8.1 Per-Unit System Fundamentals & Base Conversion Formulas

8.1 Per-Unit System Fundamentals & Base Conversion Formulas

Key Exam Takeaway: In balanced three-phase power system calculations, base impedance is derived exclusively from the three-phase power base and the line-to-line voltage base: $Z_{base} = \frac{(kV_{base,LL})^2}{MVA_{base,3\phi}}$. When converting equipment impedance to a new system base, always apply the ratio of voltages squared and the inverse ratio of power ratings: $Z_{pu,new} = Z_{pu,old} \left(\frac{V_{base,old}}{V_{base,new}}\right)^2 \left(\frac{S_{base,new}}{S_{base,old}}\right)$.

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1. Rationale of the Per-Unit System

Modern electrical power networks span multiple voltage levels—ranging from generation ($13.8\text{ kV}$ to $24\text{ kV}$), extra-high-voltage transmission ($138\text{ kV}$, $230\text{ kV}$, $500\text{ kV}$), sub-transmission ($69\text{ kV}$, $34.5\text{ kV}$), distribution ($12.47\text{ kV}$, $4.16\text{ kV}$), down to utilization ($480\text{ V}$, $208\text{ V}$, $120\text{ V}$). Analyzing these networks using actual physical quantities (Ohms, Volts, Amperes) requires reflecting impedances across transformer boundaries using the square of the turns ratio ($a^2 = (N_1/N_2)^2$). In multi-bus interconnected networks with tens or hundreds of transformers, manual impedance reflection becomes intensely error-prone.

The per-unit (pu) system normalizes electrical quantities by expressing them as dimensionless decimal fractions or ratios of carefully defined base reference parameters:

$$\text{Per-Unit Value} = \frac{\text{Actual Quantity (in Physical Units)}}{\text{Base Value (in Same Physical Units)}}$$

Primary Advantages in Power Engineering

1. Elimination of Transformer Turns Ratios: When base voltages in adjacent zones are selected in proportion to transformer turns ratios, the per-unit equivalent impedance of a transformer is identical whether calculated from the high-voltage or low-voltage winding.
2. Parameter Standardization: Apparatus of similar design and construction exhibit per-unit impedances within narrow, predictable bands regardless of physical size or MVA rating. For example, large two-pole synchronous generator subtransient reactances consistently fall between $0.12\text{ pu}$ and $0.25\text{ pu}$, and two-winding substation transformers typically have leakage reactances between $0.06\text{ pu}$ and $0.10\text{ pu}$.
3. Simplification of 3-Phase Formulations: Factors of $\sqrt{3}$ and $3$ vanish from balanced three-phase voltage, current, and complex power equations when standard three-phase base conventions are applied.
4. Immediate Insight into Operating Limits: Operating at $V = 0.94\text{ pu}$ instantly conveys a $6\%$ undervoltage condition without needing to recall whether the nominal bus voltage is $13.8\text{ kV}$ or $115\text{ kV}$.

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2. Fundamental Base Relationships in Three-Phase Networks

To establish a per-unit system for a three-phase network, two independent base quantities must be selected. By universal utility and NCEES convention, these two quantities are:

1. System Three-Phase Apparent Power Base ($S_{base,3\phi}$ or $MVA_{base,3\phi}$): A single common value selected for the entire power system (most commonly $100\text{ MVA}$ or $10\text{ MVA}$).
2. Line-to-Line Voltage Base ($V_{base,LL}$ or $kV_{base,LL}$): Specified for a reference zone and transferred across transformer boundaries according to nominal voltage ratings.

All remaining base parameters (Base Current, Base Impedance, and Base Admittance) are strictly derived from these two fundamental quantities.

                                  ┌────────────────────────┐
                                  │  Selected System Base  │
                                  │  S_base,3φ & V_base,LL │
                                  └───────────┬────────────┘
                                              │
                     ┌────────────────────────┴────────────────────────┐
                     ▼                                                 ▼
        ┌─────────────────────────┐                       ┌─────────────────────────┐
        │       Base Current      │                       │      Base Impedance     │
        │         S_base          │                       │       (V_base,LL)^2     │
        │ I_base = ─────────────  │                       │ Z_base = ─────────────  │
        │          √3 · V_base    │                       │          S_base,3φ      │
        └─────────────────────────┘                       └───────────┬─────────────┘
                                                                      │
                                                                      ▼
                                                          ┌─────────────────────────┐
                                                          │     Base Admittance     │
                                                          │ Y_base = 1 / Z_base     │
                                                          └─────────────────────────┘

Mathematical Derivations of Derived Bases

Base Current ($I_{base}$)

