18.2 Mechanical Overhead Line Design, Underground Cables & Insulators

18.2 Mechanical Overhead Line Design, Underground Cables & Insulators

Mechanical overhead line design, insulator string voltage distribution, corona phenomenon control, and underground cable engineering represent vital practical competencies tested in the PRC Registered Electrical Engineer (REE) Licensure Examination. Overhead line conductors must withstand severe mechanical tension and environmental loading (wind pressure and ice buildup) while maintaining statutory ground clearances mandated by the Philippine Electrical Code (PEC) and National Electrical Safety Code (NESC).

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1. Mechanical Overhead Line Design & Sag Calculations

Sag ($S$) is the maximum vertical distance between the line conductor and the imaginary straight line joining its two supporting tower points. Sufficient sag is necessary to prevent conductor mechanical tension ($T$) from exceeding safe working stress limits.

Conductor Sag on Equal-Level Supports

For a span length $L$ (meters) between two supports at equal height, carrying a conductor of resultant weight $w$ (N/m) under tension $T$ (N):

$S = \frac{w L^2}{8 T} \quad \implies \quad T = \frac{w L^2}{8 S}$

The total curved length of the conductor along the span is:

$S_{\text{length}} \approx L + \frac{8 S^2}{3 L}$

Conductor Sag on Unequal-Level Supports

When supporting towers have a height difference $h = h_2 - h_1$ across span $L$:

$x_1 = \frac{L}{2} - \frac{T h}{w L}, \quad x_2 = \frac{L}{2} + \frac{T h}{w L}$

where $x_1$ and $x_2$ are the horizontal distances from the lower support 1 and higher support 2 to the lowest point (vertex) of the conductor catenary. The respective sags are:

$S_1 = \frac{w x_1^2}{2 T}, \quad S_2 = \frac{w x_2^2}{2 T}$

Environmental Vector Loading (Wind & Ice Effects)

Under real environmental conditions, conductors experience three concurrent force vectors:

1. Self-Weight ($w_c$): Vertical downward force due to gravity ($w_c = m \cdot g$).
2. Ice Coating Weight ($w_i$): Radial ice layer of thickness $t$ adds vertical downward weight: $w_i = \rho_{\text{ice}} \cdot V_{\text{ice}} = \rho_{\text{ice}} \cdot \pi t (d + t) \cdot g$ where $d$ is conductor outer diameter and $\rho_{\text{ice}} \approx 915\ \text{kg/m}^3$.
3. Wind Pressure Force ($w_w$): Horizontal wind pressure $P_{\text{wind}}$ acting perpendicular to projected surface area: $w_w = P_{\text{wind}} \cdot (d + 2t) \cdot 1.0$

Total Resultant Weight ($w_R$):

$w_R = \sqrt{(w_c + w_i)^2 + w_w^2}$

• Slanted Sag ($S_{\text{slant}}$) along the resultant force plane: $S_{\text{slant}} = \frac{w_R L^2}{8 T}$
• Vertical Sag ($S_{\text{vert}}$) for ground clearance evaluation: $S_{\text{vert}} = S_{\text{slant}} \cos \theta = S_{\text{slant}} \left( \frac{w_c + w_i}{w_R} \right)$

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2. Overhead Line Insulators & String Efficiency

Insulators physically support high-voltage conductors while preventing current leakage to grounded steel towers.

Insulator Types & Applications

• Pin Insulators: Used for primary distribution lines up to $33\ \text{kV}$. Economical, but physically bulky above $33\ \text{kV}$.
• Suspension Insulators: Used for transmission lines above $33\ \text{kV}$ ($69\ \text{kV}, 138\ \text{kV}, 230\ \text{kV}, 500\ \text{kV}$). Standard porcelain or toughened glass discs linked in series strings. If one disc fails, only that disc is replaced.
• Strain Insulators: Heavy-duty suspension discs mounted horizontally at dead-end towers, river crossings, and sharp line angle turns.
• Shackle (Spool) Insulators: Used for low-voltage secondary distribution ($230\ \text{V}/400\ \text{V}$).

Potential Distribution & String Efficiency

A suspension string consisting of $n$ series disc units has a mutual capacitance $C$ across each porcelain unit. Shunt capacitance $C_g = k C$ ($k \approx 0.10 \text{ to } 0.20$) exists between the metal pin connectors and the grounded steel tower structure.

Due to current escaping through shunt capacitances, charging current is maximum at the disc closest to the line conductor and minimum at the top disc near the cross-arm. Consequently, voltage distribution across discs is non-uniform, with the bottom disc experiencing peak dielectric stress.

