Overhead and Underground Transmission

Stranded conductors

A stranded conductor is made of many thin wires twisted together in layers (e.g. 1, 7, 19, 37 ... strands), each layer twisted in the opposite direction.

Reasons:

1. Flexibility: A solid conductor of large area is stiff and hard to handle, wind on drums and string. Stranding makes it flexible.
2. Less breakage: Solid wires can crack due to vibration and bending; in a stranded conductor a broken strand does not break the whole conductor.
3. Higher strength with composite strands: Steel core strands can be combined with aluminium strands (ACSR) to give high tensile strength and good conductivity.
4. Easy manufacture and transport of long lengths.
5. Slightly larger diameter for the same area, helping to reduce surface stress.

   Stranded (7 strands)    ACSR
       o o                ooo
      o O o              o###o   ### = steel core
       o o                ooo

Bundled conductors

In EHV lines (220 kV and above), each phase uses two or more sub-conductors spaced 30–45 cm apart by spacers.

Reasons:

1. Reduced corona: The bundle behaves like a conductor of larger equivalent radius, so the surface electric field falls, raising the critical disruptive voltage and reducing corona loss, radio and audible noise.
2. Lower inductive reactance: The geometric mean radius (self-GMD) increases, so $L = 2\times10^{-7}\ln(D_m/D_s)$ falls.
3. Higher capacitance and surge impedance loading: Lower $Z_c=\sqrt{L/C}$ gives higher SIL and power capacity.
4. Better stability: Lower reactance raises $P_{max}=V_sV_r/X$.
5. Better cooling and current capacity: More surface area dissipates heat.

 2-bundle    3-bundle      4-bundle
  o---o        o          o----o
              / \         |    |
             o---o        o----o

The Murray loop test is a bridge (Wheatstone principle) method to locate a short-circuit or earth fault in an underground cable. It needs one sound core running alongside the faulty core.

Connections

• At the far end, the faulty core and a sound core are joined by a low-resistance link, forming a loop.
• At the test end, the two ends of the loop are connected to the ratio arms $P$ and $Q$ (variable resistors) with a galvanometer G between them.
• A battery is connected between the junction of $P$ and $Q$ and earth (for an earth fault) or the other core in contact (for a short-circuit fault). The fault point completes the circuit.

           P            Q
   +----/\/\/---G---/\/\/\----+
   |            |             |
   |   Test end |             |  Far end
   |   +--------+ faulty core |   (cores
   |   |  <--- x --->X(fault) |   shorted)
   |   +======================+
   |   |  sound core          |
   |   +======================+
   |                          |
   +--------[Battery]--- Earth (sheath)
            fault to earth at X

Theory

The loop forms a Wheatstone bridge: $P$, $Q$, $R_x$ (faulty core from test end to fault) and $R$ (sound core plus faulty core from far end back to fault). At balance (G reads zero):

$$\frac{P}{Q} = \frac{R}{R_x} \quad\Rightarrow\quad \frac{P+Q}{Q} = \frac{R+R_x}{R_x}$$ $$R_x = \frac{Q}{P+Q}(R+R_x)$$

$(R+R_x)$ is the total loop resistance $= 2l\times r$ (r per unit length). If both cores have the same cross-section, resistance is proportional to length:

$$x = \frac{Q}{P+Q}\times2l$$

where $x$ is the distance of the fault from the test end and $l$ the cable length.

Example

For a 500 m cable balanced at $P = 3\ \Omega$, $Q = 1\ \Omega$: $x = \frac{1}{4}\times1000 = 250$ m from the test end.

Points to note

• Fault resistance appears in the battery arm, so it does not affect the balance (only sensitivity).
• Needs a sound core of known length; works best for low-resistance faults.
• If the cores differ in area, convert to equivalent lengths before using the formula.

FALSE. The voltage across the discs of a suspension string is not uniform; the disc nearest the line conductor takes the largest share. This is due to the shunt capacitance between each metal link (cap/pin joint) and the earthed tower.

Derivation (3-disc string)

Let each disc have self-capacitance $C$ and each joint have shunt capacitance $mC$ to earth. Let $V_1, V_2, V_3$ be the voltages across discs from the cross-arm (top) to the line (bottom), and $V = V_1+V_2+V_3$.

  Cross-arm (earth)
       |
     [C] V1
       |----- mC --> tower
     [C] V2
       |----- mC --> tower
     [C] V3
       |
    Line conductor (V)

Joint A (between discs 1 and 2): the current through disc 2 equals the current through disc 1 plus the current through the shunt capacitance:

$$\begin{aligned} \omega CV_2 &= \omega CV_1 + \omega mCV_1\\ V_2 &= V_1(1+m) \end{aligned}$$

Joint B (between discs 2 and 3): the potential to earth is $V_1+V_2$:

$$\begin{aligned} \omega CV_3 &= \omega CV_2 + \omega mC(V_1+V_2)\\ V_3 &= V_2 + m(V_1+V_2) = V_1(1+3m+m^2) \end{aligned}$$

Since $m>0$: $V_3 > V_2 > V_1$.

Numerical evidence

Take $m = 0.1$ and $V_1 = 1$: $V_2 = 1.1$, $V_3 = 1.31$. Total $=3.41$, so the shares are 29.3 %, 32.3 % and 38.4 % of the string voltage, not 33.3 % each.

