Per-Unit System Calculator: Base Values, Per-Unit Impedance and Fault Level

Convert impedances to per unit on a common base and find the fault level at a point.

Calculated from standard textbook formulas, with every step shown. The formulas are fixed, so the working does not need checking: what matters is the numbers you type in. Use these calculators only if you are competent to choose the inputs (loads, material strength, dimensions, supports) and check your units and assumptions. For learning and checking your own working: design and sign-off are for a qualified engineer. Drafting Central does not store the numbers you type.

How this calculator works

Choose the base MVA and the base voltage. The calculator gives the base impedance V squared over S and the base current.

Enter the fault level of the supply, the transformer percentage impedance and its rating, and the impedance in ohms of the cable beyond it. Each is converted to per unit on your base.

The impedances are added to give the total, then the fault level S over Z and the fault current I over Z at the point.

Formulas used

  • Z_b = \frac{V_b^2}{S_b}
  • I_b = \frac{S_b}{\sqrt{3}V_b}
  • Z_{new} = Z_{old}\frac{S_{new}}{S_{old}}
  • S_{fault} = \frac{S_b}{Z_{pu}}

Worked example

With the default numbers in the calculator above:

  1. Base impedance: 1.21 ohm
  2. Base current: 5248.64 A
  3. Source impedance on the base: 0.40 pu
  4. Transformer impedance changed to the new base: 1.20 pu
  5. Cable or line impedance in per unit: 0.41 pu
  6. Total impedance to the fault (added as magnitudes: a safe, slightly high, fault level): 2.01 pu
  7. Fault level at the point: 49.67 MVA
  8. Three-phase fault current: 2.61 kA (2607.08 A)

Change the numbers in the calculator to see the full working, every step, for your own case.

Common questions

Why use per unit?

The values are the same on both sides of a transformer, and the numbers of all the equipment are similar in size, so errors are easy to spot.

How do I change base?

Multiply the per-unit value by the new base MVA over the old, and by the square of the old base voltage over the new.

Why are the impedances added as magnitudes?

It is a simple, slightly pessimistic estimate. A full study adds resistances and reactances separately.

What is an infinite source?

A supply with no impedance, which gives the highest possible fault level. Enter 0 for the fault level of the supply to use it.

Related calculators

Got a question about this? Ask an engineering question and get a plain-language answer with references.

Further reading: Grainger and Stevenson, Power System Analysis (the per-unit system); AS 3851 Calculation of short-circuit currents in three-phase a.c. systems.

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