High Stress Electric Fields

The charge coefficient (Maxwell capacitance) matrix relates the line charges of the phase conductors to their voltages: $[Q] = [C][V]$. With ground wires on the tower, the ground wires also carry charges, but their potential is zero. So they are eliminated to give a reduced $P \times P$ matrix for the phase conductors only.

Method of images

The earth is treated as a perfect conductor at zero potential. Each conductor $i$ (radius $r_i$, height $H_i$, charge $q_i$ C/m) has an image of charge $-q_i$ at depth $H_i$ below ground.

   P1 o       o P2       (phase)   ^
         o g             (gnd wire) | H
 //////////////////////////////// ground
         o -g
   -P1 o       o -P2     (images)
   A_ij = conductor i to conductor j
   I_ij = conductor i to image of j

Maxwell's potential coefficients

The potential of conductor $i$ due to all charges is

$$V_i = \sum_{j} P_{ij}\, q_j$$

where

$$\begin{aligned} P_{ii} &= \frac{1}{2\pi\varepsilon_0}\ln\frac{2H_i}{r_i} \\ P_{ij} &= \frac{1}{2\pi\varepsilon_0}\ln\frac{I_{ij}}{A_{ij}} \end{aligned}$$

with $1/(2\pi\varepsilon_0) = 18 \times 10^9$ m/F. For bundles, $r_i$ is replaced by the equivalent bundle radius.

Step 1: Full matrix for P phases and g ground wires

With $n = P + g$ conductors, partition the matrix (subscript $p$ = phases, $g$ = ground wires):

$$\begin{bmatrix} [V_p] \\ [V_g] \end{bmatrix} = \begin{bmatrix} [P_{pp}] & [P_{pg}] \\ [P_{gp}] & [P_{gg}] \end{bmatrix} \begin{bmatrix} [Q_p] \\ [Q_g] \end{bmatrix}$$

Sizes: $P_{pp}$ is $P \times P$, $P_{pg}$ is $P \times g$, $P_{gg}$ is $g \times g$.

Step 2: Use the ground-wire condition

Ground wires are at (or near) ground potential, so $[V_g] = 0$:

$$\begin{aligned} 0 &= [P_{gp}][Q_p] + [P_{gg}][Q_g] \\ [Q_g] &= -[P_{gg}]^{-1}[P_{gp}][Q_p] \end{aligned}$$

The ground wires carry induced charges of opposite sign.

Step 3: Reduced potential coefficient matrix

Substitute into the first row:

$$\begin{aligned} [V_p] &= [P_{pp}][Q_p] + [P_{pg}][Q_g] \\ &= \big([P_{pp}] - [P_{pg}][P_{gg}]^{-1}[P_{gp}]\big)[Q_p] \\ &= [P_r][Q_p] \end{aligned}$$

Step 4: Charge coefficient matrix

$$[Q_p] = [P_r]^{-1}[V_p] = [C][V_p], \qquad [C] = [P_r]^{-1}$$

$[C]$ is a $P \times P$ symmetric matrix. Its diagonal elements are positive and its off-diagonal elements are negative.

One ground wire (g = 1)

$[P_{gg}]$ is a scalar, so each element is simply

$$P'_{ij} = P_{ij} - \frac{P_{ig}\,P_{gj}}{P_{gg}}, \quad i, j = 1 \ldots P$$

Two ground wires (g, h)

$$[P_{gg}]^{-1} = \frac{1}{P_{gg}P_{hh} - P_{gh}^2}\begin{bmatrix} P_{hh} & -P_{gh} \\ -P_{gh} & P_{gg} \end{bmatrix}$$

so

$$P'_{ij} = P_{ij} - \frac{P_{ig}\left(P_{hh}P_{gj} - P_{gh}P_{hj}\right) + P_{ih}\left(P_{gg}P_{hj} - P_{gh}P_{gj}\right)}{P_{gg}P_{hh} - P_{gh}^2}$$

Effect of ground wires

• The subtracted term makes $P'_{ii} < P_{ii}$, so the self capacitance of the phase conductors increases slightly.
• For the same voltage, the phase charge (and so the conductor surface gradient) rises a little.
• The ground wires also shield the phases from direct lightning strokes.

A high gradient field is a region where the electric field (voltage gradient, kV/cm) is much higher than the average value $V/d$. Such regions start corona, partial discharge and finally breakdown, so they decide the size and shape of HV equipment. The main causes are listed below.

1. Small radius of curvature (sharp points and edges)

Near a surface of radius $r$, the field is roughly $E \approx V/[r \ln(d/r)]$, so a small radius gives a very high gradient. Examples: sharp edges of electrodes, bolt heads, conductor strands, needle points, and thin conductors on EHV lines. This is why EHV lines use large-diameter or bundled conductors and fittings are given corona rings.

2. Non-uniform electrode geometry

Rod–plane, sphere–plane and point–plane gaps, and coaxial systems produce strongly non-uniform fields. The field is highest at the electrode with the smaller radius. The field utilisation factor $\eta = E_{av}/E_{max}$ is much less than 1 for such geometries.

3. Surface irregularities on conductors

Scratches, burrs, broken strands, dust, insects and water drops on conductors locally raise the gradient. This is why corona inception falls in rain or fog (irregularity factor $m < 1$).

