8.2 Line Models (Short, Medium Nominal-π, Long Lines) & ABCD Two-Port Parameters

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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

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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

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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)

8.2 Line Models (Short, Medium Nominal-π, Long Lines) & ABCD Two-Port Parameters

Key Takeaways Explain This Topic AI

Transmission lines are classified by length at 60 Hz: Short Lines (<50 miles / <80 km) neglect shunt capacitance; Medium Lines (50–150 miles / 80–240 km) use the Nominal-π lumped circuit; Long Lines (>150 miles / >240 km) require exact distributed hyperbolic wave equations.

The Nominal-π model splits total line shunt admittance Y equally (Y/2) between sending and receiving buses, yielding ABCD parameters: A = D = 1 + Z Y/2, B = Z, and C = Y (1 + Z*Y/4).

Long distributed lines model wave propagation with propagation constant γ = α + jβ = √(z y) and characteristic impedance Zc = √(z/y), yielding exact parameters A = D = cosh(γl), B = Zc sinh(γl), and C = sinh(γl)/Zc.

All linear, passive, bilateral transmission networks satisfy the reciprocity identity AD - BC = 1, and symmetrical lines (identical looking from either terminal) satisfy A = D.

Cascaded transmission systems (such as a step-up transformer, transmission line, and step-down transformer) are analyzed by multiplying their individual ABCD transfer matrices: [T_total] = [T_1] * [T_2] * [T_3].

Last updated: August 2026

8.2 Line Models (Short, Medium Nominal-π, Long Lines) & ABCD Two-Port Parameters

Transmission lines are distributed-parameter circuits whose series resistance, series inductance, shunt conductance, and shunt capacitance are uniformly distributed along their entire physical length. In power system engineering, the mathematical complexity required to model a line depends directly on its physical length relative to the system electrical wavelength ( 60 Hz 60\text{ Hz} 60 Hz wavelength λ = v / f ≈ 3 , 000 miles ≈ 5 , 000 km \lambda = v/f \approx 3,000\text{ miles} \approx 5,000\text{ km} λ = v / f ≈ 3 , 000 miles ≈ 5 , 000 km ). +-----------------------------------------------------------------------------+

• | TRANSMISSION LINE LENGTH CLASSIFICATION |
• | |
• | Classification Length at 60 Hz Circuit Model Employed |
• | ----------------------------------------------------------------------- |
• | Short Line < 50 miles (< 80 km) Series Z only (Y neglected) |
• | Medium Line 50 - 150 miles (80-240 km)Nominal-π Lumped Network |
• | Long Line > 150 miles (> 240 km) Distributed Hyperbolic Model |
• +-----------------------------------------------------------------------------+
• 1. Short Transmission Line Model ( l < 50 miles l < 50\text{ miles} l < 50 miles )

For line lengths under 50 miles 50\text{ miles} 50 miles ( 80 km 80\text{ km} 80 km ) at 60 Hz 60\text{ Hz} 60 Hz , the total capacitive charging current ( I c = ω C l V I_c = \omega C l V I c ​ = ω Cl V ) is negligible compared to full-load current. Shunt capacitance is ignored, reducing the per-phase equivalent circuit to a simple lumped series impedance: Z = z ⋅ l = R + j X = ( r ⋅ l ) + j ( ω L ⋅ l ) [ Ω ] \mathbf{Z} = \mathbf{z} \cdot l = R + jX = (r \cdot l) + j(\omega L \cdot l) \quad [\Omega] Z = z ⋅ l = R + j X = ( r ⋅ l ) + j ( ω L ⋅ l ) [ Ω ] SHORT LINE EQUIVALENT CIRCUIT

• I_s ----> Z = R + jX ----> I_r
• +-----------------------[ Z ]-----------------------+
• | |
• + | | +
• V_s | | V_r Load
• | | -
• +-------------------------------------------------------+
• Neutral

Governing Terminal Equations: V s = V r + I r Z \mathbf{V}_s = \mathbf{V}_r + \mathbf{I}_r \mathbf{Z} V s ​ = V r ​ + I r ​ Z I s = I r \mathbf{I}_s = \mathbf{I}_r I s ​ = I r ​

ABCD Matrix Form: [ V s I s ] = [ 1 Z 0 1 ] [ V r I r ] \begin{bmatrix} \mathbf{V}_s \\ \mathbf{I}_s \end{bmatrix} = \begin{bmatrix} 1 & \mathbf{Z} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \mathbf{V}_r \\ \mathbf{I}_r \end{bmatrix} [ V s ​ I s ​ ​ ] = [ 1 0 ​ Z 1 ​ ] [ V r ​ I r ​ ​ ]

Where A = 1 A = 1 A = 1 , B = Z B = \mathbf{Z} B = Z , C = 0 C = 0 C = 0 , and D = 1 D = 1 D = 1 . Verifying reciprocity: A D − B C = ( 1 ) ( 1 ) − ( Z ) ( 0 ) = 1 AD - BC = (1)(1) - (\mathbf{Z})(0) = 1 A D − BC = ( 1 ) ( 1 ) − ( Z ) ( 0 ) = 1 .

