Line Parameter Calculations

Skin effect

Skin effect is the tendency of alternating current to crowd towards the outer surface of a conductor. The current density is highest at the surface and lowest at the centre.

Cause: Imagine the conductor as many thin filaments. A filament at the centre is linked by the flux inside the conductor and also by the flux outside, while a filament at the surface is linked only by the outside flux. So the central filaments have more inductance and a higher reactance $\omega L$. The AC current takes the easier, low-reactance path near the surface.

  DC: uniform           AC: crowded at surface
    .-------.               .-------.
   / . . . . \             / ####### \
  | . . . . . |           | ##     ## |
  | . . . . . |           | #  low  # |
   \ . . . . /             \ ####### /
    '-------'               '-------'

Effects:

• The effective area is reduced, so $R_{ac} > R_{dc}$. This increases $I^2R$ loss.
• Internal inductance is slightly reduced.

Factors: the effect increases with frequency, conductor diameter, permeability and conductivity. The depth of current flow is $\delta = \sqrt{\frac{2\rho}{\omega\mu}}$; for copper at 50 Hz, $\delta \approx 9.3$ mm.

Remedies: stranded and ACSR conductors (the steel core carries little current), hollow or tubular conductors, and bundled conductors. There is no skin effect with DC.

Proximity effect

Proximity effect is the non-uniform current distribution in a conductor caused by the alternating magnetic field of a nearby conductor carrying current.

• When two nearby conductors carry current in the same direction, current crowds to their far sides.
• When they carry current in opposite directions (go and return), current crowds to the sides facing each other.

 Same direction        Opposite directions
 (#  )   (  #)          (  #) (#  )
 crowd outward          crowd towards each other

Effects: the effective resistance rises further, and the inductance changes slightly.

Where it matters: in cables and closely spaced conductors (bus-bars, cable cores, transformer windings). It is negligible in overhead lines, where the conductors are metres apart.

Statement: the internal inductance of a straight round conductor is $L_{int} = \frac{\mu_0}{8\pi} = \frac{1}{2}\times10^{-7}$ H/m. It contains no current term, so it is independent of the current.

Derivation

Take a long round conductor of radius r carrying current I, with uniform current density (skin effect neglected) and $\mu_r = 1$.

        .---------.
      /   .-----.   \     r = conductor radius
     |   /   x   \   |    x = radius of the circular path
     |  |  <--dx  |  |    dx = thin tubular element
     |   \       /   |
      \   '-----'   /
        '---------'

1. Current enclosed by a circle of radius x < r:

$$I_x = \frac{\pi x^2}{\pi r^2}I = \frac{x^2}{r^2}I$$

2. Ampere's law around that circle:

$$H_x(2\pi x) = I_x \;\Rightarrow\; H_x = \frac{x}{2\pi r^2}I, \qquad B_x = \mu_0H_x = \frac{\mu_0 x I}{2\pi r^2}$$

3. Flux in the tube of thickness dx and length 1 m:

$$d\phi = B_x\,dx = \frac{\mu_0 x I}{2\pi r^2}dx$$

4. Flux linkage. This flux links only the fraction $x^2/r^2$ of the total current:

$$d\lambda = \frac{x^2}{r^2}d\phi = \frac{\mu_0 I x^3}{2\pi r^4}dx$$

5. Total internal flux linkage:

$$\lambda_{int} = \int_0^r \frac{\mu_0 I x^3}{2\pi r^4}dx = \frac{\mu_0 I}{2\pi r^4}\cdot\frac{r^4}{4} = \frac{\mu_0 I}{8\pi}\ \text{Wb-T/m}$$

6. Internal inductance:

$$L_{int} = \frac{\lambda_{int}}{I} = \frac{\mu_0}{8\pi} = \frac{4\pi\times10^{-7}}{8\pi} = \frac{1}{2}\times10^{-7}\ \text{H/m}$$

Conclusion

The flux linkage $\lambda_{int}$ is directly proportional to I, so the ratio $\lambda/I$ is a constant. The current cancels and $L_{int} = 0.5\times10^{-7}$ H/m for any value of current. This is true as long as μ is constant, which holds for non-magnetic conductors such as copper and aluminium. (The same result also shows that $L_{int}$ does not depend on the radius r.)

Physical reason: doubling I doubles both the field H at every point and the flux linkage, so their ratio stays the same.

What transposition is

Transposition is the regular exchange of the positions of the three phase conductors along a line. Each phase occupies each of the three positions for one-third of the length of a complete transposition cycle. The exchange is done at special transposition towers.

