18.1 Transmission Line Parameters & Performance Models

18.1 Transmission Line Parameters & Performance Models

Electric power transmission line theory and steady-state modeling constitute major core subjects on the PRC Registered Electrical Engineer (REE) Licensure Examination. Power transmission lines connect distant generating plants to distribution load centers. Precise analytical modeling of transmission line parameters—resistance ($R$), inductance ($L$), capacitance ($C$), and shunt conductance ($G$)—is essential for calculating voltage drop, power losses, transient limits, and voltage regulation across high-voltage grid networks.

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1. Electrical Transmission Line Parameters

An overhead power transmission line possesses four distributed electrical parameters evenly distributed along its physical length.

Resistance ($R$)

The DC resistance of a solid cylindrical conductor of length $l$ (meters) and cross-sectional area $A$ (square meters) is:

$$R_{\text{dc}} = \frac{\rho l}{A}$$

where $\rho$ is the material volume resistivity ($\Omega \cdot \text{m}$). Conductor resistance increases linearly with operating temperature:

$$R_{T2} = R_{T1} \left[ 1 + \alpha_{T1} (T_2 - T_1) \right] = R_{T1} \left( \frac{T_2 + T_0}{T_1 + T_0} \right)$$

where $\alpha_{T1}$ is the temperature coefficient of resistance at $T_1$, and $T_0$ is the inferred absolute zero-resistance temperature ($T_0 = 234.5^\circ\text{C}$ for $100\%$ IACS copper, $T_0 = 228.1^\circ\text{C}$ for hard-drawn aluminum).

Skin Effect & Proximity Effect

• Skin Effect: Under alternating current (AC), changing magnetic flux linkages inside the conductor generate internal eddy currents that oppose current flow at the conductor core while reinforcing current flow near the outer surface. Consequently, AC current density concentrates near the conductor surface ("skin"). This reduces the effective cross-sectional area, raising AC resistance ($R_{\text{ac}} > R_{\text{dc}}$). Skin effect increases with higher supply frequency ($f$), larger conductor diameter ($d$), higher permeability ($\mu$), and lower resistivity ($\rho$). At $60\ \text{Hz}$, $R_{\text{ac}} / R_{\text{dc}} \approx 1.02 \text{ to } 1.08$.
• Proximity Effect: Alternate magnetic fields produced by adjacent parallel conductors alter the current density distribution across conductor cross-sections, further increasing effective AC line resistance.

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2. Inductance of Transmission Lines

Inductance per phase arises from total magnetic flux linkages per unit current, comprising internal flux linkages (inside the conductor body) and external flux linkages (between conductor surface and infinity).

Single Conductor & Internal Inductance

For a solid cylindrical conductor carrying uniform current $I$, the internal inductance per meter is constant regardless of conductor radius:

$$L_{\text{int}} = \frac{\mu_0}{8 \pi} = \frac{4 \pi \times 10^{-7}}{8 \pi} = \frac{1}{2} \times 10^{-7}\ \text{H/m} = 0.05\ \text{mH/km}$$

Self Geometric Mean Radius (GMR / $D_s$) & Geometric Mean Distance (GMD / $D_m$)

Combining internal and external flux linkages yields the total inductance formula using the concept of Self GMR ($D_s$) and Mutual GMD ($D_m$).

Self GMR ($D_s$)

For a solid conductor of radius $r$, the internal inductance is mathematically absorbed into an equivalent fictitious hollow thin-walled conductor of radius $D_s$ with zero internal flux:

$$D_s = r e^{-1/4} = r e^{-0.25} \approx 0.7788 r$$

For multi-strand composite or ACSR conductors, $D_s$ is determined by taking the $n^2$-th root of all inter-strand distance products:

$$D_s = \sqrt[n^2]{\prod_{i=1}^n \prod_{j=1}^n d_{ij}}$$

For a bundled conductor consisting of $n$ sub-conductors separated by bundle spacing $d$:

• 2-conductor bundle: $D_s^b = \sqrt{D_s \cdot d}$
• 3-conductor bundle (equilateral triangle): $D_s^b = \sqrt[3]{D_s \cdot d^2}$
• 4-conductor bundle (square arrangement): $D_s^b = \sqrt[4]{D_s \cdot d^3 \cdot \sqrt{2}} = 1.091 \sqrt[4]{D_s \cdot d^3}$

Mutual GMD ($D_m$)

For a 3-phase line with phase conductor spacings $D_{ab}, D_{bc}, D_{ca}$:

$$D_m = \sqrt[3]{D_{ab} D_{bc} D_{ca}}$$

If the 3-phase conductors are symmetrically placed at the vertices of an equilateral triangle ($D_{ab} = D_{bc} = D_{ca} = D$), then $D_m = D$.

