Auburn University
1.2 Per-Unit Scaling Extended to Three-Phase Systems
1.3 Per-Unit Scaling Extended to a General Three-Phase System
In many engineering situations, it is useful to scale or normalize quantities. This is commonly done in power system analysis, and the standard method used is referred to as the per-unit system. Historically, this was done to simplify numerical calculations that were made by hand. Although this advantage has been eliminated by using the computer, other advantages remain:
• Device parameters tend to fall into a relatively narrow range, making erroneous values conspicuous.
• The method is defined in order to eliminate ideal transformers as circuit components.
• The voltage throughout the power system is normally close to unity.
Some disadvantages are that component equivalent circuits are somewhat more abstract. Sometimes phase shifts that are clearly present in the unscaled circuit are eliminated in the per-unit circuit.
It is necessary for power system engineers to become familiar with the system because of its wide industrial acceptance and use and also to take advantage of its analytical simplifications. This discussion is limited to traditional AC analysis, with voltages and currents represented as complex phasor values. Per-unit is sometimes extended to transient analysis and may include quantities other than voltage, power, current, and impedance.
The basic per-unit scaling equation is
(1.1) |
The base value always has the same units as the actual value, forcing the per-unit value to be dimension-less. Also, the base value is always a real number, whereas the actual value may be complex. Representing a complex value in polar form, the angle of the per-unit value is the same as that of the actual value.
Consider complex power
(1.2) |
or
where
V = phasor voltage, in volts
I = phasor current, in amperes
Suppose we arbitrarily pick a value Sbase, a real number with the units of volt-amperes. Dividing through by Sbase,
We further define
(1.3) |
Either Vbase or Ibase may be selected arbitrarily, but not both. Substituting Equation 1.3 into Equation 1.2, we obtain
(1.4) |
The subscript pu indicates per-unit values. Note that the form of Equation 1.4 is identical to Equation 1.2. This was not inevitable, but resulted from our decision to relate Vbase Ibase and Sbase through Equation 1.3. If we select Zbase by
(1.5) |
Convert Ohm’s law:
(1.6) |
into per-unit by dividing by Zbase.
Observe that
(1.7) |
Thus, separate bases for R and X are not necessary:
By the same logic,
Example 1.1
(a) Solve for Z, I, and S at Port ab in Figure 1.1a.
(b) Repeat (a) in per-unit on bases of Vbase = 100 V and Sbase = 1000 V. Draw the corresponding per-unit circuit.
Solution
(a)
(b) On bases Vbase and Sbase = 1000 VA:
Converting results in (b) to SI units:
The results of (a) and (b) are identical.
For power system applications, base values for Sbase and Vbase are arbitrarily selected. Actually, in practice, values are selected that force results into certain ranges. Thus, for Vbase, a value is chosen such that the normal system operating voltage is close to unity. Popular power bases used are 1, 10, 100, and 1000 MVA, depending on system size.
To understand the impact of pu scaling on transformer, consider the three-winding ideal device (see Figure 1.2).
For sinusoidal steady-state performance:
(1.8a) |
(1.8b) |
(1.8c) |
and
(1.9) |
Consider the total input complex power S.
(1.10) |
The interpretation to be made here is that the ideal transformer can neither absorb real nor reactive power. An example should clarify these properties.
Arbitrarily select two base values V1base and S1base. Require base values for windings 2 and 3 to be
(1.11a) |
(1.11b) |
and
(1.12) |
By definition,
(1.13a) |
(1.13b) |
(1.13c) |
It follows that
(1.14a) |
(1.14b) |
Recall that a per-unit value is the actual value divided by its appropriate base. Therefore:
(1.15a) |
and
(1.15b) |
or
(1.15c) |
indicates per-unit values. Similarly,
(1.16a) |
or
(1.16b) |
Summarizing:
(1.17) |
Divide Equation 1.9 by N1
Now divide through by I1base
Simplifying to
(1.18) |
Equations 1.17 and 1.18 suggest the basic scaled equivalent circuit, shown in Figure 1.3. It is cumbersome to carry the pu in the subscript past this point: no confusion should result, since all quantities will show units, including pu.
Example 1.2
The three-winding single-phase transformer of Figure 1.1 is rated at 13.8 kV/138 kV/4.157 kV and 50 MVA/40 MVA/10 MVA. Terminations are as follows:
13.8 kV winding: 13.8 kV Source
138 kV winding: 35 MVA load, pf = 0.866 lagging
4.157 kV winding: 5 MVA load, pf = 0.866 leading
Using Sbase = 10 MVA, and voltage ratings as bases,
(a) Draw the pu equivalent circuit.
(b) Solve for the primary current, power, and power, and power factor.
Solution
(a) See Figure 1.4.
