TRANSFORMER LOSSES COMPONENTS TUTORIALS
What Are Transformer Losses Components?
Transformer Losses is a natural occurrence in the Power System Cycle. Below are the different components of the Transformer Losses.
No-Load Loss and Exciting Current
When alternating voltage is applied to a transformer winding, an alternating magnetic flux is induced in the core. The alternating flux produces hysteresis and eddy currents within the electrical steel, causing heat to be generated in the core. Heating of the core due to applied voltage is called no-load loss.
Other names are iron loss or core loss. The term “no-load” is descriptive because the core is heated regardless of the amount of load on the transformer. If the applied voltage is varied, the no-load loss is very roughly proportional to the square of the peak voltage, as long as the core is not taken into saturation.
The current that flows when a winding is energized is called the “exciting current” or “magnetizing current,” consisting of a real component and a reactive component. The real component delivers power for no-load losses in the core.
The reactive current delivers no power but represents energy momentarily stored in the winding inductance. Typically, the exciting current of a distribution transformer is less than 0.5% of the rated current of the winding that is being energized.
Load Loss
A transformer supplying load has current flowing in both the primary and secondary windings that will produce heat in those windings. Load loss is divided into two parts, I2R loss and stray losses.
I2R Loss
Each transformer winding has an electrical resistance that produces heat when load current flows. Resistance of a winding is measured by passing dc current through the winding to eliminate inductive effects.
Stray Losses
When alternating current is used to measure the losses in a winding, the result is always greater than the I2R measured with dc current. The difference between dc and ac losses in a winding is called “stray loss.”
One portion of stray loss is called “eddy loss” and is created by eddy currents circulating in the winding conductors. The other portion is generated outside of the windings, in frame members, tank walls, bushing flanges, etc.
Although these are due to eddy currents also, they are often referred to as “other strays.” The generation of stray losses is sometimes called “skin effect” because induced eddy currents tend to flow close to the surfaces of the conductors.
Stray losses are proportionally greater in larger transformers because their higher currents require larger conductors. Stray losses tend to be proportional to current frequency, so they can increase dramatically when loads with high-harmonic currents are served. The effects can be reduced by subdividing large conductors and by using stainless steel or other nonferrous materials for frame parts and bushing plates.
Harmonics and DC Effects
Rectifier and discharge-lighting loads cause currents to flow in the distribution transformer that are not pure power-frequency sine waves. Using Fourier analysis, distorted load currents can be resolved into components that are integer multiples of the power frequency and thus are referred to as harmonics. Distorted load currents are expected to be high in the 3rd, 5th, 7th, and sometimes the 11th and 13th harmonics, depending on the character of the load.
Odd-Ordered Harmonics
Load currents that contain the odd-numbered harmonics will increase both the eddy losses and other stray losses within a transformer. If the harmonics are substantial, then the transformer must be derated to prevent localized and general overheating.
ANSI standards suggest that any transformer with load current containing more than 5% total harmonic distortion should be loaded according to the appropriate ANSI guide (IEEE, 1998).
Even-Ordered Harmonics
Analysis of most harmonic currents will show very low amounts of even harmonics (2nd, 4th, 6th, etc.) Components that are even multiples of the fundamental frequency generally cause the waveform to be nonsymmetrical about the zero-current axis.
The current therefore has a zeroth harmonic or dc-offset component. The cause of a dc offset is usually found to be half-wave rectification due to a defective rectifier or other component. The effect of a significant dc current offset is to drive the transformer core into saturation on alternate half-cycles.
When the core saturates, exciting current can be extremely high, which can then burn out the primary winding in a very short time. Transformers that are experiencing dc-offset problems are usually noticed because of objectionably loud noise coming from the core structure.
Industry standards are not clear regarding the limits of dc offset on a transformer. A recommended value is a dc current no larger than the normal exciting current, which is usually 1% or less of a winding’s rated current (Galloway, 1993).
MATCHING TRANSFORMERS FOR PARALLEL AND BANK OPERATIONS BASIC AND TUTORIALS
PARALLEL AND BANK OPERATIONS TRANSFORMERS MATCHING
How To Match Transformers For Banking and Parallel Operation?
The following rules must be obeyed in order to successfully connect two or more transformers in parallel with each other:
1. The turns ratios of all of the transformers must be nearly equal.
2. The phase angle displacements of all of the transformers must be identical.
3. The series impedances of all transformers must be nearly equal, when expressed as ‘‘%Z’’ using the transformer impedance base.
The first two rules are required so that the open-circuit secondary voltages of the transformers are closely matched in order to avoid excessive circulating currents when the parallel connections are made.
The last rule is based on the fact that for a given voltage rating and %Z, the ohmic impedance of a transformer is inversely proportional to its KVA rating. When transformers having the same %Z are connected in parallel, the load currents will split in proportion to the KVA ratings of the units.
Therefore, transformers with different KVA ratings can be successfully operated in parallel as long as their %Z values are all approximately the same.
