Conductor and Support Selection

When the two supports are at different heights, the lowest point of the conductor is not at mid-span. It shifts towards the lower support. Its position is found from the parabolic approximation.

Derivation

Let:

• $L$ = span (horizontal distance between supports A and B)
• $h$ = difference in level between the supports (B higher than A)
• $w$ = weight of conductor per unit length, and $T$ = horizontal tension
• $x_1$, $x_2$ = horizontal distances of the lowest point O from A (lower) and B (higher), so $x_1 + x_2 = L$

                                B
                              /|
                    S2      /  |
     A                    /    |  h
     |\               /        |
  S1 | \_____ O _____/  .......|
     |<- x1 ->|<----- x2 ---->|

Taking O as the origin, the conductor is the parabola $y = \dfrac{w x^2}{2T}$. So:

$$S_1 = \frac{w x_1^2}{2T}, \qquad S_2 = \frac{w x_2^2}{2T}$$ $$\begin{aligned} h &= S_2 - S_1 = \frac{w}{2T}(x_2^2 - x_1^2) \\ &= \frac{w}{2T}(x_2 + x_1)(x_2 - x_1) = \frac{wL}{2T}(x_2 - x_1) \\ \Rightarrow x_2 - x_1 &= \frac{2Th}{wL} \end{aligned}$$

Solving with $x_1 + x_2 = L$:

$$x_1 = \frac{L}{2} - \frac{Th}{wL}, \qquad x_2 = \frac{L}{2} + \frac{Th}{wL}$$

The lowest point (point of maximum sag) is at $x_1$ from the lower tower and $x_2$ from the higher tower. The sags are $S_1 = w x_1^2/2T$ below A and $S_2 = w x_2^2/2T$ below B. Under wind and ice, $w$ is replaced by the resultant loading $w_r$.

If the supports are at equal height ($h = 0$), $x_1 = x_2 = L/2$ and $S = wL^2/8T$.

Virtual sag

• If $h > \dfrac{wL^2}{2T}$, then $x_1 = \dfrac{L}{2} - \dfrac{Th}{wL}$ becomes negative. The lowest point O of the parabola then lies outside the span, beyond the lower support.
• The conductor rises all the way from A to B and has no real lowest point inside the span. The sag $S_1$ measured to the imaginary point O is called the virtual sag.
• This happens on steep hill slopes, which are common in Nepal. The conductor pulls upward on the lower tower (uplift). That tower must be designed as a tension tower, or guyed, and the string must be a tension string, because a suspension string would be pulled up.

A conductor for an HV line must carry the required power within safe temperature, loss, voltage-drop and corona limits. The electrical considerations are:

1. Current-carrying capacity (thermal limit / ampacity): the conductor must carry the maximum load current, including contingency and overload, without exceeding its maximum allowable temperature (about 75–85 °C for ACSR). Above that temperature the conductor anneals and loses strength, and its sag increases. Ampacity depends on ambient temperature, wind speed, solar radiation and emissivity: $I^2R =$ heat lost by convection and radiation minus solar heat gained.
2. Line losses and efficiency: $I^2R$ loss falls as the conductor area increases. Efficiency should usually be at least 94–95%. The economic choice balances the cost of losses against the capital cost (Kelvin's law and present-worth comparison).
3. Voltage regulation: the drop $IR\cos\phi + IX\sin\phi$ must keep regulation within limits (about 10–12% for transmission). $R$ falls with a larger area, while $X$ falls only slightly with diameter and more with bundling.
4. Corona: the conductor surface gradient must stay below the corona inception gradient in fair weather. Corona causes loss, radio and TV interference and audible noise. It needs a minimum diameter (about 1 cm at 132 kV, 1.75–2 cm at 220 kV) or bundled conductors at 400 kV and above. Check using Peek's formula $V_d = 21.1\, m\,\delta\, r \ln(D/r)$ kV.
5. Stability and power-transfer limit: a lower reactance (larger or bundled conductor) raises $P_{max} = V_sV_r/X$ and the SIL, which matters for long lines.
6. Skin and proximity effect: AC resistance is higher than DC resistance. ACSR places the aluminium outside the steel core, which suits the skin effect.
7. Short-circuit rating: the conductor must withstand the fault current for the fault duration without damage.
8. Radio interference and audible noise limits: these set the minimum diameter and the bundle design at EHV.
9. Resistivity of material: aluminium (or ACSR, AAAC) is chosen over copper for lower cost and weight at equal conductance.

The final conductor is the cheapest one that satisfies all these electrical checks together with the mechanical checks (tension, sag, vibration).

TRUE.

In steady state, the heat produced in the conductor equals the heat it loses:

$$I^2R + q_{solar} = q_{conv} + q_{rad}$$

where $q_{conv} \propto (\theta_c - \theta_a)$ and $q_{rad} \propto (T_c^4 - T_a^4)$.