2. Medium Transmission Line Model: Nominal- π \pi π ( 50 mi ≤ l ≤ 150 mi 50\text{ mi} \le l \le 150\text{ mi} 50 mi ≤ l ≤ 150 mi )

For lines between 50 50 50 and 150 miles 150\text{ miles} 150 miles , shunt capacitive charging current cannot be neglected. In the standard Nominal- π \pi π model , the total series impedance Z = z ⋅ l \mathbf{Z} = \mathbf{z} \cdot l Z = z ⋅ l remains in the series branch, while the total shunt admittance Y = y ⋅ l = j ω C ⋅ l \mathbf{Y} = \mathbf{y} \cdot l = j\omega C \cdot l Y = y ⋅ l = jω C ⋅ l is split into two equal halves ( Y / 2 \mathbf{Y}/2 Y /2 ) lumped at the sending and receiving buses. NOMINAL-π EQUIVALENT CIRCUIT

• I_s ----> Z = R + jX ----> I_r
• +------------+------------[ Z ]------------+------------+
• | | | |
• | [Y/2] [Y/2] |
• + | (Sending) (Receiving) | +
• V_s | | | | V_r Load
• | | | | -
• +------------+---------------------------------+------------+
• Neutral

Derivation of Nominal- π \pi π Equations:

Receiving Shunt Current: I C , r = V r ( Y 2 ) \mathbf{I}_{C,r} = \mathbf{V}_r \left(\frac{\mathbf{Y}}{2}\right) I C , r ​ = V r ​ ( 2 Y ​ )

Series Branch Current: I L = I r + I C , r = I r + V r ( Y 2 ) \mathbf{I}_L = \mathbf{I}_r + \mathbf{I}_{C,r} = \mathbf{I}_r + \mathbf{V}_r \left(\frac{\mathbf{Y}}{2}\right) I L ​ = I r ​ + I C , r ​ = I r ​ + V r ​ ( 2 Y ​ )

Sending Voltage: V s = V r + I L Z = V r + [ I r + V r ( Y 2 ) ] Z = V r ( 1 + Z Y 2 ) + I r Z \mathbf{V}_s = \mathbf{V}_r + \mathbf{I}_L \mathbf{Z} = \mathbf{V}_r + \left[\mathbf{I}_r + \mathbf{V}_r \left(\frac{\mathbf{Y}}{2}\right)\right]\mathbf{Z} = \mathbf{V}_r \left(1 + \frac{\mathbf{Z}\mathbf{Y}}{2}\right) + \mathbf{I}_r \mathbf{Z} V s ​ = V r ​ + I L ​ Z = V r ​ + [ I r ​ + V r ​ ( 2 Y ​ ) ] Z = V r ​ ( 1 + 2 ZY ​ ) + I r ​ Z

Sending Shunt Current: I C , s = V s ( Y 2 ) \mathbf{I}_{C,s} = \mathbf{V}_s \left(\frac{\mathbf{Y}}{2}\right) I C , s ​ = V s ​ ( 2 Y ​ )

Sending Current: I s = I L + I C , s = I L + V s ( Y 2 ) = V r Y ( 1 + Z Y 4 ) + I r ( 1 + Z Y 2 ) \mathbf{I}_s = \mathbf{I}_L + \mathbf{I}_{C,s} = \mathbf{I}_L + \mathbf{V}_s \left(\frac{\mathbf{Y}}{2}\right) = \mathbf{V}_r \mathbf{Y} \left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) + \mathbf{I}_r \left(1 + \frac{\mathbf{Z}\mathbf{Y}}{2}\right) I s ​ = I L ​ + I C , s ​ = I L ​ + V s ​ ( 2 Y ​ ) = V r ​ Y ( 1 + 4 ZY ​ ) + I r ​ ( 1 + 2 ZY ​ ) +-----------------------------------------------------------------------------+