Nodal Voltage Formulas for a 3-Disc String:

Let $V_1, V_2, V_3$ be disc voltages from top (cross-arm) to bottom (conductor): $V_2 = V_1 (1 + k)$ $V_3 = V_1 (1 + 3k + k^2)$ $\text{Total Line-to-Neutral Voltage: } V = V_1 + V_2 + V_3 = V_1 (3 + 4k + k^2)$

Insulator String Efficiency ($\eta_{\text{string}}$):

$\eta_{\text{string}} = \frac{\text{Total Phase Voltage Across String }(V)}{n \times \text{Voltage Across Disc Nearest Conductor }(V_n)} \times 100\%$

Methods to Improve String Efficiency:

1. Longer Cross-Arms: Reduces shunt capacitance $C_g$, lowering ratio $k$.
2. Capacitance Grading: Using larger capacitance units near the bottom conductor end.
3. Static Shielding / Guard Ring: A metal ring connected to the line conductor surrounding the bottom disc. The guard ring introduces line-to-pin capacitances that cancel shunt ground currents, equalizing disc voltage distribution towards $100%$.

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3. Corona Effect in Overhead Transmission Lines

Corona is the localized dielectric breakdown and ionization of air surrounding high-voltage conductors, accompanied by a violet glow, hissing noise, ozone ($O_3$) production, and power loss.

Physical Breakdown Mechanism

When the electric field gradient at the conductor surface exceeds the dielectric breakdown strength of air ($g_0 = 30\ \text{kV/cm peak} = 21.2\ \text{kV/cm RMS}$ at standard temperature $25^\circ\text{C}$ and pressure $76\ \text{cmHg}$), air molecules ionize into free electrons and positive ions, initiating corona discharge.

Critical Voltage Formulations

1. Disruptive Critical Voltage ($V_d$): Line-to-neutral voltage at which atmospheric air breakdown begins: $V_d = m_0 \cdot g_0 \cdot \delta \cdot r \cdot \ln\left( \frac{D}{r} \right) \quad (\text{kV/phase RMS})$ where:

   • $m_0$ = conductor surface irregularity factor ($1.0$ for smooth polished wires, $0.80\text{--}0.92$ for stranded ACSR conductors).
   • $\delta$ = air density factor: $\delta = \frac{3.92 b}{273 + T}$ (where $b$ is barometric pressure in cmHg, $T$ in $^\circ\text{C}$; $\delta = 1.0$ at STP).
   • $r$ = conductor radius (cm); $D$ = inter-conductor spacing (cm).

2. Visual Critical Voltage ($V_v$): Line-to-neutral voltage at which visible violet glow appears along the line: $V_v = m_v \cdot g_0 \cdot \delta \cdot r \left( 1 + \frac{0.3}{\sqrt{\delta r}} \right) \ln\left( \frac{D}{r} \right) \quad (\text{kV/phase RMS})$

3. Corona Power Loss (Peek's Empirical Formula): $P_c = \frac{242.2}{\delta} (f + 25) \sqrt{\frac{r}{D}} (V - V_d)^2 \times 10^{-5} \quad (\text{kW/phase/km})$ where $V$ is operating line-to-neutral voltage (kV), and $f$ is frequency (Hz).

Methods to Reduce Corona:

• Increase conductor outer diameter using hollow conductors or ACSR steel-reinforced cores.
• Utilize bundled conductors (2, 3, or 4 sub-conductors per phase), which dramatically increases effective self GMR ($D_s$), lowering surface electric stress below $g_0$.
• Increase inter-conductor spacing $D$.

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4. Underground Power Cable Engineering

Underground cables transport electric power in urban environments where overhead lines are physically impractical or unsafe.

Cable Construction Layers

1. Core / Conductor: Tinned copper or aluminum strands.
2. Insulation: Cross-linked polyethylene (XLPE) or impregnated paper.
3. Metallic Sheath: Lead or aluminum extrusion preventing moisture ingress.
4. Bedding: Jute or hessian tape protecting sheath from mechanical damage.
5. Armouring: Galvanized steel wire or tape serving as mechanical armor.
6. Serving: Fibrous outer layer protecting armouring against soil corrosion.

Insulation Resistance of Single-Core Cable ($R_{\text{ins}}$)

For a single-core cable of conductor radius $r$, internal sheath radius $R$, length $l$, and insulation resistivity $\rho$:

$R_{\text{ins}} = \frac{\rho}{2 \pi l} \ln\left( \frac{R}{r} \right) \quad (\Omega)$

CRITICAL REE EXAM FACT: Cable insulation resistance varies inversely with length ($R_{\text{ins}} \propto 1/l$). Doubling cable length halves its insulation resistance!

Cable Capacitance & Dielectric Stress

1. Single-Core Capacitance: $C = \frac{2 \pi \epsilon_0 \epsilon_r}{\ln\left( \frac{R}{r} \right)}\ \text{F/m} = \frac{\epsilon_r}{18 \ln\left( \frac{R}{r} \right)}\ \mu\text{F/km}$
2. Electrostatic Dielectric Stress ($E$) at distance $x$ from core center ($r \le x \le R$): $E(x) = \frac{V}{x \ln\left( \frac{R}{r} \right)}$
   • Maximum Stress occurs at conductor surface ($x = r$): $E_{\max} = \frac{V}{r \ln\left( \frac{R}{r} \right)}$
   • Minimum Stress occurs at inner sheath boundary ($x = R$): $E_{\min} = \frac{V}{R \ln\left( \frac{R}{r} \right)} \implies \frac{E_{\max}}{E_{\min}} = \frac{R}{r}$

Optimum Cable Dimensioning:

To minimize $E_{\max}$ for a given sheath radius $R$ and operating voltage $V$, differentiate $E_{\max}$ with respect to $r$. Setting $\frac{d E_{\max}}{dr} = 0$ yields:

$\ln\left( \frac{R}{r} \right) = 1 \implies \frac{R}{r} = e \approx 2.718$

Under optimum geometry, $E_{\max,\text{opt}} = \frac{V}{r}$.