4. Voids and cavities in solid insulation

A gas-filled void in a solid of relative permittivity $\varepsilon_r$ carries a field about $\varepsilon_r$ times that in the solid, and the gas is also weaker. This leads to partial discharges.

5. Interfaces of different permittivities (composite dielectrics)

In series dielectrics, the field divides inversely with permittivity ($E_1\varepsilon_1 = E_2\varepsilon_2$). The lower permittivity material (often air or oil) is over-stressed. Example: an air gap between a bushing and its insulation.

6. Triple junctions

The point where an electrode, a solid insulator and a gas meet has a very high local field. Discharges often start there and creep along the surface.

7. Conductive particles and contamination

Metallic particles in GIS or oil, and moisture and salt on insulator surfaces, distort the field and create local high stress, leading to flashover.

8. Space charge and terminations

Charges left by previous discharges, and the ends of cable screens and shields, concentrate the field at the termination edge. Stress cones are therefore used at cable ends.

9. Overvoltages

Lightning and switching surges raise the voltage, and therefore every local gradient, far above normal values.

Remedies (in brief)

• Use large radii, grading (corona) rings and stress cones.
• Use bundled conductors.
• Make insulation void-free and keep it clean.
• Choose permittivities that grade the field properly.

The capacitance matrix of the line is found from Maxwell's potential coefficients, using the image method and replacing each bundle by its equivalent radius: $[C] = 2\pi\varepsilon_0 [M]^{-1}$.

Data

• 3 phases in flat horizontal configuration, phase spacing $S = 11$ m
• Height $H = 15$ m
• Sub-conductor radius $r = 3.18$ cm; bundle spacing $B = 45.72$ cm; $N = 2$
• $2\pi\varepsilon_0 = \dfrac{1}{18 \times 10^9}$ F/m $= 55.56$ pF/m

    1          2          3
    o--- 11 m--o--- 11 m--o
    |                       H = 15 m
 ////////////////////////////////

Step 1: Equivalent radius of the bundle

For $N = 2$:

$$\begin{aligned} r_{eq} &= \sqrt{r\,B} = \sqrt{3.18 \times 45.72} \\ &= 12.058\ \text{cm} = 0.12058\ \text{m} \end{aligned}$$

Step 2: Maxwell coefficient matrix $[M]$ (dimensionless)

Self terms:

$$M_{11} = M_{22} = M_{33} = \ln\frac{2H}{r_{eq}} = \ln\frac{30}{0.12058} = 5.5167$$

Mutual terms ($A_{ij}$ = distance to conductor, $I_{ij}$ = distance to image):

$$\begin{aligned} M_{12} = M_{23} &= \ln\frac{\sqrt{11^2 + 30^2}}{11} = \ln\frac{31.953}{11} = 1.0664 \\ M_{13} &= \ln\frac{\sqrt{22^2 + 30^2}}{22} = \ln\frac{37.202}{22} = 0.5253 \end{aligned}$$ $$[M] = \begin{bmatrix} 5.5167 & 1.0664 & 0.5253 \\ 1.0664 & 5.5167 & 1.0664 \\ 0.5253 & 1.0664 & 5.5167 \end{bmatrix}$$

(The potential coefficient matrix is $[P] = 18 \times 10^9 [M]$ m/F, i.e. diagonal $99.30 \times 10^9$, $P_{12} = 19.20 \times 10^9$, $P_{13} = 9.46 \times 10^9$.)

Step 3: Inverse of $[M]$

Determinant of $[M]$ = 155.02.

$$[M]^{-1} = \begin{bmatrix} 0.18899 & -0.03434 & -0.01136 \\ -0.03434 & 0.19454 & -0.03434 \\ -0.01136 & -0.03434 & 0.18899 \end{bmatrix}$$

Step 4: Capacitance matrix

$$\begin{aligned} [C] &= 2\pi\varepsilon_0 [M]^{-1} = 55.56 \times [M]^{-1}\ \text{pF/m} \\[4pt] [C] &= \begin{bmatrix} 10.499 & -1.908 & -0.631 \\ -1.908 & 10.808 & -1.908 \\ -0.631 & -1.908 & 10.499 \end{bmatrix}\ \text{pF/m (= nF/km)} \end{aligned}$$

Interpretation

• Diagonal terms are the self (Maxwell) capacitances. The centre phase has the largest value, 10.81 nF/km, because it is shielded by both outer phases.
• Off-diagonal terms are negative, as expected in a Maxwell capacitance matrix.
• If the line were transposed, $M_s = 5.5167$ and $M_m = 0.8860$ (averages), giving a positive-sequence capacitance $C_1 = 55.56/(5.5167 - 0.8860) = 12.00$ nF/km.

Answer: $[C] = \begin{bmatrix} 10.50 & -1.91 & -0.63 \\ -1.91 & 10.81 & -1.91 \\ -0.63 & -1.91 & 10.50 \end{bmatrix}$ nF/km (pF/m).

The audible noise (AN) from each phase is found from the given empirical formula using the distance from that phase to the microphone. The three levels are then added on a power (energy) basis, not arithmetically.