• | NOMINAL-π ABCD PARAMETER MATRIX |
• | |
• | A = 1 + (Z * Y) / 2 (dimensionless, complex numeric) |
• | B = Z [Ohms] |
• | C = Y * [ 1 + (Z * Y) / 4 ] [Siemens] |
• | D = 1 + (Z * Y) / 2 = A (dimensionless, symmetrical) |
• | |
• | Reciprocity Check: AD - BC = [1 + ZY/2]^2 - Z * Y * [1 + ZY/4] |
• | = 1 + ZY + Z^2*Y^2/4 - ZY - Z^2*Y^2/4 = 1 |
• +-----------------------------------------------------------------------------+
• 3. Long Transmission Line Model: Distributed Parameters ( l > 150 miles l > 150\text{ miles} l > 150 miles )

For line lengths exceeding 150 miles 150\text{ miles} 150 miles , lumped models introduce unacceptable errors. Parameters must be treated as continuous differential quantities distributed along the line. DIFFERENTIAL SECTION OF DISTRIBUTED LINE

• I(x+dx) ----> z*dx = (r + jωL)*dx ----> I(x)
• +-----------------------[ z*dx ]---------------------+
• | |
• + | [y*dx] +
• V(x+dx) | V(x)
• | | -
• +-------------------------------------------------------+
• |<------------------------ dx ------------------------->|

Wave Equations & Distributed Parameter Derivation: d V ( x ) d x = z I ( x ) d I ( x ) d x = y V ( x ) \frac{d\mathbf{V}(x)}{dx} = \mathbf{z} \mathbf{I}(x) \qquad \frac{d\mathbf{I}(x)}{dx} = \mathbf{y} \mathbf{V}(x) d x d V ( x ) ​ = zI ( x ) d x d I ( x ) ​ = yV ( x )

Differentiating with respect to x x x yields the second-order wave equations: d 2 V ( x ) d x 2 = z y V ( x ) = γ 2 V ( x ) \frac{d^2\mathbf{V}(x)}{dx^2} = \mathbf{z}\mathbf{y} \mathbf{V}(x) = \gamma^2 \mathbf{V}(x) d x 2 d 2 V ( x ) ​ = zyV ( x ) = γ 2 V ( x ) d 2 I ( x ) d x 2 = z y I ( x ) = γ 2 I ( x ) \frac{d^2\mathbf{I}(x)}{dx^2} = \mathbf{z}\mathbf{y} \mathbf{I}(x) = \gamma^2 \mathbf{I}(x) d x 2 d 2 I ( x ) ​ = zyI ( x ) = γ 2 I ( x )

Propagation Constant ( γ \gamma γ ) & Characteristic Impedance ( Z c Z_c Z c ​ ): Propagation Constant: γ = z y = α + j β [ per unit length ] \text{Propagation Constant: } \gamma = \sqrt{\mathbf{z}\mathbf{y}} = \alpha + j\beta \quad [\text{per unit length}] Propagation Constant: γ = zy ​ = α + j β [ per unit length ]

α \alpha α = Attenuation constant ( Nepers/mile \text{Nepers/mile} Nepers/mile or Np/m \text{Np/m} Np/m ), representing dielectric and ohmic dissipation.

β \beta β = Phase constant ( rad/mile \text{rad/mile} rad/mile or rad/m \text{rad/m} rad/m ), representing wave phase shift along the line: β = ω L C = 2 π λ \beta = \omega \sqrt{LC} = \frac{2\pi}{\lambda} β = ω L C ​ = λ 2 π ​ Characteristic (Surge) Impedance: Z c = z y = r + j ω L g + j ω C [ Ω ] \text{Characteristic (Surge) Impedance: } Z_c = \sqrt{\frac{\mathbf{z}}{\mathbf{y}}} = \sqrt{\frac{r + j\omega L}{g + j\omega C}} \quad [\Omega] Characteristic (Surge) Impedance: Z c ​ = y z ​ ​ = g + jω C r + jω L ​ ​ [ Ω ]

Exact Hyperbolic Terminal Equations:

Evaluating the general solutions at x = l x = l x = l yields the exact relationship between sending and receiving terminals: V s = V r cosh ⁡ ( γ l ) + I r Z c sinh ⁡ ( γ l ) \mathbf{V}_s = \mathbf{V}_r \cosh(\gamma l) + \mathbf{I}_r Z_c \sinh(\gamma l) V s ​ = V r ​ cosh ( γ l ) + I r ​ Z c ​ sinh ( γ l ) I s = V r ( 1 Z c ) sinh ⁡ ( γ l ) + I r cosh ⁡ ( γ l ) \mathbf{I}_s = \mathbf{V}_r \left(\frac{1}{Z_c}\right) \sinh(\gamma l) + \mathbf{I}_r \cosh(\gamma l) I s ​ = V r ​ ( Z c ​ 1 ​ ) sinh ( γ l ) + I r ​ cosh ( γ l ) +-----------------------------------------------------------------------------+

