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

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 ($60\text{ Hz}$ wavelength $\lambda = v/f \approx 3,000\text{ miles} \approx 5,000\text{ km}$).

+-----------------------------------------------------------------------------+
|                   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 ($l < 50\text{ miles}$)

For line lengths under $50\text{ miles}$ ($80\text{ km}$) at $60\text{ Hz}$, the total capacitive charging current ($I_c = \omega C l V$) is negligible compared to full-load current. Shunt capacitance is ignored, reducing the per-phase equivalent circuit to a simple lumped series impedance:

$$\mathbf{Z} = \mathbf{z} \cdot l = R + jX = (r \cdot l) + j(\omega L \cdot l) \quad [\Omega]$$

                         SHORT LINE EQUIVALENT CIRCUIT

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

Governing Terminal Equations:

$$\mathbf{V}_s = \mathbf{V}_r + \mathbf{I}_r \mathbf{Z}$$ $$\mathbf{I}_s = \mathbf{I}_r$$

ABCD Matrix Form:

$$\begin{bmatrix} \mathbf{V}_s \\ \mathbf{I}_s \end{bmatrix} = \begin{bmatrix} 1 & \mathbf{Z} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \mathbf{V}_r \\ \mathbf{I}_r \end{bmatrix}$$

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

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

For lines between $50$ and $150\text{ miles}$, shunt capacitive charging current cannot be neglected. In the standard Nominal-$\pi$ model, the total series impedance $\mathbf{Z} = \mathbf{z} \cdot l$ remains in the series branch, while the total shunt admittance $\mathbf{Y} = \mathbf{y} \cdot l = j\omega C \cdot l$ is split into two equal halves ($\mathbf{Y}/2$) 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-$\pi$ Equations:

1. Receiving Shunt Current: $\mathbf{I}_{C,r} = \mathbf{V}_r \left(\frac{\mathbf{Y}}{2}\right)$
2. Series Branch Current: $\mathbf{I}_L = \mathbf{I}_r + \mathbf{I}_{C,r} = \mathbf{I}_r + \mathbf{V}_r \left(\frac{\mathbf{Y}}{2}\right)$
3. Sending Voltage:

$$\mathbf{V}_s = \mathbf{V}_r + \mathbf{I}_L \mathbf{Z} = \mathbf{V}_r + \left[\mathbf{I}_r + \mathbf{V}_r \left(\frac{\mathbf{Y}}{2}\right)\right]\mathbf{Z} = \mathbf{V}_r \left(1 + \frac{\mathbf{Z}\mathbf{Y}}{2}\right) + \mathbf{I}_r \mathbf{Z}$$

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

$$\mathbf{I}_s = \mathbf{I}_L + \mathbf{I}_{C,s} = \mathbf{I}_L + \mathbf{V}_s \left(\frac{\mathbf{Y}}{2}\right) = \mathbf{V}_r \mathbf{Y} \left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) + \mathbf{I}_r \left(1 + \frac{\mathbf{Z}\mathbf{Y}}{2}\right)$$

+-----------------------------------------------------------------------------+
|                     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    |
+-----------------------------------------------------------------------------+

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3. Long Transmission Line Model: Distributed Parameters ($l > 150\text{ miles}$)

For line lengths exceeding $150\text{ miles}$, 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:

$$\frac{d\mathbf{V}(x)}{dx} = \mathbf{z} \mathbf{I}(x) \qquad \frac{d\mathbf{I}(x)}{dx} = \mathbf{y} \mathbf{V}(x)$$

Differentiating with respect to $x$ yields the second-order wave equations:

$$\frac{d^2\mathbf{V}(x)}{dx^2} = \mathbf{z}\mathbf{y} \mathbf{V}(x) = \gamma^2 \mathbf{V}(x)$$ $$\frac{d^2\mathbf{I}(x)}{dx^2} = \mathbf{z}\mathbf{y} \mathbf{I}(x) = \gamma^2 \mathbf{I}(x)$$

Propagation Constant ($\gamma$) & Characteristic Impedance ($Z_c$):

$$\text{Propagation Constant: } \gamma = \sqrt{\mathbf{z}\mathbf{y}} = \alpha + j\beta \quad [\text{per unit length}]$$

• $\alpha$ = Attenuation constant ($\text{Nepers/mile}$ or $\text{Np/m}$), representing dielectric and ohmic dissipation.
• $\beta$ = Phase constant ($\text{rad/mile}$ or $\text{rad/m}$), representing wave phase shift along the line:

$$\beta = \omega \sqrt{LC} = \frac{2\pi}{\lambda}$$ $$\text{Characteristic (Surge) Impedance: } Z_c = \sqrt{\frac{\mathbf{z}}{\mathbf{y}}} = \sqrt{\frac{r + j\omega L}{g + j\omega C}} \quad [\Omega]$$