String efficiency:

$$\eta = \frac{V}{n\,V_{n}} = \frac{3.41}{3\times1.31} = 86.8\ \%$$

For uniform distribution $\eta$ would be 100 %. Only if $m = 0$ (no shunt capacitance, e.g. in the laboratory with the string far from earthed metal) would the voltage be equal.

Consequence and remedies

The bottom disc is overstressed and may flash over first. Remedies: longer cross-arms (reduce $m$), grading of discs (higher capacitance near line), and guard (grading) rings which add line-to-joint capacitance to cancel the earth currents.

Faults in underground cables (insulation breakdown to sheath, short between cores, or open conductor) cannot be seen, so they are detected and located electrically in three steps.

1. Identify the faulty core and type of fault

• Use an insulation tester (megger) between each core and earth/sheath and between cores. Very low reading = earth or short-circuit fault.
• Use a continuity test (cores shorted at far end) to find open-circuit faults.

2. Locate the fault (pre-location) – loop tests

Murray loop test (for earth and short-circuit faults): the faulty core and a sound core are joined at the far end to form a loop, which with ratio arms $P$ and $Q$ forms a Wheatstone bridge.

           P            Q
   +----/\/\/---G---/\/\/\----+
   |            |             |
   |   +--------+ faulty core |
   |   |  <--- x --->X(fault) |
   |   +======================+  far ends
   |   |  sound core          |  shorted
   |   +======================+
   |
   +--------[Battery]--- Earth (sheath)

At balance:

$$\frac{P}{Q} = \frac{R}{R_x},\qquad x = \frac{Q}{P+Q}\times2l$$

where $x$ = distance of fault from test end and $l$ = cable length (cores of equal area).

Varley loop test: Similar loop, but with fixed ratio arms $P$, $Q$ and a variable resistor $S$ in series with the faulty-core arm; the loop resistance $(R_1+R_x)$ is first measured by a separate balance. Balance gives

$$R_x = \frac{(R_1+R_x)Q - SP}{P+Q}$$

and $x = R_x/r$. It is useful when the loop resistance is not known.

Pulse echo (time domain reflectometry, TDR): A short pulse is sent into the cable; it reflects at the fault. Distance $= v t/2$, where $v$ is the pulse velocity in the cable and $t$ the echo time. Fast and suitable for open and short faults.

3. Pinpoint the exact spot

Use an acoustic thumper (surge generator, listen for the discharge sound) or tracer/audio-frequency methods along the route before excavating.

In a single-core cable, the electric field (dielectric stress) in the insulation is inversely proportional to the radius, so it is highest at the conductor surface and falls towards the sheath.

Derivation

Let conductor radius $= r$, inner sheath radius $= R$, charge $= Q$ C/m length, permittivity $\varepsilon = \varepsilon_0\varepsilon_r$. Take a coaxial cylinder of radius $x$ ($r < x < R$) and length 1 m.

By Gauss's law, flux density and field at $x$:

$$D_x = \frac{Q}{2\pi x},\qquad g_x = \frac{D_x}{\varepsilon} = \frac{Q}{2\pi\varepsilon x}$$

Potential difference between conductor and sheath:

$$V = \int_r^R g_x\,dx = \frac{Q}{2\pi\varepsilon}\ln\frac{R}{r} \quad\Rightarrow\quad \frac{Q}{2\pi\varepsilon} = \frac{V}{\ln(R/r)}$$

Substituting:

$$g_x = \frac{V}{x\ln(R/r)}$$

Conclusion

Since $V$ and $\ln(R/r)$ are constants, $g_x \propto \dfrac{1}{x}$:

• At the conductor surface ($x = r$): $g_{max} = \dfrac{V}{r\ln(R/r)}$, the maximum.
• At the sheath ($x = R$): $g_{min} = \dfrac{V}{R\ln(R/r)}$, the minimum.
• $\dfrac{g_{max}}{g_{min}} = \dfrac{R}{r}$.

 stress g
  ^
  |*
  | *
  |   *
  |      *  *   *
  +--+-------------+---> x
     r             R

Example

$V = 33$ kV, $r = 1$ cm, $R = 2.5$ cm: $\ln 2.5 = 0.916$, $g_{max} = 33/(1\times0.916) = 36.0$ kV/cm, $g_{min} = 14.4$ kV/cm.

Because the inner insulation is most stressed, cables use grading (capacitance or intersheath grading) to make the stress more uniform; also the most economical size is when $R/r = e = 2.718$.

A conductor for an overhead line should carry current efficiently and safely for many years under mechanical and weather loads. Its main requirements are:

1. High electrical conductivity (low resistivity) to reduce $I^2R$ loss and voltage drop.
2. High tensile strength to withstand its own weight, wind and ice loading and allow long spans with less sag.
3. Low specific gravity (light weight) so that towers can be lighter and sag is smaller.
4. Low cost and easy availability so the line is economical.
5. Flexibility: easy to handle, string and wind on drums (achieved by stranding).
6. Durability and corrosion resistance against weather and pollution.
7. Low coefficient of thermal expansion so that sag does not change much with temperature.
8. High melting point / ability to work hot to carry overloads and short-circuit currents.
9. Larger diameter (for EHV) to reduce corona.
10. Not brittle; should resist fatigue from vibration.

ACSR meets most of these requirements and is widely used in Nepal's transmission and distribution lines.

Data

• Line voltage 66 kV, so voltage across the string (phase to earth):

$$V = \frac{66}{\sqrt3} = 38.105\ \text{kV}$$

• Number of discs $n = 4$; shunt-to-self capacitance ratio $m = 0.1$.

