asym.SC Current Min. Value (95%) Rated (100%) Max. Value (105%)
LV Peak Current 48.60 KA 51.17 KA 53.72 KA
HV Reactance 12.78 Ω HV Inductance 33.92 mH

TAP POSITION 3 Nom.
HV Current 590.67 A 656.31 A 721.94 A
LV Current 20.38 kA 22.64 kA 24.91 kA
LV Peak Current 49.33 kA 51.94 kA 54.53 kA
TAP POSITION 5 Min.
HV Current 636.72 A 707.46 A 778.21 A
LV Current 20.87 KA 23.19 KA 25.50 KA
LV Peak Current 50.14 KA 52.78 KA 55.42 KA

 We know the %Z is expressed in Ohms per phase as
 Z (Ω) = %Z x (VL )² / 100 x MVA
 Example: %Z of a 1000kVA (1MVA) transformer at Tap
3 and @75°C = 6%
 Therefore Z (Ω) = 6 x (13.8)² / 100 x1 = 11.43Ω
 Similarly %X is expressed in Ohms per phase as
 X (Ω) = %X x (VL )² / 100 x MVA
 Example: %X of a 1000kVA (1MVA) transformer at Tap
3 = 5.932%
 Therefore X (Ω) = 5.932 x (13.8)² / 100 x1 = 11.30Ω
 Likewise %R is also expressed in Ohms per phase as
 R (Ω) = %R x (VL )² / 100 x MVA
 Example: %R of a 1000kVA (1MVA) transformer at Tap
3 and @75°C = 0.9%
 Therefore R (Ω) = 0.9 x (13.8)² / 100 x1 = 1.71Ω

 We know that the Inductive reactance XL is
given by XL = ω L
 Where “L” is the inductance “ω” is 2πf in a
resonance circuit and “f” is the resonance
frequency (60Hz).
 Therefore XL = 2πf L
 And L = XL / 2πf = 11.30/2x3.14x60 =
29.98mH
END OF PAPER

Short circuit test in brief

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