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3.3 The Per-Unit (pu) System: Bases, Conversions & System Modeling

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1 Chapter 1: Exam Overview & Power Engineering Fundamentals

1.1 PE Power Exam Structure, Official Specification & NCEES Reference Strategy 1.2 Units, Conversions, Engineering Constants & Fundamental Power Equations 1.3 Measurement & Metering Principles (Power, Energy, Power Factor) 1.4 Instrument Testing & Diagnostics (Insulation Megger & Ground Resistance)

2 Chapter 2: Circuit Analysis Fundamentals

2.1 DC Circuit Laws, Network Theorems (Thevenin, Norton, Superposition) & Transients 2.2 Single-Phase AC Steady-State Analysis, Phasors & Impedance 2.3 AC Power Relationships (Real, Reactive, Apparent Power & Power Factor Correction) 2.4 Resonance, Frequency Response & Harmonic Distortion

3 Chapter 3: Three-Phase Systems & Symmetrical Components

3.1 Three-Phase Circuit Analysis (Wye and Delta Configurations & Transformations) 3.2 Balanced Three-Phase Power Calculations & Phasor Relationships 3.3 The Per-Unit (pu) System: Bases, Conversions & System Modeling 3.4 Symmetrical Components (Fortescue Transformation, Sequence Networks & Unbalance)

4 Chapter 4: Electric Power Transformers

4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing 4.2 Transformer Voltage Regulation, Efficiency & Loss Calculations 4.3 Three-Phase Transformer Connections, Phase Shifts & Paralleling 4.4 Autotransformers, Instrument Transformers (CTs & VTs) & Special Transformers

5 Chapter 5: Rotating Machines: Induction & Synchronous

5.1 Three-Phase Induction Motors: Principles, Equivalent Circuits & Slip-Torque Characteristics 5.2 Induction Motor Starting, Speed Control & Braking Methods 5.3 Synchronous Generators: Operation, Excitation, Power-Angle Equations & V-Curves 5.4 Synchronous Motors & Condensers for Power Factor Correction

6 Chapter 6: DC Machines & Power Devices

6.1 DC Generators & Motors: Types (Shunt, Series, Compound), Characteristics & Speed Control 6.2 Power Capacitors, Shunt Reactors & Harmonic Filter Design 6.3 Battery Energy Storage Systems (BESS) & Ultra-Capacitors 6.4 Renewable Generation Integration: Solar Photovoltaics (PV) & Wind Power Systems

7 Chapter 7: Power Electronics & Industrial Controls

7.1 Power Semiconductor Switching Devices (Diodes, Thyristors, MOSFETs, IGBTs) 7.2 AC-to-DC Converters (Uncontrolled & Controlled Rectifiers) 7.3 DC-to-DC Switch-Mode Converters (Buck, Boost, Buck-Boost) & DC-AC Inverters 7.4 Variable Frequency Drives (VFDs), Motor Starters & Ladder Logic / PLC Controls

8 Chapter 8: Transmission & Distribution Systems

8.1 Transmission Line Parameters (Resistance, Inductance, Capacitance & GMD/GMR) 8.2 Line Models (Short, Medium Nominal-π, Long Lines) & ABCD Two-Port Parameters 8.3 Surge Impedance, Surge Impedance Loading (SIL) & Voltage Regulation 8.4 Distribution System Architectures, Voltage Drops & Feeder Loading 8.5 Power Flow Studies: Bus Classification, Y-Bus & Newton-Raphson Solutions 8.6 Power System Stability: Swing Equation, Equal-Area Criterion & Critical Clearing Time

9 Chapter 9: Power System Fault Analysis

9.1 Fault Analysis Fundamentals & Subtransient/Transient Machine Reactances 9.2 Three-Phase Symmetrical Fault Calculations 9.3 Single Line-to-Ground (SLG) & Line-to-Line (LL) Fault Analysis 9.4 Double Line-to-Ground (DLG) Faults & Bus Impedance Matrix (Zbus) Methods

10 Chapter 10: Power System Protection & Coordination

10.1 Protective Relaying Principles, IEEE Device Designations & Instrument Transformers 10.2 Overcurrent Protection (50/51), Time-Current Curves (TCC) & Relay Coordination 10.3 Differential Protection (87) for Transformers, Busbars & Generators 10.4 Distance / Impedance Protection (21) for Transmission Lines & Directional Relays (67)

