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 a 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 evaluation 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 $\scriptstyle \sqrt{3}$ 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
• Per unit data representation yields important information about relative magnitudes.

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.

As we have already known, the purpose of introducing per-unit system is to simplify our conversion between different transformers. Hence, I’d like to illustrate the steps for finding per-unit values for voltage and impedance. First, let the base power (Sbase) of each end of the transformers becomes the same. Once we set every S on the same base, we can get base voltage and base impedance for every transformer easily. The numbers we have till now are all base on the same unit. Next, substitute the real numbers of impedances and voltages into the per-unit calculation definition. We can get the answers for per-unit system. In other case, if we’ve known the per-unit values at first, we can get the real values by timing base values.

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. As we have already known, the purpose of introducing per-unit system is to simplify our conversion between different transformers. Hence, it is appropriate to illustrate the steps for finding per-unit values for voltage and impedance. First, let the base power (S_base) of each end of a transformer become the same. Once we set every S on the same base, we can get base voltage and base impedance for every transformer easily. The numbers we have until now are all based on the same unit. Next, substitute the real numbers of impedances and voltages into the per-unit calculation definition. We can get the answers for the per-unit system. In other case, if we have known the per-unit values at first, we can get the real values by multiplying by the base 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:

$P_{\text{base}} = 1 \mathrm{pu}$

$V_{\text{base}} = 1 \mathrm{pu}$

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

$Q_{\text{base}} = 1 \mathrm{pu}$

or

$S_{\text{base}} = 1 \mathrm{pu}$

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

$I_{\text{base}} = \frac{S_{\text{base}}}{V_{\text{base}}} = 1 \mathrm{pu}$

$Z_{\text{base}} = \frac{V_{\text{base}}}{I_{\text{base}}} = \frac{V_{\text{base}}^{2}}{I_{\text{base}}V_{\text{base}}} = \frac{V_{\text{base}}^{2}}{S_{\text{base}}} = 1 \mathrm{pu}$

$Y_{\mathrm{base}} = \frac{1}{Z_{\mathrm{base}}} = 1 \mathrm{pu}$

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 $P = S\cos(\phi)$ and $Q = S\sin(\phi)$ also hold. The apparent power $S$ now equals $S_{\mathrm{base}}= \sqrt{3}V_{\mathrm{base}} I_{\mathrm{base}}$

$I_{\mathrm{base}} = \frac{S_{\mathrm{base}}}{V_{\mathrm{base}} \times \sqrt{3}} = 1 \mathrm{pu}$

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

$Y_{\mathrm{base}} = \frac{1}{Z_{\mathrm{base}}} = 1 \mathrm{pu}$

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 $S_{\mathrm{base}} = 500\, \mathrm{MVA}$, and use the nominal voltage 138 kV as the base voltage $V_{\mathrm{base}}$. We then have:

$I_{\text{base}} = \frac{S_{\text{base}}}{V_{\text{base}} \times \sqrt{3}} = 2.09 \, \mathrm{kA}$

$Z_{\text{base}} = \frac{V_{\text{base}}}{I_{\text{base}} \times \sqrt{3}} = \frac{V_{\text{base}}^{2}}{S_{\text{base}}} = 38.1 \, \Omega$

