Study guide
8.2 Three-Phase Per-Unit Network Modeling & Equipment Impedance Conversion
PE Electrical and Computer: Power Exam
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1 Chapter 1: Exam Overview, Strategy & Reference Handbook
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2 Chapter 2: Measurement, Metering & Field Testing
2.1 Current Transformers (CTs), Potential Transformers (PTs) & Meter Connections 2.2 Instrument Transformer Burdens, Saturation Curves & Ratio Correction Factors 2.3 Insulation Resistance Testing (Megger, Polarization Index, Dielectric Absorption) 2.4 Ground Resistance Testing (Fall-of-Potential 62% Rule & Clamp-on Methods)
3 Chapter 3: General Applications — Grounding & Surge Protection
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6 Chapter 6: NEC Wiring Methods & Installation Requirements
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7 Chapter 7: Circuit Analysis — Fundamentals & Symmetrical Components
7.1 DC & Single-Phase AC Network Analysis (Thevenin, Norton, Superposition) 7.2 Balanced Three-Phase Circuits (Wye-Delta Configurations & Power Formulas) 7.3 Symmetrical Components & Fortescue Sequence Transformation Matrices 7.4 Sequence Impedances & Positive, Negative, Zero Sequence Network Construction
8 Chapter 8: Circuit Analysis — Per-Unit & Phasor Analysis
8.1 Per-Unit System Fundamentals & Base Conversion Formulas 8.2 Three-Phase Per-Unit Network Modeling & Equipment Impedance Conversion 8.3 Phasor Diagrams, Complex Power & Power Triangle Relationships
9 Chapter 9: Power Electronics & Logic Controls
9.1 Power Electronic Converters (Rectifiers, Inverters & DC-DC Topologies) 9.2 Inverter-Based Resources (IBRs), Grid-Forming & Grid-Following Controls 9.3 Variable Frequency Drives (VFDs), PWM Operation & Harmonic Mitigation 9.4 Boolean Algebra, Logic Gates, De Morgan's Laws & Ladder Logic Control Diagrams
10 Chapter 10: Rotating Machines — Induction & Synchronous
10.1 Three-Phase Induction Motors (Equivalent Circuits, Slip & Performance) 10.2 Induction Motor Torque-Speed Characteristics & NEMA Design Classes 10.3 Synchronous Machines (Operating Principles, Power-Angle Curves & V-Curves) 10.4 Motor Starting Methods (DOL, Wye-Delta, Autotransformer, Soft-Start) & Voltage Dip
11 Chapter 11: Electric Power Devices & Renewable Energy
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13 Chapter 13: Transmission & Distribution — Fault Analysis & Transformer Shifts
13.1 Symmetrical Three-Phase Bolted Fault Calculations 13.2 Unsymmetrical Fault Analysis (SLG, L-L, DLG Fault Calculations) 13.3 Transformer Connection Types, Zero-Sequence Paths & 30-Degree Phase Shifts
14 Chapter 14: Protection Systems & Device Coordination
14.1 Overcurrent Protection & Time-Current Characteristic (TCC) Curves 14.2 Protective Relaying Schemes (Differential 87, Distance 21, Undervoltage 27) 14.3 Circuit Breakers, Fuses & Automatic Reclosers 14.4 Protection Coordination, Selectivity & Coordination Time Intervals (CTI)
8.2 Three-Phase Per-Unit Network Modeling & Equipment Impedance Conversion
Key Takeaways Explain This Topic AI
Power transformers partition a transmission and distribution network into distinct voltage zones, where base voltages in adjacent zones must strictly scale by the transformer nominal line-to-line turns ratio ( V b a s e , 2 = V b a s e , 1 × N 2 / N 1 V_{base,2} = V_{base,1} \times N_2 / N_1 V ba se , 2 = V ba se , 1 × N 2 / N 1 ).
Manufacturer percent impedance is converted to per-unit by dividing by 100 ( % Z → Z p u = % Z / 100 \%Z \to Z_{pu} = \%Z / 100 % Z → Z p u = % Z /100 ), and transformer per-unit leakage reactance remains invariant whether referenced to primary or secondary windings on transformer base.
Loads are modeled in per-unit based on bus characteristics: constant power ( S p u = P p u + j Q p u S_{pu} = P_{pu} + jQ_{pu} S p u = P p u + j Q p u ), constant impedance ( Z p u = ∣ V p u ∣ 2 / S p u ∗ Z_{pu} = |V_{pu}|^2 / S_{pu}^* Z p u = ∣ V p u ∣ 2 / S p u ∗ ), or constant current ( I p u = I 0 ∠ ( θ V − θ ) I_{pu} = I_0 \angle (\theta_V - \theta) I p u = I 0 ∠ ( θ V − θ ) ).