The conductor temperature $\theta_c$ is limited to a fixed maximum (for example 75–80 °C for ACSR). When the ambient temperature $\theta_a$ rises, the difference $(\theta_c - \theta_a)$ falls. So less heat can be lost by convection and radiation, and a smaller $I^2R$ is allowed. Neglecting the solar term:

$$I \propto \sqrt{\theta_c - \theta_a}$$

Example: a conductor rated 400 A at 35 °C ambient with an 80 °C limit can carry only $400\sqrt{(80-40)/45} = 377$ A at 40 °C ambient. Hence ampacity decreases as ambient temperature increases. Line ratings are therefore given for summer conditions.

TRUE (for long lines).

The power a line can carry is limited by three criteria (St. Clair loadability curve):

The stability limit $P = \dfrac{V_sV_r}{X}\sin\delta$ (with $\delta$ kept to about 30°–35°) falls as the line length increases, because $X$ grows with length. The thermal limit $\sqrt{3}VI_{rated}$ does not depend on length.

For a long line, the stability-limited power is therefore much smaller than the thermal capacity of the chosen conductor. If the line is designed so that the stability criterion is met, the current is well below the conductor's ampacity, so the thermal criterion is automatically met.

The converse is not true. A long line that meets the thermal criterion may still fail the stability criterion.

Corona is the partial breakdown (ionisation) of the air immediately around a conductor. It happens when the electric field at the conductor surface exceeds the dielectric strength of air, about 30 kV/cm peak (21.1 kV/cm rms) at standard conditions. It shows as a violet glow and a hissing noise, produces ozone, and causes power loss and radio interference.

The corona inception (disruptive critical) voltage $V_d$ is the minimum phase-to-neutral voltage at which this ionisation starts.

Derivation

Consider a single-phase line, or a symmetrical 3-phase line taken to neutral. The conductor radius is $r$ and the spacing is $D$, with $D \gg r$.

1. With charge $q$ per metre on the conductor, the field at distance $x$ is

$$E_x = \frac{q}{2\pi\varepsilon_0 x}$$

2. The voltage between the conductor and the neutral plane is

$$V = \int_r^D E_x\,dx = \frac{q}{2\pi\varepsilon_0}\ln\frac{D}{r}$$

3. Eliminate $q$. The field is greatest at the conductor surface ($x = r$):

$$E_{max} = \frac{q}{2\pi\varepsilon_0 r} = \frac{V}{r\ln(D/r)}$$

4. Corona starts when $E_{max}$ reaches the breakdown strength of air $g_0$:

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

with $g_0 = 30$ kV/cm peak, or 21.1 kV/cm rms. 5. Air density correction: breakdown strength is proportional to air density, so $g_0$ is replaced by $g_0\delta$, where

$$\delta = \frac{3.92\,b}{273 + t}$$

($b$ in cm of Hg, $t$ in °C; $\delta = 1$ at 76 cm Hg and 25 °C). 6. Surface irregularity factor $m_0$: 1 for a polished wire, 0.98–0.93 for a rough wire, 0.87–0.8 for a stranded conductor.

$$V_d = 21.1\, m_0\, \delta\, r \ln\frac{D}{r}\ \text{kV (rms, phase)}$$

with $r$ and $D$ in the same units and $r$ in cm. For a 3-phase line with unequal spacing, $D$ is replaced by the equivalent spacing $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$.

Visual corona starts at a slightly higher voltage, $V_v = 21.1\,m_v\,\delta\, r\left(1 + \dfrac{0.3}{\sqrt{\delta r}}\right)\ln\dfrac{D}{r}$.

FALSE.

For lines of 400 kV and above, the horizontal configuration is generally adopted for single-circuit lines. The vertical configuration is used for double-circuit lines up to 220 kV, and occasionally for 400 kV double circuits.

Reasons for the horizontal configuration at EHV:

1. Lower tower height: all phases are on one level. A vertical arrangement would stack three large phase spacings (8–12 m each) plus long strings, making the tower very tall and heavy.
2. Smaller overturning moment: the conductors are lower, so the moment from wind is less and the foundations are cheaper.
3. Lightning performance: low conductors and two earth wires give better shielding and fewer strikes.
4. Galloping and ice shedding: with no conductor directly above another, a conductor jumping after ice shedding cannot clash with the phase above.
5. Maintenance: the lower height makes stringing and live-line work easier.

The disadvantage is a wider right-of-way. Where land is costly, a vertical or delta double-circuit tower is chosen instead.

Kelvin's law: the most economical conductor cross-section is the one for which the annual cost of energy lost in the conductor equals the annual interest and depreciation on the part of the capital cost that is proportional to the cross-section.

Proof

Let $a$ be the conductor cross-sectional area.

1. Capital cost of the line = $P_1 + P_2 a$.
   • $P_1$ is the fixed part: towers, insulators, erection.
   • $P_2 a$ is the part proportional to area: conductor material and the extra tower strength.
2. With a rate $r$ of annual interest plus depreciation, the annual capital charge is

$$C_1 = r(P_1 + P_2 a) = P_1' + P_2' a$$

3. Energy loss: resistance $R = \rho l/a$. With an equivalent loss current $I$ over a year of $t$ hours, the energy lost is $3I^2\rho l t / a$. At a cost of $k$ per kWh:

$$C_2 = \frac{P_3}{a}$$

4. Total annual cost:

$$C = P_1' + P_2' a + \frac{P_3}{a}$$

5. For minimum cost:

$$\begin{aligned} \frac{dC}{da} &= P_2' - \frac{P_3}{a^2} = 0 \\ \Rightarrow P_2' a &= \frac{P_3}{a} \end{aligned}$$

Also $d^2C/da^2 = 2P_3/a^3 > 0$, so this is a minimum.

