1

The Per-Unit System

Charles A. Gross

Auburn University

1.1    Impact on Transformers

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

$Per\text{-}unitvalue=\frac{actualvalue}{basevalue}.$

(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

$S=V{I}^{\star }$

(1.2)

or

$S\angle \text{θ}\text{=}\text{V}\angle \text{αI}\angle -\text{β}$

where

V = phasor voltage, in volts

I = phasor current, in amperes

Suppose we arbitrarily pick a value S_base, a real number with the units of volt-amperes. Dividing through by S_base,

$\frac{\text{S}\angle \text{θ}}{{\text{S}}_{\text{base}}}=\frac{\text{V}\angle \text{αI}\angle -\text{β}}{{\text{S}}_{\text{base}}}.$

We further define

${\text{V}}_{\text{base}}{\text{I}}_{\text{base}}={\text{S}}_{\text{base}}.$

(1.3)

Either V_base or I_base may be selected arbitrarily, but not both. Substituting Equation 1.3 into Equation 1.2, we obtain

$\begin{array}{ll}\frac{\text{S}\angle \text{θ}}{{\text{S}}_{\text{base}}}\hfill & =\frac{\text{V}\angle \text{α}\left(\text{I}\angle -\text{β}\right)}{{\text{V}}_{\text{base}}{\text{I}}_{\text{base}}}\hfill \\ {\text{S}}_{\text{pu}}\angle \text{θ}\hfill & =\left(\frac{\text{V}\angle \text{α}}{{\text{V}}_{\text{base}}}\right)\left(\frac{\text{I}\angle -\text{β}}{{\text{I}}_{\text{base}}}\right)\hfill \\ S_{\text{pu}}\hfill & ={\text{V}}_{\text{pu}}\angle \text{α}\left({\text{I}}_{\text{pu}}\angle -\text{β}\right)\hfill \\ S_{\text{pu}}\hfill & =V_{\text{pu}}{I}_{\text{pu}}\star \hfill \end{array}$

(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 V_base I_base and S_base through Equation 1.3. If we select Z_base by

${\text{Z}}_{\text{base}}=\frac{{\text{V}}_{\text{base}}}{{\text{I}}_{\text{base}}}=\frac{{\text{V}}_{\text{base}}^2}{{\text{S}}_{\text{base}}}.$

(1.5)

Convert Ohm’s law:

$Z=\frac{V}{I}$

(1.6)

into per-unit by dividing by Z_base.

$\begin{array}{ll}\frac{Z}{{\text{Z}}_{\text{base}}}\hfill & =\frac{V/I}{{\text{Z}}_{\text{base}}}\hfill \\ Z_{\text{pu}}\hfill & =\frac{V/{\text{V}}_{\text{base}}}{I/{\text{I}}_{\text{base}}}=\frac{V_{\text{pu}}}{I_{\text{pu}}}.\hfill \end{array}$

Observe that

$\begin{array}{l}{Z}_{\text{pu}}=\frac{Z}{{\text{Z}}_{\text{base}}}=\frac{\text{R}+\text{jX}}{{\text{Z}}_{\text{base}}}=\left(\frac{\text{R}}{{\text{Z}}_{\text{base}}}\right)+\text{j}\left(\frac{\text{X}}{{\text{Z}}_{\text{base}}}\right)\hfill \\ {\text{Z}}_{\text{pu}}={\text{R}}_{\text{pu}}+{\text{jX}}_{\text{pu}}\hfill \end{array}$

(1.7)

Thus, separate bases for R and X are not necessary:

${\text{Z}}_{\text{base}}={\text{R}}_{\text{base}}={\text{X}}_{\text{base}}$

By the same logic,

${\text{S}}_{\text{base}}={\text{P}}_{\text{base}}={\text{Q}}_{\text{base}}$

For power system applications, base values for S_base and V_base are arbitrarily selected. Actually, in practice, values are selected that force results into certain ranges. Thus, for V_base, 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.

1.1    Impact on Transformers

To understand the impact of pu scaling on transformer, consider the three-winding ideal device (see Figure 1.2).