11 Chapter 11: National Electrical Code (NEC) & Installation Design

11.1 NEC Fundamentals, Branch Circuit & Feeder Conductor Sizing (Ampacity & Derating) 11.2 Overcurrent Protection Sizing, Raceway Fill & Box Fill Calculations 11.3 Motor Branch Circuits, Controllers & Feeder Protection (NEC Article 430) 11.4 Grounding and Bonding Systems per NEC Article 250

12 Chapter 12: Electrical Safety, Standards & Facility Applications

12.1 Hazardous (Classified) Locations: Classes, Divisions, Zones, T-Codes & Sealing (NEC Articles 500-506) 12.2 Electrical Safety in the Workplace: NFPA 70E & Arc Flash Hazard Calculations 12.3 National Electrical Safety Code (NESC) Clearance & Utility Safety Requirements 12.4 Facility Applications: Illumination Design, Demand/Connected Loads & Lightning/Surge Protection 12.5 Special Occupancies & Systems: Health Care, Fire Pumps, Pools, Emergency & Standby (NEC Chapters 5-7)

3.3 The Per-Unit (pu) System: Bases, Conversions & System Modeling

Key Takeaways Explain This Topic AI

The per-unit system normalizes electrical quantities as dimensionless ratios: Value_pu = Actual_Value (in physical units) / Base_Value (in same physical units); selecting any two independent base quantities (typically S_base,3φ and V_base,LL) uniquely fixes all other system bases.

Derived three-phase base impedance is Z_base = (V_base,LL)^2 / S_base,3φ = (V_base,LN)^2 / S_base,1φ; when V_base is in kV and S_base is in MVA, Z_base = (V_base,kV)^2 / S_base,MVA in ohms.

Impedance base conversions across differing power or voltage bases follow the master scaling law: Z_pu,new = Z_pu,old * (S_base,new / S_base,old) * (V_base,old / V_base,new)^2.

In multi-voltage networks, base voltages in different zones are strictly dictated by transformer nominal turns ratios; an ideal transformer represented in per-unit has a 1:1 turns ratio, eliminating off-nominal voltage ideal couplings from network equations.

Converting a power system one-line diagram to a per-unit reactance diagram establishes a single unified per-phase impedance network, allowing direct application of Thevenin, nodal admittance, and short-circuit fault equations without explicit turns-ratio impedance scaling.

Last updated: August 2026

3.3 The Per-Unit (pu) System: Bases, Conversions & System Modeling

Executive Overview: Power systems encompass vast networks spanning multiple voltage levels—from generator terminals ( 13.8 kV 13.8\text{ kV} 13.8 kV ) to bulk transmission ( 115 kV − 765 kV 115\text{ kV} - 765\text{ kV} 115 kV − 765 kV ) down to industrial distribution ( 480 V − 13.8 kV 480\text{ V} - 13.8\text{ kV} 480 V − 13.8 kV ). Analyzing such systems in physical ohms requires continually reflecting impedances across transformer turns ratios ( a 2 a^2 a 2 ). The Per-Unit (pu) System normalizes all voltages, currents, powers, and impedances to dimensionless ratios, eliminating ideal transformers and condensing equipment parameters into predictable, standardized ranges. Mastery of per-unit base transformations is among the highest-yield competencies on the NCEES PE Power examination.

1. Core Principles & Advantages of the Per-Unit System

The per-unit value of any electrical quantity is defined as: Quantity pu = Actual Quantity (in physical engineering units) Base Value of Quantity (in identical physical units) \text{Quantity}_{\text{pu}} = \frac{\text{Actual Quantity (in physical engineering units)}}{\text{Base Value of Quantity (in identical physical units)}} Quantity pu ​ = Base Value of Quantity (in identical physical units) Actual Quantity (in physical engineering units) ​ Quantity % = Quantity pu × 100 % \text{Quantity}_{\%} = \text{Quantity}_{\text{pu}} \times 100\% Quantity % ​ = Quantity pu ​ × 100% +---------------------------------------------------------------------------------------------------+

