19.1 The Per-Unit System, Base Conversion & Impedance Diagrams

19.1 The Per-Unit System, Base Conversion & Impedance Diagrams

Power System Analysis is topic K of the Electrical Engineering subject in the PRC Enhanced Table of Specifications (PRBEE Resolution No. 40, s. 2024), weighted 9.00% of the whole examination and 20 of the 100 Electrical Engineering items. That makes it the largest single topic in the entire REE examination — larger than any Mathematics or ESAS topic. It entered the syllabus explicitly through CHED Memorandum Order No. 88, s. 2017, which lists Power Systems Analysis among the required BSEE professional courses.

Everything in this topic begins with the per-unit system, because no serious fault or load-flow calculation is done in volts and amperes.

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1. Why Per-Unit

Expressing every quantity as a fraction of a chosen base:

$\text{Per-unit value} = \frac{\text{Actual value}}{\text{Base value}}$

buys four things an electrical engineer actually needs:

1. Transformer turns ratios vanish — an ideal transformer becomes a plain connection, so a multi-voltage network collapses into one impedance diagram;
2. Manufacturer impedance data already arrives in per-unit or percent on the equipment rating;
3. Per-unit impedances of like equipment fall in narrow, memorable bands, so a wrong answer is obvious by inspection;
4. Three-phase and single-phase calculations use the same numbers when bases are chosen consistently.

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2. Base Quantities

Two bases are chosen — normally three-phase MVA and line-to-line kV — and the rest follow:

$I_{base} = \frac{S_{base(3\phi)}}{\sqrt{3}\,V_{base(LL)}} \qquad\qquad Z_{base} = \frac{V_{base(LL)}^{2}}{S_{base(3\phi)}}$

With $S$ in MVA and $V$ in kV, $Z_{base}$ comes out directly in ohms and $I_{base}$ in kA:

$Z_{base}(\Omega) = \frac{\left[kV_{LL}\right]^{2}}{MVA_{3\phi}} \qquad\qquad I_{base}(kA) = \frac{MVA_{3\phi}}{\sqrt{3}\;kV_{LL}}$

The base voltage changes at every transformer in proportion to the turns ratio; the base MVA is chosen once and stays constant across the whole system. Getting this backwards is the most common structural error in per-unit problems.

$S_{pu} = \frac{S_{actual}}{S_{base}}, \qquad V_{pu} = \frac{V_{actual}}{V_{base}}, \qquad I_{pu} = \frac{I_{actual}}{I_{base}}, \qquad Z_{pu} = \frac{Z_{actual}}{Z_{base}}$

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3. Change of Base

Equipment impedance is published on the equipment's own rating, so it must be converted to the common system base:

$Z_{pu,\text{new}} = Z_{pu,\text{old}} \times \frac{S_{base,\text{new}}}{S_{base,\text{old}}} \times \left(\frac{V_{base,\text{old}}}{V_{base,\text{new}}}\right)^{2}$

Note the asymmetry: the MVA ratio is direct, the voltage ratio is inverted and squared. When the base voltage matches the equipment's rated voltage — the usual case — only the MVA ratio survives:

$Z_{pu,\text{new}} = Z_{pu,\text{old}} \times \frac{S_{base,\text{new}}}{S_{base,\text{old}}}$

Typical Per-Unit Impedances

These bands are worth memorising: a computed generator subtransient reactance of 0.02 pu or 2.0 pu is a signal that a base conversion went wrong.

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4. Constructing the Impedance Diagram

1. Draw the single-line diagram and mark every rating;
2. Choose one system base MVA (100 MVA is conventional) and one base kV in a chosen zone;
3. Propagate base kV through each transformer by its turns ratio, defining a voltage zone per transformer section;
4. Convert every impedance to the common base;
5. Replace generators by an EMF behind the appropriate reactance ($X_d''$ for momentary duty, $X_d'$ for interrupting duty);
6. Omit ideal transformers entirely — per-unit has already accounted for them;
7. For balanced three-phase fault studies, draw only the positive-sequence network, with loads usually neglected as they are large impedances beside fault paths.

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5. Short-Circuit MVA and Interrupting Duty

Once the network is reduced to a single Thevenin impedance $Z_{th,pu}$ at the fault point:

$I_{f,pu} = \frac{V_{pu}}{Z_{th,pu}}, \qquad I_{f,\text{actual}} = I_{f,pu} \times I_{base}$

$\text{Short-circuit MVA} = \frac{S_{base}}{Z_{th,pu}} \qquad \text{(at 1.0 pu prefault voltage)}$

Equivalently, a utility that quotes an available fault level of $MVA_{sc}$ at a point of common coupling is telling you its source impedance:

$Z_{source,pu} = \frac{S_{base}}{MVA_{sc}}$

Circuit-breaker interrupting duty uses $X_d'$ (or $1.5X_d''$ by older practice) because the subtransient component has decayed by the time the contacts part, while momentary or close-and-latch duty uses $X_d''$. Choosing the wrong reactance under-rates the breaker, which is a safety defect rather than an arithmetic one.

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Solved Board Exam Examples

Example 1: Base Conversion of a Transformer Impedance

A 20 MVA, 69/13.8 kV transformer has an impedance of 8% on its own rating. Express it on a 100 MVA base.

Solution. Base voltages match the transformer ratings, so only the MVA ratio applies:

$Z_{pu,\text{new}} = 0.08 \times \frac{100}{20} = \boxed{0.40 \text{ pu}}$

The same impedance expressed on a five-times-larger base is five times larger in per-unit — the number changed, the physical ohms did not.

Example 2: Fault Current and Short-Circuit MVA

A 13.8 kV bus is supplied through a total Thevenin impedance of $j0.25$ pu on a 100 MVA base. Find the three-phase fault current and short-circuit MVA.

Solution.

$I_{base} = \frac{100}{\sqrt{3}(13.8)} = \frac{100}{23.902} = 4.184 \text{ kA}$

$I_{f,pu} = \frac{1.0}{0.25} = 4.0 \text{ pu} \implies I_f = 4.0 \times 4.184 = \boxed{16.74 \text{ kA}}$

$MVA_{sc} = \frac{100}{0.25} = \boxed{400 \text{ MVA}}$

A 25 kA-rated switchgear lineup is therefore adequate; a 16 kA lineup is not.

Example 3: Multi-Zone Per-Unit Network

A 30 MVA, 13.8 kV generator with $X_d'' = 0.20$ pu feeds a 30 MVA, 13.8/69 kV transformer of 10% impedance, then a line of 12 Ω. Using a 30 MVA, 13.8 kV base on the generator side, find the total per-unit impedance to a fault at the far end of the line.

Solution. Base voltage on the line side is $13.8 \times (69/13.8) = 69$ kV, so

$Z_{base,\text{line}} = \frac{69^{2}}{30} = \frac{4761}{30} = 158.7\ \Omega$

$Z_{line,pu} = \frac{12}{158.7} = 0.0756 \text{ pu}$

Generator and transformer are already on a 30 MVA base at their rated voltages, so:

$Z_{total} = j0.20 + j0.10 + j0.0756 = \boxed{j0.3756 \text{ pu}}$

$MVA_{sc} = \frac{30}{0.3756} = 79.9 \text{ MVA}$

Observe that the ideal transformer never appeared in the arithmetic — that is the whole point of the per-unit method.