Let $V_1$ (top, at cross-arm) to $V_4$ (bottom, at line) be the disc voltages.

Equations at each joint

Current through lower disc = current through upper disc + shunt current to tower:

$$\begin{aligned} V_2 &= V_1(1+m) = 1.1V_1\\ V_3 &= V_2 + m(V_1+V_2) = 1.1 + 0.1(2.1) = 1.31V_1\\ V_4 &= V_3 + m(V_1+V_2+V_3) = 1.31 + 0.1(3.41) = 1.651V_1 \end{aligned}$$

Solving

$$V = V_1(1 + 1.1 + 1.31 + 1.651) = 5.061V_1 \Rightarrow V_1 = \frac{38.105}{5.061} = 7.529\ \text{kV}$$

String efficiency

$$\eta = \frac{V}{n\times V_4} = \frac{38.105}{4\times12.431}\times100 = 76.6\ \%$$

The disc next to the line carries about 33 % of the string voltage, so it is the most stressed; guard rings or longer cross-arms would improve the distribution.

Answer: $V_1 = 7.53$, $V_2 = 8.28$, $V_3 = 9.86$, $V_4 = 12.43$ kV; string efficiency = 76.6 %.

Assumption: 25 kV is the voltage across the whole string (phase to earth). Each unit has self-capacitance $C$; each joint has capacitance $mC = 0.25C$ to earth (tower) and $nC = 0.1C$ to the line conductor.

   Tower (earth)
      |
    [C] V1
      +-- 0.25C to tower, 0.1C to line (joint A)
    [C] V2
      +-- 0.25C to tower, 0.1C to line (joint B)
    [C] V3
      |
   Line (25 kV)

Joint equations

Current flows from line towards the tower. At each joint, current entering (through lower disc + from line capacitance) = current leaving (through upper disc + to earth capacitance).

Joint A (potential $V_1$; line-to-joint voltage $V_2+V_3$):

$$V_2 + n(V_2+V_3) = V_1(1+m) \quad\Rightarrow\quad 1.1V_2 + 0.1V_3 = 1.25V_1$$

Joint B (potential $V_1+V_2$; line-to-joint voltage $V_3$):

$$V_3(1+n) = V_2 + m(V_1+V_2) \quad\Rightarrow\quad 1.1V_3 = 1.25V_2 + 0.25V_1$$

Solving (put $V_1 = 1$)

From joint B: $V_3 = (1.25V_2 + 0.25)/1.1$. Substituting in joint A:

$$\begin{aligned} 1.1V_2 + \frac{0.1(1.25V_2+0.25)}{1.1} &= 1.25\\ 1.21V_2 + 0.125V_2 + 0.025 &= 1.375\\ V_2 &= \frac{1.35}{1.335} = 1.0112,\qquad V_3 = \frac{1.25(1.0112)+0.25}{1.1} = 1.3764 \end{aligned}$$

Sum $= 1 + 1.0112 + 1.3764 = 3.3876$, so $V_1 = 25/3.3876 = 7.380$ kV.

String efficiency

$$\eta = \frac{V}{n\,V_3} = \frac{25}{3\times10.157}\times100 = 82.0\ \%$$

The line capacitance partly cancels the earth capacitance current, so the distribution is better than with earth capacitance alone (with the 0.25 earth capacitance alone, $\eta$ would be only 74.7 %). If 25 kV were the line voltage, every disc voltage would be divided by $\sqrt3$, but the efficiency stays 82.0 %.

Answer: $V_1 = 7.38$ kV, $V_2 = 7.46$ kV, $V_3 = 10.16$ kV; string efficiency ≈ 82.0 %.

A stranded conductor is one conductor made of many thin wires twisted together. A bundled conductor is one phase made of two or more separate (usually stranded) sub-conductors held apart by spacers.

 Stranded (one conductor)   Bundled (one phase)
        o o                   (o)------(o)
       o O o                      spacer
        o o

A guard ring adds capacitance between each link pin and the line conductor. This line capacitance partly cancels the charging current lost to the tower, so the voltage spreads more evenly across the discs.

Circuit and junction equations

Number the discs 1 (tower end) to 3 (line end), with voltages $V_1, V_2, V_3$. Let $V$ be the voltage of the line to earth. Each link pin has $0.2C$ to earth and $g\,C$ to the line.

 Tower ─[C]─ A ─[C]─ B ─[C]─ Line (V)
             |       |
   earth ─0.2C     0.2C
   line  ─gA·C     gB·C

Kirchhoff's current law at a junction (current entering from the line side = current leaving towards the tower + current to earth):

$$V_{k+1} = V_k + m\,(V_1+\dots+V_k) - g\,[\,V-(V_1+\dots+V_k)\,]$$

Case 1: line capacitance 0.1C at both pins

Junction A ($m=0.2,\ g=0.1$):

$$V_2 = V_1 + 0.2V_1 - 0.1(V - V_1) = 1.3V_1 - 0.1V$$

Junction B:

$$\begin{aligned} V_3 &= V_2 + 0.2(V_1+V_2) - 0.1(V - V_1 - V_2)\\ &= 1.99V_1 - 0.23V \end{aligned}$$

Since $V_1+V_2+V_3 = V$: $4.29V_1 - 0.33V = V$, so $V_1 = 0.3100V$.

$$\eta = \frac{V}{3V_3} = \frac{1}{3\times 0.38695} = 86.15\ \%$$

Case 2: guard ring raises the lower pin's line capacitance to 0.3C

Junction A is unchanged: $V_2 = 1.3V_1 - 0.1V$.

