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Electrical Power Transformers

Transformer Vector Group and Through-Fault Current

Introduction

The topic of Transformer Vector Group per IEC 60076-1 and Through-Fault transformer current duration under IEEE C57.109-1985 and the 4 types of transformer categories will only be covered here briefly to remind the electrical practitioner of the importance of this IEC and IEEE standards for transformers. We assume that the practicing electrical engineer will conduct his own thorough research on these topics to further his understanding of this important transformer standards.

Scope of IEC 60076-1

"This part of International Standard IEC 60076 applies to three-phase and single-phase power transformers (including auto-transformers) with the exception of certain categories of small and special transformers such as:
- single-phase transformers with rated power less than 1 kVA and three-phase transformers less than 5 kVA;
- instrument transformers;
- transformers for static convertors;
- traction transformers mounted on rolling stock;
- starting transformers;
- testing transformers;
- welding transformers."

Note: The above information was culled from IEC 60076-1 Edition 2.1 - 2000-04

Symbol Designation

Transformer Vector Group is International Electro-technical Commission's (IEC) categorization of the high voltage (HV) and low voltage (LV) winding. In the symbol designation of transformer vector group the high side of the transformer comes first and low voltage comes second. As an example if the HV winding of the transformer is "wye" connected,"Y" is used. If the HV winding is delta connected, "D" is used. The letters are in uppercase to designate it as the high voltage side and it always comes first.

For the low voltage side the delta or wye designation is in lower cased letter and it comes second. In Dy11, the transformer's HV side is delta and the secondary is wye connected. The "11" or "1" simply means the secondary leads the primary by 30o.

When "N" or "n" is added such as in Dyn11 or YdN11, the upper cased "N" simply means that the system neutral is connected to the high voltage side. The lower cased "n", on the other hand, means that the system neutral is connected to the low voltage side.

The Zig-zag Vector Group

Symbol "Z" or "z" indicates an interconnected wye winding. Zigzag-wye (ZY1 or Zy11) is a complex vector group used in specialized application.

It is not possible to generate a zero phase shift between primary and secondary with a combination of delta and star windings. For this, a more complex vector group the Zig-zag or inter-connected star configuration is used. In this format, half of each secondary is wound on a different leg and the  transformer secondary windings are then interconnected. An example of this is the DZN0, here the lead angle of one leg is used to balance the lag angle of the other leg. A zig-zag transformer's usual application is impedance grounding of the transformer neutral.

Summing it up, the upper cased "N" indicates that the system neutral is connected to the HV neutral. The lower cased "n" indicates that the system neutral is connected to the LV neutral, and the III or iii, indicates independent windings, that are not connected internally in the transformer but outside of the transformer.

Please take note that transformer vector groups are written "high side" first and in capital letter with the "low side" following and written in small letter without consideration of which side is the primary or secondary.

Confusion over IEC60076-1 Standard

Internationally adopted standard for phase rotation is always counterclockwise with multiples of 30o lag for LV with the HV side used as reference; which mean 1 is 30o; 2 is 60o; 3=90o; 6 is 180o and 12 is 360o. The confusion arises over a step-up transformer delta LV primary and HV wye connected secondary which is not written as dy11 but YD11.

Transformers Built to ANSI Standards

Transformers designed and built to ANSI standards usually have no "vector group" on the transformer's nameplate. ANSI shows a vector diagram to show the relationship of all windings of the transformer.

Transformer Through Fault

The transformer's I2t is a definite limitation since heat is equal to I2Req with Req defined as the total resistance of the transformer's primary and secondary windings expressed as constant 1.

Ex. To better understand this statement, let's assume that a transformer can carry 20X its rated full load current for t=1s. Now let's determine what maximum current can be allowed in 2s at 10X current rating without incurring damage to the transformer.

Since I2R = I2t;

202 X 1 =102 X t

t=400/100 = 4s

For maximum current of 2s: 202 X 1 = I2 X 2

I2 = 400/2 = (400 / 2)1/2  = 14.14Amps (The exponent form "1/2" is a square root)

I = 14.14 times full load current

I = 14.14 X full load current is the maximum current that can be allowed to flow in the transformer in 2 seconds.