Example
(This example is based on an actual event.)
Two three-phase 10,000 KVA 66,000Δ-12,470Y volt transformers were in parallel operation in a substation. The primaries of the two transformers are connected to a 66 kV transmission line through a single air break switch.
This switch is designed to interrupt magnetizing current only, which is less than 1 A. The transformers were being removed from service and the secondary loads had been removed. A switchman then started to open the air break switch, expecting to see a small arc as the magnetizing current was interrupted.
Instead, there was a loud ‘‘bang’’ and there was a ball of flame where the air break switch contacts had vaporized. Something was obviously wrong.
Upon closer inspection, it was revealed that the two transformers had been set on widely different taps: The first transformer was on the 62,700 V primary tap and the second transformer was on the 69,300 V primary tap.
Both transformers had a 7% impedance. Because the turns ratios were unequal, a circulating current was set up even without any secondary load. The opencircuit secondary voltage difference, assuming 66 kV at the transformer primaries, is calculated below.
ΔEs = 66,000 x ( 12,470/ 62,700 - 12,470/69,300 ) V = 1250 V = 0.10 per unit
The per-unit circulating current in the secondary loop is equal to ΔEs divided by the sum of the per-unit impedances of the two transformers:
Ic = 0.10/ 0.14 = 0.714 per unit
Converting Ic into amperes:
Ic = 0.714 x 10,000 KVA/(12.47 kV 1.732) = 331 A per phase
Since Ic flows in a loop in the secondary circuit, the current out of the secondary of the first transformer equals the current into the secondary of the second transformer. But since the turns ratios are not equal, Ic does not get transformed into equal and opposite currents at the primaries.
Primary current of first transformer
65.8 A per phase
Primary current of second transformer
59.6 A per phase
The net current through the air break switch, IAB, is the difference in the primary currents:
IAB 65.8 A per phase 59.6 A per phase 6.2 A per phase
The current through the air break switch supplies the I2c Xs reactive losses of both transformers and therefore lags the primary voltage by 90°. The resulting current exceeded the interrupting rating of the switch, causing it to fail.
How To Match Transformers For Banking and Parallel Operation?
The following rules must be obeyed in order to successfully connect two or more transformers in parallel with each other:
1. The turns ratios of all of the transformers must be nearly equal.
2. The phase angle displacements of all of the transformers must be identical.
3. The series impedances of all transformers must be nearly equal, when expressed as ‘‘%Z’’ using the transformer impedance base.
The first two rules are required so that the open-circuit secondary voltages of the transformers are closely matched in order to avoid excessive circulating currents when the parallel connections are made.
The last rule is based on the fact that for a given voltage rating and %Z, the ohmic impedance of a transformer is inversely proportional to its KVA rating. When transformers having the same %Z are connected in parallel, the load currents will split in proportion to the KVA ratings of the units.
Therefore, transformers with different KVA ratings can be successfully operated in parallel as long as their %Z values are all approximately the same.
Example
(This example is based on an actual event.)
Two three-phase 10,000 KVA 66,000Δ-12,470Y volt transformers were in parallel operation in a substation. The primaries of the two transformers are connected to a 66 kV transmission line through a single air break switch.
This switch is designed to interrupt magnetizing current only, which is less than 1 A. The transformers were being removed from service and the secondary loads had been removed. A switchman then started to open the air break switch, expecting to see a small arc as the magnetizing current was interrupted.
Instead, there was a loud ‘‘bang’’ and there was a ball of flame where the air break switch contacts had vaporized. Something was obviously wrong.
Upon closer inspection, it was revealed that the two transformers had been set on widely different taps: The first transformer was on the 62,700 V primary tap and the second transformer was on the 69,300 V primary tap.
Both transformers had a 7% impedance. Because the turns ratios were unequal, a circulating current was set up even without any secondary load. The opencircuit secondary voltage difference, assuming 66 kV at the transformer primaries, is calculated below.
ΔEs = 66,000 x ( 12,470/ 62,700 - 12,470/69,300 ) V = 1250 V = 0.10 per unit
The per-unit circulating current in the secondary loop is equal to ΔEs divided by the sum of the per-unit impedances of the two transformers:
Ic = 0.10/ 0.14 = 0.714 per unit
Converting Ic into amperes:
Ic = 0.714 x 10,000 KVA/(12.47 kV 1.732) = 331 A per phase
Since Ic flows in a loop in the secondary circuit, the current out of the secondary of the first transformer equals the current into the secondary of the second transformer. But since the turns ratios are not equal, Ic does not get transformed into equal and opposite currents at the primaries.
Primary current of first transformer
65.8 A per phase
Primary current of second transformer
59.6 A per phase
The net current through the air break switch, IAB, is the difference in the primary currents:
IAB 65.8 A per phase 59.6 A per phase 6.2 A per phase
The current through the air break switch supplies the I2c Xs reactive losses of both transformers and therefore lags the primary voltage by 90°. The resulting current exceeded the interrupting rating of the switch, causing it to fail.
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