   Section 1    Section 2    Section 3
 pos1  a ---------- c ---------- b
 pos2  b ---------- a ---------- c
 pos3  c ---------- b ---------- a
      |<- l/3 ->|<- l/3 ->|<- l/3 ->|

Why it is required

1. Equal inductance and capacitance per phase. With unsymmetrical spacing (for example horizontal or vertical arrangement), each phase sees different distances to the other two. So each phase has a different flux linkage, inductance and capacitance. Transposition gives every phase the same average spacing, $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$, and so equal parameters:

$$L = 2\times10^{-7}\ln\frac{D_{eq}}{r'}\ \text{H/m per phase}$$

2. Balanced voltage drops. With unequal reactances, a balanced load current produces unequal drops, and the receiving-end voltages become unbalanced. Transposition keeps the receiving end balanced.
3. No complex (imaginary) inductance terms. Without transposition the flux-linkage equations contain $a$-operator terms. These make power transfer between phases appear, and the per-phase model fails. Transposition removes them, so the line can be treated as balanced and analysed per phase.
4. Reduced interference with communication lines. The induced voltages from the three phases cancel over a cycle. This reduces electromagnetic and electrostatic interference with nearby telephone and communication circuits.
5. Reduced unbalance in charging currents and neutral (ground) currents.

In practice, lines are transposed at switching stations or every few tens of kilometres. Even when a line is not physically transposed, it is usually calculated as transposed, because the error is small.

Statement: the internal inductance of a round conductor is $L_{int} = \frac{\mu_0}{8\pi} = 0.5\times10^{-7}$ H/m. It does not contain the conductor radius or the spacing, so it is independent of the conductor geometry (size and arrangement).

Derivation

Consider a long solid round conductor of radius r carrying current I with uniform current density ($\mu_r = 1$).

1. Current enclosed within radius x (x < r):

$$I_x = \frac{x^2}{r^2}I$$

2. Field intensity by Ampere's law:

$$2\pi x H_x = I_x \;\Rightarrow\; H_x = \frac{xI}{2\pi r^2}, \qquad B_x = \frac{\mu_0 xI}{2\pi r^2}$$

3. Flux in a tubular element of thickness dx, per metre length:

$$d\phi = B_x\,dx = \frac{\mu_0 xI}{2\pi r^2}dx$$

4. Partial flux linkage. This flux links only $x^2/r^2$ of the current:

$$d\lambda = \frac{x^2}{r^2}\cdot\frac{\mu_0 xI}{2\pi r^2}dx = \frac{\mu_0 I x^3}{2\pi r^4}dx$$

5. Integrate from 0 to r:

$$\lambda_{int} = \frac{\mu_0 I}{2\pi r^4}\cdot\frac{r^4}{4} = \frac{\mu_0 I}{8\pi}$$

6. Inductance:

$$L_{int} = \frac{\lambda_{int}}{I} = \frac{\mu_0}{8\pi} = \frac{4\pi\times10^{-7}}{8\pi} = 0.5\times10^{-7}\ \text{H/m}$$

Why it is independent of geometry

• The $r^4$ in the denominator (from the field and partial-linkage terms) cancels exactly with the $r^4/4$ from the integral. So the radius drops out: a thin conductor and a thick conductor have the same internal inductance per metre.
• Internal flux exists only inside the conductor, so it does not depend on the spacing to other conductors or on the arrangement (horizontal, vertical, triangular).
• Only the permeability of the conductor material matters. For copper and aluminium $\mu = \mu_0$.

The spacing and radius affect only the external inductance, $2\times10^{-7}\ln\frac{D}{r}$. Combining both parts gives the familiar form with GMR:

$$L = 2\times10^{-7}\left(\frac14 + \ln\frac{D}{r}\right) = 2\times10^{-7}\ln\frac{D}{r'}, \quad r' = re^{-1/4} = 0.7788r$$

The line parameters are the four distributed constants that set the electrical behaviour of a transmission line: series resistance R and inductance L, and shunt capacitance C and conductance G. All are given per unit length. R and L form the series impedance $z = R + j\omega L$; G and C form the shunt admittance $y = G + j\omega C$.

Effect of line configuration

1. Spacing between phases ($D_m$, GMD). L and C depend on $\ln(D_m)$. A larger spacing increases L and decreases C. Higher-voltage lines need larger clearances, so they have more L and less C per km.
2. Arrangement (equilateral, horizontal, vertical). With unsymmetrical spacing, each phase has different L and C. Transposition is used, and the equivalent spacing $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$ replaces D. For example, horizontal spacing D gives $D_{eq} = 1.26D$.
3. Conductor size and type ($D_s$, GMR). A larger radius gives a larger GMR, so lower L and higher C. Stranded conductors use the GMR of the strands; for a solid conductor $D_s = r' = 0.7788r$. Larger cross-section also lowers R.
4. Bundled conductors. Using 2, 3 or 4 sub-conductors per phase raises the self-GMD: $D_s^b = \sqrt{D_sd}$, $\sqrt[3]{D_sd^2}$, $1.09\sqrt[4]{D_sd^3}$. This lowers L (and reactance) and raises C, which increases the surge-impedance loading. It also reduces corona loss.
5. Double-circuit lines. Parallel conductors of the same phase act like a bundle. Proper placement (for example the hexagonal arrangement a-b-c / c'-b'-a') gives a high $D_s$ and low L for each phase.
6. Height above ground. Earth acts like image conductors. It slightly increases C, and has a negligible effect on L for balanced operation.
7. Stranding and spiralling. Spiralled strands are slightly longer than the line, which raises R by about 1–2 %.