Three-phase symmetrical fault current in physical amperes is evaluated directly from per-unit Thevenin impedance via I s c = I b a s e / ∣ Z t h , p u ∣ = S b a s e / ( 3 V b a s e , L L ∣ Z t h , p u ∣ ) I_{sc} = I_{base} / |Z_{th,pu}| = S_{base} / (\sqrt{3} V_{base,LL} |Z_{th,pu}|) I sc = I ba se /∣ Z t h , p u ∣ = S ba se / ( 3 V ba se , LL ∣ Z t h , p u ∣ ) .
Last updated: August 2026
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 b a s e , 2 = V b a s e , 1 × N 2 N 1 V_{base,2} = V_{base,1} \times \frac{N_2}{N_1} V ba se , 2 = V ba se , 1 × N 1 N 2 ). When modeling loads for fault studies, constant impedance representations are calculated as Z l o a d , p u = ∣ V p u ∣ 2 S l o a d , p u ∗ Z_{load,pu} = \frac{|V_{pu}|^2}{S_{load,pu}^*} Z l o a d , p u = S l o a d , p u ∗ ∣ V p u ∣ 2 . Symmetrical three-phase fault currents are solved directly in per-unit using I s c , p u = V f Z t h , p u I_{sc,pu} = \frac{V_f}{Z_{th,pu}} I sc , p u = Z t h , p u V f and converted to physical amperes via I s c = I s c , p u × I b a s e I_{sc} = I_{sc,pu} \times I_{base} I sc = I sc , p u × I ba se .
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
Single System Power Base: A single value of S b a s e , 3 ϕ S_{base,3\phi} S ba se , 3 ϕ (such as 100 MVA 100\text{ MVA} 100 MVA ) applies uniformly to all zones across the entire network.
Voltage Base Transformation: The base voltage in Zone k + 1 k+1 k + 1 is derived from Zone k k k through the rated voltage ratio of the interconnecting transformer: V b a s e , k + 1 = V b a s e , k × ( V T , w i n d i n g 2 V T , w i n d i n g 1 ) V_{base,k+1} = V_{base,k} \times \left(\frac{V_{T,winding\ 2}}{V_{T,winding\ 1}}\right) V ba se , k + 1 = V ba se , k × ( V T , w in d in g 1 V T , w in d in g 2 )
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 pu 1.0\text{ pu} 1.0 pu voltage.
- 2. Per-Unit Modeling of Power System Components
- Synchronous Generators
Synchronous generators are modeled as an ideal internal electromotive force (EMF) E g E_g E g in series with an internal reactance:
Subtransient Reactance ( X d ′ ′ X_d'' X d ′′ ): Governs current during the first 1 1 1 to 5 5 5 cycles after fault initiation ( 0 0 0 to 80 ms 80\text{ ms} 80 ms ). Used for circuit breaker momentary ratings and fast-acting protective device sizing.
Transient Reactance ( X d ′ X_d' X d ′ ): Governs current from 5 5 5 to 30 30 30 cycles ( 0.1 0.1 0.1 to 0.5 s 0.5\text{ s} 0.5 s ). Used for transient stability studies and medium-speed relay operations.
Synchronous Reactance ( X d X_d X d ): Governs steady-state fault and power flow conditions ( t > 1.0 s t > 1.0\text{ s} t > 1.0 s ). Per-Unit Model: E g ′ ′ = V t e r m , p u + I g e n , p u ( R a + j X d ′ ′ ) \text{Per-Unit Model: } E_g'' = V_{term,pu} + I_{gen,pu}(R_a + jX_d'') Per-Unit Model: E g ′′ = V t er m , p u + I g e n , p u ( R a + j X d ′′ )
For fault studies, generator armature resistance R a R_a R a is negligible ( R a ≪ X d ′ ′ R_a \ll X_d'' R a ≪ X d ′′ ), and the prefault internal voltage is typically taken as E g ′ ′ ≈ 1.0 ∠ 0 ∘ pu E_g'' \approx 1.0 \angle 0^\circ\text{ pu} E g ′′ ≈ 1.0∠ 0 ∘ pu .