Transformer Impedance

Impedances of Single-Phase Transformers in Three-Phase Transformer Banks (Stevenson, 1982)

"For 3-phase transformer units, the nameplate specifies the impedance in percent on the three-phase KVA rating and the kV lie-to-line voltages. Where several KVA ratings are specified, the impedance of the ambient rating (without fans or pumps should be used.

For the individual single-phase transformers, that are in common use in electrical customer services, the transformer impedance is usually specified on the single-phase KVA and the rated winding voltage of the transformer. When three such units are used in three-phase systems, then the the three-phase KVA and the line-to-line kV bases are required. Thus, when three individual single-phase transformers are connected in the power system the individual nameplate percent or per unit impedance will be the leakage impedance, but the three-phase kVA, base and the system line-to-line kV."

Transformer Categories

The time-current characteristic curves shown is a great help to protection specialist to properly select the protective devices for various transformer categories.

Figure 1 - Category I Transformer Damage Curve

Category I Transformer Curve

Category Transformer ranges from 5KVA to 500KVA 1-ph; 15KVA TO 500KVA 3-ph The curve is available from IEEE/ANSI. Reference: Electric Power System Protection and Coordination by: Michael Anthony - University of Michigan.

View clear image here

Category I Transformers

As shown on Figure-1 on the left, the lower end of the thermal curve starts at t=0.5secs and is a function of the transformer impedance.

At t=50s, the transformer damage curve (see encircled point on the curve) increases and stops at an intercept (vertical line) at about 2.2X the full load current (please see the Times Normal Base Current of Figure-1) at about 1000s point of the upper limit in the curve .

Category I transformer curve does not take into consideration the occurrence of frequent faults. The through-fault protection curve was plotted without consideration to faults either occurring frequently or infrequently.

As can be gleaned from the curves shown, damage curve for Category I transformer starts at 0.5 sec., but for all other transformer categories, the curve's lower limit starts at t=2s.

Category II Transformers

There are two curves shown in the Figure-2. The left hand curve shows the thermal and mechanical damage concerns. The curve on the left is an application used in considering a high probability of faults in order to properly determine the appropriate feeder protective devices' time-current characteristic.

The right hand curve, on the other hand, is used for feeder protective devices' characteristic curve, too but for anticipated faults that rarely occur.

Figure 2 - Category II Transformer Damage Curve

Category II Transformer Curve

Category II Transformer - 501KVA to 1667KVA 1-ph; 501KVA to 5000KVA 3-ph. The curve is available from IEEE/ANSI. Reference: Electric Power System Protection and Coordination by: Michael Anthony - University of Michigan.

View clear image here

Category III Transformers

The thermal and mechanical damage is shown on the left side of the curve. It may be used for selecting, too feeder protective relays' time current characteristic for transformer installed in areas with high atmospheric pollution, such as dust and chemical elements present in the air that can cause frequent fault. The right hand curve shows only the thermal damage considerations used for selecting the proper protective relays time-current characteristic curve where the anticipated fault is low or infrequent.

When to Expect High Fault Incidence

The incidence of fault is usually determined in the environment where the transformer is installed. Transformers installed in salty atmosphere like seashores, areas with high motor vehicle exhaust emission, coal fired power plant where burned or unburned coal particles may be present and geothermal plants where of H2s gas molecules in the air is present; would surely subject the transformer to high incidence of external faults. Transformers installed in indoor substations can be expected to have rare fault incident.

Category IV Transformers

Category IV are actually dry type transformers as shown in Class AF cooling type below. These transformers are commonly used in power generating plants.

As an example, banking 3 X 1-phase, 3.75 MVA transformers will have a combined capacity of about 12MVA and can be used for power plant with about 10MW generating capacity. Power generating companies' utilizing generator step-up transformers of up to 300MVA or more also belong to Category IV transformers. These transformers are usually gas cooled utilizing either hydrogen, nitrogen or sulfur hexafluoride. These transformers usually belong to YdN11 vector group.

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Summing up

Summing it up, the Category II and Category III curves is used in selecting the protective devices time-characteristic curve for both the primary side and also for the secondary side taking into consideration the occurrence of frequent or infrequent fault while Category I does not.

The time-current characteristic curves shown is a great help to protection specialist to properly select the protective devices for various types of transformer.

Category III Transformer Damage Curve

Category III Transformers

Category III Transformer - 1668KVA to 10MVA 1-ph; 500KVA to 30MVA 3-ph. The curve is available from IEEE/ANSI. Reference: Electric Power System Protection and Coordination by: Michael Anthony - University of Michigan.