Starting from the balanced three-phase apparent power relation $S_{3\phi} = \sqrt{3} V_{LL} I_L$:

$$I_{base} = \frac{S_{base,3\phi}}{\sqrt{3} \cdot V_{base,LL}} = \frac{MVA_{base,3\phi} \times 10^3}{\sqrt{3} \cdot kV_{base,LL}} \quad [\text{in Amperes or kA}]$$

Base Impedance ($Z_{base}$)

Base impedance per phase is defined as the ratio of base line-to-neutral voltage ($V_{base,LN} = V_{base,LL} / \sqrt{3}$) to base line current ($I_{base}$):

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

Expressing line-to-line voltage in kilovolts ($kV$) and three-phase power in megavolt-amperes ($MVA$):

$$Z_{base} = \frac{(kV_{base,LL} \times 10^3)^2}{MVA_{base,3\phi} \times 10^6} = \frac{(kV_{base,LL})^2}{MVA_{base,3\phi}} \quad [\Omega]$$

Base Admittance ($Y_{base}$)

Base admittance is the exact reciprocal of base impedance:

$$Y_{base} = \frac{1}{Z_{base}} = \frac{S_{base,3\phi}}{(V_{base,LL})^2} = \frac{MVA_{base,3\phi}}{(kV_{base,LL})^2} \quad [\text{Siemens / }\mho]$$

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3. Converting Between Physical Units and Per-Unit

Converting between physical engineering units and per-unit quantities is governed by the following direct relations:

$$Z_{pu} = \frac{Z_{actual\ (\Omega)}}{Z_{base\ (\Omega)}} = Z_{actual\ (\Omega)} \times \frac{MVA_{base,3\phi}}{(kV_{base,LL})^2}$$ $$Z_{actual\ (\Omega)} = Z_{pu} \times Z_{base\ (\Omega)} = Z_{pu} \times \frac{(kV_{base,LL})^2}{MVA_{base,3\phi}}$$ $$V_{pu} = \frac{V_{actual\ (kV)}}{kV_{base,LL}}, \quad I_{pu} = \frac{I_{actual\ (A)}}{I_{base\ (A)}}, \quad S_{pu} = \frac{S_{actual\ (MVA)}}{MVA_{base,3\phi}}$$

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4. Equipment Base Impedance Conversion Formula

Electrical equipment (generators, transformers, motors, reactors) is tested and stamped by manufacturers with per-unit or percent impedance ($X''$, $X_d$, $\%Z$) based on its own nameplate ratings ($S_{rated}, V_{rated}$):

$$\%Z = Z_{pu} \times 100\% \implies Z_{pu,old} = \frac{\%Z}{100}$$

When incorporating equipment into a system-wide study with a common system power base ($S_{base,new}$) and zone voltage base ($V_{base,new}$), the per-unit impedance must be adjusted to the new base.

Derivation of the Base Conversion Formula

Because the actual ohmic impedance $Z_{actual\ (\Omega)}$ of the physical device is a constant physical property:

$$Z_{actual} = Z_{pu,old} \times Z_{base,old} = Z_{pu,new} \times Z_{base,new}$$

Solving for $Z_{pu,new}$:

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

Rearranging into the standard base conversion equation:

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

Where:

• $Z_{pu,old}$ = Nameplate per-unit impedance of the equipment
• $V_{base,old}$ = Nameplate rated voltage of the equipment (kV)
• $V_{base,new}$ = System base voltage of the specific zone where equipment resides (kV)
• $S_{base,old}$ = Nameplate rated apparent power of the equipment (MVA)
• $S_{base,new}$ = Common system base apparent power (MVA)

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5. Comprehensive Step-by-Step Worked Calculation Example

Problem Statement

A radial three-phase utility interconnection consists of:

• Generator G1: Rated $50\text{ MVA}$, $13.8\text{ kV}$, with subtransient reactance $X_d'' = 0.18\text{ pu}$.
• Step-Up Transformer T1: Rated $60\text{ MVA}$, $13.2\text{ kV} \Delta - 115\text{ kV}\text{ Y}$, with leakage reactance $X_{T1} = 0.09\text{ pu}$ ($9\%$).
• Transmission Line TL1: $115\text{ kV}$, $20\text{ miles}$ in length, with series impedance $z = 0.12 + j0.48\ \Omega/\text{mile}$.

Task: Using a system base of $S_{base,3\phi} = 100\text{ MVA}$ and $V_{base,1} = 13.8\text{ kV}$ at the generator bus (Zone 1), calculate the per-unit impedance of all components on the system base.

