The Per-Unit System in Transformers

In power systems engineering, transformers operate across different
voltage levels. This complicates calculations because impedances must be
referred to either the primary or secondary side using the square of the
turns ratio ($a^2$). The Per-Unit (p.u.) System
standardizes these values, allowing engineers to analyze complex grids
as a single, unified circuit.

1. Definition of Per-Unit Value

The per-unit value of any quantity is defined as the ratio of the
actual physical quantity to a selected base value of the same quantity.
It is a dimensionless number.

$$\text{Quantity in p.u.} = \frac{\text{Actual Value}}{\text{Base Value}}$$

2. Selection of Base Values

To define a per-unit system, two independent base values are
typically chosen (usually the rated values of the transformer):

1. Base Power ($S_{base}$): Usually the rated
   apparent power (e.g., $100 \text{ kVA}$). This remains constant
   throughout the entire system.

2. Base Voltage ($V_{base}$): Usually the rated
   line voltage of the specific section. Crucially, $V_{base}$
   changes across a transformer according to its turns
   ratio.

From these, the base current and base impedance are derived:

• Base Current ($I_{base}$):

$$I_{base} = \frac{S_{base}}{V_{base}}$$

• Base Impedance ($Z_{base}$):

$$Z_{base} = \frac{V_{base}}{I_{base}} = \frac{V_{base}^2}{S_{base}}$$

3. Key Advantages in Transformer Analysis

The primary reason for using the per-unit system is that the
per-unit impedance of a transformer is the same, whether it is
calculated from the primary or the secondary side.

Key Takeaway:

In a per-unit circuit, the ideal transformer “disappears” because the
turns ratio becomes $1:1$. This eliminates the need for manual impedance
transformation across different voltage levels.

4. Calculation Example: Single-Phase Transformer

Consider a single-phase transformer with the following
specifications:

• Rated Capacity ($S_{rated}$): $50 \text{ kVA}$

• Rated Voltage: $2400 / 240 \text{ V}$

• Equivalent Impedance (referred to the primary
  side): $Z_{eq1} = 1.44 + j2.88 \ \Omega$

Step 1: Establish Base Values

• Set $S_{base} = 50,000 \text{ VA}$

• Primary Base Voltage $V_{base1} = 2,400 \text{ V}$

• Secondary Base Voltage $V_{base2} = 240 \text{ V}$

Step 2: Calculate Primary Base Impedance
($Z_{base1}$)

$$Z_{base1} = \frac{V_{base1}^2}{S_{base}} = \frac{2400^2}{50000} = 115.2 \ \Omega$$

Step 3: Calculate the Per-Unit Impedance
($Z_{pu}$)

$$Z_{pu} = \frac{Z_{eq1}}{Z_{base1}} = \frac{1.44 + j2.88}{115.2} = \mathbf{0.0125 + j0.025 \ \text{p.u.}}$$

Verification: Calculation from the Secondary
Side

To prove the consistency of the p.u. system, let’s refer the actual
impedance to the secondary side first:

$$a = \frac{2400}{240} = 10$$

$$Z_{eq2} = \frac{Z_{eq1}}{a^2} = \frac{1.44 + j2.88}{100} = 0.0144 + j0.0288 \ \Omega$$

Now, calculate the Secondary Base Impedance ($Z_{base2}$):

$$Z_{base2} = \frac{V_{base2}^2}{S_{base}} = \frac{240^2}{50000} = 1.152 \ \Omega$$

Calculate the Per-Unit Impedance from the secondary side:

$$Z_{pu} = \frac{0.0144 + j0.0288}{1.152} = \mathbf{0.0125 + j0.025 \ \text{p.u.}}$$

Conclusion: The per-unit value is identical ($0.0125 + j0.025 \ \text{p.u.}$) regardless of which side is used for the
calculation.

Summary

The Per-Unit System is an essential tool because it:

• Simplifies Topology: Simplifies multi-voltage
  networks into a single-level equivalent circuit.

• Facilitates Comparison: Manufacturers provide
  impedance in percentage or p.u., which allows for easy comparison
  between different transformer sizes.

• Reduces Errors: Minimizes mistakes related to
  $\sqrt{3}$ in three-phase calculations and $a^2$ in impedance
  referrals.