 Larger spacing D  -> L up,   C down
 Larger radius r   -> L down, C up,  R down
 Bundling          -> L down, C up,  corona down

Why transposition is needed

When the phase spacings are unequal ($D_{12} \ne D_{23} \ne D_{31}$), each phase has a different flux linkage. Its inductance therefore differs from the other phases and even contains an imaginary part. This causes unequal voltage drops, an unbalanced receiving-end voltage, and interference with nearby communication lines. Transposition means each phase takes each position for one-third of the length. This makes the average inductance of all phases equal, so the line becomes balanced and can be analysed per phase.

          Sec I      Sec II     Sec III
 pos 1     a           c           b
 pos 2     b           a           c
 pos 3     c           b           a
        D12 between pos1-2, D23 pos2-3, D31 pos3-1

Derivation

Flux linkage of conductor a, of radius r, in a group of conductors (per metre):

$$\lambda_a = 2\times10^{-7}\left[I_a\ln\frac{1}{r'} + I_b\ln\frac{1}{D_{ab}} + I_c\ln\frac{1}{D_{ac}}\right], \quad r' = 0.7788r$$

Section I (a at 1, b at 2, c at 3):

$$\lambda_{a1} = 2\times10^{-7}\left[I_a\ln\frac{1}{r'} + I_b\ln\frac{1}{D_{12}} + I_c\ln\frac{1}{D_{31}}\right]$$

Section II (a at 2, b at 3, c at 1):

$$\lambda_{a2} = 2\times10^{-7}\left[I_a\ln\frac{1}{r'} + I_b\ln\frac{1}{D_{23}} + I_c\ln\frac{1}{D_{12}}\right]$$

Section III (a at 3, b at 1, c at 2):

$$\lambda_{a3} = 2\times10^{-7}\left[I_a\ln\frac{1}{r'} + I_b\ln\frac{1}{D_{31}} + I_c\ln\frac{1}{D_{23}}\right]$$

Average over the cycle:

$$\begin{aligned} \lambda_a &= \frac{\lambda_{a1} + \lambda_{a2} + \lambda_{a3}}{3}\\ &= \frac{2\times10^{-7}}{3}\left[3I_a\ln\frac{1}{r'} + (I_b + I_c)\ln\frac{1}{D_{12}D_{23}D_{31}}\right] \end{aligned}$$

For balanced currents, $I_b + I_c = -I_a$:

$$\begin{aligned} \lambda_a &= 2\times10^{-7}\left[I_a\ln\frac{1}{r'} - I_a\ln\frac{1}{\sqrt[3]{D_{12}D_{23}D_{31}}}\right]\\ &= 2\times10^{-7}I_a\ln\frac{\sqrt[3]{D_{12}D_{23}D_{31}}}{r'} \end{aligned}$$

Inductance per phase:

$$L_a = \frac{\lambda_a}{I_a} = 2\times10^{-7}\ln\frac{D_{eq}}{r'}\ \text{H/m}, \qquad D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}$$

By symmetry, $L_b = L_c = L_a$. The transposed line behaves like an equilaterally spaced line with spacing $D_{eq}$ (the geometric mean distance), and the inductance is real and equal in all phases.

In mH/km: $L = 0.2\ln\frac{D_{eq}}{r'}$ mH/km per phase. For stranded or bundled conductors, $r'$ is replaced by the GMR $D_s$.

Example: horizontal spacing 4 m, 4 m, 8 m gives $D_{eq} = \sqrt[3]{4\times4\times8} = 5.04$ m.

The earth is a conductor at zero potential. Its presence changes the electric field of an overhead line and so increases the line capacitance slightly.

Method of images

The earth's surface is an equipotential plane. Its effect is the same as removing the earth and placing an image conductor at the same depth below ground, with an equal and opposite charge. For a conductor at height h with charge +q, the image is at depth h with charge −q. The field above ground is then calculated from the real and image charges together.

     a (+qa)  <-- D -->  b (+qb)
     |                   |
     h                   h
 ====|======= earth =====|======
     h                   h
     |                   |
     a' (-qa)            b' (-qb)
  distance a to b' = H = sqrt(D^2 + 4h^2)

Single-phase line

Without earth:

$$C_{ab} = \frac{\pi\varepsilon_0}{\ln(D/r)}\ \text{F/m}$$

With earth effect (images included):

$$C_{ab} = \frac{\pi\varepsilon_0}{\ln\left(\dfrac{D}{r}\cdot\dfrac{2h}{\sqrt{D^2 + 4h^2}}\right)}$$

Because $\frac{2h}{\sqrt{D^2+4h^2}} < 1$, the denominator is smaller and the capacitance is larger.

Three-phase transposed line

$$C_n = \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r} - \ln\dfrac{\sqrt[3]{H_{12}H_{23}H_{31}}}{\sqrt[3]{H_1H_2H_3}}}\ \text{F/m to neutral}$$

Here $H_{ij}$ is the distance from conductor i to the image of conductor j, and $H_i = 2h$ is the distance to its own image.

Points to note

• The earth brings charge of the opposite sign closer to the conductor, which raises the capacitance.
• The effect is small for normal overhead lines, because $h \gg D$, so it is usually neglected. It becomes important for low-height lines, large spacings, and in unbalanced-fault (zero-sequence) calculations.
• Earth has a negligible effect on line inductance under balanced conditions.
• Higher capacitance means slightly more charging current and charging kVAR.