Power Transformers
Two-winding power transformers are represented by their equivalent series leakage impedance Z T = R T + j X T Z_T = R_T + jX_T Z T = R T + j X T on the system base:
Winding resistance R T R_T R T is typically 0.5 % 0.5\% 0.5% to 2 % 2\% 2% of X T X_T X T and is often neglected in short-circuit studies ( Z T ≈ j X T Z_T \approx jX_T Z T ≈ j X T ).
Shunt magnetizing branch ( R c ∥ j X m R_c \parallel jX_m R c ∥ j X m ) has high impedance ( > 100 pu > 100\text{ pu} > 100 pu ) and is omitted during fault analysis.
Standard ANSI/IEEE Δ − Y \Delta - \text{Y} Δ − 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 ∘ 30^\circ 3 0 ∘ : V H V = V L V ∠ + 30 ∘ \mathbf{V}_{HV} = \mathbf{V}_{LV} \angle +30^\circ V H V = V L V ∠ + 3 0 ∘
Transmission Lines and Cables
Transmission lines are represented by their series impedance ( Z s e r i e s = R + j X Z_{series} = R + jX Z ser i es = R + j X ) and total shunt susceptance ( j B jB j B ):
Short Lines ( < 50 miles < 50\text{ miles} < 50 miles ): Shunt capacitance is negligible; modeled as pure series impedance Z l i n e , p u = ( R + j X ) / Z b a s e , z o n e Z_{line,pu} = (R + jX) / Z_{base,zone} Z l in e , p u = ( R + j X ) / Z ba se , zo n e .
Medium Lines ( 50 − 150 miles 50 - 150\text{ miles} 50 − 150 miles ): Modeled using the nominal π \pi π -equivalent circuit with half of the line charging susceptance ( j B / 2 jB/2 j B /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 p u = 1.0 V_{pu} = 1.0 V p u = 1.0 ): Z l o a d , p u = ∣ V b u s , p u ∣ 2 S l o a d , p u ∗ = ∣ V b u s , p u ∣ 2 P l o a d , p u − j Q l o a d , p u = R l o a d , p u + j X l o a d , p u 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} Z l o a d , p u = S l o a d , p u ∗ ∣ V b u s , p u ∣ 2 = P l o a d , p u − j Q l o a d , p u ∣ V b u s , p u ∣ 2 = R l o a d , p u + j X l o a d , p u
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:
Positive-Sequence Impedance Diagram: Replace all generators and utility grid interconnections with ideal voltage sources ( 1.0 ∠ 0 ∘ pu 1.0 \angle 0^\circ\text{ pu} 1.0∠ 0 ∘ pu ) behind their subtransient reactances, tied to the reference neutral.
Thevenin Impedance ( Z t h , p u Z_{th,pu} Z t h , p u ): De-energize all ideal voltage sources (short to ground) and compute the driving-point impedance viewed from the faulted bus: Z t h , p u = R t h , p u + j X t h , p u ≈ j X t h , p u Z_{th,pu} = R_{th,pu} + jX_{th,pu} \approx jX_{th,pu} Z t h , p u = R t h , p u + j X t h , p u ≈ j X t h , p u
Per-Unit Symmetrical Fault Current ( I f , p u I_{f,pu} I f , p u ): I f , p u = V f Z t h , p u = 1.0 ∠ 0 ∘ j X t h , p u = − j 1.0 X t h , p u [ 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}] I f , p u = Z t h , p u V f = j X t h , p u 1.0∠ 0 ∘ = − j X t h , p u 1.0 [ pu ]
Three-Phase Short-Circuit Apparent Power ( S s c , 3 ϕ S_{sc,3\phi} S sc , 3 ϕ ): S s c , 3 ϕ = S b a s e , 3 ϕ ∣ Z t h , p u ∣ = S b a s e , 3 ϕ X t h , p u [ MVA ] S_{sc,3\phi} = \frac{S_{base,3\phi}}{|Z_{th,pu}|} = \frac{S_{base,3\phi}}{X_{th,pu}} \quad [\text{MVA}] S sc , 3 ϕ = ∣ Z t h , p u ∣ S ba se , 3 ϕ = X t h , p u S ba se , 3 ϕ [ MVA ]
Actual RMS Symmetrical Fault Current ( I f , a c t u a l I_{f,actual} I f , a c t u a l ): I f , a c t u a l = I f , p u × I b a s e , f a u l t _ z o n e = 1.0 X t h , p u × ( S b a s e , 3 ϕ 3 ⋅ V b a s e , f a u l t _ z o n e ) [ Amperes ] 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}] I f , a c t u a l = I f , p u × I ba se , f a u lt _ zo n e = X t h , p u 1.0 × ( 3 ⋅ V ba se , f a u lt _ zo n e S ba se , 3 ϕ ) [ Amperes ]
- 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 MVA 50\text{ MVA} 50 MVA , 13.8 kV 13.8\text{ kV} 13.8 kV , X d ′ ′ = 0.15 pu X_d'' = 0.15\text{ pu} X d ′′ = 0.15 pu .
