Per Unit Impedance Calculation Explained

The per unit system is a fundamental concept in power system analysis, used to simplify complex calculations by normalizing all quantities to a common base. This system helps in comparing the relative magnitudes of different quantities and makes computations more manageable, especially in large power grids. The impedance of a component, when expressed in per unit, gives a normalized value relative to a chosen base impedance.

Per Unit Impedance Formula

The per unit impedance (\(Z_{pu}\)) is determined by the actual ohmic impedance and the selected base values (base MVA and base voltage). The formula to convert an actual impedance to its per unit value is:

\[Z_{pu} = Z_{actual} \times \frac{MVA_{base}}{(kV_{base})^2}\]

Where:

• \(Z_{actual}\) represents the actual impedance in ohms (Ω). This is the physical resistance and reactance of the line.
• \(MVA_{base}\) is the base apparent power in megavolt-amperes (MVA).
• \(kV_{base}\) is the base line-to-line voltage in kilovolts (kV).

Analyzing Per Unit Impedance Change

Let's consider the initial conditions. We are given the initial per unit impedance as \(X\). Let \(V_{old\_base}\) be the initial base voltage and \(MVA_{old\_base}\) be the initial base MVA. The actual impedance of the line, \(Z_{actual}\), remains constant regardless of changes in the base values, as it is an inherent property of the line itself.

So, the initial per unit impedance \(X\) can be expressed as:

\[X = Z_{actual} \times \frac{MVA_{old\_base}}{(V_{old\_base})^2} \quad \text{(Equation 1)}\]

The problem states that the base voltage is tripled and the base MVA is doubled. Let's define the new base values:

• New base voltage: \(V_{new\_base} = 3 \times V_{old\_base}\)
• New base MVA: \(MVA_{new\_base} = 2 \times MVA_{old\_base}\)

Now, we need to find the new per unit impedance, let's call it \(X_{new}\), using these new base values. The formula for the new per unit impedance will be:

\[X_{new} = Z_{actual} \times \frac{MVA_{new\_base}}{(V_{new\_base})^2}\]

Step-by-Step Per Unit Calculation

To find the new per unit impedance, substitute the expressions for \(V_{new\_base}\) and \(MVA_{new\_base}\) into the equation for \(X_{new}\):

\[X_{new} = Z_{actual} \times \frac{(2 \times MVA_{old\_base})}{(3 \times V_{old\_base})^2}\]

First, calculate the square of the new base voltage in the denominator:

\[(3 \times V_{old\_base})^2 = 3^2 \times (V_{old\_base})^2 = 9 \times (V_{old\_base})^2\]

Now, substitute this back into the expression for \(X_{new}\):

\[X_{new} = Z_{actual} \times \frac{2 \times MVA_{old\_base}}{9 \times (V_{old\_base})^2}\]

To relate \(X_{new}\) back to the original per unit impedance \(X\), we can rearrange the terms:

\[X_{new} = \frac{2}{9} \times \left( Z_{actual} \times \frac{MVA_{old\_base}}{(V_{old\_base})^2} \right)\]

From Equation 1, we know that \(Z_{actual} \times \frac{MVA_{old\_base}}{(V_{old\_base})^2}\) is equal to \(X\).

Therefore, by substituting \(X\) into the equation:

\[X_{new} = \frac{2}{9} X\]

Final Per Unit Impedance

When the base voltage is tripled and the base MVA is doubled, the new per unit impedance becomes \(\frac{2}{9}\) times the original per unit impedance.