In past articles we have looked at a simplified method for calculation of short circuit problems routinely encountered by electric utility engineers. I have found the method to be handy in the field or in the office as a quick check on results obtained by traditional methods.
The method of solution, however, is only one piece of the puzzle when solving problems. Reliable results depend on reliable input data. Thus, we will dig a little deeper into the input data, starting this time with a look at transformer impedance.
To impede is to obstruct or block the way of something. The flow of alternating current is impeded and we use the term impedance to quantify the amount of restriction of electrical current caused by a device in the circuit.
As used by electrical engineers, impedance has both a resistive and a reactive component and we express it as
Z = R + jX
and it is normally expressed in ohms where R is the resistance, X the reactance, and j is the imaginary number operator (the square root of -1). The transformer impedance in ohms is converted to per unit and is a series component in our system short circuit diagram.
For substation transformers, the impedance is usually designed to be within a certain range, which depends primarily on the kilovolt rating of the primary winding. Since there is always some variation in the construction of a transformer, especially in the characteristics of the steel used in the core, there may be a variation in the tested impedance, compared to the design value. Industry standards allow a maximum variation of 7.5 percent from the design value.
The actual impedance value in percentage stamped on a transformer nameplate is determined by testing. The test setup concept is simple, and the classical method consists of measuring the exciting voltage necessary on the primary winding to circulate current in the short-circuited secondary equal to the current at rated kVA. This impressed primary voltage is then the “impedance voltage” or just impedance. It is expressed in percentage on the rated transformer power base. This value is useful for short circuit calculations because current scales with the applied voltage when the secondary is shorted. For example, if 8 percent of the normal voltage causes rated current to flow, then 100 percent of normal voltage will cause 100/8 or 12.5x normal current to flow.
It is possible to demonstrate the factory test with a small single-phase distribution transformer. These transformers usually have relatively low percentage impedance in the range of 1.5 to 2 percent so the voltages needed are relatively low. If we consider a 1.5 percent Z, 25-kVA, 7.2-kV single-phase transformer, we find that it will circulate full rated current in the short-circuited secondary winding with only 108 Volts on the primary. The primary will draw about 3.5 Amperes, well within the range of the small variable transformers commonly used in electrical testing work. Unfortunately, a three-phase test setup is much more involved and the voltages and current requirements to field test a substation transformer in this manner are not practical. Commercial test sets can calculate this automatically using scaled quantities.
Transformer reactance—You may have encountered the term “leakage inductance” related to electrical coils or transformers. The leak is magnetic flux. In other words, not all of the magnetic flux generated by the current in the primary winding is perfectly coupled from the primary coil to the secondary coil. This flux leakage is responsible for the transformer reactance. If all of the magnetic flux generated by the current flow in the primary winding could be linked perfectly to the secondary winding, then all of the magnetic energy in ampere-turns would be transferred to the secondary and the transformer would have zero reactive impedance. This is the classic “ideal transformer.”
Inductive reactance dominates the impedance of practical transformers. For a system frequency of 60 Hz, the reactance is the leakage inductance multiplied by 2•pi•f or 377 and it represents the combined effect of both windings. For most problems, it is assumed to be a constant value. Due to the characteristics of the steel core, the reactance is actually somewhat non-linear, and for in-rush calculations only the “air core” inductance is used.
Transformer resistance—The systems I work with involve primary voltages of 4.16-kV and 12.47-kV for distribution and 69-kV and 138-kV for sub-transmission. When faced with the need to solve a problem, the question arises whether to include or exclude the resistance of circuit components. For many quick short-circuit calculations, it is not necessary to include it, but for other problems where the X/R ratio is needed for sufficient accuracy or where we are solving for losses, harmonics, or voltage drops, the resistance may be important, even if relatively small.
There are several methods for determining or estimating the resistance of a substation transformer. The best method is to use the factory test report information, if available. Winding resistance is usually not stated on a factory test report unless specifically requested at the time of purchase. If you have this information for each winding, it must be converted to a common base and the winding connections accounted for.
For most problems, it is far simpler to find a reasonable estimate of the resistance by working from the full load winding loss from the factory test report. Using the full load loss (not total loss) the percent resistance is found by
%R = FLL / (MVA•10)
where FLL is the full load winding loss from the test report in kW and MVA is the transformer rated MVA.
To obtain R in per unit on a 1-MVA base for use in the simplified MVA method:
Rpu = FLL,kW / (MVA•1000)
I do not like equations constructed so that I have to remember factor-of-10 multipliers. Therefore, I do this calculation by first moving the decimal point mentally to convert the given kW loss to MW and the transformer kVA to MVA and therefore
Rpu = FLL / MVA
Once R is calculated, X can be found as:
Xpu = √ (Zpu2-Rpu2)
Note that the value of R calculated above is for the transformer operating at full load with the corresponding temperature rise. At less than full load, i.e., at lower temperatures, the resistance is lower. I usually do not make a temperature correction since the effect is comparatively small. Also, some other resistances in the network are usually ignored, such as that of the transmission system and this small over-estimation of the transformer resistance compensates for these other neglected resistances.
Finally, if no test data are available, there is a method for estimating the X/R value for a transformer based on industry standards. A graph on page 41 of General Electric Co. publication GET-3550F provides an estimate of power transformer X/R ratio versus MVA rating. The information presented in the graph is a little dated and modern transformers purchased on a loss cost evaluated basis are likely to have a somewhat higher X/R ratio.
Using the estimated X/R ratio:
X percent = Z
R percent = Z %•cos•arctan(X/R)
on the transformer MVA base.
Interested readers can download a small spreadsheet. It is a table of test data with conversion of nameplate impedances to R and X components in percent and per unit for a sampling of transformers for which I have test reports. It is interesting to note some trends. The newer transformers were purchased based on loss-cost evaluation, and one can see the lower losses, lower resistances, and higher X/R ratios that result. Also, the X/R ratio increases as the transformer size increases. The largest units are auto-transformers.
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