Data

• $N = 4$, sub-conductor diameter $d = 3.05$ cm
• Height $H = 20$ m, phase spacing $S = 14$ m (horizontal)
• Gradients: centre $E = 20$ kV/cm, outer phases $E = 18.4$ kV/cm
• Microphone at ground level, 30 m horizontally from the centre phase
• Formula: $AN_i = 120\log E_i + 55\log d - 11.4\log D_i - 115.4$ dB(A)

 P1(outer)  P2(centre)  P3(outer)
    o--14 m--o--14 m--o        H=20 m
 ///////////////////////////////////[mic]
             |<------- 30 m ------->|

Step 1: Distances from each phase to the microphone

$$\begin{aligned} D_{near\,outer} &= \sqrt{16^2 + 20^2} = 25.61\ \text{m} \\ D_{centre} &= \sqrt{30^2 + 20^2} = 36.06\ \text{m} \\ D_{far\,outer} &= \sqrt{44^2 + 20^2} = 48.33\ \text{m} \end{aligned}$$

Step 2: Common terms

$$\begin{aligned} 55\log 3.05 &= 26.64 \\ 120\log 20 &= 156.12 \\ 120\log 18.4 &= 151.78 \end{aligned}$$

Step 3: AN of each phase

$$\begin{aligned} AN_{centre} &= 156.12 + 26.64 - 11.4\log 36.06 - 115.4 \\ &= 156.12 + 26.64 - 17.75 - 115.4 = 49.61\ \text{dB(A)} \\ AN_{near} &= 151.78 + 26.64 - 11.4\log 25.61 - 115.4 \\ &= 151.78 + 26.64 - 16.06 - 115.4 = 46.96\ \text{dB(A)} \\ AN_{far} &= 151.78 + 26.64 - 11.4\log 48.33 - 115.4 \\ &= 151.78 + 26.64 - 19.20 - 115.4 = 43.81\ \text{dB(A)} \end{aligned}$$

Step 4: Total AN (power addition)

$$\begin{aligned} AN_{total} &= 10\log\left(10^{4.961} + 10^{4.696} + 10^{4.381}\right) \\ &= 10\log\left(91424 + 49640 + 24068\right) \\ &= 10\log(165132) = 52.18\ \text{dB(A)} \end{aligned}$$

The centre phase contributes most because it has the highest surface gradient, even though it is not the nearest phase.

Answer: total AN ≈ 52.2 dB(A) at 30 m from the line centre.

The audible noise from each phase is calculated with the empirical formula using its own distance to the microphone. The three contributions are then added on an energy basis.

Data and reading of the question

• Flat horizontal line, $H = 20$ m, phase spacing $S = 15$ m
• Bundle of $N = 2$ sub-conductors, diameter given as 0.30 m, so $d = 30$ cm
• All phases: $E_m = 20$ kV/cm
• Microphone at ground level, 20 m horizontally beyond the outermost phase
• The "$-11.4\log N$" term in the formula is read as $-11.4\log D_i$, since the note gives units for $D$ (in m) and the distance must enter:

$$AN_i = 120\log E_m + 55\log d - 11.4\log D_i - 115.4\ \ \text{dB(A)}$$

  P1        P2        P3
  o--15 m---o--15 m---o         H=20 m
 ///////////////////////////////[mic]
                      |<- 20 m ->|

Step 1: Distances to the microphone

Horizontal distances are 20, 35 and 50 m.

$$\begin{aligned} D_3 &= \sqrt{20^2 + 20^2} = 28.28\ \text{m} \\ D_2 &= \sqrt{35^2 + 20^2} = 40.31\ \text{m} \\ D_1 &= \sqrt{50^2 + 20^2} = 53.85\ \text{m} \end{aligned}$$

Step 2: Common part

$$\begin{aligned} &120\log 20 + 55\log 30 - 115.4 \\ &= 156.12 + 81.24 - 115.4 = 121.96 \end{aligned}$$

Step 3: AN from each phase

$$\begin{aligned} AN_3 &= 121.96 - 11.4\log 28.28 = 121.96 - 16.55 = 105.42\ \text{dB(A)} \\ AN_2 &= 121.96 - 11.4\log 40.31 = 121.96 - 18.30 = 103.66\ \text{dB(A)} \\ AN_1 &= 121.96 - 11.4\log 53.85 = 121.96 - 19.74 = 102.23\ \text{dB(A)} \end{aligned}$$

Step 4: Total AN

$$\begin{aligned} AN &= 10\log\left(10^{10.542} + 10^{10.366} + 10^{10.223}\right) \\ &= 10\log\left[(3.481 + 2.325 + 1.671) \times 10^{10}\right] \\ &= 108.74\ \text{dB(A)} \end{aligned}$$

Note on the conductor size

A 30 cm sub-conductor is unrealistically large; the intended value is probably 0.030 m (3 cm). With $d = 3$ cm the term $55\log d$ becomes 26.24 instead of 81.24, so every level drops by exactly 55 dB. The phase levels become 50.42, 48.66 and 47.23 dB(A), and the total is 53.74 dB(A), a typical value for a 750 kV line.

Answer: AN ≈ 108.7 dB(A) with $d = 30$ cm as printed (≈ 53.7 dB(A) if $d = 3$ cm is intended).

