8.2 Three-Phase Per-Unit Network Modeling & Equipment Impedance Conversion

8.2 Three-Phase Per-Unit Network Modeling & Equipment Impedance Conversion

Key Exam Takeaway: Power transformers partition networks into distinct voltage zones where the base voltage of each zone is dictated by the transformer turns ratio ($V_{base,2} = V_{base,1} \times \frac{N_2}{N_1}$). When modeling loads for fault studies, constant impedance representations are calculated as $Z_{load,pu} = \frac{|V_{pu}|^2}{S_{load,pu}^*}$. Symmetrical three-phase fault currents are solved directly in per-unit using $I_{sc,pu} = \frac{V_f}{Z_{th,pu}}$ and converted to physical amperes via $I_{sc} = I_{sc,pu} \times I_{base}$.

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1. Multi-Voltage Level Network Partitioning & Voltage Zones

A complete power system is an interconnected network of synchronous machines, transformers, overhead transmission lines, underground cables, industrial motors, and static loads. When modeling the positive-sequence network in per-unit, the fundamental rule is that power transformers serve as the structural boundaries between voltage zones.

Rules for Establishing Network Zones

1. Single System Power Base: A single value of $S_{base,3\phi}$ (such as $100\text{ MVA}$) applies uniformly to all zones across the entire network.
2. Voltage Base Transformation: The base voltage in Zone $k+1$ is derived from Zone $k$ through the rated voltage ratio of the interconnecting transformer: $V_{base,k+1} = V_{base,k} \times \left(\frac{V_{T,winding\ 2}}{V_{T,winding\ 1}}\right)$
3. Transformer Elimination: Once all equipment impedances are converted to their respective zone bases, all ideal transformers disappear from the positive-sequence circuit diagram. The multi-voltage network reduces to a single contiguous impedance network operating at a nominal $1.0\text{ pu}$ voltage.

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2. Per-Unit Modeling of Power System Components

Synchronous Generators

Synchronous generators are modeled as an ideal internal electromotive force (EMF) $E_g$ in series with an internal reactance:

• Subtransient Reactance ($X_d''$): Governs current during the first $1$ to $5$ cycles after fault initiation ($0$ to $80\text{ ms}$). Used for circuit breaker momentary ratings and fast-acting protective device sizing.
• Transient Reactance ($X_d'$): Governs current from $5$ to $30$ cycles ($0.1$ to $0.5\text{ s}$). Used for transient stability studies and medium-speed relay operations.
• Synchronous Reactance ($X_d$): Governs steady-state fault and power flow conditions ($t > 1.0\text{ s}$).

$\text{Per-Unit Model: } E_g'' = V_{term,pu} + I_{gen,pu}(R_a + jX_d'')$ For fault studies, generator armature resistance $R_a$ is negligible ($R_a \ll X_d''$), and the prefault internal voltage is typically taken as $E_g'' \approx 1.0 \angle 0^\circ\text{ pu}$.

Power Transformers

Two-winding power transformers are represented by their equivalent series leakage impedance $Z_T = R_T + jX_T$ on the system base:

• Winding resistance $R_T$ is typically $0.5%$ to $2%$ of $X_T$ and is often neglected in short-circuit studies ($Z_T \approx jX_T$).
• Shunt magnetizing branch ($R_c \ parallel jX_m$) has high impedance ($> 100\text{ pu}$) and is omitted during fault analysis.
• Standard ANSI/IEEE $\Delta - \text{Y}$ Phase Shift: Positive-sequence voltages and currents on the high-voltage (HV) side lead corresponding quantities on the low-voltage (LV) side by $30^\circ$: $\mathbf{V}_{HV} = \mathbf{V}_{LV} \angle +30^\circ$

Transmission Lines and Cables

Transmission lines are represented by their series impedance ($Z_{series} = R + jX$) and total shunt susceptance ($jB$):

• Short Lines ($< 50\text{ miles}$): Shunt capacitance is negligible; modeled as pure series impedance $Z_{line,pu} = (R + jX) / Z_{base,zone}$.
• Medium Lines ($50 - 150\text{ miles}$): Modeled using the nominal $\pi$-equivalent circuit with half of the line charging susceptance ($jB/2$) lumped at each bus terminal.