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Solved Board Exam Examples

Example 1: Conductor Sag under Combined Wind and Ice Loading

Problem: A transmission line conductor has a span length $L = 250\ \text{m}$ between equal-level supports. Conductor weight $w_c = 1.1\ \text{kg/m}$ ($10.791\ \text{N/m}$), outer diameter $d = 20\ \text{mm}$, allowable working tension $T = 20,000\ \text{N}$. The conductor is coated with a radial ice layer $t = 10\ \text{mm}$ thick (ice density $\rho_{\text{ice}} = 915\ \text{kg/m}^3$) and subjected to a horizontal wind pressure $P_{\text{wind}} = 390\ \text{N/m}^2$ of projected area. Determine: (a) Total resultant weight per meter $w_R$, (b) Slanted sag $S_{\text{slant}}$, and (c) Vertical sag $S_{\text{vert}}$.

Solution:

1. Calculate weight of ice per meter ($w_i$): $A_{\text{ice}} = \pi t (d + t) = \pi (0.01)(0.02 + 0.01) = 0.00094248\ \text{m}^2$ $w_i = 0.00094248 \times 915 \times 9.81 = 8.46\ \text{N/m}$
2. Calculate horizontal wind force per meter ($w_w$): $\text{Projected Diameter} = d + 2t = 0.02 + 0.02 = 0.04\ \text{m}$ $w_w = 390 \times 0.04 = 15.60\ \text{N/m}$
3. Calculate total vertical force ($w_V$) and resultant force ($w_R$): $w_V = w_c + w_i = 10.791 + 8.46 = 19.251\ \text{N/m}$ $w_R = \sqrt{w_V^2 + w_w^2} = \sqrt{19.251^2 + 15.60^2} = \sqrt{370.60 + 243.36} = \sqrt{613.96} = 24.778\ \text{N/m}$
4. Calculate slanted sag ($S_{\text{slant}}$) and vertical sag ($S_{\text{vert}}$): $S_{\text{slant}} = \frac{w_R L^2}{8 T} = \frac{24.778 \times 250^2}{8 \times 20,000} = \frac{1,548,625}{160,000} = 9.679\ \text{m}$ $S_{\text{vert}} = S_{\text{slant}} \times \left( \frac{w_V}{w_R} \right) = 9.679 \times \frac{19.251}{24.778} = 7.520\ \text{m}$

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Example 2: Suspension Insulator String Voltage Distribution

Problem: A 3-phase, $138\ \text{kV}$ transmission line uses a string of 4 suspension insulator discs. The ratio of shunt capacitance to ground to mutual disc capacitance is $k = 0.15$. Determine: (a) Voltage distribution across each disc ($V_1, V_2, V_3, V_4$), and (b) String Efficiency $\eta_{\text{string}}$.

Solution:

1. Express disc voltages in terms of top disc voltage $V_1$ ($k = 0.15$): $V_2 = V_1 (1 + k) = 1.15 V_1$ $V_3 = V_1 (1 + 3k + k^2) = V_1 (1 + 0.45 + 0.0225) = 1.4725 V_1$ $V_4 = V_3(1+k) + (V_1+V_2+V_3)k = 1.4725 V_1 (1.15) + (1 + 1.15 + 1.4725) V_1 (0.15) = 1.693375 V_1 + 0.543375 V_1 = 2.23675 V_1$
2. Calculate total string line-to-neutral voltage ($V_{LN}$): $V_{LN} = \frac{138,000}{\sqrt{3}} = 79,674.3\ \text{V}$ $V_{\text{total}} = V_1 + V_2 + V_3 + V_4 = (1 + 1.15 + 1.4725 + 2.23675) V_1 = 5.85925 V_1$
3. Solve individual disc voltages: $V_1 = \frac{79,674.3}{5.85925} = 13,598.1\ \text{V} = 13.60\ \text{kV}$ $V_2 = 1.15 \times 13,598.1 = 15,637.8\ \text{V} = 15.64\ \text{kV}$ $V_3 = 1.4725 \times 13,598.1 = 20,023.2\ \text{V} = 20.02\ \text{kV}$ $V_4 = 2.23675 \times 13,598.1 = 30,415.2\ \text{V} = 30.42\ \text{kV} \quad (\text{Bottom Disc Nearest Conductor})$
4. Calculate Insulator String Efficiency ($\eta_{\text{string}}$): $\eta_{\text{string}} = \frac{V_{LN}}{4 \times V_4} \times 100\% = \frac{79,674.3}{4 \times 30,415.2} \times 100\% = \frac{79,674.3}{121,660.8} \times 100\% = 65.49\%$