Junction B ($g = 0.3$):

$$\begin{aligned} V_3 &= V_2 + 0.2(V_1+V_2) - 0.3(V - V_1 - V_2)\\ &= 2.45V_1 - 0.45V \end{aligned}$$

Sum: $4.75V_1 - 0.55V = V$, so $V_1 = 0.3263V$.

$$\eta_2 = \frac{1}{3\times 0.34947} = 95.38\ \%$$

Answer: Without the larger guard ring capacitance the discs carry 31.00 %, 30.30 % and 38.69 % of $V$, and the string efficiency is 86.15 %. With 0.3C at the lower pin the voltages become 32.63 %, 32.42 % and 34.95 % of $V$, and the efficiency rises to 95.38 %. The guard ring takes stress off the line-end disc.

(For comparison, with no line capacitance at all the same string would carry 26.0 %, 31.3 % and 42.7 %, giving an efficiency of 78.0 %.)

The Murray loop test is a bridge method for locating faults in underground cables. For a short-circuit or earth fault it compares the resistance of the faulty core with the resistance of a sound core (Wheatstone bridge). For an open-circuit fault it compares capacitances.

Murray loop test (short-circuit or earth fault)

This is a Wheatstone bridge method. It needs one sound core running alongside the faulty core.

        P              Q
  A ---/\/\--- G --/\/\--- B
  |          (galv)        |
  |                        |
  faulty core (X)     sound core
  |                        |
  +====== loop at far end==+
  |
 fault --> earth
  |
 battery between G-side junction and earth

1. At the far end, connect the faulty core to the sound core with a low-resistance link, so the two cores form a loop.
2. Connect the ratio arms $P$ and $Q$ (variable resistors) across the two test-end terminals and the galvanometer between them. Connect the battery between the junction of $P$ and $Q$ and earth (for an earth fault). For a short circuit between two cores, connect the battery to the other shorted core instead of earth.
3. Adjust $P$ and $Q$ until the galvanometer reads zero.

Let $R_x$ be the resistance of the faulty core from the test end to the fault, and $R_L$ the resistance of the whole loop. At balance:

$$\frac{P}{Q} = \frac{R_L - R_x}{R_x} \;\Rightarrow\; R_x = \frac{Q}{P+Q}\,R_L$$

If both cores have the same cross-section and the cable length is $L$, resistance is proportional to length:

$$x = \frac{Q}{P+Q}\times 2L$$

Here $x$ is the distance from the test end to the fault. The fault resistance carries only battery current, so it does not affect the balance.

Murray loop test for an open-circuit (broken-conductor) fault

A broken core gives no resistance loop, so capacitance is compared instead (the Murray capacitance loop). The capacitance of a core to the sheath is proportional to its length.

1. At the far end, leave the broken core and the sound core open (insulated).
2. Connect the broken core and the sound core to the two ratio arms $P$ and $Q$. Connect an AC source (or a battery with a key and a ballistic galvanometer) between the junction of $P$ and $Q$ and the sheath, and connect a detector between the two cores at the test end.
3. Adjust $P$ and $Q$ for a null. The arm impedances are $1/\omega C_x$ (broken core up to the break, length $x$) and $1/\omega C_L$ (sound core, full length $L$). At balance:

$$\frac{P}{Q} = \frac{1/\omega C_x}{1/\omega C_L} = \frac{C_L}{C_x} \;\Rightarrow\; x = L\,\frac{C_x}{C_L} = L\,\frac{Q}{P}$$

($P$ is the arm joined to the broken core.)

Points to note

• The method needs at least one sound core of the same size.
• It is accurate when the fault resistance is low enough for a sensitive galvanometer reading.
• Fault-to-earth resistance does not affect the result, because it lies in the battery circuit.

With a guard ring, each link pin has capacitance to the tower (earth), $mC$, and to the guard ring (which is at line potential), $gC$. Here $m = 0.1$ and $g = 0.06$.

Junction equation

Number the discs 1 (tower end) to 4 (line end). Let $V$ be the line-to-earth voltage and $S_k = V_1+\dots+V_k$. Current balance at the junction below disc $k$:

$$V_{k+1} = V_k + mS_k - g(V - S_k)$$

Solving in terms of $V_1$

$$\begin{aligned} V_2 &= V_1 + 0.1V_1 - 0.06(V - V_1) = 1.16V_1 - 0.06V\\ V_3 &= V_2 + 0.16(V_1+V_2) - 0.06V = 1.5056V_1 - 0.1296V\\ V_4 &= V_3 + 0.16(V_1+V_2+V_3) - 0.06V = 2.0921V_1 - 0.2199V \end{aligned}$$

Sum $= V$:

$$5.7577V_1 - 0.4095V = V \;\Rightarrow\; V_1 = 0.2448V$$

Voltage distribution

The guard ring makes the distribution much flatter than a plain string. Disc 2 has the smallest share, because the guard ring current is largest near the top.

String efficiency

$$\eta = \frac{V}{4V_4} = \frac{1}{4\times 0.29223} = 85.55\ \%$$

Answer: The discs carry 24.48 %, 22.40 %, 23.90 % and 29.22 % of the phase voltage (tower end to line end), and the string efficiency is 85.55 %.

Underground cable faults are first found to exist with an insulation (megger) test. They are then located by measuring from the cable ends, and finally pinpointed on the route.