View clear image here

Transformer Cooling Classes

Dry Type Transformers

Class AA

The transformer body is provided with fins and designed to provide maximum heat dissipation. The "fins" dissipate the heat through the natural air circulating around the transformer.

Class AF

This is the "forced" air type transformer. Cooling of the transformer is provided by fans directing air to the transformer core windings.

In both Class AA and AF, proper ventilation must be provided for indoor installations in order not to operate the transformers beyond ambient temperatures specified by the manufacturer.

Hydrogen, Nitrogen or Sulfur Hexafluoride gas cooled transformers are large capacity transformers that are  under the category of Dry Type Transformers. The core of these transformers is placed in sealed containers where the circulating gas takes the heat out of the core and circulates i t to large radiator cooling fans.

Oil Cooled Transformers

The four classes of oil cooled transformers are classified as:

Oil Natural - Air Natural (ONAN)

The heat is dissipated through the oil circulating in and out of the transformer core and cooled "naturally" by the air. This is accomplished without pumps since rising oil temperature causes its specific gravity to decrease and move upwards to the tube inlets connected to external cooling fins (heat exchangers). The oil moves downwards due to gravity as it cools and the whole cycle repeats continuously.

Oil Natural - Air Forced (ONAF)

The oil circulation in ONAF system is the same as that of ONAN except that the air is now "forced" to the radiator cooling fins. The fans are idle but kick into action as soon as natural air cooling becomes deficient. The fans stops or starts automatically depending on the transformer load: It starts as it detects higher temperature due to increased transformer loading, and stops as the temperature decreases to a point where it can sustain natural air cooling.

ONAF cooled transformer capacity can be increased to 133% or 167% of its rating i.e. 133% when fan segments is augmented with 1 additional segment; and 167% when provided with 2.

Oil Forced - Air Forced (OFAF)

Increased heat transfer efficiency requires uses OFAF cooling system configuration. In OFAF cooled transformers, the heat dissipation is maximized when fans are made to continuously blow air to the radiators aided with pumps to force oil circulation in the transformer core windings.

Oil Forced - Oil Directed (OFOD)

A far better way to maximize transformer heat dissipations is by directing the oil towards the winding itself. In OFAF, the circulation of the oil is free, that is, the oil simply follows the pump pressure around the transformer windings. When oil, on the other hand, is directly forced through the windings, the cumulative heat dissipation of the cooling system becomes much more efficient.

Transformer 3rd Harmonic Considerations

We are all aware that the neutral current “In” in a balanced wye connected transformer is the vector sum of In = Ia + Ib + Ic = 0. A good example of an almost perfect 3-phase balanced load is a well designed 3-phase electric motor. However, for varying loads of single and three phase loads in a 3-phase panel, an unbalanced condition will always occur in the transformer feeding them. As an example, let's assume total line currents of Ia = 20.0A, Ib = 23.0A and Ic = 19.4A, the transformer neutral current "In" for these loads will result to about 3.34A (See Computation below).

For this neutral current, the designer will usually assume that a neutral conductor 1/3 the size of the phase conductor will suffice. For the fundamental frequency of 50 or 60hz yes, however, the presence of "triplens" - which is defined as odd 3rd, 9th, 15th, 21st, etc., harmonics developed - especially - by high non-linear loads in the circuits will overheat or even melt this conductor.

Why the 3rd harmonic currents or "triplens" will not cancel each other out and add up to the neutral conductor

When there are high non-linear loads, the wye transformer neutral current may rise as high or higher than the phase current load. This condition will result to overheating of the neutral conductor/s. Non-linear loads are industrial electronic loads produced by discreet devices in electronic circuitry such as rectifier diodes (like power diodes in welding machines), Silicon Controlled Rectifiers (SCRs) and other types of thyristors involved in high speed power electronic switching. One big contributor of third harmonic currents are electronic fluorescent lamp ballasts. These non-linear loads develop harmonics currents that do not cancel each other out because their shapes is not a perfect sine wave and their phase voltage and currents do not follow the 120o phase displacement of the fundamental frequency sine wave.

In the example above where only about 1/3 the size of one line conductor was used for the neutral grounding, the 3rd harmonic currents will, in most probability, overheat or possibly even melt it.