  • | KEY ENGINEERING ADVANTAGES OF THE PER-UNIT SYSTEM |
  • +---------------------------------------------------------------------------------------------------+
  • | 1. Eliminates Ideal Transformers: Transformer turns ratios become 1:1 in per-unit, reducing |
  • | multi-voltage networks to simple connected impedance diagrams without turns-ratio scaling. |
  • | 2. Standardized Parameter Ranges: Equivalent impedances of similar apparatus fall within narrow |
  • | numerical bands (e.g., power transformers typically have X_leakage = 0.06 - 0.12 pu), making |
  • | erroneous data immediately detectable. |
  • | 3. Equal High-Side and Low-Side pu Impedance: A transformer's per-unit impedance is identical |
  • | whether calculated from the primary or secondary winding terminals. |
  • | 4. Intuitive Voltage Profiles: Operating voltages near nominal are close to 1.00 pu (e.g., |
  • | 0.95 pu indicates a 5% undervoltage; 1.05 pu indicates a 5% overvoltage). |
  • +---------------------------------------------------------------------------------------------------+
  • 2. Selection & Derivation of System Base Quantities

A three-phase electrical network is defined by four fundamental interrelated variables: Apparent Power ( S S S ), Voltage ( V V V ), Current ( I I I ), and Impedance ( Z Z Z ). In standard practice, two independent base quantities are arbitrarily selected, which mathematically dictate the remaining two derived bases.

Standard Base Selection Protocol

System Three-Phase Apparent Power Base ( S base , 3 ϕ S_{\text{base},3\phi} S base , 3 ϕ ​ ): Chosen globally for the entire study (standard utility convention is 100 MVA 100\text{ MVA} 100 MVA or 10 MVA 10\text{ MVA} 10 MVA ; industrial studies often select 10 MVA 10\text{ MVA} 10 MVA or 1 MVA 1\text{ MVA} 1 MVA ).

Single-phase power base: S base , 1 ϕ = S base , 3 ϕ 3 S_{\text{base},1\phi} = \frac{S_{\text{base},3\phi}}{3} S base , 1 ϕ ​ = 3 S base , 3 ϕ ​ ​ .

System Line-to-Line Voltage Base ( V base , L L V_{\text{base},LL} V base , LL ​ ): Selected for one reference voltage zone, and then propagated across each transformer according to its rated voltage ratio.

Line-to-neutral voltage base: V base , L N = V base , L L 3 V_{\text{base},LN} = \frac{V_{\text{base},LL}}{\sqrt{3}} V base , L N ​ = 3 ​ V base , LL ​ ​ .

Derivation of Derived Base Current ( I base I_{\text{base}} I base ​ ) I base = S base , 3 ϕ 3 V base , L L = S base , 1 ϕ V base , L N [ Amperes ] I_{\text{base}} = \frac{S_{\text{base},3\phi}}{\sqrt{3} V_{\text{base},LL}} = \frac{S_{\text{base},1\phi}}{V_{\text{base},LN}} \quad [\text{Amperes}] I base ​ = 3 ​ V base , LL ​ S base , 3 ϕ ​ ​ = V base , L N ​ S base , 1 ϕ ​ ​ [ Amperes ]

Derivation of Derived Base Impedance ( Z base Z_{\text{base}} Z base ​ ) Z base = V base , L N I base = V base , L L / 3 S base , 3 ϕ / ( 3 V base , L L ) = ( V base , L L ) 2 S base , 3 ϕ [ Ω ] Z_{\text{base}} = \frac{V_{\text{base},LN}}{I_{\text{base}}} = \frac{V_{\text{base},LL} / \sqrt{3}}{S_{\text{base},3\phi} / (\sqrt{3} V_{\text{base},LL})} = \frac{(V_{\text{base},LL})^2}{S_{\text{base},3\phi}} \quad [\Omega] Z base ​ = I base ​ V base , L N ​ ​ = S base , 3 ϕ ​ / ( 3 ​ V base , LL ​ ) V base , LL ​ / 3 ​ ​ = S base , 3 ϕ ​ ( V base , LL ​ ) 2 ​ [ Ω ]

