Editing Per-unit system (section)

== 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]].<ref name="Sen 1997">{{cite book | last=Sen | first=P. C. | title=Principles of electric machines and power electronics | publisher=[[John Wiley & Sons]] | location=New York | year=1997 | isbn=978-0-471-02295-4}}</ref>{{rp|85}}

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''<sub>1pu</sub> and ''E''<sub>2pu</sub> respectively.

:<math>
\begin{align}
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{align}
</math>
(source: Alexandra von Meier Power System Lectures, UC Berkeley)

''E''<sub>1</sub> and ''E''<sub>2</sub> are the voltages of sides 1 and 2 in volts. ''N''<sub>1</sub> is the number of turns the coil on side 1 has. ''N''<sub>2</sub> is the number of turns the coil on side 2 has. ''V''<sub>base1</sub> and ''V''<sub>base2</sub> are the base voltages on sides 1 and 2.
: <math> V_\text{base1}=\frac{N_{1}}{N_{2}}V_\text{base2}</math>

For current, we can prove that the per-unit currents of the two sides are the same below.
:<math>
\begin{align}
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{align}
</math>
(source: Alexandra von Meier Power System Lectures, UC Berkeley)

where ''I''<sub>1,pu</sub> and ''I''<sub>2,pu</sub> are the per-unit currents of sides 1 and 2 respectively. In this, the base currents ''I''<sub>base1</sub> and ''I''<sub>base2</sub> are related in the opposite way that ''V''<sub>base1</sub> and V<sub>base2</sub> are related, in that

:<math>
\begin{align}
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{align}
</math>

The reason for this relation is for power conservation
:''S''<sub>base1</sub> = ''S''<sub>base2</sub>

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

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

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

Therefore, it may be more useful to express the resistance in per-unit form as it also represents the full-load copper loss.<ref name="Sen 1997" />{{rp|86}}

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''{{sub|base}}) and the base power (''S''{{sub|base}}). By convention, a single base power (''S''{{sub|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''{{sub|base1}} and ''V''{{sub|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&nbsp;kVA and 240/100&nbsp;V. The secondary side has an impedance equal to 1∠0°&nbsp;Ω. The base impedance on the secondary side is equal to:

<math>
\begin{align}
Z_\text{base,2}&=\frac{V_\text{base,2}^2}{S_\text{base}}\\
&= \frac {(100\text{ V})^2} {10000\text{ VA}}\\
&= \text{1 }\Omega\\
\end{align}
</math>

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

<math>
\begin{align}
Z_{2}&=\left(\frac{240}{100}\right)^2\times\text{1∠0° }\Omega\\
&= \text {5.76∠0° }\Omega\\
\end{align}
</math>

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

<math>
\begin{align}
Z_\text{base,1}&=\frac{V_\text{base,1}^2}{S_\text{base}}\\
&= \frac {(240\text{ V})^2} {10000\text{ VA}}\\
&= \text{5.76 }\Omega\\
\end{align}
</math>

This means that the per unit impedance is 5.76∠0°&nbsp;Ω / 5.76&nbsp;Ω = 1∠0°&nbsp;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&nbsp;kV might be connected to the primary side of another transformer whose rated voltage is 1&nbsp;kV. The formula is as shown below.

<math>
\begin{align}
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{align}
</math>