Per-unit system

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. The base value should only be a magnitude, while the per-unit value is a phasor. The phase angles of complex power, voltage, current, impedance, etc., are not affected by the conversion to per unit values.

The purpose of using a per-unit system is to simplify 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 every S is set on the same base, the base voltage and base impedance for every transformer can easily be obtained. Then, the real numbers of impedances and voltages can be substituted into the per-unit calculation definition to get the answers for the per-unit system. If the per-unit values are known, the real values can be obtained by multiplying by the base values.

By convention, the following two rules are adopted for base quantities:

• The base power value 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 the ideal transformer to be eliminated 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:

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

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

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

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

or

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

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

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

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

${\displaystyle Y_{\text{base}}=\frac{1}{Z_{\text{base}}}=1\text{ pu}}$

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 ${\displaystyle P=S\cos (\varphi )}$ and ${\displaystyle Q=S\sin (\varphi )}$ also hold. The apparent power ${\displaystyle S}$ now equals ${\displaystyle S_{\text{base}}=\sqrt{3}{V}_{\text{base}}{I}_{\text{base}}}$

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

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

${\displaystyle Y_{\text{base}}=\frac{1}{Z_{\text{base}}}=1\text{ pu}}$

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 ${\displaystyle S_{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}}=500\mathrm{M}\mathrm{V}\mathrm{A}}$, and use the nominal voltage 138 kV as the base voltage ${\displaystyle V_{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}}}$. We then have:

${\displaystyle I_{\text{base}}=\frac{S_{\text{base}}}{V_{\text{base}}\times \sqrt{3}}=2.09\mathrm{k}\mathrm{A}}$

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

${\displaystyle Y_{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}}=\frac{1}{Z_{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}}}=26.3\mathrm{m}\mathrm{S}}$

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

${\displaystyle V_{\mathrm{p}\mathrm{u}}=\frac{V}{V_{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}}}=\frac{136\mathrm{k}\mathrm{V}}{138\mathrm{k}\mathrm{V}}=0.9855\mathrm{p}\mathrm{u}}$

Per-unit system formulas

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

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.

For instance, for voltage, we can prove that the per unit voltages of two sides of the transformer, side 1 and side 2, are the same. Here, the per-unit voltages of the two sides are E_1pu and E_2pu respectively.

${\displaystyle \begin{array}{cc}{E}_{\text{1,pu}}& =\frac{E_1}{V_{\text{base1}}}\\ & =\frac{N_1{E}_2}{N_2{V}_{\text{base1}}}\\ & =\frac{N_1{E}_2}{N_2\frac{N_1}{N_2}{V}_{\text{base2}}}\\ & =\frac{E_2}{V_{\text{base2}}}\\ & =E_{\text{2,pu}}\end{array}}$

(source: Alexandra von Meier Power System Lectures, UC Berkeley)

E₁ and E₂ are the voltages of sides 1 and 2 in volts. N₁ is the number of turns the coil on side 1 has. N₂ is the number of turns the coil on side 2 has. V_base1 and V_base2 are the base voltages on sides 1 and 2.

${\displaystyle V_{\text{base1}}=\frac{N_1}{N_2}{V}_{\text{base2}}}$

For current, we can prove that the per-unit currents of the two sides are the same below.

${\displaystyle \begin{array}{cc}{I}_{\text{1,pu}}& =\frac{I_1}{I_{\text{base1}}}\\ & =\frac{N_2{I}_2}{N_1{I}_{\text{base1}}}\\ & =\frac{N_2{I}_2}{N_1\frac{N_2}{N_1}{I}_{base2}}\\ & =\frac{I_2}{I_{\text{base2}}}\\ & =I_{\text{2,pu}}\end{array}}$

(source: Alexandra von Meier Power System Lectures, UC Berkeley)

where I_1,pu and I_2,pu are the per-unit currents of sides 1 and 2 respectively. In this, the base currents I_base1 and I_base2 are related in the opposite way that V_base1 and V_base2 are related, in that