Techniques of locating faults

Murray loop test (short-circuit or earth fault)

This is a Wheatstone bridge method. It needs one sound core running alongside the faulty core.

        P              Q
  A ---/\/\--- G --/\/\--- B
  |          (galv)        |
  |                        |
  faulty core (X)     sound core
  |                        |
  +====== loop at far end==+
  |
 fault --> earth
  |
 battery between G-side junction and earth

1. At the far end, connect the faulty core to the sound core with a low-resistance link, so the two cores form a loop.
2. Connect the ratio arms $P$ and $Q$ (variable resistors) across the two test-end terminals and the galvanometer between them. Connect the battery between the junction of $P$ and $Q$ and earth (for an earth fault). For a short circuit between two cores, connect the battery to the other shorted core instead of earth.
3. Adjust $P$ and $Q$ until the galvanometer reads zero.

Let $R_x$ be the resistance of the faulty core from the test end to the fault, and $R_L$ the resistance of the whole loop. At balance:

$$\frac{P}{Q} = \frac{R_L - R_x}{R_x} \;\Rightarrow\; R_x = \frac{Q}{P+Q}\,R_L$$

If both cores have the same cross-section and the cable length is $L$, resistance is proportional to length:

$$x = \frac{Q}{P+Q}\times 2L$$

Here $x$ is the distance from the test end to the fault. The fault resistance carries only battery current, so it does not affect the balance.

Example

Take a 2 km cable (loop length 4 km) that balances at $P = 3\ \Omega$ and $Q = 1\ \Omega$:

$$x = \frac{1}{3+1}\times 4 = 1\ \text{km}$$

So the fault is 1 km from the test end.

Each link pin has capacitance to earth $mC = 0.2C$ and to the line $gC = 0.05C$, where $C$ is the self-capacitance of a unit. Number the discs 1 (tower end) to 4 (line end). Let $V$ be the phase voltage and $S_k = V_1+\dots+V_k$.

 Tower─[C]─A─[C]─B─[C]─D─[C]─Line(V)
           |     |     |
     earth 0.2C each; line 0.05C each

Junction equation

Current from the line side = current to the tower side + current to earth − current received from the line:

$$V_{k+1} = V_k + 0.2S_k - 0.05(V - S_k)$$ $$\begin{aligned} V_2 &= 1.25V_1 - 0.05V\\ V_3 &= V_2 + 0.25(V_1+V_2) - 0.05V = 1.8125V_1 - 0.1125V\\ V_4 &= V_3 + 0.25(V_1+V_2+V_3) - 0.05V = 2.8281V_1 - 0.2031V \end{aligned}$$

Sum $= V$:

$$6.8906V_1 - 0.3656V = V \;\Rightarrow\; V_1 = 0.1982V$$

String efficiency

$$\eta = \frac{V}{4V_4} = \frac{1}{4\times 0.35737} = 69.96\ \%$$

Answer: The string efficiency is 69.96 % (about 70 %).

In a single-core cable, the electric stress is greatest at the conductor surface. For a fixed sheath radius, this maximum stress is lowest when $R/r = e$.

Stress in a single-core cable

Let the conductor radius be $r$, the inside radius of the sheath $R$, and the voltage between core and sheath $V$. The charge per metre is $q$. By Gauss's law, the field at radius $x$ is:

$$g_x = \frac{q}{2\pi\varepsilon_0\varepsilon_r x}$$

Integrating from $r$ to $R$:

$$V = \frac{q}{2\pi\varepsilon_0\varepsilon_r}\ln\frac{R}{r} \;\Rightarrow\; g_x = \frac{V}{x\ln(R/r)}$$

The stress is largest at $x = r$:

$$g_{max} = \frac{V}{r\ln(R/r)}$$

Condition for minimum $g_{max}$

With $V$ and $R$ fixed, $g_{max}$ is least when the denominator $f(r) = r\ln(R/r)$ is greatest:

$$\begin{aligned} \frac{df}{dr} &= \ln\frac{R}{r} + r\cdot\left(-\frac{1}{r}\right) = \ln\frac{R}{r} - 1 = 0\\ \ln\frac{R}{r} &= 1 \;\Rightarrow\; \frac{R}{r} = e = 2.718 \end{aligned}$$

The second derivative, $d^2f/dr^2 = -1/r$, is negative, so this point is a maximum of $f$ and therefore a minimum of $g_{max}$.

Result:

$$r = \frac{R}{e}, \qquad g_{max}\big|_{min} = \frac{V}{r} = \frac{eV}{R}$$

The insulation thickness is then $R - r = r(e-1) = 1.718r$.

Practical note

The conductor radius from $r = R/e$ is often larger than the radius needed to carry the current. In that case a hollow conductor or a larger-diameter (for example aluminium) conductor is used to obtain the optimum radius.

Example

Take $V = 66$ kV (core to sheath) and $R = 2$ cm. Then $r = 2/2.718 = 0.736$ cm and $g_{max} = 66/0.736 = 89.7$ kV/cm.

In EHV lines, each phase is made of two or more sub-conductors (a bundle) held apart by spacers. This is done mainly to reduce corona and line reactance.