So at the most economical area, the variable annual capital charge equals the annual cost of energy loss, which proves the law. The economical area is

$$a = \sqrt{\frac{P_3}{P_2'}}$$

and the economical current density is $J = I/a$.

 cost
  |\                       / total
  | \ energy loss       /
  |  \  (P3/a)  ___/__
  |   \     _--   /
  |    `-.__     / capital (P2'a)
  |         ``--/--.____
  +-------------+---------> area a
            a_economical

Limitations

1. Interest and depreciation are hard to estimate exactly. Costs, interest rates and energy prices change over the life of the line.
2. The load varies, so the loss (equivalent) current is uncertain. Load growth is not considered.
3. It considers economy only. It ignores the technical limits of voltage regulation, corona, thermal ampacity, stability and mechanical strength. The economic size may be too small to meet these limits.
4. The conductor cost is not exactly proportional to area. Standard sizes are discrete, and the installation cost is not linear.
5. It does not include the cost of the extra generation capacity needed to supply the losses.
6. The capital cost of towers does not depend linearly on conductor area, especially when wind and ice loading govern the design.

For these reasons Kelvin's law is used only as a first guide. The final conductor is chosen by technical checks plus a present-worth comparison of standard conductors.

FALSE.

The height of a tower depends on several factors, not on the voltage alone:

$$H_t = h_g + S_{max} + \text{(vertical spacing between conductors)} + h_{ew}$$

The voltage affects only the clearances and the string length. Sag, which depends on span, conductor and temperature, often adds more height than the voltage terms do. The configuration (single or double circuit, vertical or horizontal) also has a large effect.

For example, at the same 132 kV, a 350 m span needs a taller tower than a 250 m span, because the sag is larger. A double-circuit vertical tower is taller than a single-circuit horizontal tower.

FALSE.

Corona starts when the phase voltage exceeds the disruptive critical voltage

$$V_d = 21.1\,m_0\,\delta\,r\ln\frac{D}{r}\ \text{kV}$$

• On LV and MV lines (up to 33 kV), the operating voltage is far below $V_d$ even for small conductors. For example, a 10 mm conductor at 1 m spacing has $V_d \approx 21.1 \times 0.85 \times 0.5\ln(200) \approx 47.5$ kV per phase, which is well above 33 kV/$\sqrt{3}$. So corona is practically absent.
• Corona becomes a governing design criterion only for EHV/UHV lines (220 kV and above). There it sets:
  • the minimum conductor diameter,
  • the use of bundled conductors (400 kV and above),
  • limits on radio interference and audible noise.

LV line design is governed instead by voltage drop, current-carrying capacity and mechanical strength.

Method. For each span, the transverse load at the tower comes from two sources:

• wind on the conductors and earth wires: $F = p \times d \times S$, taking the wind span equal to the span $S$ and using the full projected area;
• the resultant of the conductor tensions at a line deviation $\theta$: $F_d = 2T\sin(\theta/2)$.

These loads act at the conductor heights. Their overturning moment at ground level is

$$M = n_c\,(F_c + F_{dc})(h_1 + h_2 + h_3) + n_e\,(F_e + F_{de})\,H_t$$

where $n_c$ is the number of conductors at each level and $n_e$ the number of earth wires. The tower weight is found from Ryle's formula

$$W = K\,H_t\sqrt{M}$$

$K$ is the same for all spans, so it cancels in the comparison. Tower cost is proportional to tower weight, so the span that gives the smallest total tower weight is the most economical span (with the conductor already fixed). Wind on the tower body is neglected.

Data and deviation loads

• Double circuit: 2 power conductors at each of $h_1$, $h_2$, $h_3$, so $n_c = 2$. Two earth wires at $H_t$, so $n_e = 2$.
• Wind pressure $p = 100$ kg/m², $d_c = 0.02597$ m, $d_e = 0.0146$ m.
• All towers are designed for $\theta = 5°$:

$$\begin{aligned} F_{dc} &= 2 \times 6890 \times \sin 2.5° = 601.08\ \text{kg} \\ F_{de} &= 2 \times 3332 \times \sin 2.5° = 290.68\ \text{kg} \end{aligned}$$

Wind loads per span

Overturning moments and tower weight index

Sample calculation (250 m):

$$\begin{aligned} M_c &= 2(649.25 + 601.08)(9.70 + 14.52 + 19.35) = 108{,}953.3\ \text{kg·m} \\ M_e &= 2(365.0 + 290.68)(24.84) = 32{,}574.2\ \text{kg·m} \\ M &= 141{,}527.5\ \text{kg·m}, \quad H_t\sqrt{M} = 24.84\sqrt{141527.5} = 9{,}344.8 \end{aligned}$$

Tower steel per km

Towers per km $= 1000/S$, so the relative cost per km $= (1000/S) \times H_t\sqrt{M}$ (in units of $K$):

Moving from 250 m to 300 m saves more on the number of towers than it adds in tower weight. Beyond 300 m, the heavier and taller towers outweigh the saving.