When line-to-line base voltage is expressed in kilovolts ( V base , kV V_{\text{base},\text{kV}} V base , kV ​ ) and three-phase base power in megavolt-amperes ( S base , MVA S_{\text{base},\text{MVA}} S base , MVA ​ ): Z base = ( V base , kV × 10 3 ) 2 S base , MVA × 10 6 = ( V base , kV ) 2 S base , MVA [ Ω ] Z_{\text{base}} = \frac{(V_{\text{base},\text{kV}} \times 10^3)^2}{S_{\text{base},\text{MVA}} \times 10^6} = \frac{(V_{\text{base},\text{kV}})^2}{S_{\text{base},\text{MVA}}} \quad [\Omega] Z base ​ = S base , MVA ​ × 1 0 6 ( V base , kV ​ × 1 0 3 ) 2 ​ = S base , MVA ​ ( V base , kV ​ ) 2 ​ [ Ω ]

Derivation of Derived Base Admittance ( Y base Y_{\text{base}} Y base ​ ) Y base = 1 Z base = S base , MVA ( V base , kV ) 2 [ Siemens ] Y_{\text{base}} = \frac{1}{Z_{\text{base}}} = \frac{S_{\text{base},\text{MVA}}}{(V_{\text{base},\text{kV}})^2} \quad [\text{Siemens}] Y base ​ = Z base ​ 1 ​ = ( V base , kV ​ ) 2 S base , MVA ​ ​ [ Siemens ] +---------------------------------------------------------------------------------------------------+

  • | PER-UNIT SYSTEM FUNDAMENTAL BASE EQUATIONS (PE FORMULA MATRIX) |
  • +---------------------------------------------------------------------------------------------------+
  • | Base Current: | I_base = S_base,3φ / (sqrt(3) * V_base,LL) [A] |
  • | Base Impedance: | Z_base = (V_base,LL_kV)^2 / S_base,3φ_MVA [Ω] |
  • | Physical Ohms: | Z_pu = Z_actual_ohms / Z_base = Z_actual_ohms * (S_base,MVA / V_base,kV^2) |
  • | Actual Current: | I_actual = I_pu * I_base [A] |
  • | Actual Voltage: | V_actual = V_pu * V_base [V] |
  • +---------------------------------------------------------------------------------------------------+
  • 3. Changing Per-Unit Impedance Bases (The Master Scaling Law)

Manufacturer nameplates state equipment impedance in per-unit on the equipment's own nominal ratings ( S base , old S_{\text{base},\text{old}} S base , old ​ and V base , old V_{\text{base},\text{old}} V base , old ​ ). Before assembling a system-wide reactance network on a common system base ( S base , new S_{\text{base},\text{new}} S base , new ​ and V base , new V_{\text{base},\text{new}} V base , new ​ ), every per-unit impedance must be rescaled using the Master Base Conversion Formula : Z pu , new = Z pu , old × ( S base , new S base , old ) × ( V base , old V base , new ) 2 Z_{\text{pu},\text{new}} = Z_{\text{pu},\text{old}} \times \left(\frac{S_{\text{base},\text{new}}}{S_{\text{base},\text{old}}}\right) \times \left(\frac{V_{\text{base},\text{old}}}{V_{\text{base},\text{new}}}\right)^2 Z pu , new ​ = Z pu , old ​ × ( S base , old ​ S base , new ​ ​ ) × ( V base , new ​ V base , old ​ ​ ) 2

Analytical Derivation

Convert the old per-unit impedance to physical ohms ( Z Ω Z_\Omega Z Ω ​ ): Z Ω = Z pu , old × Z base , old = Z pu , old × ( V base , old ) 2 S base , old Z_\Omega = Z_{\text{pu},\text{old}} \times Z_{\text{base},\text{old}} = Z_{\text{pu},\text{old}} \times \frac{(V_{\text{base},\text{old}})^2}{S_{\text{base},\text{old}}} Z Ω ​ = Z pu , old ​ × Z base , old ​ = Z pu , old ​ × S base , old ​ ( V base , old ​ ) 2 ​