Each conductor above a perfectly conducting earth produces a potential at the other conductor. With the method of images, these effects are written in matrix form: $[V] = [P][Q]$. The off-diagonal terms show the proximity effect of the nearby conductor.

Assumptions

• Long, straight, parallel conductors of radii $r_1$ and $r_2$, with charges $+q_1$ and $+q_2$ C/m.
• The earth is a perfect conductor at zero potential. It is replaced by images $-q_1$ and $-q_2$ at depths $H_1$ and $H_2$ below ground.
• $A_{12}$ = distance between conductors 1 and 2; $I_{12}$ = distance from conductor 1 to the image of conductor 2 (equal to that from 2 to the image of 1).

        +q1 o . . A12 . . o +q2
            |  .         .|
         H1 |    .  I12   | H2
  //////////////////////////////// ground
         H1 |        .    | H2
        -q1 o             o -q2
           (images)

Step 1: Potential due to a line charge

The field of a line charge $q$ at distance $x$ is $E = \dfrac{q}{2\pi\varepsilon_0 x}$. The potential difference between distances $x_1$ and $x_2$ is

$$V_{x_1} - V_{x_2} = \frac{q}{2\pi\varepsilon_0}\ln\frac{x_2}{x_1}$$

For a charge $+q$ and its image $-q$, the potential at a point is

$$V = \frac{q}{2\pi\varepsilon_0}\ln\frac{\text{distance from image}}{\text{distance from charge}}$$

This is zero on the ground plane, as required.

Step 2: Potential of conductor 1

• Due to $+q_1$ and its image (distances $r_1$ and $2H_1$): $\dfrac{q_1}{2\pi\varepsilon_0}\ln\dfrac{2H_1}{r_1}$
• Due to $+q_2$ and its image (distances $A_{12}$ and $I_{12}$): $\dfrac{q_2}{2\pi\varepsilon_0}\ln\dfrac{I_{12}}{A_{12}}$

$$V_1 = \frac{q_1}{2\pi\varepsilon_0}\ln\frac{2H_1}{r_1} + \frac{q_2}{2\pi\varepsilon_0}\ln\frac{I_{12}}{A_{12}}$$

Step 3: Potential of conductor 2

$$V_2 = \frac{q_1}{2\pi\varepsilon_0}\ln\frac{I_{12}}{A_{12}} + \frac{q_2}{2\pi\varepsilon_0}\ln\frac{2H_2}{r_2}$$

Step 4: Voltage matrix

$$\begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \frac{1}{2\pi\varepsilon_0} \begin{bmatrix} \ln\dfrac{2H_1}{r_1} & \ln\dfrac{I_{12}}{A_{12}} \\[8pt] \ln\dfrac{I_{12}}{A_{12}} & \ln\dfrac{2H_2}{r_2} \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix}$$

or $[V] = [P][Q]$, with Maxwell's potential coefficients

$$P_{11} = \frac{1}{2\pi\varepsilon_0}\ln\frac{2H_1}{r_1},\quad P_{22} = \frac{1}{2\pi\varepsilon_0}\ln\frac{2H_2}{r_2},\quad P_{12} = P_{21} = \frac{1}{2\pi\varepsilon_0}\ln\frac{I_{12}}{A_{12}}$$

Here $1/(2\pi\varepsilon_0) = 18 \times 10^9$ m/F.

Interpretation (proximity effect)

• $P_{11}$ and $P_{22}$ are the self terms: the potential a conductor would have alone above ground.
• $P_{12}$ is the mutual term. Because $I_{12} > A_{12}$, it is positive. The nearby positively charged conductor raises the potential of conductor 1, and the effect grows as the conductors come closer (smaller $A_{12}$).
• Inverting gives the capacitance (charge coefficient) matrix: $[Q] = [P]^{-1}[V] = [C][V]$. This is used to find conductor charges and surface gradients. For $n$ conductors, the same rule gives an $n \times n$ matrix.

Corona starts when the voltage to neutral reaches the disruptive critical voltage $V_d = m\,\delta\, g_0\, r \ln(D/r)$. Setting $V_d$ equal to the limiting voltage gives the smallest equilateral spacing $D$ for which the line voltage does not exceed the critical value.

Data

• Conductor diameter = 3 cm, so $r = 1.5$ cm
• Irregularity factor $m = 0.96$; air density factor $\delta = 0.95$
• Dielectric strength of air $= 30$ kV(peak)/cm, so $g_0 = 30/\sqrt{2} = 21.21$ kV(rms)/cm
• Disruptive critical voltage (line) $\le 230$ kV, so per phase

$$V_d = \frac{230}{\sqrt{3}} = 132.79\ \text{kV (rms)}$$

• Equilateral spacing, so $D$ is the equivalent spacing.