Exact Hyperbolic Terminal Equations:

Evaluating the general solutions at $x = l$ yields the exact relationship between sending and receiving terminals:

$$\mathbf{V}_s = \mathbf{V}_r \cosh(\gamma l) + \mathbf{I}_r Z_c \sinh(\gamma l)$$ $$\mathbf{I}_s = \mathbf{V}_r \left(\frac{1}{Z_c}\right) \sinh(\gamma l) + \mathbf{I}_r \cosh(\gamma l)$$

+-----------------------------------------------------------------------------+
|                   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-$\pi$ Model for Long Lines

A lumped $\pi$-circuit can represent a long distributed line identically at its terminal nodes if its series impedance $\mathbf{Z}'$ and shunt admittance $\mathbf{Y}'/2$ are corrected using hyperbolic correction factors:

$$\mathbf{Z}' = \mathbf{Z} \left[\frac{\sinh(\gamma l)}{\gamma l}\right] = Z_c \sinh(\gamma l)$$ $$\frac{\mathbf{Y}'}{2} = \frac{\mathbf{Y}}{2} \left[\frac{\tanh(\gamma l / 2)}{\gamma l / 2}\right] = \frac{1}{Z_c} \tanh\left(\frac{\gamma l}{2}\right)$$

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4. ABCD Two-Port Transmission Matrix Properties

A two-port transmission network models the relationship between input (sending) and output (receiving) quantities:

$$\begin{bmatrix} \mathbf{V}_s \\ \mathbf{I}_s \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} \mathbf{V}_r \\ \mathbf{I}_r \end{bmatrix}$$

Comprehensive Line Model Comparison Table

Network Properties:

1. Reciprocity: Any linear, passive, bilateral network satisfies $AD - BC = 1$.
2. Symmetry: If the network is physically symmetrical from either terminal, $A = D$.

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 ]

$$\begin{bmatrix} \mathbf{V}_s \\ \mathbf{I}_s \end{bmatrix} = \begin{bmatrix} A_1 & B_1 \\ C_1 & D_1 \end{bmatrix} \begin{bmatrix} A_2 & B_2 \\ C_2 & D_2 \end{bmatrix} \begin{bmatrix} A_3 & B_3 \\ C_3 & D_3 \end{bmatrix} \begin{bmatrix} \mathbf{V}_r \\ \mathbf{I}_r \end{bmatrix} = \begin{bmatrix} A_{eq} & B_{eq} \\ C_{eq} & D_{eq} \end{bmatrix} \begin{bmatrix} \mathbf{V}_r \\ \mathbf{I}_r \end{bmatrix}$$ $$\text{For two cascaded networks: } \begin{bmatrix} A_{eq} & B_{eq} \\ C_{eq} & D_{eq} \end{bmatrix} = \begin{bmatrix} A_1 A_2 + B_1 C_2 & A_1 B_2 + B_1 D_2 \\ C_1 A_2 + D_1 C_2 & C_1 B_2 + D_1 D_2 \end{bmatrix}$$

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5. Step-by-Step Worked Mathematical Example

Problem Statement:

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

• Series impedance: $\mathbf{z} = 0.15 + j0.80\ \Omega/\text{mile}$
• Shunt admittance: $\mathbf{y} = j5.0 \times 10^{-6}\text{ S/mile}$

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

Calculate:

1. Total series impedance $\mathbf{Z}$ and shunt admittance $\mathbf{Y}$.
2. Nominal-$\pi$ ABCD parameters ($A, B, C, D$).
3. Sending end line-to-neutral voltage ($V_{s,LN}$), line-to-line voltage ($V_{s,LL}$), and sending end current ($I_s$).
4. Sending end real power ($P_s$) and line transmission efficiency ($\eta$).