Inductance Equations

1. Single-Phase Two-Wire Line: $$L_{\text{loop}} = 4 \times 10^{-7} \ln\left( \frac{D}{D_s} \right)\ \text{H/m} = 0.921 \log_{10}\left( \frac{D}{D_s} \right)\ \text{mH/km}$$
2. Three-Phase Line (Transposed Conductor Configuration): $$L = 2 \times 10^{-7} \ln\left( \frac{D_m}{D_s} \right)\ \text{H/m/phase} = 0.4605 \log_{10}\left( \frac{D_m}{D_s} \right)\ \text{mH/km/phase}$$

$$\text{Inductive Reactance: } X_L = 2 \pi f L \quad (\Omega/\text{km})$$

REE Exam Rule: Transposition of line conductors at regular physical intervals along the line route equalizes mutual inductances among phases, preventing unbalanced induced phase voltage drops.

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3. Capacitance of Transmission Lines

Electric potential differences between parallel energized conductors establish an electrostatic field. Transmission line capacitance per unit length is defined by electric charge per unit potential difference ($C = q / V$).

Single-Phase & Three-Phase Line-to-Neutral Capacitance

For a 3-phase transposed line with mutual GMD $D_m$ and conductor outer radius $r$:

$$C_n = \frac{2 \pi \epsilon_0}{\ln\left( \frac{D_m}{r} \right)} = \frac{2 \pi (8.854 \times 10^{-12})}{\ln\left( \frac{D_m}{r} \right)}\ \text{F/m/phase} = \frac{0.0556}{\ln\left( \frac{D_m}{r} \right)}\ \mu\text{F/km/phase}$$

For bundled conductors, the capacitive self GMR is $D_s^c = r$ (using physical radius $r$, not $0.7788r$!):

• 2-conductor bundle: $D_s^{cb} = \sqrt{r \cdot d}$
• 4-conductor bundle: $D_s^{cb} = 1.091 \sqrt[4]{r \cdot d^3}$

$$\text{Line Charging Current: } I_c = \omega C_n V_{\text{LN}} = 2 \pi f C_n \left( \frac{V_{L-L}}{\sqrt{3}} \right) \quad (\text{A/phase})$$

Effect of Earth / Ground (Method of Image Charges)

The presence of the conducting earth plane increases line-to-neutral capacitance. Using fictitious image conductors placed at depth $h$ below ground level:

$$C_{n,\text{earth}} = \frac{2 \pi \epsilon_0}{\ln\left( \frac{D_m}{r \cdot \sqrt{1 + \left( \frac{D_m}{2h} \right)^2}} \right)}$$

Because tower heights $h \gg D_m$, earth proximity increases capacitance by less than $1\%\text{--}2\%$, which is often neglected in standard overhead line calculations.

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4. Two-Port ABCD Parameter Representation & Transmission Line Models

A transmission line is modeled as a passive linear two-port network relating sending-end voltage/current ($V_S, I_S$) to receiving-end voltage/current ($V_R, I_R$):

$$\begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} V_R \\ I_R \end{bmatrix}$$ $$\begin{aligned} V_S &= A V_R + B I_R \\ I_S &= C V_R + D I_R \end{aligned}$$

Fundamental Properties of ABCD Matrix:

1. Reciprocity: $A D - B C = 1$
2. Symmetry (for identical sending and receiving ends): $A = D$
3. Units: $A$ and $D$ are dimensionless; $B$ has units of Ohms ($\Omega$); $C$ has units of Siemens ($\text{S}$).