Harmonics is defined as "a wave whose frequency is an integer multiple of the fundamental frequency."  French mathematician Joseph Fourier in his study of sound waves in the 1800s used the term "harmonics" to describe the sets of complex repeating waveforms as sums of simple sinusoidal waveforms.

Thus for the above example, the appropriate neutral conductor will be about 100% the size of one line conductor or a bit more. This practice is sometimes called "hardening the circuit".

Triplens will not pass through delta-delta connected transformer and as such, are used to feed power to critical electronic loads. When delta-delta transformers are used to eliminate 3rd harmonics, the ratio is usually 1:1. That is Vin= Vout

Mitigation of Harmonics

There are various ways of harmonics mitigation. The most popular of which are the use Harmonic Mitigation Transformers (HTM) which include the use of zigzag transformers  ZY1 or Zy11 vector group, over sizing of cables (costly option), passive and active filters, AC/DC/AC converters etc. This subject is wide ranging and falls beyond the scope of this article. For those who may want some good study materials on the subject. The following external sites are recommended:

Some Excellent Transformer Mitigation References:

- General Electric's Harmonic Design Considerations

- Schneider Electric's Harmonics Mitigation and Solutions

- IEEE Control of AC-DC-AC Converters with Minimized DC Link Capacitance under Grid Distortion

Transformer Inrush Current

Power transformer inrush current or switch-on surge current during energization can be as high as 2 to 5 times rated load current, and is caused by part cycle saturation of the magnetic core. Oscillation damping gradually reduces the inrush current magnitude due to impedance, winding and magnetizing resistances of the transformer until it recedes to the steady state condition.

Protective relays can mistake transformer inrush current as a short circuit condition. It is for this reason that value of current magnitude, its duration, and other parameters should be properly defined in order for the protective relays to properly discriminate inrush current from short circuit current. Inrush current wave-shape for a large power transformer are usually calculated during the first 5cycles. The design of the power transformer including its installation can greatly influence the magnitude of the inrush current.

For your reference and further study, download ABB'S Power Transformer Characteristics and Their Effect on Protective Relays, which is also shown at the Suggested Reading column at the bottom of this page.

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Electrical transformer parameters
1. How to compute transformer reactance X and resistance R when only the transformer impedance Z is known.

Use of short circuit software requires you to enter the reactance and the resistance of the transformer. The formula below shows the equation. You can use a default value of X/R = 30 for medium voltage transformer.

R = Z / { 1+ (X/R)2 }1/2
Where R = reactance; Z = impedance

Note 1: The exponent form "1/2" is a square root. The square root symbol is not recognized by web browsers unless it is in image form. Please see logarithm refresher at the bottom of this page.

Note 2: Download  ANSI 37.010 for typical transformer X/R values shown below for your reference.

General Electrica Transformer X/R Ratio Transformer X/R Ratio per ANSI 37.010

2. Simple transformer bolted fault calculations.

The symmetrical fault current on the secondary load side circuit breaker of a 3-phase transformer with bolted secondary is easily calculated using the transformer nameplate percent impedance.

For example, if a 3-phase, 34.5kV/4.16kV, 5MVA transformer with an impedance of 5.75% is bolted at its secondary:

Solution:

The primary feeder voltage required for the full load current to flow on the secondary will be: 34,500V x 0.0575 = 1,984 Volts.

Since the Iflsec= 5 X 106/4160 X 31/2 = 2082Amps, a rated voltage of 34.5 kV rated primary voltage will cause 100% / 5.75% = 17.4 times the rated secondary current. Thus, 17.4 X 2082 = 36,209*Amps

Or alternately, 2082Amps / 0.0575 = 36,209 Amps, which is the symmetrical fault current on the bolted secondary.

*Note: A line-to-line fault will develop about 87% of bolted 3-phase fault current. and is equal  to 15.4% phase shift in the secondary current curve.

3. How to compute three phase transformer unbalanced load neutral current

Computation:

Find In when Ia = 20.0A, Ib = 23.0A and Ic = 19.4A
Equation: In = [ ( Ia2 + Ib2 + Ic2 ) - ( Ia x Ib ) - (Ib x Ic) - (Ia x Ic) ]1/2
= [ (202 + 232 + 19.42) - (20 x 23) - (23; x 19.4) - (20 x 19.4) ]1/2
= [ (400 + 529 + 376.36) - (460) - (446.2) - (388) ]1/2
= [ (1305.36) - 1294.2]1/2 = [11.6 ]1/2 = 3.34Amps

*Note the exponent "1/2" is the logarithmic equivalent of the square root symbol.