Answer: the most economical span is 300 m, which has the lowest relative tower cost per km (36,578.1 K). Insulator and foundation costs per km also fall with longer spans, but the spread between 250 m and 350 m is small, so 300 m is a sound choice.

TRUE.

The corona inception (disruptive critical) voltage is

$$V_d = 21.1\,m_0\,\delta\,r\ln\frac{D}{r}\ \text{kV (rms per phase)}$$

• $V_d$ is almost proportional to the radius $r$. The log term changes only slowly: as $r$ rises, $\ln(D/r)$ falls a little, but the product $r\ln(D/r)$ still increases strongly.
• Increasing the spacing $D$ raises $V_d$ only through the logarithm, so it is much less effective and makes the tower bigger.
• A larger diameter therefore lowers the surface gradient $E = V/(r\ln(D/r))$ for the same voltage, which is the most effective way to raise $V_d$.

In practice, the larger effective radius is obtained with:

• ACSR with more aluminium strands,
• expanded (hollow) conductors,
• bundled conductors (2–4 sub-conductors at 400 kV and above), which give a large equivalent radius without a very heavy single conductor.

Smoother surfaces (higher $m_0$) also help.

FALSE.

The tower height is

$$H_t = h_g + S_{max} + \text{(vertical spacing between conductors)} + h_{ew}$$

The maximum sag at mid-span between supports at equal level is

$$S_{max} = \frac{w\,L^2}{8T}$$

It grows with the square of the span $L$. A longer span therefore needs a taller tower to keep the same minimum ground clearance.

For example, in a typical design table, increasing the span from 250 m to 350 m raises the maximum sag from about 3.1 m to 5.5 m. The tower height rises by about 2.4 m (28.8 m to 31.2 m).

This is why fewer but taller and heavier towers are traded against more, shorter towers when finding the most economical span.

For supports at different levels, the conductor is approximated by a parabola with its lowest point O. The "maximum sag" from each support is measured down to O.

Let $L$ be the span, $h$ the difference in support levels (B higher), $w$ the weight per metre and $T$ the horizontal tension. Let $x_1$ and $x_2$ be the horizontal distances of O from the lower support A and the higher support B.

Expressions for maximum sag

Taking O as the origin, $y = \dfrac{w x^2}{2T}$, so

$$S_1 = \frac{w x_1^2}{2T}\ \text{(from A)}, \qquad S_2 = \frac{w x_2^2}{2T}\ \text{(from B)}$$

Location of maximum sag

The difference of the sags is the difference in support levels:

$$\begin{aligned} h &= S_2 - S_1 = \frac{w}{2T}\left(x_2^2 - x_1^2\right) \\ &= \frac{w}{2T}(x_2 + x_1)(x_2 - x_1) \end{aligned}$$

Since $x_1 + x_2 = L$:

$$x_2 - x_1 = \frac{2Th}{wL}$$

Adding and subtracting with $x_1 + x_2 = L$:

$$x_1 = \frac{L}{2} - \frac{Th}{wL}, \qquad x_2 = \frac{L}{2} + \frac{Th}{wL}$$

The point of maximum sag therefore lies $\dfrac{Th}{wL}$ from mid-span, towards the lower support.

• With equal supports ($h = 0$): $x_1 = x_2 = L/2$.
• If $h > wL^2/2T$, then $x_1 < 0$ and the lowest point falls outside the span (virtual sag). The lower tower then has uplift.
• Under wind and ice, $w$ is replaced by the resultant load per metre.

Ampacity follows from the heat balance of the conductor. Neglecting solar heating, and with the resistance taken at the maximum temperature as constant:

$$I^2R = h\,A_s(\theta_c - \theta_a) \quad\Rightarrow\quad I \propto \sqrt{\theta_c - \theta_a}$$

New current-carrying capacity

Original: $\theta_c - \theta_a = 80 - 35 = 45\ °\text{C}$ at $I_1 = 400$ A.

New: $\theta_c - \theta_a = 75 - 40 = 35\ °\text{C}$.

$$\begin{aligned} I_2 &= I_1\sqrt{\frac{35}{45}} \\ &= 400 \times 0.8819 \\ &= 352.8\ \text{A} \end{aligned}$$

Conductor temperature at 320 A

The temperature rise is proportional to $I^2$. Using the new rating (35 °C rise at 352.8 A) with ambient 40 °C:

$$\begin{aligned} \theta_c - 40 &= 35\left(\frac{320}{352.77}\right)^2 = 45\left(\frac{320}{400}\right)^2 = 28.8\ °\text{C} \\ \theta_c &= 40 + 28.8 = 68.8\ °\text{C} \end{aligned}$$

Answer: ampacity = 352.8 A (about 353 A); conductor temperature at 320 A = 68.8 °C (below the 75 °C limit).

The small change of resistance with temperature and the solar heat gain are neglected.