(b)
All values in Per-Unit Equivalent Circuit:
1.2 Per-Unit Scaling Extended to Three-Phase Systems
The extension to three-phase systems has been complicated to some extent by the use of traditional terminology and jargon, and a desire to normalize phase-to-phase and phase-to-neutral voltage simultaneously. The problem with this practice is that it renders Kirchhoff’s voltage and current laws invalid in some circuits. Consider the general three-phase situation in Figure 1.5, with all quantities in SI units.
Define the complex operator:
The system is said to be balanced, with sequence abc, if
and
Likewise:
If the load consists of wye-connected impedance:
The equivalent delta element is
To convert to per-unit, define the following bases:
S3ϕbase = The three-phase apparent base at a specific location in a three-phase system, in VA.
VLbase = The line (phase-to-phase) rms voltage base at a specific location in a three-phase system, in V.
From the above, define:
(1.19) |
(1.20) |
It follows that
(1.21) |
(1.22) |
An example will be useful.
Example 1.3
Consider a balanced three-phase 60 MVA 0.8 pf lagging load, sequence abc operating from a 13.8 kV (line voltage) bus. On bases of S3ϕbase = 100 MVA and VLbase = 13.8 kV:
(a) Determine all bases.
(b) Determine all voltages, currents, and impedances in SI units and per-unit.
Solution
(a)
(b)
Converting voltages and currents to symmetrical components:
Inclusion of transformers demonstrates the advantages of per-unit scaling.
Example 1.4
A 3ϕ 240 kV <Insert Symbol1>:15 kV <Insert Symbol2> transformer supplies a 13.8 kV 60 MVA pf = 0.8 lagging load, and is connected to a 230 kV source on the high-voltage (HV) side, as shown in Figure 1.6.
(a) Determine all base values on both sides for S3ϕbase = 100 MVA. At the low-voltage (LV) bus, VLbase = 13.8 kV.
(b) Draw the positive sequence circuit in per-unit, modeling the transformer as ideal.
(c) Determine all currents and voltages in SI and per-unit.
Solution
(a) Base values on the LV side are the same as in Example 1.3. The turns ratio may be derived from the voltage ratings ratios:
Results are presented in the following chart.
(b)
The positive sequence circuit is shown as Figure 1.7.
(c) All values determined in pu are valid on both sides of the transformer! To determine SI values on the HV side, use HV bases. For example:
Example 1.5
Repeat the previous example using a 3ϕ 240 kV:15 kV <Insert Symbol1> Δ
Solution
All results are the same as before. The reasoning is as follows.
The voltage ratings are interpreted as line (phase-to-phase) values independent of connection (wye or delta). Therefore the turns ratio remains:
As before:
However, Van is no longer in phase on both sides. This is a consequence of the transformer model, and not due to the scaling procedure. Whether this is important depends on the details of the analysis.
1.3 Per-Unit Scaling Extended to a General Three-Phase System
The ideas presented are extended to a three-phase system using the following procedure.
1. Select a three-phase apparent power base (S3ph base), which is typically 1, 10, 100, or 1000 MVA. This base is valid at every bus in the system.
2. Select a line voltage base (VL base), user defined, but usually the nominal rms line-to-line voltage at a user-defined bus (call this the “reference bus”).
3. Compute
(1.23) |
4. At the reference bus:
(1.24) |
(1.25) |
(1.26) |
5. To determine the bases at the remaining busses in the system, start at the reference bus, which we will call the “from” bus, and execute the following procedure: Trace a path to the next nearest bus, called the “to” bus. You reach the “to” bus by either passing over (a) a line or (b) a transformer.
a. The “line” case: VL base is the same at the “to” bus as it was at the “from” bus. Use Equations 1.2 through 1.4 to compute the “to” bus bases.
b. The “transformer” case: Apply VL base at the “from” bus, and treat the transformer as ideal. Calculate the line voltage that appears at the “to” bus. This is now the new VL base at the “to” bus. Use Equations 1.2 through 1.4 to compute the “to” bus bases.
Rename the bus at which you are located, the “from” bus. Repeat the above procedure until you have processed every bus in the system.
6. We now have a set of bases for every bus in the system, which are to be used for every element terminated at that corresponding bus. Values are scaled according to
where actual value = the actual complex value of S, V, Z, or I, in SI units (VA, V, Ω, A); base value = the (user-defined) base value (real) of S, V, Z, or I, in SI units (VA, V, Ω, A); per-unit value = the per-unit complex value of S, V, Z, or I, in per-unit (dimensionless).
Finally, the reader is advised that there are many scaling systems used in engineering analysis, and, in fact, several variations of per-unit scaling have been used in electric power engineering applications. There is no standard system to which everyone conforms in every detail. The key to successfully using any scaling procedure is to understand how all base values are selected at every location within the power system. If one receives data in per-unit, one must be in a position to convert all quantities to SI units. If this cannot be done, the analyst must return to the data source for clarification on what base values were used.