• Less corona: A bundle acts like one conductor of much larger effective radius. This lowers the surface voltage gradient, so the corona loss, radio interference and audible noise are reduced.
• Lower inductance: The GMR of the bundle, $D_s^b$ (e.g. $\sqrt{D_s d}$ for two sub-conductors), is much larger than that of one conductor. This lowers the series reactance, so more power can be transmitted ($P \propto 1/X$) with less voltage drop.
• Higher capacitance and SIL: The capacitance rises, so the surge impedance $\sqrt{L/C}$ falls and the surge impedance loading rises.
• Better current capacity and cooling: Several smaller conductors have more surface area than one large conductor and less skin effect.

Example: 400 kV lines usually use twin or quad bundles of ACSR "Moose" conductor.

An overhead line insulator must keep the energised conductor isolated from the earthed tower. It must also carry the mechanical load of the conductor in all weather.

Electrical requirements

• High dielectric strength, so it does not puncture at operating and surge voltages
• High insulation resistance, with very low leakage current over the surface
• A high ratio of puncture strength to flashover voltage (safety factor), so that it flashes over through air rather than breaking down internally
• A non-porous, glazed surface free from impurities and cracks, so that moisture and dirt do not form conducting paths
• Good performance under contamination, with enough creepage distance

Mechanical requirements

• High mechanical (tensile and compressive) strength to carry the conductor weight plus wind and ice loads
• Ability to withstand sudden temperature changes without cracking
• Stable properties over time, with no ageing
• Light weight and easy handling and installation

Porcelain, toughened glass and composite (silicone rubber) insulators are used. Glass shatters visibly when it fails, which makes inspection easy.

With no guard ring, the voltage across the discs increases towards the line, because each link pin leaks charging current to the tower through its shunt capacitance $C_1 = mC$.

Data

Phase voltage:

$$V = \frac{66}{\sqrt 3} = 38.105\ \text{kV}$$

Number the discs 1 (tower end) to 5 (line end). At each junction, the current through the lower disc equals the current through the upper disc plus the current to earth:

$$V_{k+1} = V_k + m(V_1+\dots+V_k)$$

Case 1: $C = 5C_1$, so $m = 0.2$

$$\begin{aligned} V_2 &= 1.2V_1\\ V_3 &= V_2 + 0.2(2.2V_1) = 1.64V_1\\ V_4 &= V_3 + 0.2(3.84V_1) = 2.408V_1\\ V_5 &= V_4 + 0.2(6.248V_1) = 3.6576V_1 \end{aligned}$$

$V = 9.9056V_1$, so $V_1 = 38.105/9.9056 = 3.847$ kV.

$$\eta_1 = \frac{V}{5V_5} = \frac{38.105}{5\times 14.070} = 54.16\ \%$$

Case 2: $C = 6C_1$, so $m = 1/6$

$$\begin{aligned} V_2 &= 1.1667V_1,\quad V_3 = 1.5278V_1\\ V_4 &= 2.1435V_1,\quad V_5 = 3.1165V_1 \end{aligned}$$

$V = 8.9545V_1$, so $V_1 = 4.255$ kV and $V_5 = 13.262$ kV.

$$\eta_2 = \frac{38.105}{5\times 13.262} = 57.47\ \%$$

Answer: The string efficiency is 54.16 % when $C = 5C_1$. Moving the insulators further from the tower lowers the shunt capacitance so that $C = 6C_1$. The efficiency then rises to 57.47 %, and the line-end disc voltage falls from 14.07 kV to 13.26 kV. A lower ratio $m = C_1/C$ always gives a more uniform distribution.

In capacitance grading, each disc gets a different capacitance. The disc nearer the line, which carries more charging current, gets a larger capacitance, so that every disc has the same voltage $v$.

Condition at each junction

Number the discs 1 (top, at the cross-arm) to 4 (line end), with capacitances $C_1 \dots C_4$. Let the pin-to-earth capacitance be $C$.

The junction below disc $k$ is at voltage $kv$ above earth. Current through disc $(k+1)$ = current through disc $k$ + current to earth:

$$\omega C_{k+1}v = \omega C_k v + \omega C(kv) \;\Rightarrow\; C_{k+1} = C_k + kC$$

Calculation (given $C_1 = 7C$)

$$\begin{aligned} C_2 &= C_1 + 1\cdot C = 7C + C = 8C\\ C_3 &= C_2 + 2C = 8C + 2C = 10C\\ C_4 &= C_3 + 3C = 10C + 3C = 13C \end{aligned}$$

Answer: The remaining discs need 8C, 10C and 13C. The string efficiency is then 100 %.

Grading needs discs of different sizes, which makes spares and maintenance difficult. In practice, guard rings are preferred.

A guard ring is a metal ring at line potential placed around the bottom of the string. It gives each link pin a capacitance to the line that partly offsets the pin-to-tower (earth) capacitance. Here the pin-to-tower capacitance is $mC = 0.1C$ and the pin-to-guard-ring capacitance is $gC = 0.05C$.

 Tower ─[C]─ A ─[C]─ B ─[C]─ D ─[C]─ Line (V)
             |       |       |
  to tower  0.1C    0.1C    0.1C
  to ring   0.05C   0.05C   0.05C

Junction equation

Number the discs 1 (tower) to 4 (line end). Let $V$ be the phase voltage and $S_k = V_1+\dots+V_k$. At the junction below disc $k$:

$$\omega C V_{k+1} + \omega gC(V - S_k) = \omega C V_k + \omega mC S_k$$ $$\Rightarrow V_{k+1} = V_k + 0.1S_k - 0.05(V - S_k)$$

Solution

$$\begin{aligned} V_2 &= 1.15V_1 - 0.05V\\ V_3 &= V_2 + 0.15(V_1+V_2) - 0.05V = 1.4725V_1 - 0.1075V\\ V_4 &= V_3 + 0.15(V_1+V_2+V_3) - 0.05V = 2.0159V_1 - 0.1811V \end{aligned}$$

Adding: $5.6384V_1 - 0.3386V = V$, so $V_1 = 0.2374V$.