Skin effect is the tendency of alternating current to flow mostly near the outer surface ("skin") of a conductor, instead of being spread evenly over its cross-section.

With DC

• The current is steady, so there is no changing flux and no induced emf inside the conductor.
• The current divides according to resistance only. Every filament has the same resistance, so the current density is uniform over the whole area.
• The full area is used, and the resistance is $R_{dc} = \rho l/A$.

With AC

• The changing current produces alternating flux inside the conductor. A filament at the centre is linked by more flux (internal plus external) than a filament at the surface.
• So the central filaments have higher inductance and reactance, and larger back-emf. The current is pushed towards the surface, where the impedance is lower.
• The effective conducting area is reduced, so $R_{ac} > R_{dc}$. The effect grows with frequency, diameter and permeability. The skin depth is $\delta = \sqrt{2\rho/(\omega\mu)}$, about 9.3 mm for copper at 50 Hz.

 DC (uniform J)          AC (J higher near surface)
   .-------.               .-------.
  / ::::::: \             / ####### \
 | ::::::::: |           | ##.....## |
 | ::::::::: |           | #.......# |
  \ ::::::: /             \ ####### /
   '-------'               '-------'

Effects and remedies: higher losses and lower efficiency. It is reduced by using stranded, ACSR (current flows mainly in the outer aluminium strands), hollow or bundled conductors.

For a bundled, transposed line the inductance per phase is

$$L = 2\times10^{-7}\ln\frac{D_m}{D_s^b}\ \text{H/m}$$

Here $D_s^b$ is the self-GMD (GMR) of the bundle and $D_m$ is the mutual GMD between phases. The bundle spacing (0.305 m) is much smaller than the phase spacing, so distances between phases are taken centre to centre. The line is assumed to be fully transposed.

Data

• Sub-conductor radius: $r = 2.54/2 = 1.27$ cm $= 0.0127$ m
• GMR of a solid sub-conductor: $r' = 0.7788r = 0.7788\times0.0127 = 0.009891$ m
• Sub-conductor spacing: $d = 0.305$ m (equilateral triangle)

Self-GMD of the three-conductor bundle

$$\begin{aligned} D_s^b &= \sqrt[9]{(r'\,d\,d)^3} = \sqrt[3]{r'd^2}\\ &= \sqrt[3]{0.009891\times0.305^2}\\ &= 0.09726\ \text{m} \end{aligned}$$

Mutual GMD

Horizontal arrangement: $D_{ab} = D_{bc} = 6.1$ m and $D_{ca} = 12.2$ m.

$$D_m = \sqrt[3]{6.1\times6.1\times12.2} = 6.1\times\sqrt[3]{2} = 7.686\ \text{m}$$

Inductance

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{7.686}{0.09726}\ \text{H/m}\\ &= 2\times10^{-7}\times\ln(79.02) = 2\times10^{-7}\times4.3697\\ &= 0.8739\times10^{-6}\ \text{H/m} = 0.8739\ \text{mH/km} \end{aligned}$$

If the frequency is 50 Hz, the reactance is $X_L = 2\pi\times50\times0.8739\times10^{-3} = 0.2746\ \Omega$/km per phase.

Answer: $L \approx 0.874$ mH per phase per km ($X_L \approx 0.275\ \Omega$/km at 50 Hz).

The line length is not given, so all results are per km. The ACSR conductor is treated as a solid round conductor of the given effective diameter, with GMR $r' = 0.7788r$ (no strand data is given).

Data

• $V_{LL} = 132$ kV, so $V_{ph} = 132/\sqrt3 = 76.21$ kV; f = 50 Hz, $\omega = 314.16$ rad/s
• $r = 39.8/2 = 19.9$ mm $= 1.99$ cm; D = 8 m = 800 cm (equilateral)
• Copper-equivalent area $A = 1.5$ cm²; $\rho = 1.73\times10^{-6}\ \Omega$-cm

(i) Line parameters

Resistance (l = 1 km = $10^5$ cm):

$$R = \frac{\rho l}{A} = \frac{1.73\times10^{-6}\times10^5}{1.5} = 0.1153\ \Omega/\text{km}$$

Inductance: $r' = 0.7788\times1.99 = 1.5498$ cm.