Step 1: Disruptive critical voltage expression

$$V_d = m\,\delta\, g_0\, r\, \ln\frac{D}{r}\ \ \text{kV (rms, phase)}$$

Step 2: Substitute

$$\begin{aligned} m\,\delta\, g_0\, r &= 0.96 \times 0.95 \times 21.21 \times 1.5 = 29.02\ \text{kV} \\ \ln\frac{D}{r} &= \frac{132.79}{29.02} = 4.576 \\ \frac{D}{r} &= e^{4.576} = 97.11 \\ D &= 97.11 \times 1.5 = 145.67\ \text{cm} \end{aligned}$$

Step 3: Check

$$V_{d(line)} = \sqrt{3} \times 29.02 \times \ln\frac{145.67}{1.5} = \sqrt{3} \times 132.79 = 230\ \text{kV}$$

Result

The spacing at which the disruptive critical voltage equals 230 kV (line) is about 1.46 m. Because $V_d$ increases with $D$ (logarithmically), a smaller spacing would make corona start below 230 kV, and a larger spacing raises the corona threshold above it.

Answer: minimum spacing $D \approx 145.7$ cm ≈ 1.46 m.

Corona starts when the phase voltage reaches the disruptive critical voltage $V_d = m\,\delta\,g_0\,r\ln(D/r)$. Setting $V_d$ equal to the limiting value gives the required equilateral spacing $D$.

Data

• Conductor diameter = 30 mm, so $r = 1.5$ cm
• Air density factor $\delta = 0.95$; irregularity factor $m = 0.95$
• Breakdown strength of air = 30 kV(peak)/cm, so $g_0 = 30/\sqrt{2} = 21.21$ kV(rms)/cm
• Line voltage limit 230 kV, so per phase

$$V_d = \frac{230}{\sqrt{3}} = 132.79\ \text{kV (rms)}$$

• Equilateral arrangement, so the spacing $D$ is also the equivalent spacing.

Step 1: Formula

$$V_d = m\,\delta\,g_0\,r\,\ln\frac{D}{r}$$

Step 2: Substitute and solve

$$\begin{aligned} m\,\delta\,g_0\,r &= 0.95 \times 0.95 \times 21.21 \times 1.5 = 28.72\ \text{kV} \\ \ln\frac{D}{r} &= \frac{132.79}{28.72} = 4.624 \\ \frac{D}{r} &= e^{4.624} = 101.91 \\ D &= 101.91 \times 1.5 = 152.86\ \text{cm} \end{aligned}$$

Step 3: Check

$$\sqrt{3} \times 28.72 \times \ln\frac{152.86}{1.5} = \sqrt{3} \times 132.79 = 230\ \text{kV}$$

Comment

$V_d$ increases with $D$, so at this spacing the corona threshold is exactly 230 kV (line). A closer spacing would let corona start below 230 kV. Compared with $m = 0.96$ (which gives about 1.46 m), the rougher conductor surface ($m = 0.95$) needs a slightly larger spacing.

Answer: minimum spacing $D \approx 152.9$ cm ≈ 1.53 m.

Radio interference (RI) from each phase is found with the CIGRE formula using its distance to the measuring point. The line RI is then obtained by the CIGRE addition rule.

Data

• $N = 4$, sub-conductor diameter $d = 3.05$ cm, $B = 45.72$ cm
• Height $H = 20$ m, phase spacing $S = 14$ m (horizontal)
• $g_m = 20$ kV/cm (centre) and 18.4 kV/cm (outer phases)
• Measuring point at ground level, 15 m horizontally outside one outer phase
• CIGRE formula (0.5 MHz, average fair weather), $D$ in m, $d$ in cm:

$$RI_i = 3.5\,g_m + 6d - 33\log_{10}\frac{D_i}{20} - 30\ \ \text{dB above 1 μV/m}$$

  P1        P2        P3
  o--14 m---o--14 m---o          H=20 m
 ///////////////////////////////[P]
                      |<- 15 m ->|

Step 1: Distances

Horizontal distances are 15, 29 and 43 m.

$$\begin{aligned} D_3 &= \sqrt{15^2 + 20^2} = 25.00\ \text{m} \\ D_2 &= \sqrt{29^2 + 20^2} = 35.23\ \text{m} \\ D_1 &= \sqrt{43^2 + 20^2} = 47.42\ \text{m} \end{aligned}$$

Step 2: RI of each phase

$6d = 6 \times 3.05 = 18.3$; $3.5 \times 20 = 70$; $3.5 \times 18.4 = 64.4$.

$$\begin{aligned} RI_3 &= 64.4 + 18.3 - 33\log\frac{25}{20} - 30 = 64.4 + 18.3 - 3.20 - 30 = 49.50\ \text{dB} \\ RI_2 &= 70 + 18.3 - 33\log\frac{35.23}{20} - 30 = 70 + 18.3 - 8.11 - 30 = 50.19\ \text{dB} \\ RI_1 &= 64.4 + 18.3 - 33\log\frac{47.42}{20} - 30 = 64.4 + 18.3 - 12.37 - 30 = 40.33\ \text{dB} \end{aligned}$$

Step 3: RI of the line (CIGRE rule)

• If the highest phase value exceeds the others by 3 dB or more, the line RI equals that highest value.
• Otherwise, line RI = (average of the two highest) + 1.5 dB.

Here the two highest values (50.19 and 49.50 dB) differ by only 0.69 dB, so

$$RI = \frac{50.19 + 49.50}{2} + 1.5 = 51.34\ \text{dB}$$

This is the average fair-weather value at 0.5 MHz. In heavy rain it would be about 17 dB higher (≈ 68.3 dB).