GSU Transformer T1 (Buses 1–2): Rated 60 MVA 60\text{ MVA} 60 MVA , 13.8 kV / 138 kV 13.8\text{ kV} / 138\text{ kV} 13.8 kV /138 kV , X T 1 = 0.09 pu X_{T1} = 0.09\text{ pu} X T 1 = 0.09 pu .
Transmission Line TL1 (Buses 2–3): 138 kV 138\text{ kV} 138 kV , 15 miles 15\text{ miles} 15 miles , total series reactance X l i n e = 19.044 Ω X_{line} = 19.044\ \Omega X l in e = 19.044 Ω (resistance neglected).
Substation Transformer T2 (Buses 3–4): Rated 25 MVA 25\text{ MVA} 25 MVA , 138 kV / 4.16 kV 138\text{ kV} / 4.16\text{ kV} 138 kV /4.16 kV , X T 2 = 0.0625 pu X_{T2} = 0.0625\text{ pu} X T 2 = 0.0625 pu ( 6.25 % 6.25\% 6.25% ).
Plant Induction Motor Load M1 (Bus 4): Totaling 20 MVA 20\text{ MVA} 20 MVA , rated 4.16 kV 4.16\text{ kV} 4.16 kV , with subtransient reactance X m ′ ′ = 0.20 pu X_m'' = 0.20\text{ pu} X m ′′ = 0.20 pu on motor base.
Objective: Using a system base of S b a s e = 100 MVA S_{base} = 100\text{ MVA} S ba se = 100 MVA and V b a s e , 1 = 13.8 kV V_{base,1} = 13.8\text{ kV} V ba se , 1 = 13.8 kV , calculate:
Base voltages and base currents in all zones.
All component per-unit reactances on the common 100 MVA 100\text{ MVA} 100 MVA system base.
The total symmetrical subtransient fault current ( I s c I_{sc} I sc ) in physical Amperes for a solid 3-phase fault at the 4.16 kV 4.16\text{ kV} 4.16 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 kV 13.8\text{ kV} 13.8 kV ): V b a s e , 1 = 13.8 kV , I b a s e , 1 = 100 MVA 3 × 13.8 kV = 4 , 183.7 A 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} V ba se , 1 = 13.8 kV , I ba se , 1 = 3 × 13.8 kV 100 MVA = 4 , 183.7 A
Zone 2 ( 138 kV 138\text{ kV} 138 kV ): V b a s e , 2 = 13.8 kV × ( 138 kV 13.8 kV ) = 138 kV , I b a s e , 2 = 100 MVA 3 × 138 kV = 418.37 A 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} V ba se , 2 = 13.8 kV × ( 13.8 kV 138 kV ) = 138 kV , I ba se , 2 = 3 × 138 kV 100 MVA = 418.37 A Z b a s e , 2 = ( 138 kV ) 2 100 MVA = 190.44 Ω Z_{base,2} = \frac{(138\text{ kV})^2}{100\text{ MVA}} = 190.44\ \Omega Z ba se , 2 = 100 MVA ( 138 kV ) 2 = 190.44 Ω
Zone 3 ( 4.16 kV 4.16\text{ kV} 4.16 kV ): V b a s e , 3 = 138 kV × ( 4.16 kV 138 kV ) = 4.16 kV , I b a s e , 3 = 100 MVA 3 × 4.16 kV = 13 , 878.6 A 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} V ba se , 3 = 138 kV × ( 138 kV 4.16 kV ) = 4.16 kV , I ba se , 3 = 3 × 4.16 kV 100 MVA = 13 , 878.6 A
Step 2: Convert Component Reactances to 100 MVA Base
Generator G1: X G 1 , p u = 0.15 × ( 13.8 13.8 ) 2 × ( 100 50 ) = 0.3000 pu X_{G1,pu} = 0.15 \times \left(\frac{13.8}{13.8}\right)^2 \times \left(\frac{100}{50}\right) = 0.3000\text{ pu} X G 1 , p u = 0.15 × ( 13.8 13.8 ) 2 × ( 50 100 ) = 0.3000 pu