An overhead line conductor should have:

1. High electrical conductivity (low resistivity), for low losses and voltage drop.
2. High tensile strength, to support long spans and wind and ice loads.
3. Low weight per unit length (low density), for less sag and lighter towers.
4. Low cost and easy availability.
5. Low coefficient of thermal expansion, so that sag changes little with temperature.
6. Resistance to corrosion and weather.
7. A large enough diameter to limit corona at high voltage.
8. Easy jointing, and good resistance to fatigue and vibration.

Aluminium vs copper

Aluminium (as ACSR or AAAC) is preferred for overhead transmission. It is lighter and cheaper, and its larger diameter reduces corona. The steel core of ACSR makes up for its low strength. Copper is used mainly where space is limited (cables, substations).

The tower dimensions follow from the minimum air clearance, the insulator string length and the 45° swing. They are taken from the given appendix rules and common IS practice.

Data assumed: maximum system voltage $V_m = 245$ kV. Tower width at cross-arm level is 1.5 m. Discs are 254 × 154 mm.

Minimum air clearance $a$

Using 6.5 inch per 10 kV of maximum phase rms voltage plus 8 inch:

$$\begin{aligned} V_{ph} &= \frac{245}{\sqrt{3}} = 141.45\ \text{kV} \\ a &= 6.5 \times 14.145 + 8 = 99.94\ \text{in} = 2.54\ \text{m} \end{aligned}$$

As a check, 1 cm per kV of phase peak voltage plus 30 cm gives $200 + 30 = 230$ cm. The larger value, $a = 2.54$ m, is used.

(ii) Insulator string length

For 220 kV, 10 discs meet the wet (395 kV) and impulse (900 kV) withstand levels. One spare disc is added, giving 11 discs:

$$l = 11 \times 0.154 + 0.30\ (\text{hardware}) \approx 2.0\ \text{m}$$

(i) Cross-arm length

When the string swings by $\theta = 45°$, the conductor must stay at least $a$ from the tower:

$$\begin{aligned} L_{ca} &= a + l\sin\theta \\ &= 2.54 + 2.0 \times 0.707 = 3.95\ \text{m} \end{aligned}$$

Take 4.0 m from the tower face.

(iii) Conductor separation

• Vertical (between cross-arms): the conductor hanging from the upper arm must keep clearance $a$ from the arm below.

$$V = l + a = 2.0 + 2.54 = 4.54\ \text{m} \approx 4.6\ \text{m}$$

• Horizontal (conductors on opposite sides of the tower):

$$H = 2L_{ca} + w = 2 \times 4.0 + 1.5 = 9.5\ \text{m}$$

This is well above the phase-to-phase clearance (about $1.15a \approx 2.9$ m).

(iv) Number of earth wires

A 220 kV single-circuit line in a vertical or delta configuration is narrow enough for one earth wire at the peak to give a 30° shielding angle.

(v) Height of earth wire above the top conductor

For a shielding angle $\alpha = 30°$, the horizontal distance from the tower centre to the outermost conductor is $4.0 + 0.75 = 4.75$ m. So:

$$y = \frac{4.75}{\tan 30°} = 8.23\ \text{m} \approx 8.3\ \text{m}$$

The line is 130 km long, so it is a medium line and the nominal-π model is used. The two circuits are in parallel.

Assumptions: power factor 0.9 lagging (not given in the question); receiving-end voltage held at 220 kV.

Line parameters (per phase)

Per circuit:

$$\begin{aligned} R &= 0.0544 \times 130 = 7.072\ \Omega \\ X &= 2\pi(50)(0.494\times10^{-3})(130) = 20.175\ \Omega \\ Y &= 2\pi(50)(23.1\times10^{-9})(130) = 9.434\times10^{-4}\ \text{S} \end{aligned}$$

Two circuits in parallel:

$$Z = \frac{7.072 + j20.175}{2} = 3.536 + j10.088\ \Omega, \qquad Y = j1.887\times10^{-3}\ \text{S}$$

Constants and currents

$$\begin{aligned} A &= 1 + \frac{ZY}{2} = 0.9905\angle 0.19° \\ V_R &= \frac{220}{\sqrt{3}} = 127.02\angle 0°\ \text{kV} \\ I_R &= \frac{144\times10^6}{\sqrt{3}\times220\times10^3\times0.9} = 419.9\angle -25.84°\ \text{A} \end{aligned}$$ $$\begin{aligned} V_S &= AV_R + ZI_R = 129.04\angle 1.59°\ \text{kV/phase} \\ &= 223.51\ \text{kV (line)} \\ I_S &= CV_R + AI_R = 379.1\angle 8.88°\ \text{A} \end{aligned}$$

a) Efficiency

$$\begin{aligned} P_S &= \text{Re}(3V_SI_S^*) = 145.56\ \text{MW} \\ \text{Loss} &= 1.56\ \text{MW} \\ \eta &= \frac{144}{145.56}\times100 = 98.93\% \end{aligned}$$

b) Voltage regulation

At no load, $V_{R0} = V_S/|A|$:

$$\begin{aligned} V_{R0} &= \frac{223.51}{0.9905} = 225.65\ \text{kV (line)} \\ \%VR &= \frac{225.65 - 220}{220}\times100 = 2.57\% \end{aligned}$$

c) Receiving-end voltage rise at no load (Ferranti effect)

With $V_S$ held at 223.51 kV and the load removed:

$$\Delta V = V_{R0} - V_S = 225.65 - 223.51 = 2.15\ \text{kV}$$

The no-load voltage is 5.65 kV above the rated 220 kV.