${\displaystyle \begin{array}{cc}{I}_{\text{base1}}& =\frac{S_{\text{base1}}}{V_{\text{base1}}}\\ S_{\text{base1}}& =S_{\text{base2}}\\ V_{\text{base2}}& =\frac{N_2}{N_1}{V}_{\text{base1}}\\ I_{\text{base2}}& =\frac{S_{\text{base2}}}{V_{\text{base2}}}\\ I_{\text{base1}}& =\frac{S_{\text{base2}}}{\frac{N_1}{N_2}{V}_{\text{base2}}}\\ & =\frac{N_2}{N_1}{I}_{\text{base2}}\end{array}}$

The reason for this relation is for power conservation

S_base1 = S_base2

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

${\displaystyle \begin{array}{cc}{P}_{\text{cu,FL}}& =\text{full-load copper loss}\\ & =I_{R1}^2{R}_{eq1}\end{array}}$

${\displaystyle \begin{array}{cc}{P}_{\text{cu,FL,pu}}& =\frac{P_{\text{cu,FL}}}{P_{\text{base}}}\\ & =\frac{I_{R1}^2{R}_{eq1}}{V_{R1}{I}_{R1}}\\ & =\frac{R_{\text{eq1}}}{V_{R1}/I_{R1}}\\ & =\frac{R_{\text{eq1}}}{Z_{B1}}\\ & =R_{\text{eq1,pu}}\end{array}}$

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

As stated above, there are two degrees of freedom within the per unit system that allow the engineer to specify any per unit system. The degrees of freedom are the choice of the base voltage (V_base) and the base power (S_base). By convention, a single base power (S_base) is chosen for both sides of the transformer and its value is equal to the rated power of the transformer. By convention, there are actually two different base voltages that are chosen, V_base1 and V_base2 which are equal to the rated voltages for either side of the transformer. By choosing the base quantities in this manner, the transformer can be effectively removed from the circuit as described above. For example:

Take a transformer that is rated at 10 kVA and 240/100 V. The secondary side has an impedance equal to 1∠0° Ω. The base impedance on the secondary side is equal to:

${\displaystyle \begin{array}{cc}{Z}_{\text{base,2}}& =\frac{V_{\text{base,2}}^2}{S_{\text{base}}}\\ & =\frac{(100\text{ V}{)}^2}{10000\text{ VA}}\\ & =\text{1 }\mathrm{\Omega }\end{array}}$

This means that the per unit impedance on the secondary side is 1∠0° Ω / 1 Ω = 1∠0° pu When this impedance is referred to the other side, the impedance becomes:

${\displaystyle \begin{array}{cc}{Z}_2& ={\left(\frac{240}{100}\right)}^2\times \text{1∠0° }\mathrm{\Omega }\\ & =\text{5.76∠0° }\mathrm{\Omega }\end{array}}$

The base impedance for the primary side is calculated the same way as the secondary:

${\displaystyle \begin{array}{cc}{Z}_{\text{base,1}}& =\frac{V_{\text{base,1}}^2}{S_{\text{base}}}\\ & =\frac{(240\text{ V}{)}^2}{10000\text{ VA}}\\ & =\text{5.76 }\mathrm{\Omega }\end{array}}$

This means that the per unit impedance is 5.76∠0° Ω / 5.76 Ω = 1∠0° pu, which is the same as when calculated from the other side of the transformer, as would be expected.

Another useful tool for analyzing transformers is to have the base change formula that allows the engineer to go from a base impedance with one set of a base voltage and base power to another base impedance for a different set of a base voltage and base power. This becomes especially useful in real life applications where a transformer with a secondary side voltage of 1.2 kV might be connected to the primary side of another transformer whose rated voltage is 1 kV. The formula is as shown below.

${\displaystyle \begin{array}{cc}{Z}_{\text{pu,new}}& =Z_{\text{pu,old}}\times \frac{Z_{\text{base,old}}}{Z_{\text{base,new}}}=Z_{\text{pu,old}}\times {\left(\frac{V_{\text{base,old}}}{V_{\text{base,new}}}\right)}^2\times \left(\frac{S_{\text{base,new}}}{S_{\text{base,old}}}\right)\end{array}}$