Load Modeling in Per-Unit

In power system analysis, electrical loads are represented under three primary modeling conventions:

┌───────────────────────────────────────────────────────────────────────────────┐
│                            PER-UNIT LOAD MODELS                               │
├───────────────────────┬───────────────────────────────┬───────────────────────┤
│ Constant Power (PQ)   │ Constant Current (I)          │ Constant Impedance (Z)│
│ S_pu = P_pu + jQ_pu   │ |I_pu| = Constant             │ Z_pu = |V_pu|² / S_pu*│
│ I_pu = (S_pu/V_pu)*   │ P_pu ∝ |V_pu|, Q_pu ∝ |V_pu|  │ P, Q ∝ |V_pu|²        │
│ (Power Flow Analysis) │ (Arc Furnaces, Discharge)     │ (Fault Studies)       │
└───────────────────────┴───────────────────────────────┴───────────────────────┘

For short-circuit analysis and linear network reduction, loads are converted to constant per-unit impedance at nominal voltage ($V_{pu} = 1.0$):

$Z_{load,pu} = \frac{|V_{bus,pu}|^2}{S_{load,pu}^*} = \frac{|V_{bus,pu}|^2}{P_{load,pu} - jQ_{load,pu}} = R_{load,pu} + jX_{load,pu}$

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3. End-to-End Three-Phase Symmetrical Fault Analysis

The per-unit system reduces complex three-phase short-circuit calculations to basic single-phase Thevenin equivalent circuits:

1. Positive-Sequence Impedance Diagram: Replace all generators and utility grid interconnections with ideal voltage sources ($1.0 \angle 0^\circ\text{ pu}$) behind their subtransient reactances, tied to the reference neutral.
2. Thevenin Impedance ($Z_{th,pu}$): De-energize all ideal voltage sources (short to ground) and compute the driving-point impedance viewed from the faulted bus: $Z_{th,pu} = R_{th,pu} + jX_{th,pu} \approx jX_{th,pu}$
3. Per-Unit Symmetrical Fault Current ($I_{f,pu}$): $I_{f,pu} = \frac{V_f}{Z_{th,pu}} = \frac{1.0 \angle 0^\circ}{jX_{th,pu}} = -j \frac{1.0}{X_{th,pu}} \quad [\text{pu}]$
4. Three-Phase Short-Circuit Apparent Power ($S_{sc,3\phi}$): $S_{sc,3\phi} = \frac{S_{base,3\phi}}{|Z_{th,pu}|} = \frac{S_{base,3\phi}}{X_{th,pu}} \quad [\text{MVA}]$
5. Actual RMS Symmetrical Fault Current ($I_{f,actual}$): $I_{f,actual} = I_{f,pu} \times I_{base,fault\_zone} = \frac{1.0}{X_{th,pu}} \times \left(\frac{S_{base,3\phi}}{\sqrt{3} \cdot V_{base,fault\_zone}}\right) \quad [\text{Amperes}]$

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4. Comprehensive Multi-Bus Industrial System Worked Example

System Configuration & Parameters

An industrial manufacturing complex is supplied by a utility generation source through a transmission line and plant substation:

• Utility Generator G1 (Bus 1): Rated $50\text{ MVA}$, $13.8\text{ kV}$, $X_d'' = 0.15\text{ pu}$.
• GSU Transformer T1 (Buses 1–2): Rated $60\text{ MVA}$, $13.8\text{ kV} / 138\text{ kV}$, $X_{T1} = 0.09\text{ pu}$.
• Transmission Line TL1 (Buses 2–3): $138\text{ kV}$, $15\text{ miles}$, total series reactance $X_{line} = 19.044\ \Omega$ (resistance neglected).
• Substation Transformer T2 (Buses 3–4): Rated $25\text{ MVA}$, $138\text{ kV} / 4.16\text{ kV}$, $X_{T2} = 0.0625\text{ pu}$ ($6.25%$).
• Plant Induction Motor Load M1 (Bus 4): Totaling $20\text{ MVA}$, rated $4.16\text{ kV}$, with subtransient reactance $X_m'' = 0.20\text{ pu}$ on motor base.

Objective: Using a system base of $S_{base} = 100\text{ MVA}$ and $V_{base,1} = 13.8\text{ kV}$, calculate:

1. Base voltages and base currents in all zones.
2. All component per-unit reactances on the common $100\text{ MVA}$ system base.
3. The total symmetrical subtransient fault current ($I_{sc}$) in physical Amperes for a solid 3-phase fault at the $4.16\text{ kV}$ industrial bus (Bus 4).

  Bus 1 (13.8 kV)     Bus 2 (138 kV)               Bus 3 (138 kV)    Bus 4 (4.16 kV)
 ┌──────────────┐     ┌───────────┐                ┌───────────┐     ┌─────────────┐
 │ Generator G1 ├──┬──┤ T1 (GSU)  ├───[TL1: 15 mi]─┤ T2 (Plant)├──┬──┤   FAULT F1  │
 │   50 MVA     │  │  │ 60 MVA    │   X=19.044 Ω   │ 25 MVA    │  │  │      X      │
 │   13.8 kV    │  │  │ 13.8/138  │                │ 138/4.16  │  │  ├─────────────┤
 │  X'' = 0.15  │  │  │ X = 0.09  │                │ X = 0.0625│  │  │  Motor M1   │
 └──────────────┘  │  └───────────┘                └───────────┘  │  │  20 MVA     │
                   │                                              │  │  X'' = 0.20 │
                 Zone 1                                         Zone 2 └─────────────┘
                (13.8 kV)                                      (138 kV)     Zone 3
                                                                           (4.16 kV)