Transformer T1: X T 1 , p u = 0.09 × ( 13.8 13.8 ) 2 × ( 100 60 ) = 0.1500 pu X_{T1,pu} = 0.09 \times \left(\frac{13.8}{13.8}\right)^2 \times \left(\frac{100}{60}\right) = 0.1500\text{ pu} X T 1 , p u = 0.09 × ( 13.8 13.8 ) 2 × ( 60 100 ) = 0.1500 pu
Transmission Line TL1: X T L 1 , p u = X a c t u a l Z b a s e , 2 = 19.044 Ω 190.44 Ω = 0.1000 pu X_{TL1,pu} = \frac{X_{actual}}{Z_{base,2}} = \frac{19.044\ \Omega}{190.44\ \Omega} = 0.1000\text{ pu} X T L 1 , p u = Z ba se , 2 X a c t u a l = 190.44 Ω 19.044 Ω = 0.1000 pu
Transformer T2: X T 2 , p u = 0.0625 × ( 138 138 ) 2 × ( 100 25 ) = 0.2500 pu X_{T2,pu} = 0.0625 \times \left(\frac{138}{138}\right)^2 \times \left(\frac{100}{25}\right) = 0.2500\text{ pu} X T 2 , p u = 0.0625 × ( 138 138 ) 2 × ( 25 100 ) = 0.2500 pu
Induction Motor Load M1: X M 1 , p u = 0.20 × ( 4.16 4.16 ) 2 × ( 100 20 ) = 1.0000 pu X_{M1,pu} = 0.20 \times \left(\frac{4.16}{4.16}\right)^2 \times \left(\frac{100}{20}\right) = 1.0000\text{ pu} X M 1 , p u = 0.20 × ( 4.16 4.16 ) 2 × ( 20 100 ) = 1.0000 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 s y s X_{sys} X sys ): X s y s = X G 1 + X T 1 + X T L 1 + X T 2 = 0.30 + 0.15 + 0.10 + 0.25 = 0.8000 pu X_{sys} = X_{G1} + X_{T1} + X_{TL1} + X_{T2} = 0.30 + 0.15 + 0.10 + 0.25 = 0.8000\text{ pu} X sys = X G 1 + X T 1 + X T L 1 + X T 2 = 0.30 + 0.15 + 0.10 + 0.25 = 0.8000 pu
Motor Branch Impedance ( X m o t o r X_{motor} X m o t or ): X m o t o r = X M 1 = 1.0000 pu X_{motor} = X_{M1} = 1.0000\text{ pu} X m o t or = X M 1 = 1.0000 pu
Parallel Equivalent Thevenin Reactance ( X t h X_{th} X t h ): X t h , p u = X s y s × X m o t o r X s y s + X m o t o r = 0.80 × 1.00 0.80 + 1.00 = 0.80 1.80 = 0.4444 pu 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} X t h , p u = X sys + X m o t or X sys × X m o t or = 0.80 + 1.00 0.80 × 1.00 = 1.80 0.80 = 0.4444 pu
Step 4: Compute Fault Quantities
Per-Unit Symmetrical Fault Current: I f , p u = V f X t h , p u = 1.0 pu 0.4444 pu = 2.2500 pu I_{f,pu} = \frac{V_f}{X_{th,pu}} = \frac{1.0\text{ pu}}{0.4444\text{ pu}} = 2.2500\text{ pu} I f , p u = X t h , p u V f = 0.4444 pu 1.0 pu = 2.2500 pu
Short-Circuit MVA: S s c , 3 ϕ = S b a s e X t h , p u = 100 MVA 0.4444 = 225.0 MVA S_{sc,3\phi} = \frac{S_{base}}{X_{th,pu}} = \frac{100\text{ MVA}}{0.4444} = 225.0\text{ MVA} S sc , 3 ϕ = X t h , p u S ba se = 0.4444 100 MVA = 225.0 MVA
Actual Physical RMS Symmetrical Short-Circuit Current: I f , a c t u a l = I f , p u × I b a s e , 3 = 2.2500 × 13 , 878.6 A = 31 , 227 A = 31.23 kA 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} I f , a c t u a l = I f , p u × I ba se , 3 = 2.2500 × 13 , 878.6 A = 31 , 227 A = 31.23 kA
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 b a s e , 3 I_{base,3} I ba se , 3 at the 4.16 kV 4.16\text{ kV} 4.16 kV bus), not the generator bus or transmission line base current.