• | EXACT DISTRIBUTED LINE ABCD PARAMETERS |
• | |
• | A = cosh(γl) (dimensionless) |
• | B = Zc * sinh(γl) [Ohms] |
• | C = (1 / Zc) * sinh(γl) [Siemens] |
• | D = cosh(γl) = A (dimensionless) |
• | |
• | Reciprocity: AD - BC = cosh^2(γl) - sinh^2(γl) = 1.0 |
• +-----------------------------------------------------------------------------+
• Equivalent- π \pi π Model for Long Lines

A lumped π \pi π -circuit can represent a long distributed line identically at its terminal nodes if its series impedance Z ′ \mathbf{Z}' Z ′ and shunt admittance Y ′ / 2 \mathbf{Y}'/2 Y ′ /2 are corrected using hyperbolic correction factors: Z ′ = Z [ sinh ⁡ ( γ l ) γ l ] = Z c sinh ⁡ ( γ l ) \mathbf{Z}' = \mathbf{Z} \left[\frac{\sinh(\gamma l)}{\gamma l}\right] = Z_c \sinh(\gamma l) Z ′ = Z [ γ l sinh ( γ l ) ​ ] = Z c ​ sinh ( γ l ) Y ′ 2 = Y 2 [ tanh ⁡ ( γ l / 2 ) γ l / 2 ] = 1 Z c tanh ⁡ ( γ l 2 ) \frac{\mathbf{Y}'}{2} = \frac{\mathbf{Y}}{2} \left[\frac{\tanh(\gamma l / 2)}{\gamma l / 2}\right] = \frac{1}{Z_c} \tanh\left(\frac{\gamma l}{2}\right) 2 Y ′ ​ = 2 Y ​ [ γ l /2 tanh ( γ l /2 ) ​ ] = Z c ​ 1 ​ tanh ( 2 γ l ​ )

4. ABCD Two-Port Transmission Matrix Properties

A two-port transmission network models the relationship between input (sending) and output (receiving) quantities: [ V s I s ] = [ A B C D ] [ V r I r ] \begin{bmatrix} \mathbf{V}_s \\ \mathbf{I}_s \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} \mathbf{V}_r \\ \mathbf{I}_r \end{bmatrix} [ V s ​ I s ​ ​ ] = [ A C ​ B D ​ ] [ V r ​ I r ​ ​ ]

Comprehensive Line Model Comparison Table

Line Model Length Range Parameter A A A Parameter B B B ( Ω \Omega Ω ) Parameter C C C ( S \text{S} S ) Parameter D D D

Short Line < 50 mi < 50\text{ mi} < 50 mi 1.0 1.0 1.0 Z \mathbf{Z} Z 0 0 0 1.0 1.0 1.0

Nominal- π \pi π 50 − 150 mi 50 - 150\text{ mi} 50 − 150 mi 1 + Z Y 2 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} 1 + 2 ZY ​ Z \mathbf{Z} Z Y ( 1 + Z Y 4 ) \mathbf{Y}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) Y ( 1 + 4 ZY ​ ) 1 + Z Y 2 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} 1 + 2 ZY ​

Nominal-T 50 − 150 mi 50 - 150\text{ mi} 50 − 150 mi 1 + Z Y 2 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} 1 + 2 ZY ​ Z ( 1 + Z Y 4 ) \mathbf{Z}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) Z ( 1 + 4 ZY ​ ) Y \mathbf{Y} Y 1 + Z Y 2 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} 1 + 2 ZY ​

Long Distributed > 150 mi > 150\text{ mi} > 150 mi cosh ⁡ ( γ l ) \cosh(\gamma l) cosh ( γ l ) Z c sinh ⁡ ( γ l ) Z_c \sinh(\gamma l) Z c ​ sinh ( γ l ) sinh ⁡ ( γ l ) Z c \frac{\sinh(\gamma l)}{Z_c} Z c ​ s i n h ( γ l ) ​ cosh ⁡ ( γ l ) \cosh(\gamma l) cosh ( γ l )

Network Properties:

Reciprocity: Any linear, passive, bilateral network satisfies A D − B C = 1 AD - BC = 1 A D − BC = 1 .

Symmetry: If the network is physically symmetrical from either terminal, A = D A = D A = D .