Answer: η = 98.93%, VR = 2.57%, no-load rise = 2.15 kV above $V_S$ (receiving end 225.65 kV).

The line already meets η > 94% and VR < 12% by a wide margin.

Options if the limits were not met

1. Larger conductor (lower $R$) for higher efficiency, or bundled conductors (lower $X$) for better regulation.
2. Raise the power factor with shunt capacitors or SVCs at the load end. At unity pf the regulation falls to about 1.1%.
3. Series capacitor compensation to cut $X$ and the voltage drop.
4. More circuits or a higher voltage (for example 400 kV), which reduces the current and therefore $I^2R$ and $IX$.
5. Shunt reactors at the receiving end to limit the no-load (Ferranti) rise.
6. On-load tap-changing transformers to correct the receiving-end voltage.

The most economical conductor is the one with the lowest total annual cost per km. This is the annual charge on the capital (towers plus conductor), spread over the project life at the given interest rate, plus the annual cost of energy lost.

Capital recovery factor

With $i = 14\%$ and $n = 25$ years:

$$\begin{aligned} CRF &= \frac{i(1+i)^n}{(1+i)^n - 1} \\ &= \frac{0.14(1.14)^{25}}{(1.14)^{25} - 1} = 0.14550 \end{aligned}$$

Annual cost of energy loss

Column C is in thousands of kWh per km per year, and the energy rate is Rs 9 per kWh. So the loss cost in thousand Rs is $9C$. For Goat: $32 \times 9 = 288$ thousand Rs.

Comparison (per km)

Sample calculation for Deer: $(910 + 1663) \times 0.14550 = 374.37$, and $374.37 + 20 \times 9 = 554.37$ thousand Rs.

Answer: Deer is the most economical conductor, with a total annual cost of about Rs 554.37 thousand per km. Its higher capital cost is more than recovered by its much lower energy loss.

If the energy loss is instead worked as a present worth (annual loss cost ÷ CRF, added to the capital), the ranking is the same, because both methods differ only by the factor CRF.

Method. For each span, the transverse load at the tower comes from two sources:

• wind on the conductors and earth wires: $F = p \times d \times S$, taking the wind span equal to the span $S$ and using the full projected area;
• the resultant of the conductor tensions at a line deviation $\theta$: $F_d = 2T\sin(\theta/2)$.

These loads act at the conductor heights. Their overturning moment at ground level is

$$M = n_c\,(F_c + F_{dc})(h_1 + h_2 + h_3) + n_e\,(F_e + F_{de})\,H_t$$

where $n_c$ is the number of conductors at each level and $n_e$ the number of earth wires. The tower weight is found from Ryle's formula

$$W = K\,H_t\sqrt{M}$$

$K$ is the same for all spans, so it cancels in the comparison. Tower cost is proportional to tower weight, so the span that gives the smallest total tower weight is the most economical span (with the conductor already fixed). Wind on the tower body is neglected.

Assumptions

• Single circuit: one conductor at each of $H_1$, $H_2$, $H_3$ ($n_c = 1$), and one earth wire at $H_t$ ($n_e = 1$).
• Standard tower types (IS 802 practice): type A is a suspension tower for 0°–2° deviation and type B is a small-angle tower for 2°–15°. Each type is designed for its maximum angle.
• Number of towers for 180 km: $N = \lceil 180000/S \rceil + 1$, with 80% type A and 20% type B.

Deviation loads

Wind loads ($p = 100$ kg/m²)

Moments and tower weight index

Sample calculation (250 m, type A):

$$\begin{aligned} M_c &= (630 + 239.45)(10.22 + 13.60 + 16.98) = 35{,}473.4\ \text{kg·m} \\ M_e &= (410 + 99.13)(28.84) = 14{,}683.3\ \text{kg·m} \\ H_t\sqrt{M} &= 28.84\sqrt{50156.7} = 6{,}458.9 \end{aligned}$$

Total tower steel for the line

Weighted index $= 0.8(H_t\sqrt{M})_A + 0.2(H_t\sqrt{M})_B$, multiplied by the number of towers $N$:

The total tower weight keeps falling up to the largest span tabulated.

Answer: the most economical span is 350 m (lowest total tower weight index, 4.846 × 10⁶ K, with 516 towers).

FALSE.

Tower strength must be enough for the worst (maximum) loading the tower may face during its life. So the maximum design wind speed or pressure of the zone is used (for example the basic wind speed with a return period of 50 years or more, as in IS 802), together with the minimum temperature and ice where applicable.

• Wind on the conductors, earth wires, insulators and tower body gives the transverse load and the overturning moment. Both rise with the square of the wind speed.
• Minimum wind speed would underestimate the loads and the tower could collapse in a storm.

Minimum (low) wind speed is used for a different purpose: ampacity (thermal rating) calculations, where low wind means poor cooling and is the conservative case. It is also used in the still-air, maximum-temperature condition that gives the maximum sag for ground clearance.