$Y_{\mathrm{base}} = \frac{1}{Z_{\mathrm{base}}} = 26.3 \, \mathrm{mS}$

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

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

Per-unit system formulas

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

Equation
$\text{Base number selection}$
	$\text{Arbitrarily selecting from ohm's law the two base numbers: base voltage and base current}$
$1$	$\text{We have, Z}=\frac{E}{I}$
$2$	$\text{Base ohms}=\frac{\text{base volts}}{\text{base amperes}}$
$3$	$\text{Per-unit volts}=\frac{\text{volts}}{\text{base volts}}$
$4$	$\text{Per-unit amperes}=\frac{\text{amperes}}{\text{base amperes}}$
$5$	$\text{Per-unit ohms}=\frac{\text{ohms}}{\text{base ohms}}$
	$\text{Alternatively, choosing base volts and base kva values, we have,}$
	$\text{in single-phase systems:}$
$6$	$\text{Base amperes }=\frac{\text{base kva * 1000}}{\text{base volts}}$
$7$	$\text{Base amperes }=\frac{\text{base kva}}{\text{base kv}_{L-L}}$
$8$	$\text{Base ohms }=\frac{\text{base volts}}{\text{base amperes}}$
	$\text{and in three-phase systems:}$
$9$	$\text{Base amperes }=\frac{\text{base kva * 1000}}{\sqrt{3 } * \text{base volts}}$
$10$	$\text{Base amperes }=\frac{\text{base kva}}{\sqrt{3 } * \text{base kv}_{L-L}}$
$11$	$\text{Base ohms }=\frac{\text{base volts}}{\sqrt{3 } * \text{base amperes}}$
	$\text{Working out for convenience per-unit ohms directly, we have}$
	$\text{for single-phase and three-phase systems:}$
$12$	$\text{Base ohms }=\frac{\text{ohms * base kva}}{kv_{L-L}^2 * 1000}$
$\text{Short-Circuit Calculation Formulas}$
	$\text{Ohms conversions:}$
$13$	$\text{Per-unit ohms reactance} =\frac{\text{ohms reactance * }\text{kva base}}{kv_{L-L}^2*1000}$
$14$	$\text{Ohms reactance} =\frac{%\text{ reactance}*kv_{L-L}^2 * 10}{\text{kva base}}$
$15$	$\text{Per-unit ohms reactance} = \frac{\text{per cent ohms reactance}}{100}$
	$\text{Changing ohms from one kva base to another:}$
$16$	$\%\text{ ohms reactance on kva base}_2=\frac{\text{kva base}_2}{\text{kva base}_1} * \%\text{ ohms reactance on base}_1$
$17$	$\text{0/1 ohms reactance on kva base}_2=\frac{\text{kva base}_2}{\text{kva base}_1}\text{ * 0/1 ohms reactance on base}_1$
	$\text{Changing incoming system reactance:}$
	$\text{a. If system reactance is given in percent, use Eq. 16 to change from one kva base to another.}$
	$\text{b. If system reactance is given in short-circuit symmetrical rms kva or current, convert to per-unit as follows:}$
$18$	$\text{0/1 reactance} = \frac{\text{kva base used in reactance in studied calculation}}{\text{system short-circuit kva}}$
$19$	$\text{0/1 reactance} = \frac{\text{kva base used in reactance in studied calculation}}{\text{system short-circuit current * }\sqrt{3} \text{ * system kv}_{L-L}}$
	$\text{Calculating approximate motor kva base:}$
	$\text{a. For induction motors and 0.8 power factor synchronous motors}$
$20$	$\text{kva base} \approx\text{ horsepower rating}$
	$\text{b. For unity power factor synchronous motors}$
$21$	$\text{kva base} \approx\text{ 0.8 * horsepower rating}$
	$\text{Converting ohms from one voltage to another:}$
$22$	$\text{Ohms on basis of voltage}_1 =(\frac{\text{voltage}_1}{\text{voltage}_2})^2 \text{ * ohms on basis of voltage}_2$
	$\text{Short-circuit kva and current calculations}$
	$\text{Symmetrical short circuit kva:}$
$23$	$=\frac{\text{100 * kva base}}{%\text{ X}}$
$24$	$=\frac{\text{kva base}}{\text{0/1 X}}$
$25$	$=3 * \frac{\text{Voltage}_{L-N}^2}{\text{ohms reactance}\text{ * 1000}}$
$26$	$=\frac{\text{kv}_{L-L}^2 \text{ * 1000}}{\text{ohms reactance}}$
	$\text{Symmetrical short circuit current:}$
$27$	$=\frac{\text{100 * kva base}}{%\text{ X}* \sqrt{3} *\text{kv}_{L-L}}$
$28$	$=\frac{\text{kva base}}{\text{0/1 X}* \sqrt{3} *\text{kv}_{L-L}}$
$29$	$=\frac{\text{kv}_{L-L} \text{ * 1000}}{\sqrt{3} * \text{ohms reactance}}$
	$\text{Asymmetrical short-circuit current and kva:}$
$30$	$\text{Asymmetrical short-circuit current = symmetrical current * X/R factor }$
$31$	$\text{Asymmetrical short-circuit kva = symmetrical kva * X/R factor }$

In transformers

It can be shown that voltages, currents, and impedances in a per-unit system will have the same values whether they are referred to primary or secondary of a transformer.

The full load copper loss of a transformer in per-unit form is equal to the per-unit value of its resistance:

$\begin{align} P_{cu,FL}&=\text{full-load copper loss}\\ &= I_{R1}^2R_{eq1}\\ \end{align}$

$\begin{align} P_{cu,FL,pu}&=\frac{P_{cu,FL}}{P_{base}}\\ &= \frac {I_{R1}^2R_{eq1}} {V_{R1}I_{R1}}\\ &= \frac {R_{eq1}} {V_{R1}/I_{R1}}\\ &= \frac {R_{eq1}} {Z_{B1}}\\ &= R_{eq1,pu}\\ \end{align}$

Therefore, it may be more useful to express the resistance in per-unit form as it also represents the full-load copper loss.