Cascaded Two-Port Networks

When multiple power system components (such as a step-up transformer, transmission line, and step-down transformer) are connected in series, the composite ABCD matrix is obtained via matrix multiplication in order of power flow : CASCADED TWO-PORT POWER SYSTEM

• +-------------+ +-------------+ +-------------+
• -->| [ T_XF1 ] |------->| [ T_LINE ] |------->| [ T_XF2 ] |--->
• Vs | Transformer | V1 | Line | V2 | Transformer | Vr
• Is | (T1) | I1 | (T2) | I2 | (T3) | Ir
• +-------------+ +-------------+ +-------------+

[ T_total ] = [ T_1 ] * [ T_2 ] * [ T_3 ] [ V s I s ] = [ A 1 B 1 C 1 D 1 ] [ A 2 B 2 C 2 D 2 ] [ A 3 B 3 C 3 D 3 ] [ V r I r ] = [ A e q B e q C e q D e q ] [ V r I r ] \begin{bmatrix} \mathbf{V}_s \\ \mathbf{I}_s \end{bmatrix} = \begin{bmatrix} A_1 & B_1 \\ C_1 & D_1 \end{bmatrix} \begin{bmatrix} A_2 & B_2 \\ C_2 & D_2 \end{bmatrix} \begin{bmatrix} A_3 & B_3 \\ C_3 & D_3 \end{bmatrix} \begin{bmatrix} \mathbf{V}_r \\ \mathbf{I}_r \end{bmatrix} = \begin{bmatrix} A_{eq} & B_{eq} \\ C_{eq} & D_{eq} \end{bmatrix} \begin{bmatrix} \mathbf{V}_r \\ \mathbf{I}_r \end{bmatrix} [ V s ​ I s ​ ​ ] = [ A 1 ​ C 1 ​ ​ B 1 ​ D 1 ​ ​ ] [ A 2 ​ C 2 ​ ​ B 2 ​ D 2 ​ ​ ] [ A 3 ​ C 3 ​ ​ B 3 ​ D 3 ​ ​ ] [ V r ​ I r ​ ​ ] = [ A e q ​ C e q ​ ​ B e q ​ D e q ​ ​ ] [ V r ​ I r ​ ​ ] For two cascaded networks: [ A e q B e q C e q D e q ] = [ A 1 A 2 + B 1 C 2 A 1 B 2 + B 1 D 2 C 1 A 2 + D 1 C 2 C 1 B 2 + D 1 D 2 ] \text{For two cascaded networks: } \begin{bmatrix} A_{eq} & B_{eq} \\ C_{eq} & D_{eq} \end{bmatrix} = \begin{bmatrix} A_1 A_2 + B_1 C_2 & A_1 B_2 + B_1 D_2 \\ C_1 A_2 + D_1 C_2 & C_1 B_2 + D_1 D_2 \end{bmatrix} For two cascaded networks: [ A e q ​ C e q ​ ​ B e q ​ D e q ​ ​ ] = [ A 1 ​ A 2 ​ + B 1 ​ C 2 ​ C 1 ​ A 2 ​ + D 1 ​ C 2 ​ ​ A 1 ​ B 2 ​ + B 1 ​ D 2 ​ C 1 ​ B 2 ​ + D 1 ​ D 2 ​ ​ ]

5. Step-by-Step Worked Mathematical Example

Problem Statement:

A three-phase, 60 Hz 60\text{ Hz} 60 Hz , 230 kV 230\text{ kV} 230 kV transmission line is 120 miles 120\text{ miles} 120 miles long (Medium Line). The per-phase distributed line parameters are:

Series impedance: z = 0.15 + j 0.80 Ω / mile \mathbf{z} = 0.15 + j0.80\ \Omega/\text{mile} z = 0.15 + j 0.80 Ω/ mile

Shunt admittance: y = j 5.0 × 10 − 6 S/mile \mathbf{y} = j5.0 \times 10^{-6}\text{ S/mile} y = j 5.0 × 1 0 − 6 S/mile

The line delivers a full load of 150 MVA 150\text{ MVA} 150 MVA at 220 kV 220\text{ kV} 220 kV (line-to-line) at 0.85 0.85 0.85 power factor lagging to the receiving end substation.

Calculate:

Total series impedance Z \mathbf{Z} Z and shunt admittance Y \mathbf{Y} Y .

Nominal- π \pi π ABCD parameters ( A , B , C , D A, B, C, D A , B , C , D ).

Sending end line-to-neutral voltage ( V s , L N V_{s,LN} V s , L N ​ ), line-to-line voltage ( V s , L L V_{s,LL} V s , LL ​ ), and sending end current ( I s I_s I s ​ ).

Sending end real power ( P s P_s P s ​ ) and line transmission efficiency ( η \eta η ).