For sinusoidal steady-state performance:

$V_1=\frac{{\text{N}}_1}{{\text{N}}_2}{V}_2$

(1.8a)

$V_2=\frac{{\text{N}}_2}{{\text{N}}_3}{V}_3$

(1.8b)

$V_3=\frac{{\text{N}}_3}{{\text{N}}_1}{V}_1$

(1.8c)

and

${\text{N}}_1{I}_1+{\text{N}}_2{I}_2+{\text{N}}_3{I}_3=0$

(1.9)

FIGURE 1.2  The three-winding ideal transformer.

Consider the total input complex power S.

$\begin{array}{ll}S\hfill & =V_1{I}_1\star +V_2{I}_2\star +V_3{I}_3\star \hfill \\ \hfill & =V_1{I}_1\star +\frac{{\text{N}}_2}{{\text{N}}_1}{V}_1{I}_2\star +\frac{{\text{N}}_3}{{\text{N}}_1}{V}_1{I}_3\star \hfill \\ \hfill & =\frac{V_1}{{\text{N}}_1}{\left[{\text{N}}_1{I}_1+{\text{N}}_2{I}_2+{\text{N}}_3{I}_3\right]}^{\star }\hfill \\ \hfill & =0\hfill \end{array}$

(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 V_1base and S_1base. Require base values for windings 2 and 3 to be

${\text{V}}_{2\text{base}}=\frac{{\text{N}}_2}{{\text{N}}_1}{\text{V}}_{1\text{base}}$

(1.11a)

${\text{V}}_{3\text{base}}=\frac{{\text{N}}_3}{{\text{N}}_1}{\text{V}}_{1\text{base}}$

(1.11b)

and

${\text{S}}_{1\text{base}}={\text{S}}_{2\text{base}}={\text{S}}_{3\text{base}}={\text{S}}_{\text{base}}$

(1.12)

By definition,

${\text{I}}_{1\text{base}}=\frac{{\text{S}}_{\text{base}}}{{\text{V}}_{1\text{base}}}$

(1.13a)

${\text{I}}_{2\text{base}}=\frac{{\text{S}}_{\text{base}}}{{\text{V}}_{2\text{base}}}$

(1.13b)

${\text{I}}_{3\text{base}}=\frac{{\text{S}}_{\text{base}}}{{\text{V}}_{3\text{base}}}$

(1.13c)

It follows that

${\text{I}}_{2\text{base}}=\frac{{\text{N}}_1}{{\text{N}}_2}{\text{I}}_{1\text{base}}$

(1.14a)

${\text{I}}_{3\text{base}}=\frac{{\text{N}}_1}{{\text{N}}_3}{\text{I}}_{1\text{base}}$

(1.14b)

Recall that a per-unit value is the actual value divided by its appropriate base. Therefore:

$\frac{V_1}{{\text{V}}_{1\text{base}}}=\frac{\left({\text{N}}_1/{\text{N}}_2\right)V_2}{{\text{V}}_{1\text{base}}}$

(1.15a)

and

$\frac{V_1}{{\text{V}}_{1\text{base}}}=\frac{\left({\text{N}}_1/{\text{N}}_2\right)V_2}{\left({\text{N}}_1/{\text{N}}_2\right){\text{V}}_{2\text{base}}}$

(1.15b)

or

$V_{1\text{pu}}=V_{2\text{pu}}$

(1.15c)

indicates per-unit values. Similarly,

$\frac{V_1}{{\text{V}}_{1\text{base}}}=\frac{\left({\text{N}}_1/{\text{N}}_3\right)V_3}{\left({\text{N}}_1/{\text{N}}_3\right){\text{V}}_{3\text{base}}}$

(1.16a)

or

$V_{1\text{pu}}=V_{3\text{pu}}$

(1.16b)

Summarizing:

${\text{V}}_{1\text{pu}}={\text{V}}_{2\text{pu}}={\text{V}}_{3\text{pu}}$

(1.17)

Divide Equation 1.9 by N₁

$I_1+\frac{{\text{N}}_2}{{\text{N}}_1}{I}_2+\frac{{\text{N}}_3}{{\text{N}}_1}{I}_3=0$

