Per-Unit System Representation: Complete Beginner's Guide

The per-unit system expresses every voltage, current, impedance and power as a fraction of a chosen base value: per-unit quantity = actual quantity / base quantity. A transformer rated 100 MVA with 8% impedance is "0.08 per unit" whether it sits on the 11 kV side or the 132 kV side — the awkward voltage ratio simply disappears from the math.

This matters because real power systems span many voltage levels, and carrying √3 factors and transformer ratios through every calculation is slow and error-prone. In per unit, a 0.95 pu voltage means 95% of normal everywhere in the system, equipment of wildly different sizes can be compared at a glance, and computer programs for load flow and fault analysis take per-unit data as their standard input.

Key Concepts

• Base quantities: pick S_base (usually 100 MVA) and V_base (the nominal line voltage at each voltage level); every other base follows from these two.
• One-line bases: I_base = S_base / (√3 × V_base) in amperes, Z_base = V_base² / S_base in ohms per phase, using line-to-line volts and three-phase MVA.
• Actual from per unit: actual value = per-unit value × base value — the system never hides the real units, it just postpones them.
• Base changing: manufacturer data given on the equipment's own base must be converted to the study base before use.
• Transformer advantage: a transformer's per-unit impedance is identical referred to either side, so ideal-transformer ratios vanish from the network model.

Formulas

• Per-unit definition: X_pu = X_actual / X_base.
• Base impedance: Z_base = V_base² / S_base, with V_base in volts (line-to-line) and S_base in volt-amperes, giving ohms per phase.
• Base current: I_base = S_base / (√3 × V_base) amperes.
• Changing bases: Z_pu,new = Z_pu,old × (S_base,new / S_base,old) × (V_base,old / V_base,new)².
• Per-unit power: S_pu = P_pu + jQ_pu, V_pu × I_pu* with angles in the usual complex form.

Worked Example

Take S_base = 100 MVA and V_base = 33 kV. Then Z_base = 33²/100 = 10.89 Ω, and I_base = 100×10⁶ / (√3 × 33,000) ≈ 1,750 A. A feeder with 10.89 Ω actual impedance is exactly 1.0 pu. A load drawing 50 MW at 0.8 power factor lagging has S = 50/0.8 = 62.5 MVA, so S_pu = 0.625 pu, and its current is I_pu = S_pu / V_pu = 0.625∠−36.87° pu (taking V_pu = 1.0∠0°). Converting back, 0.625 × 1750 ≈ 1,094 A — the same as the direct calculation 50×10⁶/(√3 × 33,000 × 0.8). The per-unit route used smaller, cleaner numbers throughout.

Common Misconceptions

• "Per-unit values are absolute." They are not: change the base and every per-unit number changes, while the physical volts and amps stay the same. Always state your bases.
• "Each voltage level needs its own separate per-unit system." The opposite — one S_base with a V_base per voltage level ties the whole system into a single consistent model.
• "Per-unit hides the units." It postpones them: multiply by the base at the end and every unit reappears exactly.

Real-World Applications

• Load-flow and short-circuit software accept network data only in per unit on a common MVA base.
• Transformer and generator test certificates quote impedances in percent (per unit × 100), letting engineers compare a 5 MVA and a 500 MVA unit instantly.
• Protection engineers compute fault currents in per unit, then convert once at the end to set relay pickups in amperes.
• Manufacturers publish equipment data on the machine's own base, and every system study begins by converting it to the study base.

The per-unit system is bookkeeping that pays for itself: two base choices turn kilovolts, mega-amperes and transformer ratios into small, comparable decimals. Learn the four base formulas and the base-change rule, and every power-system calculation — load flow, faults, stability — becomes dramatically simpler.