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{D}{r'} = 2\times10^{-7}\ln\frac{800}{1.5498}\\ &= 2\times10^{-7}\times6.2465 = 1.2493\ \text{mH/km}\\ X_L &= 314.16\times1.2493\times10^{-3} = 0.3925\ \Omega/\text{km} \end{aligned}$$

Capacitance to neutral:

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D/r)} = \frac{5.563\times10^{-11}}{\ln(800/1.99)} = \frac{5.563\times10^{-11}}{5.9965}\\ &= 9.277\times10^{-12}\ \text{F/m} = 9.277\times10^{-3}\ \mu\text{F/km}\\ X_C &= \frac{1}{\omega C} = \frac{1}{314.16\times9.277\times10^{-9}} = 3.431\times10^5\ \Omega\cdot\text{km} \end{aligned}$$

(ii) Charging current and charging MVA

$$\begin{aligned} I_c &= \frac{V_{ph}}{X_C} = 76210\times314.16\times9.277\times10^{-9} = 0.2221\ \text{A/km}\\ Q_c &= 3V_{ph}I_c = \sqrt3\times132\times10^3\times0.2221\\ &= 50.78\ \text{kVA/km} = 0.0508\ \text{MVA/km} \end{aligned}$$

Answer: R = 0.115 Ω/km, L = 1.249 mH/km, C = 0.00928 μF/km; charging current 0.222 A/km and charging 0.0508 MVA/km. For example, a 100 km line would draw 22.2 A and 5.08 MVAr.

For a transposed double-circuit line the inductance per phase is

$$L = 2\times10^{-7}\ln\frac{D_m}{D_s}\ \text{H/m}$$

Here $D_m$ is the equivalent mutual GMD between phase groups and $D_s$ is the self-GMD of each phase (its two parallel conductors). Both are worked out for the section shown. By transposition, the other sections give the same result.

Geometry (section shown)

Coordinates in metres, with c at the origin:

   a (0,6) ------ 6 m ------ c' (6,6)
   |3 m
   b (0,3) ------ 7 m ------ b' (7,3)
   |3 m
   c (0,0) ------ 6 m ------ a' (6,0)

Self-GMD

$r' = 0.7788\times0.009 = 0.007009$ m (solid copper).

$$\begin{aligned} D_{sa} &= \sqrt{r'\,D_{aa'}} = \sqrt{0.007009\times8.485} = 0.2439\ \text{m}\\ D_{sb} &= \sqrt{r'\,D_{bb'}} = \sqrt{0.007009\times7} = 0.2215\ \text{m}\\ D_{sc} &= \sqrt{r'\,D_{cc'}} = 0.2439\ \text{m}\\ D_s &= \sqrt[3]{0.2439\times0.2215\times0.2439} = 0.2362\ \text{m} \end{aligned}$$

Mutual GMD

$$\begin{aligned} D_{AB} &= \sqrt[4]{D_{ab}D_{ab'}D_{a'b}D_{a'b'}} = \sqrt[4]{3\times7.616\times6.708\times3.162} = 4.692\ \text{m}\\ D_{BC} &= \sqrt[4]{D_{bc}D_{bc'}D_{b'c}D_{b'c'}} = \sqrt[4]{3\times6.708\times7.616\times3.162} = 4.692\ \text{m}\\ D_{CA} &= \sqrt[4]{6\times6\times6\times6} = 6\ \text{m}\\ D_m &= \sqrt[3]{4.692\times4.692\times6} = 5.093\ \text{m} \end{aligned}$$

Inductance

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{5.093}{0.2362} = 2\times10^{-7}\times3.0710\\ &= 0.6142\times10^{-6}\ \text{H/m} = 0.6142\ \text{mH/km per phase} \end{aligned}$$

At 50 Hz this gives $X_L = 314.16\times0.6142\times10^{-3} = 0.193\ \Omega$/km per phase.

Answer: $D_s = 0.236$ m, $D_m = 5.093$ m, and the inductance is 0.614 mH per phase per km.

The capacitance to neutral of a transposed three-phase line is $C_n = \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)}$. It contains only the conductor radius r and the spacings (through $D_{eq}$), apart from the constant $\varepsilon_0$. The derivation below shows this.

Potential difference between two conductors

For a long conductor with charge q C/m, the potential difference between two points at distances $D_1$ and $D_2$ is $V_{12} = \frac{q}{2\pi\varepsilon_0}\ln\frac{D_2}{D_1}$. For a group of conductors, using superposition:

$$V_{ab} = \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D_{ab}}{r} + q_b\ln\frac{r}{D_{ab}} + q_c\ln\frac{D_{bc}}{D_{ca}}\right]$$

Equilateral spacing D

$$\begin{aligned} V_{ab} &= \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D}{r} + q_b\ln\frac{r}{D}\right]\\ V_{ac} &= \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D}{r} + q_c\ln\frac{r}{D}\right] \end{aligned}$$

Adding and using $q_b + q_c = -q_a$ (balanced charges):

$$V_{ab} + V_{ac} = \frac{3q_a}{2\pi\varepsilon_0}\ln\frac{D}{r}$$

For a balanced system, $V_{ab} + V_{ac} = 3V_{an}$. So

$$V_{an} = \frac{q_a}{2\pi\varepsilon_0}\ln\frac{D}{r}, \qquad C_n = \frac{q_a}{V_{an}} = \frac{2\pi\varepsilon_0}{\ln(D/r)}\ \text{F/m}$$

Unsymmetrical spacing (transposed)

Averaging $V_{ab}$ over the three transposition sections gives the same form, with D replaced by the geometric mean spacing:

$$C_n = \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r}}, \qquad D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$$

For bundled conductors, r is replaced by the bundle's self-GMD for capacitance, for example $\sqrt{rd}$.