Checking the above example using polar coordinate system:

In = Ia+ Ib + Ic = 0 for balanced load
In = Ia^0o+ Ib^120o + Ic^240o

Note: The "^" denotes an "angle symbol" since it is not recreated correctly by web browsers.

= 20(Cos0o + JSin0o) + 23(Cos120o + JSin120o) + 19.4(Cos240o + JSin240o)
= 20(1 + J0) + 23(-0.5 + J0.866) + 19.4(-0.5 + J-0.866)
= (20 + J0) + (-11.5 + J19.918) + (-9.7 + J-16.8)
Adding real values (X-axis) = 20 - 11.5 - 9.7 = -1.2
Adding imaginary values (Y-axis) = J0 + J19.918 + J-16.8 = 3.118
Therefore, the In vector representation of the current is:
-1.2 + J3.118 with a displacement of tan-1(3.118 / -1.2)
= -68.95o = 111.05o

Note: tan-1(3.118 / -1.2) is = arctan(3.118 / -1.2)
In = [ (-1.2)2 + (3.118)2 ]1/2 = 11.1621/2 = 3.34^111° Amps.
Checking:

In = 3.118/Sin111.05o = 3.118/0.933 = 3.34Amps
In = -1.2/Cos111.05o = -1.2/-0.3591 = 3.34Amps

Note: Logarithm form in lieu of the square root symbol:

31/2 = 1/2Log3 = 0.477121 / 2 = 0.2385606
Antilog of 0.2385606 = 1.732 which is the square root of 3

Memory Refresher

Phasors

A rotating vector or phasor is sinusoidal function defined by IEEE Dictionary "...as a complex number expressing the magnitude and phase of a time-varying quantity. Unless otherwise specified, it is used only within the context of steady-state alternating linear systems."

.

Phasors are written in polar form: Y = |Y|^?; and in rectangular form: Ycos? + j |Y|sin?; where Y is the phasor, |Y| is its magnitude; and ? the phase angle.

Cartesian Coordinate System

Complex a and j operators can be plotted on a Cartesian coordinates by single point p, where a is the abscissa on the x-axis (real quantities), and b the imaginary quantity.

Polar Form

Note: The multiplication and division of complex numbers is best carried out using the polar form.
Multiplication

Note: the "^" denotes an "angle symbol" since it is not recreated correctly by web browsers.

Multiply A and B where A = 10^25o and B = 5^50o

Solution: A x B = (10^25o)(5^50o) = (10 x 5)^(35o + 45o) = 50^75o

Division

Divide A and B where A = 21^35o and B = 3^45o

Solution: A/B = (21^35o) / (3^45o) = (21/3)^(35o - 45o) = 7^(-10o)

As can be seen from the above, multiplying and dividing complex numbers can be easily done using polar coordinate system.

Addition

Note: The addition and subtraction of complex numbers can be best carried out using rectangular coordinate system.

Please see above example for "Three Phase Transformer Unbalanced Load Neutral Current Calculation".

For further review download:

Cartesian Components of Vectors Cartesian Components of Vectors

About the absolute symbol |Y|

The symbol |Y| denotes an absolute value and describes the distance of a number on a line number from zero axis without consideration of the direction from zero the number lies. An absolute number is never negative.

As an example, the absolute value of 7 is 7 and its distance from 0 is 7 units The absolute value of -7 is 7. Distance from 0 is 7 units

The absolute value of 7 + (-9)  is 2; distance of sum from 0 is 2 units

The absolute value of 0 is 0. It is for this reason an absolute value of a number can never be always positive since zero is neither positive nor negative. :-)

References:

- Electric Power System Protection and Coordination by: Michael Anthony - University of Michigan

- IEC 60076-1 International Standard - Edition 2.

- Schneider Electric Low Voltage Transformer Through-Fault Protection: A System Approach

- Power Transformers by Bulox Power

- An Introduction to Symmetrical Components, System Modeling and Fault Calculation by: By Stephen Marx, and Dean Bender Bonneville Power Administration

- Fourier Series by: Prof. Bill Lionheart School of Mathematics The University of Manchester


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