Step-by-Step Solution

Step 1: Establish Zone Voltage and Current Bases

• Zone 1 ($13.8\text{ kV}$): $V_{base,1} = 13.8\text{ kV}, \quad I_{base,1} = \frac{100\text{ MVA}}{\sqrt{3} \times 13.8\text{ kV}} = 4,183.7\text{ A}$
• Zone 2 ($138\text{ kV}$): $V_{base,2} = 13.8\text{ kV} \times \left(\frac{138\text{ kV}}{13.8\text{ kV}}\right) = 138\text{ kV}, \quad I_{base,2} = \frac{100\text{ MVA}}{\sqrt{3} \times 138\text{ kV}} = 418.37\text{ A}$ $Z_{base,2} = \frac{(138\text{ kV})^2}{100\text{ MVA}} = 190.44\ \Omega$
• Zone 3 ($4.16\text{ kV}$): $V_{base,3} = 138\text{ kV} \times \left(\frac{4.16\text{ kV}}{138\text{ kV}}\right) = 4.16\text{ kV}, \quad I_{base,3} = \frac{100\text{ MVA}}{\sqrt{3} \times 4.16\text{ kV}} = 13,878.6\text{ A}$

Step 2: Convert Component Reactances to 100 MVA Base

• Generator G1: $X_{G1,pu} = 0.15 \times \left(\frac{13.8}{13.8}\right)^2 \times \left(\frac{100}{50}\right) = 0.3000\text{ pu}$
• Transformer T1: $X_{T1,pu} = 0.09 \times \left(\frac{13.8}{13.8}\right)^2 \times \left(\frac{100}{60}\right) = 0.1500\text{ pu}$
• Transmission Line TL1: $X_{TL1,pu} = \frac{X_{actual}}{Z_{base,2}} = \frac{19.044\ \Omega}{190.44\ \Omega} = 0.1000\text{ pu}$
• Transformer T2: $X_{T2,pu} = 0.0625 \times \left(\frac{138}{138}\right)^2 \times \left(\frac{100}{25}\right) = 0.2500\text{ pu}$
• Induction Motor Load M1: $X_{M1,pu} = 0.20 \times \left(\frac{4.16}{4.16}\right)^2 \times \left(\frac{100}{20}\right) = 1.0000\text{ pu}$

Step 3: Calculate Thevenin Equivalent Impedance at Faulted Bus 4

The total impedance is formed by the utility supply branch in parallel with the motor back-feed branch:

• Utility Supply Branch Impedance ($X_{sys}$): $X_{sys} = X_{G1} + X_{T1} + X_{TL1} + X_{T2} = 0.30 + 0.15 + 0.10 + 0.25 = 0.8000\text{ pu}$
• Motor Branch Impedance ($X_{motor}$): $X_{motor} = X_{M1} = 1.0000\text{ pu}$
• Parallel Equivalent Thevenin Reactance ($X_{th}$): $X_{th,pu} = \frac{X_{sys} \times X_{motor}}{X_{sys} + X_{motor}} = \frac{0.80 \times 1.00}{0.80 + 1.00} = \frac{0.80}{1.80} = 0.4444\text{ pu}$

Step 4: Compute Fault Quantities

• Per-Unit Symmetrical Fault Current: $I_{f,pu} = \frac{V_f}{X_{th,pu}} = \frac{1.0\text{ pu}}{0.4444\text{ pu}} = 2.2500\text{ pu}$
• Short-Circuit MVA: $S_{sc,3\phi} = \frac{S_{base}}{X_{th,pu}} = \frac{100\text{ MVA}}{0.4444} = 225.0\text{ MVA}$
• Actual Physical RMS Symmetrical Short-Circuit Current: $I_{f,actual} = I_{f,pu} \times I_{base,3} = 2.2500 \times 13,878.6\text{ A} = 31,227\text{ A} = 31.23\text{ kA}$

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5. Critical Exam Tips & Common Traps

[!WARNING] Exam Trap 1: Neglecting Motor Contribution to Faults. Induction and synchronous motors act as generators during the first few cycles of a fault due to trapped rotor magnetic flux. In short-circuit duty calculations, always place motor subtransient reactances in parallel with the utility grid supply.

[!CAUTION] Exam Trap 2: Using the Wrong Base Current for Physical Conversion. When converting per-unit fault current to physical Amperes, you must multiply by the base current of the specific zone where the fault occurs ($I_{base,3}$ at the $4.16\text{ kV}$ bus), not the generator bus or transmission line base current.