Conclusion

• $\varepsilon_0$ is a constant (air). So C depends only on the ratio $D_{eq}/r$, that is, on conductor size and spacing.
• Larger r (or bundling) gives higher C. Larger spacing gives lower C. The dependence is logarithmic, so a big change in geometry causes only a modest change in C.
• Current, voltage and frequency do not appear in C. The earth's effect (images) is small for normal conductor heights and is neglected. That is why size and spacing are the primary factors.

Example: r = 1 cm and D = 4 m give $C_n = \frac{5.563\times10^{-11}}{\ln 400} = 9.29$ nF/km. Doubling D to 8 m gives 8.32 nF/km.

The voltage and frequency are not printed. They are assumed as 400 kV (line), 50 Hz, the usual data for this two-conductor-bundle configuration. Reactance per km does not depend on the voltage; only the charging current and VAR scale with it.

Data

• Sub-conductor radius $r = 9$ mm $= 0.009$ m. Bundle spacing $d = 0.40$ m.
• Phase spacing: 10 m, 10 m, 20 m (horizontal), transposed.
• $V_{ph} = 400/\sqrt3 = 230.94$ kV; $\omega = 314.16$ rad/s.

Self-GMD for capacitance

For capacitance the actual radius r is used, not $r'$:

$$D_{sc} = \sqrt{r\,d} = \sqrt{0.009\times0.40} = 0.06\ \text{m}$$

Mutual GMD

$$D_{eq} = \sqrt[3]{10\times10\times20} = 12.599\ \text{m}$$

Capacitance and capacitive reactance per phase per km

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_{eq}/D_{sc})} = \frac{5.563\times10^{-11}}{\ln(12.599/0.06)} = \frac{5.563\times10^{-11}}{5.3470}\\ &= 1.0404\times10^{-11}\ \text{F/m} = 0.010404\ \mu\text{F/km}\\ X_C &= \frac{1}{\omega C} = \frac{1}{314.16\times1.0404\times10^{-8}} = 3.059\times10^5\ \Omega\cdot\text{km} \end{aligned}$$

Charging current and charging VAR (per km)

$$\begin{aligned} I_c &= \frac{V_{ph}}{X_C} = \frac{230940}{3.059\times10^5} = 0.7548\ \text{A/km}\\ Q_{c,ph} &= V_{ph}I_c = 230940\times0.7548 = 174.32\ \text{kVAR/km per phase}\\ Q_{c,3\phi} &= 3\times174.32 = 522.97\ \text{kVAR/km} \end{aligned}$$

Answer: $X_C \approx 3.06\times10^5\ \Omega$·km per phase. At 400 kV, $I_c \approx 0.755$ A/km and the charging VAR is about 174.3 kVAR per phase per km. (At another line voltage V, scale $I_c$ by $V/400$ and the VAR by $(V/400)^2$.)

The geometric mean radius (GMR) of a conductor is the radius of a fictitious thin tube that has the same internal plus external flux linkage as the real conductor. For a solid round conductor of radius $R$, $GMR = r' = e^{-1/4}R = 0.7788R$. A stranded conductor of the same overall diameter contains air gaps between strands, so its current is spread over less metal and its effective GMR is smaller.

Example: 7-strand conductor

Take a 7-strand conductor, each strand of radius $r$: one central strand and six touching strands around it. Overall radius $R = 3r$.

        (2)   (3)
     (7)   (1)   (4)      1 = centre strand
        (6)   (5)         ring radius = 2r

Distances between strand centres:

With $n = 7$ strands there are $n^2 = 49$ terms:

$$\begin{aligned} D_s &= \left[(0.7788r)^7 (2r)^{24} (2\sqrt{3}r)^{12} (4r)^6\right]^{1/49} \\ &= 2.1767\,r \end{aligned}$$

Solid conductor of the same overall diameter

$$GMR_{solid} = 0.7788 \times 3r = 2.3364\,r$$

Comparison

$$\frac{D_{s,\,stranded}}{GMR_{solid}} = \frac{2.1767r}{2.3364r} = 0.932$$

So the 7-strand conductor has a GMR about 7% smaller than a solid conductor of the same diameter.

Why: the GMR depends on the actual cross-section that carries current. Stranding leaves air gaps, so the current-carrying material sits in a smaller effective area. A smaller GMR means the stranded conductor has a slightly higher inductance ($L = 2\times10^{-7}\ln(D/D_s)$) than a solid conductor of the same outer diameter.

Answer: $D_s = 2.1767r$ (stranded) < $2.3364r$ (solid).

Each phase is a 4-conductor square bundle, so we find the bundle GMR (self-GMD) and the mutual GMD between phase centres, then use the standard formulas for a transposed line.

   o o        o o        o o
   o o        o o        o o
    A    6 m   B    6 m   C
   |<------- 12 m ------->|
 square side d = 0.45 m, r = 12 mm

Data: $r = 0.012$ m, $d = 0.45$ m, $D_{AB} = D_{BC} = 6$ m, $D_{CA} = 12$ m, length 200 km.