String efficiency

$$\eta = \frac{V}{4V_4} = \frac{1}{4\times 0.29747} = 84.04\ \%$$

Answer: The voltage distribution is 23.74 %, 22.30 %, 24.21 % and 29.75 % of the phase voltage, and the string efficiency is 84.04 %. (Without the guard ring, with $m = 0.1$, the line-end disc would take 32.6 % and the efficiency would be about 76.6 %.)

The guard ring adds a capacitance $C_x$ between the ring (at line potential) and the link pin at the top of the line-end unit. This capacitance is chosen so that the two units nearest the line share the voltage equally.

 Tower ─[C]─ A ─[C]─ B ─[C]─ Line (20 kV)
             |       |  \
          C/5       C/5  Cx (ring to pin B)
             |       |
           earth   earth

Units: 1 (tower end), 2, 3 (line end), with voltages $V_1, V_2, V_3$. The shunt capacitance is $C/5 = 0.2C$, and the condition is $V_2 = V_3$.

Junction A (no guard ring capacitance)

$$\omega CV_2 = \omega CV_1 + \omega(0.2C)V_1 \;\Rightarrow\; V_2 = 1.2V_1$$

Junction B (guard ring capacitance $C_x$ to line)

The voltage of B to earth is $V_1+V_2$, and to the line it is $V_3$:

$$\omega CV_3 + \omega C_xV_3 = \omega CV_2 + \omega(0.2C)(V_1+V_2)$$

With $V_3 = V_2$:

$$C_x V_2 = 0.2C(V_1 + V_2) \;\Rightarrow\; C_x = 0.2C\left(\frac{1}{1.2} + 1\right) = 0.3667C$$

(i) Voltage on the line-end unit

$$V = V_1 + V_2 + V_3 = V_1 + 1.2V_1 + 1.2V_1 = 3.4V_1 = 20\ \text{kV}$$ $$V_1 = 5.882\ \text{kV},\qquad V_2 = V_3 = 7.059\ \text{kV}$$

String efficiency $= 20/(3\times 7.059) = 94.44\ \%$.

(ii) Capacitance between ring and pin

$$C_x = 0.3667C \;(= 11C/30)$$

Answer: (i) The line-end unit carries 7.06 kV (top unit 5.88 kV, middle unit 7.06 kV). (ii) The ring-to-pin capacitance is $0.367C$.

For a single-core cable, the insulation resistance and the capacitance both depend on $\ln(R/r)$, where $R$ is the outer radius of the insulation.

Data

• $r = 13$ mm
• $R = 13 + 4.6 = 17.6$ mm
• $l = 2$ km $= 2000$ m
• $\rho = 7\times10^{12}\ \Omega$m
• $\varepsilon_r = 3.2$

$$\ln\frac{R}{r} = \ln\frac{17.6}{13} = 0.30295$$

Insulation resistance

$$\begin{aligned} R_{ins} &= \frac{\rho}{2\pi l}\ln\frac{R}{r}\\ &= \frac{7\times10^{12}}{2\pi\times 2000}\times 0.30295\\ &= 1.688\times10^{8}\ \Omega \approx 168.8\ \text{M}\Omega \end{aligned}$$

Capacitance per metre

$$\begin{aligned} C &= \frac{2\pi\varepsilon_0\varepsilon_r}{\ln(R/r)} = \frac{2\pi\times 8.854\times10^{-12}\times 3.2}{0.30295}\\ &= 5.876\times10^{-10}\ \text{F/m} = 0.588\ \text{nF/m} \end{aligned}$$

Answer: The insulation resistance of the 2 km cable is about 168.8 MΩ, and the capacitance is 0.588 nF per metre (about 1.175 µF for the whole 2 km).

The line-end disc carries the largest voltage, so the string is limited when that disc reaches 17.5 kV.

Data

Shunt capacitance $C_1 = 0.125C$, so $m = 0.125$. Number the discs 1 (tower end) to 3 (line end).

Voltage distribution

At junction A:

$$V_2 = V_1 + mV_1 = 1.125V_1$$

At junction B:

$$V_3 = V_2 + m(V_1+V_2) = 1.125V_1 + 0.125(2.125V_1) = 1.3906V_1$$

The line-end disc has the highest voltage. Setting $V_3 = 17.5$ kV:

$$V_1 = \frac{17.5}{1.3906} = 12.584\ \text{kV},\qquad V_2 = 1.125\times12.584 = 14.157\ \text{kV}$$

Maximum voltage of the string

$$V = V_1+V_2+V_3 = 12.584 + 14.157 + 17.5 = 44.24\ \text{kV}$$

This is the phase (line-to-earth) voltage. The corresponding three-phase line voltage is:

$$V_L = \sqrt3\times 44.24 = 76.63\ \text{kV}$$

String efficiency

$$\eta = \frac{V}{3V_3} = \frac{44.24}{3\times17.5} = 84.27\ \%$$

Answer: The string can withstand 44.24 kV to earth (a line voltage of 76.63 kV), with a string efficiency of 84.27 %.

A phase-to-phase (short-circuit) fault between two cores of a multi-core cable can be located by the Murray loop test, a Wheatstone bridge method that needs one sound core.

