Per-Unit System Simulator — Power System p.u. Value Conversion
Choose base voltage V_base and apparent-power base S_base for a three-phase grid to derive Z_base and I_base. Normalize V and Z into per-unit values and watch the bar chart and single-line diagram update live.
Ohm's law in p.u.: $$V_{\mathrm{pu}} = Z_{\mathrm{pu}} \cdot I_{\mathrm{pu}}$$
$V_{\mathrm{base}}$ is the line-to-line voltage (kV) and $S_{\mathrm{base}}$ is the three-phase apparent power (MVA). The factor $\sqrt{3}$ comes from converting between line and phase quantities.
What is the Per-Unit System Simulator
🙋
Professor, every power-system textbook uses "p.u. values" instead of real volts and ohms. Why bother normalising? Working with actual values seems more intuitive.
🎓
Great question. A grid spans many voltage levels — 13.8 kV at the generator, 230 kV on transmission, 13.8 kV again at distribution — and every transformer scales the numbers by its turns ratio. p.u. values absorb those ratios so the whole grid becomes a single equivalent circuit. Run this simulator with V_base = 400 kV and S_base = 100 MVA. Z_base comes out to 1600 Ω, so a real 50 Ω line becomes Z_pu ≈ 0.0313, a tiny dimensionless number.
🙋
OK Z_pu = 0.0313, but how should I read that? Is 50 Ω large or small for a transmission line?
🎓
The beauty of p.u. is that it reads as "fraction of the rated operating point". Z_pu = 0.0313 means a voltage drop of about 3.13% at rated current. In practice, a synchronous machine has X_pu = 1.0 - 2.0, a transformer leakage reactance is X_pu = 0.08 - 0.15, and transmission lines are well under 1.0 per unit. Once you internalise these scales, you can eyeball a system data card and tell whether a number is plausible or wildly wrong.
🙋
How is the base MVA chosen? The simulator defaults to 100 MVA.
🎓
Utilities almost universally use S_base = 100 MVA as their planning standard. Equipment vendors publish p.u. impedance referenced to the device rating, so engineers must convert it to the system 100-MVA base using Z_pu_new = Z_pu_old × (S_base_new / S_base_old). Drag the S_base slider in this tool and watch Z_pu scale exactly linearly — that is the same arithmetic you do by hand when you build a per-unit single-line diagram.
Physical model and key equations
For a three-phase system, two base quantities are chosen independently: the line-to-line base voltage $V_{\mathrm{base}}$ (kV) and the three-phase apparent-power base $S_{\mathrm{base}}$ (MVA). The remaining bases follow:
Actual values $V$, $I$, $Z$ are divided by the corresponding base to yield p.u. values: $V_{\mathrm{pu}} = V/V_{\mathrm{base}}$, $I_{\mathrm{pu}} = I/I_{\mathrm{base}}$, $Z_{\mathrm{pu}} = Z/Z_{\mathrm{base}}$. The key feature is that in the per-unit system Ohm's law collapses to the simple form $V_{\mathrm{pu}} = Z_{\mathrm{pu}}\,I_{\mathrm{pu}}$, with no $\sqrt{3}$ factor and no transformer turns ratio. This is what makes the multi-voltage-level grid solvable as one homogeneous circuit.
Real-world applications
Power-flow studies: Every commercial power-flow solver — PSS/E, PowerWorld, PSCAD, DIgSILENT — works in per unit. 100 MVA is the de-facto common base. Knowing how Z_pu maps back to ohms makes input data sanity checks much easier, which is the everyday job of a planning engineer.
Short-circuit calculation: Three-phase fault current is concise in p.u.: $I_{\mathrm{sc,pu}} = 1/Z_{\mathrm{Thev,pu}}$. The result is rescaled to amperes via $I_{\mathrm{sc}} = I_{\mathrm{sc,pu}} \times I_{\mathrm{base}}$ — the I_base output of this tool is exactly what you need for breaker rating studies.
Transformer specification: Name-plate %Z is a per-unit value referenced to the transformer's own rating. A 20-MVA, %Z = 10% transformer has X_pu = 0.10 on its self base; on a 100-MVA system base it becomes X_pu = 0.50.
Stability analysis: Transient and voltage-stability software accepts only p.u. data. Generator inertia constants, governor gains, AVR limits — everything is per unit. Fluency with the system is non-negotiable for system planners.
Common misconceptions
First, p.u. is not simply "percent divided by 100". The shortcut "%Z = 10 means X_pu = 0.10" is true only when the base is the equipment's own rating. Once you change to the system common base, every per-unit number changes. Drag the S_base slider in this tool and observe how Z_pu scales proportionally.
Second, line-to-line vs. phase voltage. The relation $Z_{\mathrm{base}} = V_{\mathrm{base}}^2/S_{\mathrm{base}}$ assumes $V_{\mathrm{base}}$ is the line-to-line value (the power-system convention used here). If someone has chosen the phase voltage instead, the formulas pick up factor-of-three differences. Always confirm the convention when reading a foreign model.
Third, each voltage level needs its own V_base. The two sides of a transformer are assigned V_base values in the ratio of their nominal voltages. With this choice the transformer's own leakage reactance has the same per-unit value on both sides and the turns ratio drops out of the equivalent circuit. S_base, in contrast, is single and shared by the whole grid.
Frequently asked questions
Yes. With V_base = 400 kV and S_base = 100 MVA, Z_base = V_base² / S_base = (4×10⁵)² / 10⁸ = 1.6×10¹¹ / 10⁸ = 1600 Ω. Similarly I_base = S_base / (√3 · V_base) = 10⁸ / (√3 · 4×10⁵) ≈ 144.3 A. At 230 kV the base impedance drops to ≈ 529 Ω and at 69 kV to ≈ 47.6 Ω — two orders of magnitude across normal voltage levels.
V_pu = 1.05 means the actual voltage is 105% of the rated voltage. Most utilities enforce 0.95 - 1.05 p.u. (±5%) on the bus. Customer service entrances in IEC and ANSI codes have similar narrow bands. Above 1.10 p.u. overvoltage protection trips, and below 0.85 p.u. under-voltage load shedding is expected to act.
Z_pu is directly proportional to S_base. Scaling S_base from 100 MVA to 1000 MVA multiplies Z_pu of the same 50-Ω line from 0.0313 to 0.313 — a factor of ten. Try it with the S_base slider. This is precisely why a transformer rated at 20 MVA must have its X_pu rescaled before being mixed with 100-MVA-base system data.
Apparent power simplifies to $S_{\mathrm{pu}} = V_{\mathrm{pu}}\,I_{\mathrm{pu}}^*$ (with complex conjugate). The $\sqrt{3}$ and $\cos\varphi$ factors that usually decorate three-phase formulas disappear. This compact form is one reason power-flow Jacobian equations are so symmetric and easy to program.