Now divide through by I_1base

$\begin{array}{l}\frac{I_1}{{\text{I}}_{1\text{base}}}+\frac{\left({\text{N}}_2/{\text{N}}_1\right)I_2}{{\text{I}}_{1\text{base}}}+\frac{\left({\text{N}}_3/{\text{N}}_1\right)I_3}{{\text{I}}_{1\text{base}}}=0\hfill \\ \frac{I_1}{{\text{I}}_{1\text{base}}}+\frac{\left({\text{N}}_2/{\text{N}}_1\right)I_2}{\left({\text{N}}_2/{\text{N}}_1\right){\text{I}}_{2\text{base}}}+\frac{\left({\text{N}}_3/{\text{N}}_1\right)I_3}{\left({\text{N}}_3/{\text{N}}_1\right){\text{I}}_{3\text{base}}}=0\hfill \end{array}$

Simplifying to

$I_{1\text{pu}}+I_{2\text{pu}}+I_{3\text{pu}}=0$

(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.

FIGURE 1.3  Single-phase ideal transformer.

FIGURE 1.4  Per-unit 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:

$a=1\angle 120°$

The system is said to be balanced, with sequence abc, if

$\begin{array}{l}{V}_{\text{bn}}=a^2{V}_{\text{an}}\hfill \\ V_{\text{cn}}=a{V}_{\text{an}}\hfill \end{array}$

and

$\begin{array}{ll}{I}_{\text{b}}\hfill & =a^2{I}_{\text{a}}\hfill \\ I_{\text{c}}\hfill & =a{I}_{\text{a}}\hfill \\ -I_{\text{n}}\hfill & =I_{\text{a}}+I_{\text{b}}+I_{\text{c}}=0\hfill \end{array}$

Likewise:

$\begin{array}{l}{V}_{\text{ab}}=V_{\text{an}}-V_{\text{bn}}\hfill \\ V_{\text{bc}}=V_{\text{bn}}-V_{\text{cn}}=a^2{V}_{\text{ab}}\hfill \\ V_{\text{ca}}=V_{\text{cn}}-V_{\text{an}}=a{V}_{\text{ab}}\hfill \end{array}$

If the load consists of wye-connected impedance:

$Z_{\text{y}}=\frac{V_{\text{an}}}{I_{\text{a}}}=\frac{V_{\text{bn}}}{I_{\text{b}}}=\frac{V_{\text{cn}}}{I_{\text{c}}}$

The equivalent delta element is

$Z_{\Delta }=3Z_{\text{Y}}$

FIGURE 1.5  General three-phase system.

To convert to per-unit, define the following bases:

S_3ϕbase = The three-phase apparent base at a specific location in a three-phase system, in VA.

V_Lbase = The line (phase-to-phase) rms voltage base at a specific location in a three-phase system, in V.

From the above, define:

${\text{S}}_{\text{base}}=\frac{{\text{S}}_{3\varphi \text{base}}}{3}$

(1.19)

${\text{V}}_{\text{base}}=\frac{{\text{V}}_{\text{Lbase}}}{\sqrt{3}}$

(1.20)

It follows that

${\text{I}}_{\text{base}}=\frac{{\text{S}}_{\text{base}}}{{\text{V}}_{\text{base}}}$

(1.21)

${\text{Z}}_{\text{base}}=\frac{{\text{V}}_{\text{base}}}{{\text{I}}_{\text{base}}}$

(1.22)

An example will be useful.

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 (S_{3ph 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 (V_{L base}), user defined, but usually the nominal rms line-to-line voltage at a user-defined bus (call this the “reference bus”).

3.  Compute

${\text{S}}_{\text{base}}=\frac{\left({\text{S}}_{3\text{ph}\text{base}}\right)}{3}\left(\text{Valid}\text{at}\text{every bus}\right)$

(1.23)

4.  At the reference bus:

${\text{V}}_{\text{base}}=\frac{{\text{V}}_{\text{L}\text{base}}}{\sqrt{3}}$

(1.24)

${\text{I}}_{\text{base}}=\frac{{\text{S}}_{\text{base}}}{{\text{V}}_{\text{base}}}$

(1.25)

${\text{Z}}_{\text{base}}=\frac{{\text{V}}_{\text{base}}}{I_{\text{base}}}=\frac{{\text{V}}_{\text{base}}^2}{{\text{S}}_{\text{base}}}$

(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: V_{L 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 V_{L 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 V_{L 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

$per\text{-}unitvalue=actualvalue/basevalue$

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.