Convert physical ohms to the new per-unit base: Z pu , new = Z Ω Z base , new = Z pu , old × ( V base , old ) 2 S base , old ( V base , new ) 2 S base , new = Z pu , old × ( S base , new S base , old ) × ( V base , old V base , new ) 2 Z_{\text{pu},\text{new}} = \frac{Z_\Omega}{Z_{\text{base},\text{new}}} = \frac{Z_{\text{pu},\text{old}} \times \frac{(V_{\text{base},\text{old}})^2}{S_{\text{base},\text{old}}}}{\frac{(V_{\text{base},\text{new}})^2}{S_{\text{base},\text{new}}}} = Z_{\text{pu},\text{old}} \times \left(\frac{S_{\text{base},\text{new}}}{S_{\text{base},\text{old}}}\right) \times \left(\frac{V_{\text{base},\text{old}}}{V_{\text{base},\text{new}}}\right)^2 Z pu , new ​ = Z base , new ​ Z Ω ​ ​ = S base , new ​ ( V base , new ​ ) 2 ​ Z pu , old ​ × S base , old ​ ( V base , old ​ ) 2 ​ ​ = Z pu , old ​ × ( S base , old ​ S base , new ​ ​ ) × ( V base , new ​ V base , old ​ ​ ) 2

Exam Warning on Voltage Squaring: In the base conversion formula, power scales linearly ( S new S old \frac{S_{\text{new}}}{S_{\text{old}}} S old ​ S new ​ ​ ), whereas voltage scales with the square of the inverse ratio ( ( V old V new ) 2 \left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2 ( V new ​ V old ​ ​ ) 2 ). Omitting the square exponent is one of the most common calculation mistakes on the PE Power exam!

4. Multi-Voltage Zone Power System Partitioning

In a multi-voltage power network, transformers establish distinct voltage zones . Once a base voltage is chosen for any one zone, the base voltages for all other zones are strictly locked by the nominal rated turns ratios of the intervening transformers . MULTI-ZONE VOLTAGE PARTITIONING:

  • [Zone 1: Generation] [Zone 2: Transmission] [Zone 3: Distribution]
  • V_base1 = 13.8 kV V_base2 = 138 kV V_base3 = 4.16 kV
  • +---------------------+ +---------------------+ +---------------------+
  • | (~) Generator | T1 | Transmission | T2 | Motor Load |
  • | 13.8 kV Nominal |==#==| Line |==#==| 4.16 kV Nominal |
  • +---------------------+ +---------------------+ +---------------------+

T1 Ratio: T2 Ratio:

  • 13.8 kV / 138 kV 138 kV / 4.16 kV
  • Step-by-Step Zone Propagation Rules

Set V base 1 V_{\text{base}1} V base 1 ​ in Zone 1 (e.g., V base 1 = 13.8 kV V_{\text{base}1} = 13.8\text{ kV} V base 1 ​ = 13.8 kV ).

Across Transformer T 1 T_1 T 1 ​ (ratio N 1 : N 2 = 13.8 kV : 138 kV N_1 : N_2 = 13.8\text{ kV} : 138\text{ kV} N 1 ​ : N 2 ​ = 13.8 kV : 138 kV ): V base 2 = V base 1 × ( V T 1 , sec V T 1 , pri ) = 13.8 kV × ( 138 kV 13.8 kV ) = 138 kV V_{\text{base}2} = V_{\text{base}1} \times \left(\frac{V_{T1,\text{sec}}}{V_{T1,\text{pri}}}\right) = 13.8\text{ kV} \times \left(\frac{138\text{ kV}}{13.8\text{ kV}}\right) = 138\text{ kV} V base 2 ​ = V base 1 ​ × ( V T 1 , pri ​ V T 1 , sec ​ ​ ) = 13.8 kV × ( 13.8 kV 138 kV ​ ) = 138 kV

Across Transformer T 2 T_2 T 2 ​ (ratio N 2 : N 3 = 138 kV : 4.16 kV N_2 : N_3 = 138\text{ kV} : 4.16\text{ kV} N 2 ​ : N 3 ​ = 138 kV : 4.16 kV ): V base 3 = V base 2 × ( V T 2 , sec V T 2 , pri ) = 138 kV × ( 4.16 kV 138 kV ) = 4.16 kV V_{\text{base}3} = V_{\text{base}2} \times \left(\frac{V_{T2,\text{sec}}}{V_{T2,\text{pri}}}\right) = 138\text{ kV} \times \left(\frac{4.16\text{ kV}}{138\text{ kV}}\right) = 4.16\text{ kV} V base 3 ​ = V base 2 ​ × ( V T 2 , pri ​ V T 2 , sec ​ ​ ) = 138 kV × ( 138 kV 4.16 kV ​ ) = 4.16 kV

Off-Nominal Transformer Voltage Ratings: If a motor or generator operates at a nominal nameplate voltage that differs from the zone base voltage (e.g., a 4.0 kV 4.0\text{ kV} 4.0 kV motor installed in a 4.16 kV 4.16\text{ kV} 4.16 kV zone), the voltage correction term ( V old V new ) 2 = ( 4.0 4.16 ) 2 = 0.9245 \left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2 = \left(\frac{4.0}{4.16}\right)^2 = 0.9245 ( V new ​ V old ​ ​ ) 2 = ( 4.16 4.0 ​ ) 2 = 0.9245 must be included in that equipment's base transformation.