Step-by-Step Solution:

Step 1: Compute Total Line Parameters Z = z ⋅ l = 120 × ( 0.15 + j 0.80 ) = 18.0 + j 96.0 Ω = 97.6729 ∠ 79.380 ∘ Ω \mathbf{Z} = \mathbf{z} \cdot l = 120 \times (0.15 + j0.80) = 18.0 + j96.0\ \Omega = 97.6729 \angle 79.380^\circ\ \Omega Z = z ⋅ l = 120 × ( 0.15 + j 0.80 ) = 18.0 + j 96.0 Ω = 97.6729∠79.38 0 ∘ Ω Y = y ⋅ l = 120 × ( j 5.0 × 10 − 6 ) = j 0.00060 S = 0.00060 ∠ 90.0 ∘ S \mathbf{Y} = \mathbf{y} \cdot l = 120 \times (j5.0 \times 10^{-6}) = j0.00060\text{ S} = 0.00060 \angle 90.0^\circ\text{ S} Y = y ⋅ l = 120 × ( j 5.0 × 1 0 − 6 ) = j 0.00060 S = 0.00060∠90. 0 ∘ S Z Y = ( 97.6729 ∠ 79.380 ∘ ) ( 0.00060 ∠ 90.0 ∘ ) = 0.058604 ∠ 169.380 ∘ = − 0.05760 + j 0.01079 \mathbf{Z}\mathbf{Y} = (97.6729 \angle 79.380^\circ)(0.00060 \angle 90.0^\circ) = 0.058604 \angle 169.380^\circ = -0.05760 + j0.01079 ZY = ( 97.6729∠79.38 0 ∘ ) ( 0.00060∠90. 0 ∘ ) = 0.058604∠169.38 0 ∘ = − 0.05760 + j 0.01079

Step 2: Calculate Nominal- π \pi π ABCD Parameters A = D = 1 + Z Y 2 = 1 + − 0.05760 + j 0.01079 2 = 1 − 0.02880 + j 0.005395 = 0.97120 + j 0.005395 = 0.971215 ∠ 0.318 ∘ A = D = 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} = 1 + \frac{-0.05760 + j0.01079}{2} = 1 - 0.02880 + j0.005395 = 0.97120 + j0.005395 = 0.971215 \angle 0.318^\circ A = D = 1 + 2 ZY ​ = 1 + 2 − 0.05760 + j 0.01079 ​ = 1 − 0.02880 + j 0.005395 = 0.97120 + j 0.005395 = 0.971215∠0.31 8 ∘ B = Z = 18.0 + j 96.0 Ω = 97.6729 ∠ 79.380 ∘ Ω B = \mathbf{Z} = 18.0 + j96.0\ \Omega = 97.6729 \angle 79.380^\circ\ \Omega B = Z = 18.0 + j 96.0 Ω = 97.6729∠79.38 0 ∘ Ω C = Y ( 1 + Z Y 4 ) = ( j 0.00060 ) ( 1 − 0.01440 + j 0.002698 ) = ( j 0.00060 ) ( 0.98560 + j 0.00270 ) C = \mathbf{Y}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) = (j0.00060)\left(1 - 0.01440 + j0.002698\right) = (j0.00060)(0.98560 + j0.00270) C = Y ( 1 + 4 ZY ​ ) = ( j 0.00060 ) ( 1 − 0.01440 + j 0.002698 ) = ( j 0.00060 ) ( 0.98560 + j 0.00270 ) C = − 0.00000162 + j 0.00059136 S = 0.00059136 ∠ 90.157 ∘ S C = -0.00000162 + j0.00059136\text{ S} = 0.00059136 \angle 90.157^\circ\text{ S} C = − 0.00000162 + j 0.00059136 S = 0.00059136∠90.15 7 ∘ S

Step 3: Calculate Receiving End Operating Quantities Receiving line-to-neutral reference phasor: V r = 220 , 000 V 3 ∠ 0 ∘ = 127 , 017.06 ∠ 0 ∘ V \mathbf{V}_r = \frac{220,000\text{ V}}{\sqrt{3}} \angle 0^\circ = 127,017.06 \angle 0^\circ\text{ V} V r ​ = 3 ​ 220 , 000 V ​ ∠ 0 ∘ = 127 , 017.06∠ 0 ∘ V

