# Per-unit system

In the power systems analysis field of electrical engineering, a per-unit system is the expression of system quantities as fractions of a defined base unit quantity. Calculations are simplified because quantities expressed as per-unit do not change when they are referred from one side of transformer to the other. This can be a pronounced advantage in power system analysis where large numbers of transformers may be encountered. Moreover, similar types of apparatus will have the impedances lying within a narrow numerical range when expressed as a per-unit fraction of the equipment rating, even if the unit size varies widely. Conversion of per-unit quantities to volts, ohms, or amperes requires a knowledge of the base that the per-unit quantities were referenced to.

The main idea of a per unit system is to absorb large difference in absolute values into base relationships. Thus, representations of elements in the system with per unit values become more uniform.

A per-unit system provides units for; power, voltage, current, impedance, and admittance. Except impedance and admittance, any two of these are independent and can be arbitrarily selected as base values, usually power and voltage. All quantities are specified as multiples of selected base values. For example, the base power might be the rated power of a transformer, or perhaps an arbitrarily selected power which makes power quantities in the system more convenient. The base voltage might be the nominal voltage of a bus. Different types of quantities are labeled with the same symbol (pu); it should be clear from context whether the quantity is a voltage, current, etc.

Per-unit being used in power flow, short-circuit and motor starting studies, it is important for all power engineers to be familiar with the concept.

## Purpose

There are several reasons for using a per-unit system:

• Similar apparatus (generators, transformers, lines) will have similar per-unit impedances and losses expressed on their own rating, regardless of their absolute size. Because of this, per-unit data can be checked rapidly for gross errors. A per unit value out of normal range is worth looking into for potential errors.
• Manufacturers usually specify the impedance of apparati in per unit values.
• Use of the constant ${\displaystyle \scriptstyle {\sqrt {3}}}$ is reduced in three-phase calculations.
• Per-unit quantities are the same on either side of a transformer, independent of voltage level
• By normalizing quantities to a common base, both hand and automatic calculations are simplified.
• It improves numerical stability of automatic calculation methods

The per-unit system was developed to make manual analysis of power systems easier. Although power-system analysis is now done by computer, results are often expressed as per-unit values on a convenient system-wide base.

## Base quantities

Generally base values of power and voltage are chosen. The base power may be the rating of a single piece of apparatus such as a motor or generator. If a system is being studied, the base power is usually chosen as a convenient round number such as 10 MVA or 100 MVA. The base voltage is chosen as the nominal rated voltage of the system. All other base quantities are derived from these two base quantities. Once the base power and the base voltage are chosen, the base current and the base impedance are determined by the natural laws of electrical circuits. Note the base value should only be magnitudes, while the per-unit values are phasors. The phase angles of complex power, voltage, current, impedance etc. are not affected by the conversion to per unit values.

By convention, we adopt the following two rules for base quantities:

• The value of base power is the same for the entire power system of concern.
• The ratio of the voltage bases on either side of a transformer is selected to be the same as the ratio of the transformer voltage ratings.

With these two rules, a per-unit impedance remains unchanged when referred from one side of a transformer to the other. This allows us to eliminate ideal transformer from a transformer model.

## Relationship between units

The relationship between units in a per-unit system depends on whether the system is single-phase or three-phase.

### Single-phase

Assuming that the independent base values are power and voltage, we have:

${\displaystyle P_{base}=1pu}$
${\displaystyle V_{base}=1pu}$

Alternatively, the base value for power may be given in terms of reactive or apparent power, in which case we have, respectively,

${\displaystyle Q_{base}=1pu}$

or

${\displaystyle S_{base}=1pu}$

The rest of the units can be derived from power and voltage using the equations ${\displaystyle S=IV}$, ${\displaystyle P=Scos(\phi )}$, ${\displaystyle Q=Ssin(\phi )}$ and ${\displaystyle {\underline {V}}={\underline {I}}{\underline {Z}}}$ (Ohm's law), ${\displaystyle Z}$ being represented by ${\displaystyle {\underline {Z}}=R+jX=Zcos(\phi )+jZsin(\phi )}$. We have:

${\displaystyle I_{base}={\frac {S_{base}}{V_{base}}}=1pu}$
${\displaystyle Z_{base}={\frac {V_{base}}{I_{base}}}={\frac {V_{base}^{2}}{I_{base}V_{base}}}={\frac {V_{base}^{2}}{S_{base}}}=1pu}$
${\displaystyle Y_{base}={\frac {1}{Z_{base}}}=1pu}$

### Three-phase

Power and voltage are specified in the same way as single-phase systems. However, due to differences in what these terms usually represent in three-phase systems, the relationships for the derived units are different. Specifically, power is given as total (not per-phase) power, and voltage is line-to-line voltage. In three-phase systems the equations ${\displaystyle P=Scos(\phi )}$ and ${\displaystyle Q=Ssin(\phi )}$ also hold. The apparent power S now equals ${\displaystyle S_{base}={\sqrt {3}}V_{base}I_{base}}$

${\displaystyle I_{base}={\frac {S_{base}}{V_{base}\times {\sqrt {3}}}}=1pu}$
${\displaystyle Z_{base}={\frac {V_{base}}{I_{base}\times {\sqrt {3}}}}={\frac {V_{base}^{2}}{S_{base}}}=1pu}$
${\displaystyle Y_{base}={\frac {1}{Z_{base}}}=1pu}$

## Example of per-unit

As an example of how per-unit is used, consider a three-phase power transmission system that deals with powers of the order of 500 MW and uses a nominal voltage of 138 kV for transmission. We arbitrarily select ${\displaystyle S_{base}=500}$ MVA, and use the nominal voltage 138 kV as the base voltage ${\displaystyle V_{base}}$. We then have:

${\displaystyle I_{base}={\frac {S_{base}}{V_{base}\times {\sqrt {3}}}}=2.09\,{\mathrm {kA} }}$
${\displaystyle Z_{base}={\frac {V_{base}}{I_{base}\times {\sqrt {3}}}}={\frac {V_{base}^{2}}{S_{base}}}=38.1\,\Omega }$
${\displaystyle Y_{base}={\frac {1}{Z_{base}}}=26.3\,{\mathrm {mS} }}$

If, for example, the actual voltage at one of the buses is measured to be 136 kV, we have:

${\displaystyle V_{pu}={\frac {V}{V_{base}}}={\frac {136\,{\mathrm {kV} }}{138\,{\mathrm {kV} }}}=0.9855\,{\mathrm {pu} }}$

## Per-unit system formulas

The following tabulation of per-unit system formulas is adapted from Beeman's Industrial Power Systems Handbook.

## References

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