Connections

        P             Q
 a ---/\/\--- (G) ---/\/\--- c
 |                           |
 faulty core 1 ......... sound core 3
 |           x                |
 |      fault(1-2)            |
 +======= link at far end ====+
 battery: between P-Q junction and core 2

1. Core 1 and core 2 are shorted together at the fault. Core 3 is sound.
2. At the far end, link core 1 to sound core 3 to form a loop.
3. Connect the ratio arms $P$ and $Q$ and the galvanometer across the test ends of cores 1 and 3.
4. Connect the battery between the junction of $P$ and $Q$ and core 2 (the other shorted core). Core 2 takes the place of earth in the earth-fault test. Current then flows through the fault into the loop.
5. Adjust $P$ and $Q$ for zero galvanometer deflection.

Calculation

Let $R_x$ be the resistance of core 1 from the test end to the fault, and $R_L$ the whole loop resistance (core 1 + core 3). At balance:

$$\frac{P}{Q} = \frac{R_L - R_x}{R_x} \;\Rightarrow\; R_x = \frac{Q}{P+Q}R_L$$

For cores of equal size and cable length $L$:

$$x = \frac{Q}{P+Q}\times 2L$$

Example

A 3 km cable balances at $P = 2.5\ \Omega$ and $Q = 1\ \Omega$:

$$x = \frac{1}{3.5}\times 6 = 1.714\ \text{km}$$

The fault resistance lies in the battery circuit, so it does not affect the balance. The Varley loop test can be used in the same way when fine adjustment by a series resistor is preferred.

The greatest working voltage is the voltage at which the line-end disc (the most stressed disc) just reaches its 20 kV limit.

Data

The shunt capacitance is $C/8$, so $m = 0.125$. Number the discs 1 (tower end) to 3 (line end).

Voltage distribution

At junction A:

$$V_2 = V_1(1+m) = 1.125V_1$$

At junction B:

$$V_3 = V_2 + m(V_1+V_2) = 1.125V_1 + 0.125(2.125V_1) = 1.3906V_1$$

Setting $V_3 = 20$ kV:

$$V_1 = \frac{20}{1.3906} = 14.382\ \text{kV},\qquad V_2 = 16.180\ \text{kV}$$

Greatest working voltage

Voltage across the string (phase to earth):

$$V = 14.382 + 16.180 + 20 = 50.56\ \text{kV}$$

For a three-phase line, the line voltage is:

$$V_L = \sqrt3\times 50.56 = 87.58\ \text{kV}$$

String efficiency

$$\eta = \frac{V}{3V_3} = \frac{50.56}{3\times 20} = 84.27\ \%$$

Answer: The greatest working voltage is 50.56 kV across the string (87.58 kV line voltage), and the string efficiency is 84.27 %.

Stranded conductor

A stranded conductor is made of many thin wires twisted together in concentric layers. Each layer is laid in the opposite direction to the one below it. For $n$ layers around a central wire, the number of strands is $3n^2+3n+1$ (7, 19, 37, ...). ACSR is the common example: aluminium strands around a steel core for strength.

• It is flexible and easy to wind on drums, string and sag.
• It does not break easily from vibration.
• A broken strand does not cause the whole conductor to fail.

Bundled conductor

A bundled conductor uses two, three or four sub-conductors per phase, held a short distance apart (30–45 cm) by spacers. Together they act as one phase conductor.

 twin       triple       quad
 o   o        o         o   o
             o o        o   o

Reasons for bundling in EHV lines

• Reduced corona: The effective radius is larger, so the surface voltage gradient falls. Corona loss, radio interference and audible noise are lower.
• Lower reactance: The GMR of the bundle is large ($D_s^b = \sqrt{D_s d}$ for twin, $\sqrt[3]{D_s d^2}$ for triple, $1.09\sqrt[4]{D_s d^3}$ for quad), so the inductive reactance is reduced.
• Higher power capacity: Lower $X$ and higher $C$ reduce the surge impedance, so the SIL and the steady-state stability limit $P_{max} = V_SV_R/X$ increase.
• Better current rating: Several conductors have a larger surface area for cooling and less skin effect.
• Less voltage drop and better regulation, because of the lower reactance.

Typical practice is a twin bundle at 220–400 kV and a quad bundle at 400–765 kV.

Data

• Phase voltage $V = 66/\sqrt3 = 38.105$ kV
• Self-capacitance $C = 10C_1$, so $m = C_1/C = 0.1$
• Discs numbered 1 (tower end) to 4 (line end)

(i) Voltage distribution

At each junction, the current through the lower disc equals the current through the upper disc plus the current to earth:

$$V_{k+1} = V_k + m(V_1 + \dots + V_k)$$ $$\begin{aligned} V_2 &= 1.1V_1\\ V_3 &= V_2 + 0.1(2.1V_1) = 1.31V_1\\ V_4 &= V_3 + 0.1(3.41V_1) = 1.651V_1 \end{aligned}$$ $$V = (1 + 1.1 + 1.31 + 1.651)V_1 = 5.061V_1 \;\Rightarrow\; V_1 = \frac{38.105}{5.061} = 7.529\ \text{kV}$$

(ii) String efficiency

$$\eta = \frac{V}{4V_4} = \frac{38.105}{4\times 12.431} = 76.64\ \%$$

Answer: The disc voltages are 7.53, 8.28, 9.86 and 12.43 kV (tower to line end), and the string efficiency is 76.64 %.