Answer: phase RI = 49.50, 50.19, 40.33 dB; line RI ≈ 51.3 dB above 1 μV/m (fair weather, 0.5 MHz).

The capacitance matrix is found from Maxwell's coefficients, using the average height over the span and the equivalent bundle radius. The charge matrix then follows from $[Q] = [C][V]$.

Data

• 400 kV, flat horizontal, $S = 11.3$ m, $N = 2$
• Sub-conductor radius $r = 3.17$ cm (as given); bundle spacing $B = 45.72$ cm
• Height at tower 11.875 m, at midspan 8.85 m
• $2\pi\varepsilon_0 = 1/(18 \times 10^9) = 55.56$ pF/m

Step 1: Average height (parabolic sag)

$$\begin{aligned} H_{av} &= H_{min} + \frac{1}{3}(H_{tower} - H_{min}) \\ &= 8.85 + \frac{11.875 - 8.85}{3} = 9.858\ \text{m} \end{aligned}$$

Step 2: Equivalent radius

$$r_{eq} = \sqrt{r B} = \sqrt{3.17 \times 45.72} = 12.04\ \text{cm}$$

Step 3: Maxwell coefficients

$$\begin{aligned} M_s &= \ln\frac{2H}{r_{eq}} = \ln\frac{19.717}{0.1204} = 5.0985 \\ M_{12} &= \ln\frac{\sqrt{11.3^2 + 19.717^2}}{11.3} = 0.6987 \\ M_{13} &= \ln\frac{\sqrt{22.6^2 + 19.717^2}}{22.6} = 0.2830 \end{aligned}$$

Untransposed line

$$[M] = \begin{bmatrix} 5.0985 & 0.6987 & 0.2830 \\ 0.6987 & 5.0985 & 0.6987 \\ 0.2830 & 0.6987 & 5.0985 \end{bmatrix}, \quad \det = 127.42$$ $$[C] = 55.56\,[M]^{-1} = \begin{bmatrix} 11.121 & -1.467 & -0.416 \\ -1.467 & 11.298 & -1.467 \\ -0.416 & -1.467 & 11.121 \end{bmatrix}\ \text{pF/m}$$

Charge matrix. Take balanced phase voltages $V = 400/\sqrt{3} = 230.94$ kV rms ($\angle 0°, \angle -120°, \angle 120°$). Then $[Q] = [C][V]$:

$$[Q] = \begin{bmatrix} 2.794\angle 4.31° \\ 2.948\angle -120° \\ 2.794\angle 115.69° \end{bmatrix}\ \mu\text{C/m (rms)}$$

Peak values are 3.95, 4.17 and 3.95 μC/m. The centre phase carries the most charge, so it has the highest surface gradient.

Transposed line

Average the coefficients:

$$\begin{aligned} M_s &= 5.0985 \\ M_m &= \frac{0.6987 + 0.6987 + 0.2830}{3} = 0.5601 \end{aligned}$$ $$[C_t] = \begin{bmatrix} 11.139 & -1.103 & -1.103 \\ -1.103 & 11.139 & -1.103 \\ -1.103 & -1.103 & 11.139 \end{bmatrix}\ \text{pF/m}$$

The positive-sequence capacitance is

$$C_1 = \frac{55.56}{M_s - M_m} = \frac{55.56}{4.5384} = 12.24\ \text{pF/m (nF/km)}$$

and each phase charge is equal in magnitude:

$$q = C_1 V = 12.24 \times 10^{-12} \times 230.94 \times 10^3 = 2.827\ \mu\text{C/m (rms)} \ (4.00\ \mu\text{C/m peak})$$ $$[Q_t] = \begin{bmatrix} 2.827\angle 0° \\ 2.827\angle -120° \\ 2.827\angle 120° \end{bmatrix}\ \mu\text{C/m}$$

Answer: untransposed $C_{11} = 11.12$, $C_{22} = 11.30$, $C_{12} = -1.47$, $C_{13} = -0.42$ pF/m, with charges 2.79/2.95/2.79 μC/m rms; transposed $C_s = 11.14$, $C_m = -1.10$ pF/m, $C_1 = 12.24$ nF/km, with $q = 2.83$ μC/m rms per phase.

RI of each phase is calculated with the CIGRE formula and combined by the CIGRE rule. Rain and frequency corrections are then added.

Data

• $H = 14$ m, $S = 15$ m, bundle $4 \times 3.18$ cm, so $d = 3.18$ cm
• $g_m$ = 16.2 kV/cm (outer) and 17.3 kV/cm (centre)
• Point at ground level, 30 m horizontally from an outer phase (outside the line)
• $RI_i = 3.5g_m + 6d - 33\log_{10}(D_i/20) - 30$ dB (0.5 MHz, average fair weather)

(i) Fair weather at 0.5 MHz

Horizontal distances are 30, 45 and 60 m.

$$\begin{aligned} D_{near} &= \sqrt{30^2 + 14^2} = 33.11\ \text{m} \\ D_{centre} &= \sqrt{45^2 + 14^2} = 47.13\ \text{m} \\ D_{far} &= \sqrt{60^2 + 14^2} = 61.61\ \text{m} \end{aligned}$$

