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In electrical engineering in the field of power transmission 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 are the same regardless of the voltage level. Similar types of apparatus will have impedances, voltage drops and losses that are the same 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. ## Purpose

## Relationship between units

### Single phase

### Three phase

## Example of per-unit

## References

A per-unit system provides units for power, voltage, current, impedance, and admittance. Only two of these are independent, 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 is used primarily in power flow studies; however, because parameters of transformers and machines (electric motors and electrical generators) are often specified in terms of per-unit, it is important for all power engineers to be familiar with the concept.

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

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

- $P\_\{base\}\; =\; 1\; pu$

- $V\_\{base\}\; =\; 1\; pu$

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

- $Q\_\{base\}\; =\; 1\; pu$

or

- $S\_\{base\}\; =\; 1\; pu$

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

- $I\_\{base\}\; =\; frac\{S\_\{base\}\}\{V\_\{base\}\}\; =\; 1\; pu$

- $Z\_\{base\}\; =\; frac\{V\_\{base\}\}\{I\_\{base\}\}\; =\; frac\{V\_\{base\}^\{2\}\}\{I\_\{base\}V\_\{base\}\}\; =\; frac\{V\_\{base\}^\{2\}\}\{S\_\{base\}\}\; =\; 1\; pu$

- $Y\_\{base\}\; =\; frac\{1\}\{Z\_\{base\}\}\; =\; 1\; pu$

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 $P\; =\; Scos(phi)$ and $Q\; =\; Ssin(phi)$ also hold. The apparent power S now equals $S\_\{base\}=\; sqrt\{3\}V\_\{base\}\; I\_\{base\}$

- $I\_\{base\}\; =\; frac\{S\_\{base\}\}\{V\_\{base\}\; times\; sqrt\{3\}\}\; =\; 1\; pu$

- $Z\_\{base\}\; =\; frac\{V\_\{base\}\}\{I\_\{base\}\; times\; sqrt\{3\}\}\; =\; frac\{V\_\{base\}^2\}\{S\_\{base\}\}\; =\; 1\; pu$

- $Y\_\{base\}\; =\; frac\{1\}\{Z\_\{base\}\}\; =\; 1\; pu$

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

- $Z\_\{base\}\; =\; frac\{V\_\{base\}^\{2\}\}\{S\_\{base\}\}\; =\; 38.1\; Omega$

- $I\_\{base\}\; =\; frac\{S\_\{base\}\}\{V\_\{base\}\; times\; sqrt\{3\}\}\; =\; 2.09\; kA$

- $Z\_\{base\}\; =\; frac\{V\_\{base\}\}\{I\_\{base\}\; times\; sqrt\{3\}\}\; =\; 38.1\; Omega$

- $Y\_\{base\}\; =\; frac\{1\}\{Z\_\{base\}\}\; =\; 26.3\; mS$

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

- $V\_\{pu\}\; =\; frac\{V\}\{V\_\{base\}\}\; =\; frac\{136\; kV\}\{138\; kV\}\; =\; 0.9855\; pu$

- William D. Stevenson, Jr. Elements of Power System Analysis Third Edition,McGraw-Hill, New York (1975) ISBN 0-07-061285-4
- B. M. Weedy, Electric Power Systems Second Edition, John Wiley and Sons, London, 1972, ISBN 0-471-92445-8

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Last updated on Wednesday September 24, 2008 at 00:21:32 PDT (GMT -0700)

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Last updated on Wednesday September 24, 2008 at 00:21:32 PDT (GMT -0700)

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