The conductor must pass three checks: the thermal limit, efficiency ≥ 94% and regulation ≤ 12%. The smallest standard ACSR that passes all three is chosen.

Assumptions (not given in the question):

• Power factor 0.9 lagging.
• Equivalent spacing $D_{eq} = 5.0$ m, typical for a 132 kV single-circuit tower.
• Resistance at 80 °C: $R_{80} = R_{20}[1 + 0.004(80 - 20)] = 1.24R_{20}$.
• GMR $= 0.7788r$.
• 70 km is treated as a medium line using the nominal-π model.
• Ambient 50 °C and conductor 80 °C, as in the table, so the tabulated current rating is used directly.

(i) Thermal limit

$$I = \frac{P}{\sqrt{3}V\cos\phi} = \frac{100\times10^6}{\sqrt{3}\times132\times10^3\times0.9} = 486.0\ \text{A}$$

The rating must be at least 486 A. Bear (480 A) and smaller conductors are rejected. Goat (543 A) is the smallest conductor that passes.

Line constants (per km)

$$\begin{aligned} X &= 2\pi f\left(2\times10^{-7}\ln\frac{D}{0.7788r}\right)\times1000\ \Omega/\text{km} \\ B &= 2\pi f\,\frac{2\pi\varepsilon_0}{\ln(D/r)}\times1000\ \text{S/km} \end{aligned}$$

For Deer ($d = 29.89$ mm, $r = 1.4945$ cm):

• $R_{80} = 0.0673 \times 1.24 = 0.0835\ \Omega$/km
• $X = 0.3809\ \Omega$/km
• $B = 3.007\ \mu$S/km

(ii) and (iii) Efficiency and regulation

For each conductor:

• $Z = (R + jX)\times70$, $Y = jB\times70$, $A = 1 + ZY/2$
• $V_S = AV_R + ZI_R$, $I_S = CV_R + AI_R$, with $V_R = 132/\sqrt{3}$ kV
• $\eta = P_R/P_S$ and $VR = (V_S/|A| - V_R)/V_R$

For Deer: $V_S = 146.96$ kV (line), loss = 4.081 MW, so $\eta = 100/(100 + 4.081) = 96.08\%$ and $VR = 11.65\%$.

Selection

• Goat passes the thermal and efficiency checks but fails regulation (12.88% > 12%).
• Sheep and Bison are marginal, at about 12.2%.
• Deer passes all three: rating 643 A > 486 A, η = 96.08% > 94%, VR = 11.65% < 12%.

Answer: ACSR Deer (30/4.27 + 7/4.27 mm) is the suitable conductor.

If a smaller spacing (about 4 m) is used, Sheep and Bison would just meet the 12% limit (VR about 11.9%), and they could be compared with Deer on cost.

FALSE.

Ampacity is the largest current the conductor can carry continuously without exceeding its maximum allowable temperature. It comes from the heat balance:

$$I^2R + q_{solar} = q_{convection} + q_{radiation}$$

Convective cooling rises strongly with wind speed. If the maximum wind speed were assumed, the cooling would be overestimated. The rating would be too high, and on calm days the conductor would overheat, anneal and sag below the safe ground clearance.

So, for a safe (conservative) rating, a low wind speed is assumed, typically 0.5–0.6 m/s (as in the given ACSR table, 0.6 m/s). It is combined with high ambient temperature and full solar radiation.

The maximum wind speed is used instead for mechanical design: tower loads, conductor tension and insulator swing.

The tower height is built up from the ground clearance, the sag and the conductor spacing. The tower weight is then found from the overturning moment of the wind load with Ryle's formula, and the total tower steel per km is compared.

Tower heights

With 6 m minimum ground clearance and conductors 2 m apart vertically (from the figure), and the ground wire neglected:

Working tension $T = \text{UTS}/\text{FOS} = 15910/2 = 7955$ kg. All towers are straight-line towers, so there is no deviation load. $T$ is needed only to check that the given sags are consistent with it. The 8.7 m horizontal width is the same for every span, so it does not affect the comparison.

Method. For each span, the transverse load at the tower comes from two sources:

• wind on the conductors and earth wires: $F = p \times d \times S$, taking the wind span equal to the span $S$ and using the full projected area;
• the resultant of the conductor tensions at a line deviation $\theta$: $F_d = 2T\sin(\theta/2)$.

These loads act at the conductor heights. Their overturning moment at ground level is

$$M = n_c\,(F_c + F_{dc})(h_1 + h_2 + h_3) + n_e\,(F_e + F_{de})\,H_t$$

where $n_c$ is the number of conductors at each level and $n_e$ the number of earth wires. The tower weight is found from Ryle's formula

$$W = K\,H_t\sqrt{M}$$

$K$ is the same for all spans, so it cancels in the comparison. Tower cost is proportional to tower weight, so the span that gives the smallest total tower weight is the most economical span (with the conductor already fixed). Wind on the tower body is neglected.

Wind load on each conductor ($p = 100$ kg/m², $d = 0.028$ m)

Overturning moment and tower weight index

$M = F_c(h_1 + h_2 + h_3)$. For example, at 250 m: $M = 700 \times (9.24 + 11.24 + 13.24) = 23{,}604$ kg·m and $H_t\sqrt{M} = 13.24\sqrt{23604} = 2{,}034.1$.