5. One-Line Diagram to Per-Unit Reactance Diagram Conversion

To perform fault analysis or load flow on a power system, the physical one-line diagram is systematically transformed into a single-phase per-unit reactance network: PHYSICAL ONE-LINE DIAGRAM:

[Gen G1]----(Bus 1)====[Trans T1]====(Bus 2)----[Trans Line]----(Bus 3)====[Trans T2]====(Bus 4)----[Motor M1]

PER-UNIT REACTANCE DIAGRAM (ALL VALUES ON COMMON 100 MVA BASE):

  • jX_G1 jX_T1 Z_line,pu jX_T2 jX_M1
  • +-[ZZZZ]- Bus 1 -[ZZZZ]- Bus 2 --[ZZZZZZZZ]-- Bus 3 -[ZZZZ]- Bus 4 -[ZZZZ]-+
  • | |
  • (+)[E_G1] (+)[E_M1]
  • | |
  • === Reference Bus (Ground Neutral) ==========================================
  • Systematic 5-Step Network Modeling Algorithm

Select Global System Bases: Establish a common three-phase power base S base , 3 ϕ S_{\text{base},3\phi} S base , 3 ϕ ​ (e.g., 100 MVA 100\text{ MVA} 100 MVA ) and a voltage base for one designated zone.

Determine Base Voltages for All Zones: Calculate V base V_{\text{base}} V base ​ for each zone using transformer nominal turns ratios.

Compute Base Impedances ( Z base Z_{\text{base}} Z base ​ ) and Base Currents ( I base I_{\text{base}} I base ​ ): Determine Z base = ( V base , kV ) 2 / S base , MVA Z_{\text{base}} = (V_{\text{base},\text{kV}})^2 / S_{\text{base},\text{MVA}} Z base ​ = ( V base , kV ​ ) 2 / S base , MVA ​ for every zone.

Convert All Component Impedances to System Base:

Generators & Motors: Rescale per-unit subtransient reactances X d ′ ′ X_d'' X d ′′ ​ via the Master Base Conversion Formula.

Transformers: Rescale per-unit leakage reactances X T X_T X T ​ .

Transmission Lines & Cables: Convert physical series ohms ( Z Ω Z_\Omega Z Ω ​ ) to per-unit: Z pu = Z Ω / Z base , zone Z_{\text{pu}} = Z_\Omega / Z_{\text{base},\text{zone}} Z pu ​ = Z Ω ​ / Z base , zone ​ .

Assemble the Reactance Diagram & Solve: Combine all per-unit impedances into a single interconnected network. Solve for per-unit fault currents or bus voltages, then multiply by zone base quantities to recover actual physical values.

  • 6. Comprehensive Step-by-Step Worked System Example
  • Problem Statement

A balanced three-phase power system has the following equipment specifications:

Generator G 1 G_1 G 1 ​ : 50 MVA 50\text{ MVA} 50 MVA , 13.8 kV 13.8\text{ kV} 13.8 kV , subtransient reactance X d ′ ′ = 0.15 pu X_d'' = 0.15\text{ pu} X d ′′ ​ = 0.15 pu .

Step-Up Transformer T 1 T_1 T 1 ​ : 60 MVA 60\text{ MVA} 60 MVA , 13.8 kV ( Δ ) − 115 kV (Y) 13.8\text{ kV (}\Delta\text{)} - 115\text{ kV (Y)} 13.8 kV ( Δ ) − 115 kV (Y) , leakage reactance X T 1 = 0.10 pu X_{T1} = 0.10\text{ pu} X T 1 ​ = 0.10 pu .

Transmission Line: 20 miles 20\text{ miles} 20 miles in length, series impedance z = 0.15 + j 0.80 Ω / mile z = 0.15 + j0.80\,\Omega/\text{mile} z = 0.15 + j 0.80 Ω/ mile , operating at 115 kV 115\text{ kV} 115 kV .