Receiving current phasor ( 0.85 0.85 0.85 lagging ⟹ θ = − arccos ⁡ ( 0.85 ) = − 31.788 ∘ \implies \theta = -\arccos(0.85) = -31.788^\circ ⟹ θ = − arccos ( 0.85 ) = − 31.78 8 ∘ ): ∣ I r ∣ = S 3 ϕ 3 V L L = 150 × 10 6 VA 3 × 220 , 000 V = 393.648 A |I_r| = \frac{S_{3\phi}}{\sqrt{3} V_{LL}} = \frac{150 \times 10^6\text{ VA}}{\sqrt{3} \times 220,000\text{ V}} = 393.648\text{ A} ∣ I r ​ ∣ = 3 ​ V LL ​ S 3 ϕ ​ ​ = 3 ​ × 220 , 000 V 150 × 1 0 6 VA ​ = 393.648 A I r = 393.648 ∠ − 31.788 ∘ A = 334.601 − j 207.370 A \mathbf{I}_r = 393.648 \angle -31.788^\circ\text{ A} = 334.601 - j207.370\text{ A} I r ​ = 393.648∠ − 31.78 8 ∘ A = 334.601 − j 207.370 A

Step 4: Compute Sending End Voltage ( V s V_s V s ​ ) V s = A V r + B I r \mathbf{V}_s = A \mathbf{V}_r + B \mathbf{I}_r V s ​ = A V r ​ + B I r ​ A V r = ( 0.971215 ∠ 0.318 ∘ ) ( 127 , 017.06 ∠ 0 ∘ ) = 123 , 360.9 + j 685.3 V A \mathbf{V}_r = (0.971215 \angle 0.318^\circ)(127,017.06 \angle 0^\circ) = 123,360.9 + j685.3\text{ V} A V r ​ = ( 0.971215∠0.31 8 ∘ ) ( 127 , 017.06∠ 0 ∘ ) = 123 , 360.9 + j 685.3 V B I r = ( 97.6729 ∠ 79.380 ∘ ) ( 393.648 ∠ − 31.788 ∘ ) = 38 , 448.6 ∠ 47.592 ∘ V = 25 , 930.2 + j 28 , 388.4 V B \mathbf{I}_r = (97.6729 \angle 79.380^\circ)(393.648 \angle -31.788^\circ) = 38,448.6 \angle 47.592^\circ\text{ V} = 25,930.2 + j28,388.4\text{ V} B I r ​ = ( 97.6729∠79.38 0 ∘ ) ( 393.648∠ − 31.78 8 ∘ ) = 38 , 448.6∠47.59 2 ∘ V = 25 , 930.2 + j 28 , 388.4 V V s = ( 123 , 360.9 + 25 , 930.2 ) + j ( 685.3 + 28 , 388.4 ) = 149 , 291.1 + j 29 , 073.7 V \mathbf{V}_s = (123,360.9 + 25,930.2) + j(685.3 + 28,388.4) = 149,291.1 + j29,073.7\text{ V} V s ​ = ( 123 , 360.9 + 25 , 930.2 ) + j ( 685.3 + 28 , 388.4 ) = 149 , 291.1 + j 29 , 073.7 V ∣ V s ∣ = 149 , 291.1 2 + 29 , 073.7 2 = 22 , 287 , 832 , 519 + 845 , 280 , 032 = 152 , 095.8 V = 152.10 kV (L-N) |\mathbf{V}_s| = \sqrt{149,291.1^2 + 29,073.7^2} = \sqrt{22,287,832,519 + 845,280,032} = 152,095.8\text{ V} = 152.10\text{ kV (L-N)} ∣ V s ​ ∣ = 149 , 291. 1 2 + 29 , 073. 7 2 ​ = 22 , 287 , 832 , 519 + 845 , 280 , 032 ​ = 152 , 095.8 V = 152.10 kV (L-N) θ V s = arctan ⁡ ( 29 , 073.7 149 , 291.1 ) = 11.018 ∘ \theta_{Vs} = \arctan\left(\frac{29,073.7}{149,291.1}\right) = 11.018^\circ θ V s ​ = arctan ( 149 , 291.1 29 , 073.7 ​ ) = 11.01 8 ∘ V s , L L = 3 × 152.0958 kV = 263.44 kV V_{s,LL} = \sqrt{3} \times 152.0958\text{ kV} = 263.44\text{ kV} V s , LL ​ = 3 ​ × 152.0958 kV = 263.44 kV

