Transmission Voltage Level and Number of Circuits

The aim is to choose the standard voltage and number of circuits that carry the given power $P$ over length $L$ with acceptable technical performance and the least total annual cost (capital charges + cost of losses).

Steps

1. Collect data: power to be transmitted $P$ (MW), power factor $\cos\phi$, line length $L$ (km), future load growth, reliability requirement, voltage of the grid bus to be connected.

2. Find the most economical voltage for single and double circuit using the empirical formula

$$V_{econ} = 5.5\sqrt{\frac{L}{1.6} + \frac{P \times 1000}{\cos\phi \times N_c \times 150}}\ \text{kV}$$

with $N_c = 1, 2$.

3. Select nearby standard voltages (66, 132, 220, 400 kV in Nepal/India) above and below each $V_{econ}$. This gives a set of candidate options, e.g. 132 kV D/C, 220 kV S/C, 220 kV D/C.

4. Check technical capability of each option:

   • Surge impedance loading $SIL = V^2/Z_c$ (MW), with $Z_c \approx 400\ \Omega$ for a single circuit.
   • From the line capability curve, read the multiplying factor (m.f.) for length $L$ (m.f. falls with length because of voltage-drop and stability limits).
   • Capability $= \text{m.f.} \times SIL \times N_c$; it must be $\ge P$ (with some margin for growth).

5. Check other technical limits: thermal rating of available conductor, voltage regulation (about 10%), corona, and N-1 security (for double circuit, how much one circuit can carry).

6. Economic comparison of the feasible options: estimate capital cost of conductors, insulators, towers, right-of-way and substations, and the capitalised cost of $I^2R$ and corona losses. Higher voltage means lower current and loss but higher insulation and tower cost.

7. Consider practical factors: voltage of existing grid (avoid extra transformation), future expansion, reliability (D/C preferred for important evacuation lines), standardisation of equipment, right-of-way and environment.

8. Final choice: the option with lowest total cost that meets all technical and reliability requirements.

 Data (P, L, pf)
      |
 V_econ for Nc = 1, 2
      |
 Nearby standard voltages
      |
 Capability = m.f. x SIL x Nc >= P ?
      |  no -> reject option
      | yes
 Thermal, regulation, N-1 checks
      |
 Cost comparison -> final V and Nc

FALSE. Power transfer capability decreases as line length increases.

• Maximum power across a line: $P_{max} = \frac{V_s V_r}{X}$, and $X = xL$ grows with length, so $P_{max}$ falls roughly as $1/L$ (stability limit).
• Voltage drop also grows with length, so the voltage-regulation limit is reached at lower power.
• Only short lines (below about 80 km) are limited by the thermal rating, which does not depend on length.

The line capability curve shows this: power capability in multiples of SIL is about $2.75 \times SIL$ at 80 km, $1.75 \times SIL$ at 240 km and only $0.75 \times SIL$ at 640 km.

Given: $P = 190$ MW, $L = 240$ km, $\cos\phi = 0.96$, $Z_c = 400\ \Omega$.

Step 1: Most economical voltage

$$V = 5.5\sqrt{\frac{L}{1.6} + \frac{P \times 1000}{\cos\phi \times N_c \times 150}}$$

Single circuit ($N_c = 1$):

$$\begin{aligned} V &= 5.5\sqrt{\frac{240}{1.6} + \frac{190 \times 1000}{0.96 \times 1 \times 150}} \\ &= 5.5\sqrt{150 + 1319.44} = 5.5\sqrt{1469.44} \\ &= 5.5 \times 38.333 = 210.83\ \text{kV} \end{aligned}$$

Nearest standard: 220 kV.

Double circuit ($N_c = 2$):

$$\begin{aligned} V &= 5.5\sqrt{150 + 659.72} = 5.5\sqrt{809.72} \\ &= 5.5 \times 28.456 = 156.51\ \text{kV} \end{aligned}$$

Nearest standard: 132 kV (220 kV also checked).

Step 2: Capability check

For $L = 240$ km, Table A-1 gives m.f. $= 1.75$.

$SIL = V^2/400$; capability $= \text{m.f.} \times SIL \times N_c$.

Step 3: Selection

• 132 kV (even double circuit) cannot carry 190 MW over 240 km, so it is rejected.
• 220 kV single circuit matches the economic voltage (210.8 kV) and has capability 211.75 MW (about 11% margin).
• 220 kV double circuit is technically adequate but more than twice the required capacity; it is justified only if N-1 reliability or large future growth is required.

Answer: 220 kV, single circuit (most economical and technically adequate; capability 211.75 MW > 190 MW).