Step-Down Transformer T 2 T_2 T 2 ​ : 40 MVA 40\text{ MVA} 40 MVA , 115 kV (Y) − 13.8 kV ( Δ ) 115\text{ kV (Y)} - 13.8\text{ kV (}\Delta\text{)} 115 kV (Y) − 13.8 kV ( Δ ) , leakage reactance X T 2 = 0.08 pu X_{T2} = 0.08\text{ pu} X T 2 ​ = 0.08 pu .

Motor M 1 M_1 M 1 ​ : 30 MVA 30\text{ MVA} 30 MVA , 13.2 kV 13.2\text{ kV} 13.2 kV , subtransient reactance X m ′ ′ = 0.20 pu X_m'' = 0.20\text{ pu} X m ′′ ​ = 0.20 pu .

System Base Specifications: Select a common system power base of S base , 3 ϕ = 100 MVA S_{\text{base},3\phi} = 100\text{ MVA} S base , 3 ϕ ​ = 100 MVA and a generator zone voltage base of V base 1 = 13.8 kV V_{\text{base}1} = 13.8\text{ kV} V base 1 ​ = 13.8 kV .

Calculate:

The voltage base ( V base V_{\text{base}} V base ​ ) and base impedance ( Z base Z_{\text{base}} Z base ​ ) in all three zones.

The per-unit impedance of each system component on the common 100 MVA 100\text{ MVA} 100 MVA base.

The total series Thevenin impedance between the internal generator EMF and motor EMF.

The symmetrical three-phase short-circuit current in Amperes at the motor bus (Bus 4) for a solid three-phase fault, assuming a prefault voltage of V f = 1.0 ∠ 0 ∘ pu V_f = 1.0\angle 0^\circ\text{ pu} V f ​ = 1.0∠ 0 ∘ pu and neglecting prefault load current and line resistance. =========================================================================================

CALCULATION WORKFLOW & DETAILED STEP-BY-STEP SOLUTION:

=========================================================================================

  • Step 1: Determine Voltage and Impedance Bases for All Zones
  • System Power Base: S_base,3ph = 100 MVA

Zone 1 (Generator Zone):

  • V_base1 = 13.8 kV
  • Z_base1 = (V_base1)^2 / S_base = (13.8)^2 / 100 = 190.44 / 100 = 1.9044 ohms
  • I_base1 = S_base / (sqrt(3) * V_base1) = 100,000 / (1.73205 * 13.8) = 4,183.7 A

Zone 2 (Transmission Line Zone):

  • V_base2 = V_base1 * (115 / 13.8) = 13.8 * (115 / 13.8) = 115.0 kV
  • Z_base2 = (V_base2)^2 / S_base = (115.0)^2 / 100 = 13225.0 / 100 = 132.25 ohms
  • I_base2 = S_base / (sqrt(3) * V_base2) = 100,000 / (1.73205 * 115.0) = 502.04 A

Zone 3 (Motor Zone):

  • V_base3 = V_base2 * (13.8 / 115.0) = 115.0 * (13.8 / 115.0) = 13.8 kV
  • Z_base3 = (V_base3)^2 / S_base = (13.8)^2 / 100 = 1.9044 ohms
  • I_base3 = S_base / (sqrt(3) * V_base3) = 4,183.7 A

Step 2: Convert Component Impedances to Common 100 MVA Base

1. Generator G1 (50 MVA, 13.8 kV, X" = 0.15 pu):

  • X_G1,new = X_G1,old * (S_base,new / S_base,old) * (V_base,old / V_base,new)^2
  • = 0.15 * (100 / 50) * (13.8 / 13.8)^2
  • = 0.15 * 2.0 * 1.0
  • = 0.3000 pu

2. Step-Up Transformer T1 (60 MVA, 13.8/115 kV, X = 0.10 pu):

  • X_T1,new = 0.10 * (100 / 60) * (115 / 115)^2
  • = 0.10 * 1.6667 * 1.0
  • = 0.1667 pu

3. Transmission Line (20 miles * (0.15 + j0.80 ohms/mi) = 3.0 + j16.0 ohms):

  • Z_line,pu = Z_actual / Z_base2
  • = (3.0 + j16.0) / 132.25
  • = (3.0 / 132.25) + j(16.0 / 132.25)
  • = 0.02268 + j0.12098 pu

4. Step-Down Transformer T2 (40 MVA, 115/13.8 kV, X = 0.08 pu):

  • X_T2,new = 0.08 * (100 / 40) * (115 / 115)^2
  • = 0.08 * 2.50 * 1.0
  • = 0.2000 pu

5. Motor M1 (30 MVA, 13.2 kV, X" = 0.20 pu):

Notice: Motor nominal voltage (13.2 kV) differs from Zone 3 base voltage (13.8 kV)!