Step 5: Compute Sending End Current ( I s I_s I s ​ ) I s = C V r + D I r \mathbf{I}_s = C \mathbf{V}_r + D \mathbf{I}_r I s ​ = C V r ​ + D I r ​ C V r = ( 0.00059136 ∠ 90.157 ∘ ) ( 127 , 017.06 ∠ 0 ∘ ) = 75.113 ∠ 90.157 ∘ A = − 0.206 + j 75.113 A C \mathbf{V}_r = (0.00059136 \angle 90.157^\circ)(127,017.06 \angle 0^\circ) = 75.113 \angle 90.157^\circ\text{ A} = -0.206 + j75.113\text{ A} C V r ​ = ( 0.00059136∠90.15 7 ∘ ) ( 127 , 017.06∠ 0 ∘ ) = 75.113∠90.15 7 ∘ A = − 0.206 + j 75.113 A D I r = ( 0.971215 ∠ 0.318 ∘ ) ( 393.648 ∠ − 31.788 ∘ ) = 382.317 ∠ − 31.470 ∘ A = 326.079 − j 199.587 A D \mathbf{I}_r = (0.971215 \angle 0.318^\circ)(393.648 \angle -31.788^\circ) = 382.317 \angle -31.470^\circ\text{ A} = 326.079 - j199.587\text{ A} D I r ​ = ( 0.971215∠0.31 8 ∘ ) ( 393.648∠ − 31.78 8 ∘ ) = 382.317∠ − 31.47 0 ∘ A = 326.079 − j 199.587 A I s = ( − 0.206 + 326.079 ) + j ( 75.113 − 199.587 ) = 325.873 − j 124.474 A \mathbf{I}_s = (-0.206 + 326.079) + j(75.113 - 199.587) = 325.873 - j124.474\text{ A} I s ​ = ( − 0.206 + 326.079 ) + j ( 75.113 − 199.587 ) = 325.873 − j 124.474 A ∣ I s ∣ = 325.873 2 + ( − 124.474 ) 2 = 348.835 A |\mathbf{I}_s| = \sqrt{325.873^2 + (-124.474)^2} = 348.835\text{ A} ∣ I s ​ ∣ = 325.87 3 2 + ( − 124.474 ) 2 ​ = 348.835 A θ I s = arctan ⁡ ( − 124.474 325.873 ) = − 20.908 ∘ \theta_{Is} = \arctan\left(\frac{-124.474}{325.873}\right) = -20.908^\circ θ I s ​ = arctan ( 325.873 − 124.474 ​ ) = − 20.90 8 ∘

Step 6: Real Power and Efficiency Sending power factor angle: ϕ s = θ V s − θ I s = 11.018 ∘ − ( − 20.908 ∘ ) = 31.926 ∘ \phi_s = \theta_{Vs} - \theta_{Is} = 11.018^\circ - (-20.908^\circ) = 31.926^\circ ϕ s ​ = θ V s ​ − θ I s ​ = 11.01 8 ∘ − ( − 20.90 8 ∘ ) = 31.92 6 ∘ P s = 3 ⋅ V s , L N ⋅ I s ⋅ cos ⁡ ( ϕ s ) = 3 × 152.0958 kV × 348.835 A × cos ⁡ ( 31.926 ∘ ) = 135.08 MW P_s = 3 \cdot V_{s,LN} \cdot I_s \cdot \cos(\phi_s) = 3 \times 152.0958\text{ kV} \times 348.835\text{ A} \times \cos(31.926^\circ) = 135.08\text{ MW} P s ​ = 3 ⋅ V s , L N ​ ⋅ I s ​ ⋅ cos ( ϕ s ​ ) = 3 × 152.0958 kV × 348.835 A × cos ( 31.92 6 ∘ ) = 135.08 MW P r = 150 MVA × 0.85 = 127.50 MW P_r = 150\text{ MVA} \times 0.85 = 127.50\text{ MW} P r ​ = 150 MVA × 0.85 = 127.50 MW Line Efficiency: η = P r P s × 100 % = 127.50 MW 135.08 MW × 100 % = 94.39 % \text{Line Efficiency: } \eta = \frac{P_r}{P_s} \times 100\% = \frac{127.50\text{ MW}}{135.08\text{ MW}} \times 100\% = 94.39\% Line Efficiency: η = P s ​ P r ​ ​ × 100% = 135.08 MW 127.50 MW ​ × 100% = 94.39%

6. Common Exam Traps & Pitfalls +-----------------------------------------------------------------------------+

• | LINE MODELING TRAPS |
• | |
• | [!] Forgetting the Y/2 Split in Nominal-π: |
• | Parameter A is 1 + Z*Y/2, NOT 1 + Z*Y. Omitting the factor of 1/2 |
• | overestimates shunt capacitive effects by 100%. |
• | |
• | [!] Matrix Multiplication Sequence: |
• | Cascaded ABCD parameters are non-commutative: [T1][T2] ≠ [T2][T1]. |
• | Always multiply in the strict direction of power flow from source to |
• | load. |
• | |
• | [!] Line-to-Line vs Line-to-Neutral Voltages in ABCD Equations: |
• | ABCD matrix equations MUST be evaluated per-phase using Line-to- |
• | Neutral voltages. Multiply by sqrt(3) only after finding Vs,LN. |
• +-----------------------------------------------------------------------------+