The per-unit system on the PE Power exam

A per-unit question on the PE Power exam asks you to put every piece of equipment on one common base, combine the impedances, and convert the result back to amperes, volts or MVA at a named bus. The NCEES handbook covers the topic in the Per Unit System section of its Circuits chapter. That section holds the base current and base impedance relations for single-phase and three-phase systems, and the change-of-base equation. It does not state the defining ratio, so carry that one yourself: a per-unit value is the actual quantity divided by the base quantity of the same kind.

What you actually need to own

• The definition. Per unit = actual ÷ base. The units cancel, so the result is dimensionless. For a voltage, put both numbers on the same reference: both line-to-line or both line-to-neutral.
• Choose two bases, derive the rest. A three-phase study fixes a base MVA and a base line-to-line kV. Then I_base (A) = base kVA / (√3 × base kV_LL), and Z_base (Ω) = (base kV_LL)² / base MVA. The impedance form has no √3, because line-to-line kV squared over three-phase MVA equals line-to-neutral kV squared over per-phase MVA. A single-phase system uses base kVA / base kV with no √3.
• One base power, one base voltage per zone. The base MVA is the same everywhere in the study. The base voltage changes at each transformer in that transformer's nameplate voltage ratio, starting from the bus where the base was declared. Z_base changes with the square of the base voltage. Convert an ohmic impedance with the Z_base of the zone it sits in.
• Nameplate percent impedance. A transformer's percent impedance is its per-unit impedance on its own kVA and voltage rating, and it is the same number from either winding. When base voltages follow the nameplate ratio, the ideal transformer becomes 1:1 in per unit and leaves the diagram.
• Change of base. Z_new = Z_old × (base kV_old / base kV_new)² × (base kVA_new / base kVA_old). When the voltage bases match the nameplate, only the power ratio is left. An 8% transformer rated 15 MVA is 0.08 × 50/15 = 0.267 per unit on a 50 MVA base. The same hardware is a larger per-unit number on a larger base.
• Sources and faults. A utility given as a short-circuit duty converts directly when the base kV equals the nominal kV the duty is stated at: Z_pu = base MVA / S_sc, or I_base / I_sc. With 1.0 per-unit prefault voltage, a bolted three-phase fault is I_pu = 1 / Z_pu, and the fault MVA is base MVA / Z_pu. Impedances in one current path add. Two sources feeding the same bus combine in parallel. Finish with the base current of the zone the question asks about.

Where people lose points

• Using an equipment rating as the study base. A nameplate is the base only for that equipment's own impedance.
• Taking the base voltage from the winding rating. If the declared base voltage differs from the transformer's winding rating on that side, the base on the other side is the declared value scaled by the nameplate ratio, which no longer equals the other winding's rating.
• Building Z_base wrong. The usual three: kV / MVA without the square, line-to-neutral kV squared over three-phase MVA, and base MVA in the numerator.
• Using the three-phase base current for a single-phase transformer.
• Stopping at the base current. Base current is the current at 1.0 per unit. A fault current is that divided by the per-unit impedance.
• Quoting amperes on the wrong side. Use the base current of the bus named in the question.
• Combining before reading the one-line. Averaging two parallel branches, adding them in series, or including a branch behind an open breaker each give a tidy wrong answer. A parallel impedance must be lower than either branch impedance.

How to study it

Draw a three-zone system: a utility given as a fault duty, two transformers with different ratings and percent impedances, and one cable given in ohms. Choose a base MVA that matches none of the nameplates. Write the base kV, I_base and Z_base for each zone before touching an impedance. Convert every element, find the three-phase fault current at each bus in amperes, then repeat on a different base MVA. The ampere answers must not change. The same bookkeeping carries into fault current analysis and into the sequence networks in symmetrical components and unbalanced faults.

Part of the Circuit Analysis area of the PE Power exam. → Practice it on CastorPrep