Given: $P = 220$ MW, $\cos\phi = 0.9$, $L = 175$ km, $Z_c = 400\ \Omega$. Economic voltage formula (standard appendix):

$$V = 5.5\sqrt{\frac{L}{1.6} + \frac{P \times 1000}{\cos\phi \times N_c \times 150}}\ \text{kV}$$

Most economical voltage

$N_c = 1$:

$$\begin{aligned} V &= 5.5\sqrt{109.375 + \frac{220000}{0.9 \times 150}} = 5.5\sqrt{109.375 + 1629.63} \\ &= 5.5\sqrt{1739.00} = 5.5 \times 41.701 = 229.36\ \text{kV} \end{aligned}$$

$N_c = 2$:

$$V = 5.5\sqrt{109.375 + 814.81} = 5.5\sqrt{924.19} = 167.20\ \text{kV}$$

Capability check

m.f. at 175 km by linear interpolation between 160 km (2.25) and 240 km (1.75):

$$\text{m.f.} = 2.25 - \frac{175-160}{80}(0.50) = 2.156$$

Selection

• 132 kV double circuit fails (187.85 MW < 220 MW).
• 220 kV single circuit is close to the economical voltage (229.4 kV) and carries 260.9 MW (about 19% margin).

Answer: 220 kV, single circuit. A 220 kV double circuit would only be chosen if N-1 reliability is demanded.

TRUE.

• Steady-state power transfer: $P = \frac{V_s V_r}{X_L}\sin\delta$.
• A series capacitor of reactance $X_C$ reduces the net reactance to $X_L - X_C = X_L(1-k)$, where $k = X_C/X_L$ is the degree of compensation (normally 25–70%).
• Hence

$$P_{max} = \frac{V_s V_r}{X_L(1-k)}$$

e.g. 50% compensation doubles the steady-state limit.

• It also improves transient stability, reduces voltage drop and allows better load sharing between parallel lines.

Limits: risk of sub-synchronous resonance with turbine-generator shafts, and need for over-voltage protection (MOV, spark gap) across capacitors. Series compensation does not raise the thermal limit, so it is useful mainly on long EHV lines that are stability-limited.

For a given power $P$ and distance $L$, the total cost of a line has parts that fall with voltage and parts that rise with voltage. Their sum has a minimum; the voltage at this minimum is the most economical voltage.

Costs that decrease with voltage

• Line current $I = \frac{P}{\sqrt{3}V\cos\phi}$ falls as $V$ rises.
• For a given loss or current density, conductor cross-section $a \propto 1/V$ (and for a fixed area, loss $\propto 1/V^2$).
• So conductor cost and cost of $I^2R$ energy loss fall with voltage:

$$C_1 \propto \frac{1}{V}\ (\text{or}\ \frac{1}{V^2})$$

Costs that increase with voltage

• Insulator strings need more discs, cross-arms longer, phase spacing larger, towers taller and heavier, ground clearance higher, right-of-way wider.
• Transformers, circuit breakers, lightning arresters and substation equipment become costlier.
• Corona loss rises.

$$C_2 \propto V$$

Total annual cost

$$C = \frac{A}{V} + B\,V + C_0$$ $$\frac{dC}{dV} = -\frac{A}{V^2} + B = 0 \Rightarrow V_{econ} = \sqrt{\frac{A}{B}}$$

Because $\frac{d^2C}{dV^2} = \frac{2A}{V^3} > 0$, this is a minimum.

 Cost
  |\                       /
  | \ conductor+loss      / insulation+
  |  \                  /   towers
  |   \    total      /
  |    \  ___.___   /
  |     \/   |   \/
  |     /\   |   /\
  |    /  \__|__/  \__
  +----------|----------> Voltage
          V_econ

Effect of P and L

$A$ increases with power (more current), and insulation cost per km is fixed by voltage, while some costs scale with length. Hence $V_{econ}$ rises with both $P$ and $L$, which is what the empirical formula shows:

$$V_{econ} = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{N_c\cos\phi \times 150}}\ \text{kV}$$

In practice the computed value is rounded to a nearby standard voltage (66, 132, 220, 400 kV) because equipment is made only for these values, and the choice is checked against capability, regulation and reliability.

Given: $P = 200$ MW, $L = 160$ km. Power factor not given, assumed $\cos\phi = 0.9$. The economical voltage formula is used with the usual constant 5.5:

$$V = 5.5\sqrt{\frac{L}{1.6} + \frac{P \times 1000}{\cos\phi \times N_c \times 150}}\ \text{kV}$$

i) Standard voltage

$N_c = 1$:

$$\begin{aligned} V &= 5.5\sqrt{\frac{160}{1.6} + \frac{200000}{0.9 \times 150}} = 5.5\sqrt{100 + 1481.48} \\ &= 5.5 \times 39.768 = 218.72\ \text{kV} \end{aligned}$$

$N_c = 2$:

$$V = 5.5\sqrt{100 + 740.74} = 5.5 \times 28.996 = 159.48\ \text{kV}$$

Candidate standard voltages: 220 kV (S/C) and 132 kV or 220 kV (D/C).

ii) Number of circuits (capability check)

At 160 km, m.f. = 2.25 (Table A-1); $SIL = V^2/400$.