---

Step-by-Step Solution:

Step 1: Compute Total Line Parameters

$$\mathbf{Z} = \mathbf{z} \cdot l = 120 \times (0.15 + j0.80) = 18.0 + j96.0\ \Omega = 97.6729 \angle 79.380^\circ\ \Omega$$ $$\mathbf{Y} = \mathbf{y} \cdot l = 120 \times (j5.0 \times 10^{-6}) = j0.00060\text{ S} = 0.00060 \angle 90.0^\circ\text{ S}$$ $$\mathbf{Z}\mathbf{Y} = (97.6729 \angle 79.380^\circ)(0.00060 \angle 90.0^\circ) = 0.058604 \angle 169.380^\circ = -0.05760 + j0.01079$$

Step 2: Calculate Nominal-$\pi$ ABCD Parameters

$$A = D = 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} = 1 + \frac{-0.05760 + j0.01079}{2} = 1 - 0.02880 + j0.005395 = 0.97120 + j0.005395 = 0.971215 \angle 0.318^\circ$$ $$B = \mathbf{Z} = 18.0 + j96.0\ \Omega = 97.6729 \angle 79.380^\circ\ \Omega$$ $$C = \mathbf{Y}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) = (j0.00060)\left(1 - 0.01440 + j0.002698\right) = (j0.00060)(0.98560 + j0.00270)$$ $$C = -0.00000162 + j0.00059136\text{ S} = 0.00059136 \angle 90.157^\circ\text{ S}$$

Step 3: Calculate Receiving End Operating Quantities Receiving line-to-neutral reference phasor:

$$\mathbf{V}_r = \frac{220,000\text{ V}}{\sqrt{3}} \angle 0^\circ = 127,017.06 \angle 0^\circ\text{ V}$$

Receiving current phasor ($0.85$ lagging $\implies \theta = -\arccos(0.85) = -31.788^\circ$):

$$|I_r| = \frac{S_{3\phi}}{\sqrt{3} V_{LL}} = \frac{150 \times 10^6\text{ VA}}{\sqrt{3} \times 220,000\text{ V}} = 393.648\text{ A}$$ $$\mathbf{I}_r = 393.648 \angle -31.788^\circ\text{ A} = 334.601 - j207.370\text{ A}$$

Step 4: Compute Sending End Voltage ($V_s$)

$$\mathbf{V}_s = A \mathbf{V}_r + B \mathbf{I}_r$$ $$A \mathbf{V}_r = (0.971215 \angle 0.318^\circ)(127,017.06 \angle 0^\circ) = 123,360.9 + j685.3\text{ V}$$ $$B \mathbf{I}_r = (97.6729 \angle 79.380^\circ)(393.648 \angle -31.788^\circ) = 38,448.6 \angle 47.592^\circ\text{ V} = 25,930.2 + j28,388.4\text{ V}$$ $$\mathbf{V}_s = (123,360.9 + 25,930.2) + j(685.3 + 28,388.4) = 149,291.1 + j29,073.7\text{ V}$$ $$|\mathbf{V}_s| = \sqrt{149,291.1^2 + 29,073.7^2} = \sqrt{22,287,832,519 + 845,280,032} = 152,095.8\text{ V} = 152.10\text{ kV (L-N)}$$ $$\theta_{Vs} = \arctan\left(\frac{29,073.7}{149,291.1}\right) = 11.018^\circ$$ $$V_{s,LL} = \sqrt{3} \times 152.0958\text{ kV} = 263.44\text{ kV}$$

Step 5: Compute Sending End Current ($I_s$)

$$\mathbf{I}_s = C \mathbf{V}_r + D \mathbf{I}_r$$ $$C \mathbf{V}_r = (0.00059136 \angle 90.157^\circ)(127,017.06 \angle 0^\circ) = 75.113 \angle 90.157^\circ\text{ A} = -0.206 + j75.113\text{ A}$$ $$D \mathbf{I}_r = (0.971215 \angle 0.318^\circ)(393.648 \angle -31.788^\circ) = 382.317 \angle -31.470^\circ\text{ A} = 326.079 - j199.587\text{ A}$$ $$\mathbf{I}_s = (-0.206 + 326.079) + j(75.113 - 199.587) = 325.873 - j124.474\text{ A}$$ $$|\mathbf{I}_s| = \sqrt{325.873^2 + (-124.474)^2} = 348.835\text{ A}$$ $$\theta_{Is} = \arctan\left(\frac{-124.474}{325.873}\right) = -20.908^\circ$$

Step 6: Real Power and Efficiency Sending power factor angle: $\phi_s = \theta_{Vs} - \theta_{Is} = 11.018^\circ - (-20.908^\circ) = 31.926^\circ$

$$P_s = 3 \cdot V_{s,LN} \cdot I_s \cdot \cos(\phi_s) = 3 \times 152.0958\text{ kV} \times 348.835\text{ A} \times \cos(31.926^\circ) = 135.08\text{ MW}$$ $$P_r = 150\text{ MVA} \times 0.85 = 127.50\text{ MW}$$ $$\text{Line Efficiency: } \eta = \frac{P_r}{P_s} \times 100\% = \frac{127.50\text{ MW}}{135.08\text{ MW}} \times 100\% = 94.39\%$$

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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.       |
+-----------------------------------------------------------------------------+