Intro - In a perfect world, transformers convert from one voltage level to another without any impact on the system. Unfortunately, this isn't the way things actually work. In a real transformer, there will be a voltage drop across the transformer.
At first, this may seem confusing. For instance, if you step up in voltage from 13.8kV to 345kV, how can there be a voltage drop? The answer is that the drop is in per unit. If your input voltage to your transformer is 13.8kV (100% of nominal), then the output voltage may only be 341.6kV (99% of nominal). Even though you have increased your voltage, you have still lost voltage on a per unit basis.
Transformer Impedance - This voltage drop across a transformer is generally driven by the transformer's impedance. The impedance of any real-world transformer is provided as a percentage referenced to the transformer nameplate power and voltages. The impedance is usually dominated by a reactive (inductive) component. This makes sense since transformers are magnetic devices relying on magnetic flux linkages to do their job. The impedance of the transformer usually falls within a standard range of values that correspond to the transformer's power rating. Very large transformers (e.g. 50+ MVA) generally have an impedance on the order of 10% at their base nameplate rating. Smaller transformers may be somewhere from 3-6%. Transformer impedances aren't all bad-they play an important role in limiting fault currents. At the secondary of a transformer, fault currents are primarily driven down by the transformer impedance. If transformers had NO impedance, there could be outrageously large short circuit currents in your system.
Figure 1: A Large Transformer in a Substation-Detailed Nameplate Information is Likely Visible on the Transformer Exterior
The impedance of the transformer can be converted from a nameplate percentage value to an Ohmic value using the following relationship:
Where:
Z_Ohms is the ohmic impedance value
Z_% is the percent impedance of the transformer (nameplate)
V_Line is the line-to-line voltage of the transformer on one of its windings, usually the secondary
S_T is the 3-Phase apparent power rating of the transformer
Notice that V_Line in the equation above can be referenced to either the primary or secondary winding voltage. Results can be converted back and forth but making use of the secondary winding is a more physically meaningful approach.
When X/R ratio is available, this impedance can be broken down into a resistance R and a reactance X for more accurate results. The new effective impedance, based on information from NEC Chapter 9 Table 9 is:
Where:
R is the transformer's resistance in Ohms
X is the transformer's reactance in Ohms
Θ is the power factor angle of the load
Note that the value of Θ is somewhat of an assumption. The load power factor is the standard assumed value. However, the power factor at the transformer primary may be worse than the load power factor due to resistive and reactive losses. If input power factor information is available, it is recommended to consider the drop using this information as well.
Calculating Transformer Voltage Drop - We can use the effective impedance of the transformer to calculate a voltage drop across it. The voltage drop on the secondary of the transformer is then:
Where I is the load current magnitude on the transformer secondary and V_D is the voltage drop in terms of volts on the transformer secondary. If the voltage drop needs to be normalized (put into %), the results should be divided by the transformer secondary voltage.
When the X/R ratio is not known for the transformer, the impedance Z_Ohms can be used in place of the effective impedance (Rcos(Θ)+Xsin(Θ)). This approach may be necessary in some cases to get an understanding of the upper bound for transformer voltage drop. However, the impact of power factor of the load on transformer voltage drop is generally nontrivial and can lead to significantly lower results.
Additional Considerations - The results in the above section are based on a transformer operating at its nominal turns ratio. This means that the transformer is being used exactly at the voltage transformation ratio it is stated to perform (e.g. 13.8kV to 480V). In many cases, transformers have taps, the ability to alter a winding to improve voltage regulation during heavy loading. The effect of a winding tap change is approximately equal to an additional % increase or decrease on the resultant % voltage drop. For example, if a transformer produces a voltage drop of 3% on its secondary, but the secondary winding is tapped up 2.5%, the realized voltage drop would be ~0.5%.
An additional consideration for voltage drop calculations is transformer impedance tolerances. Per IEEE standards, two-winding transformers have a tolerance of 7.5% and three-winding transformers have a tolerance of 10% on their impedances during design. This means that preliminary models should account for this impact when performing voltage drop (e.g. impedance is 7.5% or 10% higher than what the nameplate shows). When tested information is available from the as-built transformer, this tolerance is not necessary.
The model presented in this article is only an approximation. Detailed load flow calculations using iterative numerical methods can provide more accurate results, usually achieved using commercial power systems software like ETAP, EasyPower, or similar.
Example: A transformer serves a load of 2MW at 480V and at a power factor of .85. The transformer has the following nameplate information:
Apparent Power Rating: 3.33 MVA
Impedance: 6% @ 3.33 MVA
X/R Ratio: 20
What is the voltage drop caused by the transformer? Assume no taps are present and that the impedance has no tolerance.
Solution: First, determine the impedance of the system in Ohms:
Z_Ohms = 6% * (480 V^2 / 3.33 MVA) = .00415 Ohms
Next, use the X/R ratio of the system to determine the resistive and reactive components of the transformer impedance. The results are determined using the Power Triangle: R = .000207 Ohms
X = .00414 Ohms
Third, we need to determine the load current, based on the power factor, voltage, and power required:
I = 2 MW / .85 / 480 V / √(3) = 2830 A
Lastly, determine the voltage drop using the load power factor:
V_D = √(3) * 2830 A * (.000207*.85 + .00414* sin( acos(.85) ) ) = 11.6 V
Or, normalizing into percent, we get the following:
V_D% = 11.6 V / 480 V = 2.42%