All Exams Test series for 1 year @ ₹349 only
Question

The relation between the old and new per unit impedance values is given by:

The correct answer is \(Z_{pu}^{new} = Z_{pu}^{old}\frac{{S_B^{new}}}{{S_B^{old}}}{\left( {\frac{{V_B^{old}}}{{V_B^{new}}}} \right)^2}\)

Per Unit Impedance Relation Between Different Base Values

In power systems analysis, the per unit (pu) system is commonly used to simplify calculations. When the base values for voltage and apparent power are changed, the per unit values of impedance also change. We need to find the relation between the old per unit impedance value and the new per unit impedance value.

The per unit impedance is defined as the ratio of the actual impedance to the base impedance:

\(Z_{pu} = \frac{Z_{actual}}{Z_B}\)

The base impedance \(Z_B\) is related to the base voltage \(V_B\) and base apparent power \(S_B\) by the formula:

\(Z_B = \frac{V_B^2}{S_B}\)

Let the old base values be \(V_B^{old}\) and \(S_B^{old}\), and the new base values be \(V_B^{new}\) and \(S_B^{new}\).

The old per unit impedance is given by:

\(Z_{pu}^{old} = \frac{Z_{actual}}{Z_B^{old}}\)

Where \(Z_B^{old} = \frac{(V_B^{old})^2}{S_B^{old}}\).

So, \(Z_{actual} = Z_{pu}^{old} \times Z_B^{old} = Z_{pu}^{old} \times \frac{(V_B^{old})^2}{S_B^{old}}\).

The new per unit impedance is given by:

\(Z_{pu}^{new} = \frac{Z_{actual}}{Z_B^{new}}\)

Where \(Z_B^{new} = \frac{(V_B^{new})^2}{S_B^{new}}\).

Substitute the expression for \(Z_{actual}\) into the formula for \(Z_{pu}^{new}\):

\(Z_{pu}^{new} = \frac{Z_{pu}^{old} \times Z_B^{old}}{Z_B^{new}}\)

Now, substitute the formulas for \(Z_B^{old}\) and \(Z_B^{new}\):

\(Z_{pu}^{new} = Z_{pu}^{old} \times \frac{\frac{(V_B^{old})^2}{S_B^{old}}}{\frac{(V_B^{new})^2}{S_B^{new}}}\)

\(Z_{pu}^{new} = Z_{pu}^{old} \times \frac{(V_B^{old})^2}{S_B^{old}} \times \frac{S_B^{new}}{(V_B^{new})^2}\)

\(Z_{pu}^{new} = Z_{pu}^{old} \times \frac{S_B^{new}}{S_B^{old}} \times \left(\frac{V_B^{old}}{V_B^{new}}\right)^2\)

This equation gives the relation between the old and new per unit impedance values when the base values are changed.

Comparing this derived formula with the given options:

  • Option 1: \(Z_{pu}^{new} = Z_{pu}^{old}\frac{{S_B^{old}}}{{S_B^{new}}}{\left( {\frac{{V_B^{old}}}{{V_B^{new}}}} \right)^2}\) (Incorrect - \(S_B\) ratio is inverted)
  • Option 2: \(Z_{pu}^{new} = Z_{pu}^{old}\frac{{S_B^{new}}}{{S_B^{old}}}{\left( {\frac{{V_B^{old}}}{{V_B^{new}}}} \right)^2}\) (Correct)
  • Option 3: \(Z_{pu}^{new} = Z_{pu}^{old}\frac{{S_B^{new}}}{{S_B^{old}}}{\left( {\frac{{V_B^{new}}}{{V_B^{old}}}} \right)^2}\) (Incorrect - \(V_B\) ratio is inverted)
  • Option 4: \(z_{pu}^{old} = Z_{pu}^{new}\frac{{S_B^{new}}}{{S_B^{old}}}{\left( {\frac{{V_B^{old}}}{{V_B^{new}}}} \right)^2}\) (Incorrect - Equation is rearranged incorrectly, plus uses lowercase z)

The derived relation matches option 2.

Was this answer helpful?

Important Questions from Per Unit System

  1. A synchronous generator is rated at 40 MVA, 14.6 kV and 50 Hz. The base impedance of the generator will be

  2. The per unit impedance Z (Pu) in 3 - phase system is -
  3. A synchronous generator is rated at 40 MVA, 10 kV and 50 Hz. The base impedance of the generator will be:
  4. The per unit impedance of a line is X p.u. If base voltage is tripled and base MVA is doubled, the new per unit impedance is:

  5. In symmetrical fault calculations, percentage reactance at base kVA is equal to

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App