  Zone 1 (13.8 kV)              Zone 2 (120.23 kV or 115 kV)
 ┌───────────────┐           ┌────────────────────────────────┐
 │  Generator G1 │    T1     │         Line TL1 (20 mi)       │
 │   50 MVA      ├───[88]───┬┼───────────────────────────────■ Load Bus
 │   13.8 kV     │ 13.2/115 ││     z = 0.12 + j0.48 Ω/mi      │
 │  X'' = 0.18   │   60 MVA ││                                │
 └───────────────┘  X = 0.09│└────────────────────────────────┘
                            │
                     Zone Boundary

Step-by-Step Solution

Step 1: Establish System Voltage Zones

• Zone 1 (Generator Zone): Given $V_{base,1} = 13.8\text{ kV}$.
• Zone 2 (Transmission Line Zone): Determined by transformer T1 turns ratio ($13.2\text{ kV} : 115\text{ kV}$): $$V_{base,2} = V_{base,1} \times \left(\frac{V_{T1,sec}}{V_{T1,pri}}\right) = 13.8\text{ kV} \times \left(\frac{115\text{ kV}}{13.2\text{ kV}}\right) = 120.227\text{ kV}$$

Step 2: Convert Generator G1 Reactance

• $S_{base,old} = 50\text{ MVA}$, $V_{base,old} = 13.8\text{ kV}$, $X_{old} = 0.18\text{ pu}$
• $S_{base,new} = 100\text{ MVA}$, $V_{base,new} = 13.8\text{ kV}$

$$X_{G1,pu} = 0.18 \times \left(\frac{13.8\text{ kV}}{13.8\text{ kV}}\right)^2 \times \left(\frac{100\text{ MVA}}{50\text{ MVA}}\right) = 0.18 \times (1.0)^2 \times 2.0 = 0.3600\text{ pu}$$

Step 3: Convert Transformer T1 Reactance

Let us evaluate the conversion from both the primary (LV) and secondary (HV) sides to prove mathematical invariance:

• From Primary (LV) Side: $V_{base,old} = 13.2\text{ kV}$, $V_{base,new} = 13.8\text{ kV}$, $S_{old} = 60\text{ MVA}$, $S_{new} = 100\text{ MVA}$: $$X_{T1,pu} = 0.09 \times \left(\frac{13.2\text{ kV}}{13.8\text{ kV}}\right)^2 \times \left(\frac{100\text{ MVA}}{60\text{ MVA}}\right) = 0.09 \times 0.91493 \times 1.6667 = 0.1372\text{ pu}$$
• From Secondary (HV) Side: $V_{base,old} = 115\text{ kV}$, $V_{base,new} = 120.227\text{ kV}$, $S_{old} = 60\text{ MVA}$, $S_{new} = 100\text{ MVA}$: $$X_{T1,pu} = 0.09 \times \left(\frac{115\text{ kV}}{120.227\text{ kV}}\right)^2 \times \left(\frac{100\text{ MVA}}{60\text{ MVA}}\right) = 0.09 \times 0.91493 \times 1.6667 = 0.1372\text{ pu}$$

Both perspectives yield the exact same per-unit impedance ($0.1372\text{ pu}$).

Step 4: Convert Transmission Line TL1 Impedance

• Calculate total actual ohmic impedance: $$Z_{line,actual} = 20\text{ miles} \times (0.12 + j0.48\ \Omega/\text{mile}) = 2.40 + j9.60\ \Omega$$
• Calculate Zone 2 Base Impedance: $$Z_{base,2} = \frac{(V_{base,2})^2}{S_{base,3\phi}} = \frac{(120.227\text{ kV})^2}{100\text{ MVA}} = \frac{14454.53}{100} = 144.545\ \Omega$$
• Convert actual ohms to per-unit: $$R_{line,pu} = \frac{2.40\ \Omega}{144.545\ \Omega} = 0.0166\text{ pu}$$ $$X_{line,pu} = \frac{9.60\ \Omega}{144.545\ \Omega} = 0.0664\text{ pu}$$ $$Z_{line,pu} = 0.0166 + j0.0664\text{ pu}$$

Summary of Network Impedances on 100 MVA Base

• $X_{G1} = j0.3600\text{ pu}$
• $X_{T1} = j0.1372\text{ pu}$
• $Z_{TL1} = 0.0166 + j0.0664\text{ pu}$
• Total Series Impedance $Z_{total} = 0.0166 + j(0.3600 + 0.1372 + 0.0664) = 0.0166 + j0.5636\text{ pu}$

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6. Common NCEES Exam Pitfalls & Traps