Classification of Transmission Lines by Length

where:

• $Z = z \cdot l = (r + j \omega l) l$ is total line series impedance ($\Omega$).
• $Y = y \cdot l = (g + j \omega c) l$ is total line shunt admittance ($\text{S}$).
• $\gamma = \sqrt{z y} = \alpha + j \beta$ is the complex wave propagation constant ($\alpha$ attenuation constant in nepers/km, $\beta$ phase constant in rad/km).
• $Z_c = \sqrt{\frac{z}{y}} \approx \sqrt{\frac{L}{C}}$ is characteristic impedance / surge impedance ($\Omega$).

Voltage Regulation & Transmission Efficiency

• Percentage Voltage Regulation (%VR): $$\% \text{VR} = \frac{\left| \frac{V_S}{A} \right| - |V_R|}{|V_R|} \times 100\%$$
• Transmission Efficiency ($\eta$): $$\eta = \frac{P_{R,\text{3-phase}}}{P_{S,\text{3-phase}}} \times 100\% = \frac{3 |V_R| |I_R| \cos \phi_R}{3 |V_S| |I_S| \cos \phi_S} \times 100\%$$

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5. Ferranti Effect & Surge Impedance Loading (SIL)

Ferranti Effect

When a medium or long transmission line operates under no-load ($I_R = 0$) or very light load, the receiving-end voltage rises significantly above the sending-end voltage ($V_R > V_S$). This voltage rise is called the Ferranti Effect.

Physical Cause:

The line's distributed charging current $I_c = Y V_R$ flows through the line's series inductive reactance $X_L$. The voltage drop across series inductance leads receiving-end voltage by $90^\circ$, creating a boosting voltage component in phase with $V_S$.

Mathematical Derivation:

For an un-loaded long line ($I_R = 0$), $V_S = V_R \cosh(\gamma l)$. For a lossless line ($\alpha = 0, \gamma = j\beta$):

$$V_S = V_R \cos(\beta l) \implies V_R = \frac{V_S}{\cos(\beta l)}$$

Expanding $\cos(\beta l) \approx 1 - \frac{\beta^2 l^2}{2}$ yields the approximate receiving-end voltage rise:

$$V_R \approx V_S \left( 1 + \frac{\omega^2 l^2 L C}{2} \right) \implies \Delta V = V_R - V_S \approx \frac{V_S \omega^2 l^2 L C}{2}$$

Mitigation: Connect shunt reactors (inductive coils) at receiving-end substations to absorb leading capacitive VARs under light-load conditions.

Surge Impedance Loading (SIL)

Surge Impedance Loading (SIL) is the active power delivered by a transmission line when its load impedance equals its characteristic surge impedance ($Z_L = Z_c$):

$$\text{SIL} = \frac{\left( V_{L-L,\text{rated}} \right)^2}{Z_c} \quad (\text{MW})$$

For typical overhead AC transmission lines, $Z_c \approx 377\ \Omega \text{ to } 400\ \Omega$. For underground power cables, $Z_c \approx 40\ \Omega \text{ to } 60\ \Omega$.

Operational Significance of SIL:

• Load $= \text{SIL}$: Reactive power generated by line capacitance ($Q_C = I^2 X_C$) exactly balances reactive power absorbed by line inductance ($Q_L = I^2 X_L$). The line has a flat voltage profile ($|V(x)| = V_R = V_S$), unity power factor everywhere, and zero net reactive power exchange with grid terminals.
• Load $> \text{SIL}$: Line acts as net inductor ($Q_L > Q_C$), requiring reactive power support; receiving voltage drops ($V_R < V_S$).
• Load $< \text{SIL}$: Line acts as net capacitor ($Q_C > Q_L$), generating excess reactive power; receiving voltage rises ($V_R > V_S$, Ferranti effect).

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Solved Board Exam Examples

Example 1: Inductance & Reactance of a 3-Phase Bundled Line

Problem: A 3-phase, $60\ \text{Hz}$ transposed transmission line rated at $230\ \text{kV}$ has horizontal conductor spacing $D_{ab} = 6.0\ \text{m}$, $D_{bc} = 6.0\ \text{m}$, and $D_{ca} = 12.0\ \text{m}$. Each phase consists of a 2-conductor bundle spaced $d = 0.45\ \text{m}$ apart. Each sub-conductor has an outside radius $r = 1.5\ \text{cm}$. Calculate: (a) Mutual GMD $D_m$, (b) Bundled self GMR $D_s^b$, (c) Line inductance per phase per km, and (d) Total inductive reactance per phase for a $100\ \text{km}$ line.