With $6d = 19.08$, $3.5 \times 16.2 = 56.7$ and $3.5 \times 17.3 = 60.55$:

$$\begin{aligned} RI_{near} &= 56.7 + 19.08 - 7.22 - 30 = 38.56\ \text{dB} \\ RI_{centre} &= 60.55 + 19.08 - 12.28 - 30 = 37.35\ \text{dB} \\ RI_{far} &= 56.7 + 19.08 - 16.12 - 30 = 29.66\ \text{dB} \end{aligned}$$

The two highest (38.56 and 37.35) differ by less than 3 dB, so

$$RI_{fair} = \frac{38.56 + 37.35}{2} + 1.5 = 39.45\ \text{dB}$$

Rainy weather. Average heavy-rain RI is about 17 dB above average fair weather (CIGRE):

(ii) Values at 1 MHz

CIGRE frequency correction ($f$ in MHz):

$$\begin{aligned} \Delta RI &= 5\left[1 - 2(\log_{10} 10f)^2\right] \\ f = 1:\ \Delta RI &= 5[1 - 2(1)^2] = -5\ \text{dB} \end{aligned}$$

(At 0.5 MHz the correction is only +0.11 dB, so the formula is taken as the 0.5 MHz value.)

RI falls at higher frequency because the corona pulse spectrum falls off and the line attenuates higher frequencies more.

Answer: fair weather ≈ 39.5 dB and rain ≈ 56.5 dB at 0.5 MHz; at 1 MHz ≈ 34.5 dB (fair) and 51.5 dB (rain), all above 1 μV/m.

The charge matrix is $[Q] = [C][V] = 2\pi\varepsilon_0 [M]^{-1}[V]$. Here $[M]$ is Maxwell's coefficient matrix, built from the average height of the sagging conductor and the equivalent bundle radius.

Data

• 400 kV, flat horizontal, $S = 11.3$ m
• Bundle $2 \times 3.17$ cm diameter, so $r = 1.585$ cm; $B = 45.72$ cm
• Heights 11.875 m (tower) and 8.85 m (midspan)
• $2\pi\varepsilon_0 = 55.56$ pF/m

Step 1: Average height

$$H_{av} = 8.85 + \frac{11.875 - 8.85}{3} = 9.858\ \text{m}, \quad 2H = 19.717\ \text{m}$$

Step 2: Equivalent radius

$$r_{eq} = \sqrt{rB} = \sqrt{1.585 \times 45.72} = 8.513\ \text{cm}$$

Step 3: Maxwell coefficient matrix

$$\begin{aligned} M_{ii} &= \ln\frac{19.717}{0.08513} = 5.4451 \\ M_{12} = M_{23} &= \ln\frac{\sqrt{11.3^2 + 19.717^2}}{11.3} = 0.6987 \\ M_{13} &= \ln\frac{\sqrt{22.6^2 + 19.717^2}}{22.6} = 0.2830 \end{aligned}$$ $$[M] = \begin{bmatrix} 5.4451 & 0.6987 & 0.2830 \\ 0.6987 & 5.4451 & 0.6987 \\ 0.2830 & 0.6987 & 5.4451 \end{bmatrix}, \quad \det = 155.96$$

Step 4: Capacitance (charge-coefficient) matrix

$$[M]^{-1} = \begin{bmatrix} 0.18697 & -0.02312 & -0.00675 \\ -0.02312 & 0.18959 & -0.02312 \\ -0.00675 & -0.02312 & 0.18697 \end{bmatrix}$$ $$[C] = 55.56\,[M]^{-1} = \begin{bmatrix} 10.387 & -1.285 & -0.375 \\ -1.285 & 10.533 & -1.285 \\ -0.375 & -1.285 & 10.387 \end{bmatrix}\ \text{pF/m}$$

Step 5: Charge matrix

Balanced phase voltages: $V = 400/\sqrt{3} = 230.94$ kV rms at $0°, -120°, +120°$.

$$\begin{aligned} q_1 &= 230.94\,(10.387 - 1.285\angle{-120°} - 0.375\angle{120°}) \times 10^{-3} \\ &= 2.597\angle 4.02°\ \mu\text{C/m} \end{aligned}$$

Similarly:

$$[Q] = \begin{bmatrix} 2.597\angle 4.02° \\ 2.729\angle -120° \\ 2.597\angle 115.98° \end{bmatrix}\ \mu\text{C/m (rms)}$$

Peak values: 3.67, 3.86 and 3.67 μC/m.

The centre phase carries about 5% more charge than the outer phases, so its conductor surface gradient is the highest.

Answer: $[C]$ as above (diagonal 10.39/10.53 pF/m); charges ≈ 2.60, 2.73, 2.60 μC/m rms (3.67, 3.86, 3.67 μC/m peak).

The audible noise of each phase is found from the given empirical formula and the phase-to-microphone distance. The three levels are then added on an energy basis.