  • X_M1,new = X_M1,old * (S_base,new / S_base,old) * (V_base,old / V_base,new)^2
  • = 0.20 * (100 / 30) * (13.2 / 13.8)^2
  • = 0.20 * 3.3333 * (0.95652)^2
  • = 0.66667 * 0.91493
  • = 0.60995 pu ≈ 0.6100 pu
  • Step 3: Total Series Impedance Calculation (Generator to Motor)
  • Z_series = jX_G1 + jX_T1 + Z_line + jX_T2 + jX_M1
  • = j0.3000 + j0.1667 + (0.0227 + j0.1210) + j0.2000 + j0.6100
  • = 0.0227 + j(0.3000 + 0.1667 + 0.1210 + 0.2000 + 0.6100)
  • = 0.0227 + j1.3977 pu

Step 4: Symmetrical Short-Circuit Fault Current Calculation at Motor Bus (Bus 4)

For a solid 3-phase fault at Bus 4 (upstream Thevenin reactance excluding motor):

  • X_th,upstream = X_G1 + X_T1 + X_line + X_T2
  • = 0.3000 + 0.1667 + 0.1210 + 0.2000
  • = 0.7877 pu

Per-unit fault current contributed from utility/generator source:

  • I_f,pu = V_f / X_th,upstream
  • = 1.0 / 0.7877
  • = 1.2695 pu

Convert per-unit fault current to actual Amperes in Zone 3:

  • I_f,actual = I_f,pu * I_base3
  • = 1.2695 * 4,183.7 A
  • = 5,311.2 A = 5.311 kA
  • =========================================================================================
  • 7. Common Exam Traps & Tactical Pitfalls

Omitting the Square on the Voltage Conversion Ratio: Using V old V new \frac{V_{\text{old}}}{V_{\text{new}}} V new ​ V old ​ ​ instead of ( V old V new ) 2 \left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2 ( V new ​ V old ​ ​ ) 2 . Because impedance is proportional to voltage squared ( Z ∝ V 2 Z \propto V^2 Z ∝ V 2 ), the voltage adjustment ratio must always be squared.

Inverting the Voltage Ratio: Multiplying by ( V new V old ) 2 \left(\frac{V_{\text{new}}}{V_{\text{old}}}\right)^2 ( V old ​ V new ​ ​ ) 2 instead of ( V old V new ) 2 \left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2 ( V new ​ V old ​ ​ ) 2 . An apparatus with a lower rated voltage operating in a higher base voltage zone will have a smaller per-unit impedance on the new base.

Single-Phase vs. Three-Phase Base Power Mismatch: Using single-phase MVA in Z base = V L L 2 S Z_{\text{base}} = \frac{V_{LL}^2}{S} Z base ​ = S V LL 2 ​ ​ with line-to-line voltage. Remember: Z base = ( V base , L L ) 2 S base , 3 ϕ = ( V base , L N ) 2 S base , 1 ϕ Z_{\text{base}} = \frac{(V_{\text{base},LL})^2}{S_{\text{base},3\phi}} = \frac{(V_{\text{base},LN})^2}{S_{\text{base},1\phi}} Z base ​ = S base , 3 ϕ ​ ( V base , LL ​ ) 2 ​ = S base , 1 ϕ ​ ( V base , L N ​ ) 2 ​ . Mixing V L L V_{LL} V LL ​ with S 1 ϕ S_{1\phi} S 1 ϕ ​ causes a factor of 3 error!

Neglecting Off-Nominal Equipment Nameplates: Assuming equipment rated voltage always matches the zone base voltage. Always check the motor and transformer nameplates against the zone base.

Loading diagram... One-Line Diagram to Per-Unit Reactance Network Conversion