SHORT CIRCUIT TEST-IN BRIEF

Presented byS.M.G. MUTHAR HUSSAIN

Transformer Engineer

CONTENTS 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?

CONTENTS CONTINUED…

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) 

WHAT IS SHORT CIRCUIT?

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.

WHAT IS SHORT CIRCUIT CURRENT?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).

WHAT IS SHORT CIRCUIT CURRENT? CONTINUED…•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.

WHAT IS DIFFERENCE BETWEEN SHORT CIRCUIT AND OVERLOAD? 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.

WHAT CAUSES 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…

WHAT IS THE EFFECT OF SHORT CIRCUIT? 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.

WHAT IS THE EFFECT OF SHORT CIRCUIT? CONTINUED… 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.

WHAT IS ABILITY TO WITHSTAND SHORT CIRCUIT TEST? 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…

WHY IS SHORT CIRCUIT TEST DONE ON DISTRIBUTION AND POWER TRANSFORMERS? 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.

TRANSFORMER WITH TWO SEPARATE WINDINGS

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

RECOGNIZED MINIMUM VALUES OF SHORT CIRCUIT IMPEDANCE FOR TRANSFORMERS WITH TWO SEPARATE WINDINGS (REFER TABLE 1 OF IEC 60076-5)

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

SHORT CIRCUIT APPARENT POWER OF THE SYSTEM

Highest voltage for equipment, Um (kV) Short circuit apparent power (MVA)7.2, 12, 17.5, 24 500

36 100052 and 72.5 3000100 and 123 6000145 and 170 10000

245 20000300 30000362 35000420 40000525 60000765 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

WHAT IS SYMMETRICAL SHORT CIRCUIT CURRENT? 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 CALCULATION 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.

WHAT IS ASYMMETRICAL SHORT CIRCUIT CURRENT? 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.

ASYMMETRICAL SHORT CIRCUIT CALCULATION

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 √2To 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

SHORT CIRCUIT TESTING PROCEDURE FOR TRANSFORMERS WITH TWO WINDINGS 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.

SHORT CIRCUIT TESTING PROCEDURE FOR TRANSFORMERS WITH TWO WINDINGS 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.

SHORT CIRCUIT TESTING PROCEDURE FOR TRANSFORMERS WITH TWO WINDINGS 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.

SHORT CIRCUIT TESTING PROCEDURE FOR TRANSFORMERS WITH TWO WINDINGS 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

SC TEST SETUP IN HI-POWER LAB 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.

DETECTION OF FAULTS 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.

DETECTION OF FAULTS CONTINUED…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

EVALUATION OF COMPLIANCE TO STANDARD

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. 

FOOT NOTES 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. Z=5% means, the transformer can withstand 20 times greater short

circuits. Z=6% means, the transformer can withstand 17 times greater short

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.

FOOT NOTES CONTINUED… 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.

REFERENCES International Standard IEC 60076-5

Second edition 2000-07 Fundamentals of High Voltage

Engineering by Abdulrhman Al-Arainy, Mohammad Iqbal Qureshi, Nazar Malik

ANNEXURE A (SAMPLE CALCULATION)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

ANNEXURE A (SAMPLE CALCULATION)

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

ANNEXURE A (SAMPLE CALCULATION)

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.

ANNEXURE A (SAMPLE CALCULATION)

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.

ANNEXURE B (FINAL CALCULATION SHEET SHORT CIRCUIT CURRENT)

System Power 500MVA System Voltage 13.8kVTap 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

ANNEXURE B (FINAL CALCULATION SHEET SHORT CIRCUIT CURRENT)TAP POSITION 3 Nom.

Supply Voltage 13.8 kV Terminal Voltage 13.8 kVSym.SC Current Min. Value (90%) Rated (100%) Max. Value (110%)

HV Current 590.67 A 656.31 A 721.94 ALV Current 20.38 kA 22.64 kA 24.91 kA

Asym.SC Current Min. Value (95%) Rated (100%) Max. Value (105%)LV Peak Current 49.33 kA 51.94 kA 54.53 kAHV Reactance 11.30 Ω HV Inductance 29.98 mH

TAP POSITION 5 Min.Supply Voltage 13.11 kV Terminal Voltage 13.11 kV

Sym.SC Current Min. Value (90%) Rated (100%) Max. Value (110%)HV Current 636.72 A 707.46 A 778.21 ALV Current 20.87 KA 23.19 KA 25.50 KA

Asym.SC Current Min. Value (95%) Rated (100%) Max. Value (105%)LV Peak Current 50.14 KA 52.78 KA 55.42 KAHV Reactance 9.93 Ω HV Inductance 26.35 mH

CALCULATION OF HV REACTANCE (X) IN OHMS PER PHASE 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

CALCULATION OF HV INDUCTANCE (L) IN MILLE HENRY 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