Mutual GMD

$$D_m = \sqrt[3]{6 \times 6 \times 12} = 7.5595\ \text{m}$$

Self GMD of a 4-conductor bundle

For a square bundle, $D_s = (r_x \cdot d \cdot d \cdot \sqrt{2}d)^{1/4} = 1.09\,(r_x d^3)^{1/4}$.

• Inductance: $r' = 0.7788 \times 0.012 = 0.009346$ m

$$D_{sL} = 1.09\,(0.009346 \times 0.45^3)^{1/4} = 0.18629\ \text{m}$$

• Capacitance (uses actual radius $r$):

$$D_{sC} = 1.09\,(0.012 \times 0.45^3)^{1/4} = 0.19830\ \text{m}$$

Inductance per phase

$$\begin{aligned} L &= 2\times10^{-7}\ln\frac{D_m}{D_{sL}} = 2\times10^{-7}\ln\frac{7.5595}{0.18629} \\ &= 2\times10^{-7}\times 3.7033 = 7.4065\times10^{-7}\ \text{H/m} \\ &= 0.7407\ \text{mH/km} \end{aligned}$$

For 200 km: $L = 0.7407 \times 200 = 148.13$ mH.

Capacitance per phase (to neutral)

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_m/D_{sC})} = \frac{2\pi \times 8.854\times10^{-12}}{\ln(7.5595/0.19830)} \\ &= \frac{55.63\times10^{-12}}{3.6408} = 15.28\times10^{-12}\ \text{F/m} \\ &= 15.28\ \text{nF/km} \end{aligned}$$

For 200 km: $C = 15.28 \times 200 = 3056$ nF $= 3.056\ \mu$F.

Answer: $L = 0.7407$ mH/km per phase (148.13 mH for 200 km); $C = 15.28$ nF/km per phase to neutral (3.056 µF for 200 km).

For a transposed double-circuit line, the two conductors of a phase are treated as a two-conductor bundle. The capacitance per phase is $C = 2\pi\varepsilon_0/\ln(GMD/GMR)$, where GMD is the equivalent mutual distance between phase groups and GMR is the self-GMD of each phase group (using actual radius $r$).

   a o------- 5 m -------o c'    --
                                  3 m
  b o-------- 6 m --------o b'   --
                                  3 m
   c o------- 5 m -------o a'    --

Data: $r = 6$ mm $= 0.006$ m, vertical spacing 3 m, $f = 50$ Hz, 132 kV, 100 km.

Distances (from geometry)

Horizontal offset between a and b is $(6-5)/2 = 0.5$ m.

Mutual GMD

$$\begin{aligned} D_{AB} &= (D_{ab}D_{ab'}D_{a'b}D_{a'b'})^{1/4} = (3.0414^2 \times 6.2650^2)^{1/4} = 4.3651\ \text{m} \\ D_{BC} &= 4.3651\ \text{m} \quad (\text{by symmetry}) \\ D_{CA} &= (6 \times 5 \times 5 \times 6)^{1/4} = 5.4772\ \text{m} \\ D_m &= \sqrt[3]{4.3651 \times 4.3651 \times 5.4772} = 4.7082\ \text{m} \end{aligned}$$

Self GMD of each phase (for capacitance)

$$\begin{aligned} D_{sA} &= \sqrt{r \cdot D_{aa'}} = \sqrt{0.006 \times 7.8102} = 0.21648\ \text{m} \\ D_{sB} &= \sqrt{0.006 \times 6} = 0.18974\ \text{m} \\ D_{sC} &= D_{sA} = 0.21648\ \text{m} \\ D_s &= \sqrt[3]{0.21648 \times 0.18974 \times 0.21648} = 0.20717\ \text{m} \end{aligned}$$

Capacitance per phase per km

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_m/D_s)} = \frac{55.63\times10^{-12}}{\ln(4.7082/0.20717)} = \frac{55.63\times10^{-12}}{3.1235} \\ &= 17.81\times10^{-12}\ \text{F/m} = 17.81\ \text{nF/km} \end{aligned}$$

Capacitive reactance for 100 km

$$\begin{aligned} C_{100} &= 17.81 \times 100 = 1781\ \text{nF} = 1.781\ \mu\text{F} \\ X_C &= \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 1.781\times10^{-6}} = 1787.2\ \Omega \end{aligned}$$

Total VAR generated

Phase voltage $V_p = 132/\sqrt{3} = 76.21$ kV.

$$\begin{aligned} I_c &= \frac{V_p}{X_C} = \frac{76\,210}{1787.2} = 42.64\ \text{A} \\ Q &= 3V_p I_c = \frac{V_L^2}{X_C} = \frac{(132\times10^3)^2}{1787.2} = 9.749\ \text{MVAR} \end{aligned}$$

Answer: $C = 17.81$ nF/km per phase; $X_C = 1787.2\ \Omega$ per phase (100 km); total charging = 9.75 MVAR.

The method of images is a technique for finding the electric field (and hence capacitance) of charged conductors above a conducting plane. The earth is treated as a perfectly conducting, zero-potential plane. Its effect is replaced by imaginary "image" conductors placed as far below the ground as the real conductors are above it, carrying equal and opposite charges.