Data

• $N = 4$, $d = 3.05$ cm, $H = 20$ m, $S = 14$ m
• $E_m$ = 20 kV/cm (centre) and 18 kV/cm (outer phases)
• Microphone at ground, 30 m from the centre phase
• Formula (as given):

$$AN_i = 120\log E_i + 55\log d - 10.4\log D_i + 26.4\log N - 128.4$$

Step 1: Distances

The outer phases are 16 m and 44 m away horizontally.

$$\begin{aligned} D_{near} &= \sqrt{16^2 + 20^2} = 25.61\ \text{m} \\ D_{centre} &= \sqrt{30^2 + 20^2} = 36.06\ \text{m} \\ D_{far} &= \sqrt{44^2 + 20^2} = 48.33\ \text{m} \end{aligned}$$

Step 2: Constant terms

$$\begin{aligned} 55\log 3.05 &= 26.64 \\ 26.4\log 4 &= 15.89 \\ 120\log 20 &= 156.12 \\ 120\log 18 &= 150.63 \end{aligned}$$

Step 3: AN of each phase

$$\begin{aligned} AN_{centre} &= 156.12 + 26.64 - 10.4\log 36.06 + 15.89 - 128.4 \\ &= 156.12 + 26.64 - 16.19 + 15.89 - 128.4 = 54.06\ \text{dB(A)} \\ AN_{near} &= 150.63 + 26.64 - 14.65 + 15.89 - 128.4 = 50.12\ \text{dB(A)} \\ AN_{far} &= 150.63 + 26.64 - 17.52 + 15.89 - 128.4 = 47.25\ \text{dB(A)} \end{aligned}$$

Step 4: Total AN

$$\begin{aligned} AN &= 10\log\left(10^{5.406} + 10^{5.012} + 10^{4.725}\right) \\ &= 10\log\left[(2.547 + 1.027 + 0.531) \times 10^{5}\right] \\ &= 56.13\ \text{dB(A)} \end{aligned}$$

The centre phase dominates because of its higher gradient. (The standard BPA/Begamudre formula has $-11.4\log D_i$; the printed $-10.4$ has been used as given. With $-11.4$ and the same data, the total is about 54.6 dB(A).)

Answer: total AN ≈ 56.1 dB(A) at 30 m from the centre phase.

The capacitance matrix is $[C] = 2\pi\varepsilon_0[M]^{-1}$, where $[M]$ is Maxwell's coefficient matrix built with the image method and the equivalent bundle radius. The charge matrix is $[Q] = [C][V]$.

Data

• 400 kV, flat horizontal, $S = 11.3$ m, $H = 9.81$ m (average)
• Sub-conductor radius $r = 3.17$ cm; $B = 45.72$ cm; $N = 2$
• $2\pi\varepsilon_0 = 1/(18 \times 10^9) = 55.56$ pF/m

Step 1: Equivalent radius

$$r_{eq} = \sqrt{rB} = \sqrt{3.17 \times 45.72} = 12.04\ \text{cm} = 0.1204\ \text{m}$$

Step 2: Maxwell coefficients ($2H = 19.62$ m)

$$\begin{aligned} M_{ii} &= \ln\frac{19.62}{0.1204} = 5.0936 \\ M_{12} = M_{23} &= \ln\frac{\sqrt{11.3^2 + 19.62^2}}{11.3} = \ln\frac{22.641}{11.3} = 0.6950 \\ M_{13} &= \ln\frac{\sqrt{22.6^2 + 19.62^2}}{22.6} = \ln\frac{29.928}{22.6} = 0.2809 \end{aligned}$$ $$[M] = \begin{bmatrix} 5.0936 & 0.6950 & 0.2809 \\ 0.6950 & 5.0936 & 0.6950 \\ 0.2809 & 0.6950 & 5.0936 \end{bmatrix}, \quad \det = 127.10$$

Step 3: Inverse and capacitance matrix

$$[M]^{-1} = \begin{bmatrix} 0.20033 & -0.02632 & -0.00746 \\ -0.02632 & 0.20351 & -0.02632 \\ -0.00746 & -0.02632 & 0.20033 \end{bmatrix}$$ $$[C] = 55.56\,[M]^{-1} = \begin{bmatrix} 11.129 & -1.462 & -0.414 \\ -1.462 & 11.306 & -1.462 \\ -0.414 & -1.462 & 11.129 \end{bmatrix}\ \text{pF/m (nF/km)}$$

Step 4: Charge matrix

Balanced phase voltages $V = 400/\sqrt{3} = 230.94$ kV rms at $0°, -120°, 120°$; $[Q] = [C][V]$:

$$\begin{aligned} q_1 &= 230.94\,\big(11.129 - 1.462\angle{-120°} - 0.414\angle{120°}\big) \times 10^{-3} \\ &= 2.795\angle 4.30°\ \mu\text{C/m} \end{aligned}$$ $$[Q] = \begin{bmatrix} 2.795\angle 4.30° \\ 2.949\angle -120° \\ 2.795\angle 115.70° \end{bmatrix}\ \mu\text{C/m (rms)}$$

Peak values: 3.95, 4.17 and 3.95 μC/m.

If the line is transposed (for reference)

$$M_s = 5.0936,\quad M_m = \frac{2(0.6950) + 0.2809}{3} = 0.5569$$ $$C_1 = \frac{55.56}{5.0936 - 0.5569} = 12.25\ \text{nF/km}, \quad q = 12.25 \times 230.94 \times 10^{-3} = 2.83\ \mu\text{C/m (rms)}$$

Answer: $[C]$ as above; charges ≈ 2.79∠4.3°, 2.95∠−120°, 2.79∠115.7° μC/m rms.