Solution:

1. Calculate mutual GMD ($D_m$): $$D_m = \sqrt[3]{D_{ab} D_{bc} D_{ca}} = \sqrt[3]{6.0 \times 6.0 \times 12.0} = \sqrt[3]{432} = 7.5595\ \text{m}$$
2. Calculate single conductor self GMR ($D_s$): $$D_s = 0.7788 \times r = 0.7788 \times 0.015\ \text{m} = 0.011682\ \text{m}$$
3. Calculate bundled conductor self GMR ($D_s^b$): $$D_s^b = \sqrt{D_s \cdot d} = \sqrt{0.011682 \times 0.45} = \sqrt{0.0052569} = 0.072504\ \text{m}$$
4. Calculate line inductance per km ($L$): $$L = 2 \times 10^{-7} \ln\left( \frac{D_m}{D_s^b} \right) \times 1000 = 2 \times 10^{-4} \ln\left( \frac{7.5595}{0.072504} \right) = 2 \times 10^{-4} \ln(104.263) = 0.9294\ \text{mH/km/phase}$$
5. Calculate total inductive reactance for $100\ \text{km}$ ($X_L$): $$X_L = 2 \pi f L \times 100 = 2 \pi \times 60 \times (0.9294 \times 10^{-3}) \times 100 = 35.04\ \Omega$$

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Example 2: Medium Transmission Line Nominal-$\pi$ Performance

Problem: A 3-phase, $138\ \text{kV}$, $60\ \text{Hz}$, $150\ \text{km}$ transmission line delivers $50\ \text{MW}$ at $0.85$ power factor lagging to a $138\ \text{kV}$ load. Line series impedance per phase is $Z = 20 + j 60\ \Omega$, and line shunt admittance per phase is $Y = j 5.0 \times 10^{-4}\ \text{S}$. Using the Nominal-$\pi$ model, determine: (a) ABCD parameters, (b) Sending-end line-to-line voltage $V_{S,L-L}$, and (c) Percentage voltage regulation.

Solution:

1. Calculate Nominal-$\pi$ ABCD parameters: $$A = D = 1 + \frac{Y Z}{2} = 1 + \frac{(j 5.0 \times 10^{-4})(20 + j 60)}{2} = 1 + \frac{-0.03 + j 0.01}{2} = 0.985 + j 0.005 = 0.98501 \angle 0.291^\circ$$ $$B = Z = 20 + j 60\ \Omega = 63.2455 \angle 71.565^\circ\ \Omega$$
2. Receiving-end line-to-neutral voltage and current: $$V_R = \frac{138,000}{\sqrt{3}} = 79,674.3 \angle 0^\circ\ \text{V}$$ $$I_R = \frac{50 \times 10^6}{\sqrt{3} \times 138,000 \times 0.85} = 246.06 \angle -31.788^\circ\ \text{A} = (209.15 - j 129.62)\ \text{A}$$
3. Calculate sending-end voltage ($V_S = A V_R + B I_R$): $$A V_R = (0.985 + j 0.005)(79,674.3) = 78,479.2 + j 398.37\ \text{V}$$ $$B I_R = (63.2455 \angle 71.565^\circ)(246.06 \angle -31.788^\circ) = 15,562.2 \angle 39.777^\circ = 11,960.0 + j 9,956.8\ \text{V}$$ $$V_S = (78,479.2 + 11,960.0) + j (398.37 + 9,956.8) = 90,439.2 + j 10,355.2\ \text{V} = 91,029.7 \angle 6.536^\circ\ \text{V}$$ $$V_{S,L-L} = 91,029.7 \times \sqrt{3} = 157,668.2\ \text{V} = 157.67\ \text{kV}$$
4. Calculate Voltage Regulation (%VR): $$\% \text{VR} = \frac{\frac{|V_S|}{|A|} - |V_R|}{|V_R|} \times 100\% = \frac{\frac{91,029.7}{0.98501} - 79,674.3}{79,674.3} \times 100\% = \frac{92,415.0 - 79,674.3}{79,674.3} \times 100\% = 15.99\%$$