Basis

• The earth surface is an equipotential (zero potential) surface.
• A charge $+q$ at height $h$ and a charge $-q$ at depth $h$ below the surface together produce zero potential everywhere on the plane midway between them.
• So, for the region above ground, the field of "conductor + earth" is the same as the field of "conductor + image conductor" with the earth removed.

     +q o  a           +q o  b      (real conductors)
        |  h              |
 ====================================  earth (V = 0)
        |  h              |
     -q o  a'          -q o  b'     (images)

Use in capacitance calculation

1. Replace the earth by image conductors: for each conductor of charge $q$ at height $h$, add an image of charge $-q$ at depth $h$.
2. Write the voltage of each conductor due to all real and image charges, using $V = \dfrac{q}{2\pi\varepsilon_0}\ln\dfrac{D_2}{D_1}$ for each charge.
3. For a single-phase line (conductors a, b, spacing $D$, height $h$), this gives

$$V_{ab} = \frac{q}{\pi\varepsilon_0}\ln\frac{2hD}{r\sqrt{4h^2+D^2}}$$ $$C_{ab} = \frac{\pi\varepsilon_0}{\ln\left(\dfrac{D}{r\sqrt{1+D^2/4h^2}}\right)}\ \text{F/m}$$

4. For a transposed three-phase line, average over the transposition cycle:

$$C_n = \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r} - \ln\dfrac{\sqrt[3]{H_{12}H_{23}H_{31}}}{\sqrt[3]{H_1H_2H_3}}}\ \text{F/m}$$

where $H_1, H_2, H_3$ are distances from each conductor to its own image and $H_{12}$ etc. are distances to the images of the other conductors.

Effect

The extra term in the denominator is positive, so the ground increases the line capacitance slightly. The effect is small when conductors are high above ground compared to their spacing, so it is usually neglected except for low lines or accurate studies.

The GMR (self-GMD) used in line parameter calculations is different for inductance and capacitance because inductance depends on the magnetic flux, which exists inside the conductor too, while capacitance depends on the electric field, which exists only outside the conductor.

GMR for inductance

• Current flows through the whole cross-section, so there is magnetic flux inside the conductor as well as outside.
• The internal flux linkage ($\mu_0 I/8\pi$ per metre) is included by replacing the actual radius $r$ with a smaller fictitious radius:

$$r' = r\,e^{-1/4} = 0.7788\,r$$

• For a stranded or bundled conductor, the self-GMD uses $r'$ for each strand: $D_{sL} = (r' \cdot D_{12} \cdots D_{1n} \cdots)^{1/n^2}$; for a 2-bundle, $D_{sL} = \sqrt{r'd}$.

GMR for capacitance

• Charge resides on the surface of a conductor; the electric field inside a conductor is zero.
• So there is no "internal" term and the actual outer radius $r$ is used: $D_{sC} = (r \cdot D_{12} \cdots)^{1/n^2}$; for a 2-bundle, $D_{sC} = \sqrt{rd}$.

Comparison

Example: for $r = 1$ cm, GMR for inductance $= 0.7788$ cm, while for capacitance it is 1 cm. The mutual GMD between phases is the same for both calculations.

Each side of the line is a composite conductor. The inductance of side X is $L_X = 2\times10^{-7}\ln(D_m/D_{sX})$, and similarly for side Y; the loop inductance is $L = L_X + L_Y$.

  side X           side Y
   a o------9 m------o d
     |6 m
   b o------9 m------o e
     |6 m
   c o

Data: side X: $r = 0.25$ cm, $r' = 0.7788 \times 0.0025 = 0.001947$ m. Side Y: $r = 0.5$ cm, $r' = 0.003894$ m.

Distances

Mutual GMD (6 distances)

$$\begin{aligned} D_m &= (D_{ad}D_{ae}D_{bd}D_{be}D_{cd}D_{ce})^{1/6} \\ &= (9 \times 10.817 \times 10.817 \times 9 \times 15 \times 10.817)^{1/6} = 10.743\ \text{m} \end{aligned}$$

Self GMD of side X ($3^2 = 9$ terms)

$$\begin{aligned} D_{sX} &= \left[(r')^3 \times 6^4 \times 12^2\right]^{1/9} \\ &= \left[0.001947^3 \times 6^4 \times 12^2\right]^{1/9} = 0.4810\ \text{m} \end{aligned}$$

Self GMD of side Y ($2^2 = 4$ terms)

$$D_{sY} = \left[(0.003894)^2 \times 6^2\right]^{1/4} = \sqrt{0.003894 \times 6} = 0.15285\ \text{m}$$

Inductances

$$\begin{aligned} L_X &= 2\times10^{-7}\ln\frac{10.743}{0.4810} = 6.212\times10^{-7}\ \text{H/m} \\ L_Y &= 2\times10^{-7}\ln\frac{10.743}{0.15285} = 8.505\times10^{-7}\ \text{H/m} \\ L &= L_X + L_Y = 14.718\times10^{-7}\ \text{H/m} \end{aligned}$$

Answer: $L_X = 6.212\times10^{-7}$ H/m, $L_Y = 8.505\times10^{-7}$ H/m, complete line $L = 1.472\times10^{-6}$ H/m.