Tower steel per km

Here the sag rises quickly with span (3.24 m to 6.12 m), so both tower height and moment grow faster than the number of towers falls.

Answer: the most economical span is 250 m (lowest relative tower cost, 8,136.6 K per km).

The dimensions follow from the minimum phase-to-earth air clearance, the insulator string length and the insulator swing.

Assumptions: maximum system voltage 145 kV; swing 45°; disc 254 × 154 mm; tower width at cross-arm level 1.2 m.

Minimum air clearance

$$\begin{aligned} \text{I:}\quad & V_{ph,peak} = \frac{145\sqrt{2}}{\sqrt{3}} = 118.4\ \text{kV} \Rightarrow a = 118.4 + 30 = 148.4\ \text{cm} = 1.48\ \text{m} \\ \text{II:}\quad & V_{ph,rms} = \frac{145}{\sqrt{3}} = 83.72\ \text{kV} \Rightarrow a = 6.5 \times 8.372 + 8 = 62.4\ \text{in} = 1.585\ \text{m} \end{aligned}$$

Take the larger value: a ≈ 1.6 m.

ii) Insulator string length

For 145 kV, Table A-3 gives:

• wet withstand 230 kV needs 6 discs (250 kV),
• impulse 550 kV needs 6 discs (610 kV),
• dry withstand 265 kV needs 4 discs.

Adding one spare disc gives 7 discs:

$$l = 7 \times 0.154 + 0.30\ (\text{hardware}) = 1.38 \approx 1.4\ \text{m}$$

Common practice uses 9 discs (about 1.6 m) where pollution is a concern.

i) Cross-arm length

$$L_{ca} = a + l\sin45° = 1.6 + 1.4 \times 0.707 = 2.59 \approx 2.6\ \text{m}$$

This keeps clearance $a$ from the tower body when the string swings by 45°.

iii) Separation of conductors

• Vertical (between cross-arms on the same side):

$$V = l + a = 1.4 + 1.6 = 3.0\ \text{m}$$

The hanging conductor keeps clearance $a$ from the cross-arm below.

• Horizontal (between the two circuits across the tower):

$$H = 2L_{ca} + w = 2 \times 2.6 + 1.2 = 6.4\ \text{m}$$

This is well above the phase-to-phase clearance of about $1.15a = 1.8$ m.

            ew (earth wire)
             |
   o---------+---------o   top     \
             |                      | 3.0 m
   o---------+---------o   middle  /
             |
   o---------+---------o   bottom
   |<-2.6->|1.2|<-2.6->|

iv) Height of earth wire above the top conductor

One earth wire at the peak is enough for a 132 kV double-circuit vertical tower. For a 30° shielding angle, the horizontal distance to the outer conductor is $2.6 + 0.6 = 3.2$ m:

$$y = \frac{3.2}{\tan 30°} = 5.54 \approx 5.6\ \text{m}$$

Method. For each span, the transverse load at the tower comes from two sources:

• wind on the conductors and earth wires: $F = p \times d \times S$, taking the wind span equal to the span $S$ and using the full projected area;
• the resultant of the conductor tensions at a line deviation $\theta$: $F_d = 2T\sin(\theta/2)$.

These loads act at the conductor heights. Their overturning moment at ground level is

$$M = n_c\,(F_c + F_{dc})(h_1 + h_2 + h_3) + n_e\,(F_e + F_{de})\,H_t$$

where $n_c$ is the number of conductors at each level and $n_e$ the number of earth wires. The tower weight is found from Ryle's formula

$$W = K\,H_t\sqrt{M}$$

$K$ is the same for all spans, so it cancels in the comparison. Tower cost is proportional to tower weight, so the span that gives the smallest total tower weight is the most economical span (with the conductor already fixed). Wind on the tower body is neglected.

Assumptions

• Single circuit with a single earth wire: $n_c = 1$, $n_e = 1$.
• Tower types and the deviation each is designed for (IS 802 practice): A is 0°–2°, B is 2°–15°, C is 15°–30°. Each is designed for its upper angle.
• Number of towers: $N = \lceil 200000/S \rceil + 1$, split 80% A, 15% B and 5% C.

Deviation loads

Wind loads ($p = 100$ kg/m², $d_c = 0.02597$ m, $d_e = 0.01652$ m)

Moments and tower weight index

Sample calculation (300 m, type B):

$$\begin{aligned} M_c &= (779.1 + 1798.65)(11.383 + 14.693 + 18.003) = 113{,}624.7\ \text{kg·m} \\ M_e &= (495.6 + 745.57)(29.953) = 37{,}176.6\ \text{kg·m} \\ H_t\sqrt{M} &= 29.953\sqrt{150801.3} = 11{,}631.7 \end{aligned}$$

Total tower steel for 200 km

Weighted index $= 0.80\,A + 0.15\,B + 0.05\,C$:

The total keeps decreasing as the span increases, though the saving becomes small (about 1.2% from 325 m to 350 m).

Answer: the most economical span is 350 m, needing 573 towers (about 458 type A, 86 type B and 29 type C).