Per unit advantages and choice of bases

Advantages of the per-unit system

1. The p.u. impedance of a transformer is the same on the HV and LV sides, so ideal transformers vanish from the equivalent circuit.
2. The p.u. impedances of similar equipment lie in a narrow band, whatever the rating. This makes it easy to check data and to assume typical values.
3. Manufacturers give impedances in % or p.u. on the rating of the equipment, and these can be used directly.
4. Line and phase quantities have the same p.u. value, and $\sqrt3$ factors disappear. Single-phase and three-phase calculations look alike.
5. The numbers are near 1, which reduces arithmetic errors. It is easy to see abnormal conditions, for example $V = 0.9$ p.u.
6. It is ideal for computer studies (load flow, fault and stability analysis).

Choosing base power and base voltages

1. Base power ($S_b$, MVA): Choose one value for the whole system. It is usually the rating of the largest machine or a round figure such as 100 MVA.
2. Base voltage ($V_b$, kV): Choose it in one section, usually the generator or line section, often equal to that section's rated voltage.
3. Other sections: Carry the base voltage through each transformer by its line-to-line voltage ratio:

$$V_{b,new} = V_{b,old}\times\frac{V_{rated,\ new\ side}}{V_{rated,\ old\ side}}$$

For single-phase units in a Y–Δ bank, use the line voltage ratio ($\sqrt3 V$ on the Y side).

4. Derived bases:

$$I_b = \frac{S_b}{\sqrt3V_b},\qquad Z_b = \frac{V_b^2}{S_b}$$

5. Change of base: Convert each equipment value to the common base:

$$Z_{pu,new} = Z_{pu,old}\times\frac{S_{b,new}}{S_{b,old}}\times\left(\frac{V_{b,old}}{V_{b,new}}\right)^2$$

Example: Take $S_b = 100$ MVA and 11 kV in the generator circuit, with an 11/132 kV transformer. The line base is 132 kV and $Z_b = 132^2/100 = 174.24\ \Omega$.