132 kV D/C falls just short (196 MW < 200 MW). 220 kV S/C matches $V_{econ} = 218.7$ kV and has 36% margin.

Selected: 220 kV, single circuit.

iii) Number of disc insulators (lightning over-voltage)

Insulator design for lightning (from Tables A-2 and A-3), 220 kV system, maximum system voltage 245 kV row:

• Impulse withstand voltage (BIL) required by Table A-2: 900 kV.
• Multiply by flashover withstand ratio, non-atmospheric condition factor and factor of safety:

$$\begin{aligned} V_{FO,req} &= 900 \times 1.15 \times 1.1 \times 1.1 \\ &= 900 \times 1.3915 = 1252.35\ \text{kV} \end{aligned}$$

• From Table A-3, impulse FOV: 13 discs = 1185 kV (< 1252.35), 14 discs = 1265 kV (> 1252.35).

Answer: 220 kV, single circuit, 14 standard discs (254 × 154 mm) per suspension string. (In practice one extra disc is often added for tension strings or to allow for a damaged disc.)

Line current is $I = \dfrac{P}{\sqrt{3}\,V\cos\phi\,N_c}$ per circuit. Conductor size depends mainly on current, while insulator size depends only on voltage.

(i) Power to be transmitted and length of line

• Conductor cost: larger $P$ means larger current, so a larger conductor area (for thermal limit and to keep losses low); conductor cost per km rises roughly in proportion to $P$. Longer $L$ increases voltage drop and loss, so for long lines a larger conductor (or higher voltage) is needed, again raising conductor cost per km.
• Insulator cost: per km it does not depend directly on $P$ or $L$, because insulation depends only on voltage. Indirectly, large $P$ and long $L$ push the design to a higher voltage, which increases insulator cost per km.

(ii) Voltage level

• Conductor cost: current $\propto 1/V$, so conductor area and conductor cost per km fall as voltage rises (loss $\propto 1/V^2$ for the same conductor). At EHV, however, minimum size or bundling may be fixed by corona, so the fall stops.
• Insulator cost: number of discs, string length, cross-arm length, tower height and clearances all increase with voltage; insulator cost per km rises roughly in proportion to $V$ (or faster at EHV due to switching-surge design).

(iii) Number of circuits

• Conductor cost: for the same total power, each circuit carries $1/N_c$ of the current, so each conductor is smaller, but there are $N_c$ times as many. Total conductor cost per km is roughly the same or a little higher (smaller conductors cost more per unit current), while losses for the same total aluminium are similar.
• Insulator cost: number of strings is proportional to $N_c$, so insulator cost per km roughly doubles from single to double circuit. Double-circuit towers are taller and heavier, but cheaper than two separate S/C lines.

These opposite trends give an optimum (most economical) voltage and number of circuits for each $P$ and $L$.

FALSE. The most economical voltage decreases when the number of circuits increases.

$$V_{econ} = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{\cos\phi \times N_c \times 150}}$$

$N_c$ is in the denominator: with more circuits each circuit carries less power (less current), so a lower voltage is enough for that circuit.

Example: $P = 200$ MW, $L = 160$ km, $\cos\phi = 0.9$ gives $V_{econ} = 218.7$ kV for one circuit but only $159.5$ kV for two circuits.

Appropriateness of the empirical formula

$$V_{econ} = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{\cos\phi \times N_c \times 150}}\ \text{kV}\quad (L\ \text{in km},\ P\ \text{in MW})$$

• It is derived from the balance of costs: conductor and loss cost fall with voltage, insulation and tower cost rise with voltage. The square root reflects $V_{econ} = \sqrt{A/B}$ from minimising $C = A/V + BV$.
• Length term $L/1.6$ ($L$ converted to miles): longer lines need higher voltage to keep drop and loss low.
• Power term $1000P/(\cos\phi\,N_c\,150)$: proportional to kVA per circuit; more power per circuit needs higher voltage. Increasing $N_c$ shares the power and lowers the voltage required.
• Merits: quick, needs only $P$, $L$, pf, $N_c$; gives a good starting point; matches common practice.
• Limitations: it is an old empirical (Still's type) formula; it ignores actual costs of conductor, energy and land, does not check stability, regulation, corona or thermal limits, and gives non-standard values. So the result must be rounded to a standard voltage and checked with the SIL-based capability curve.

Selection for 150 MW over 100 km

$\cos\phi$ not given, assumed 0.9.

$N_c = 1$:

$$\begin{aligned} V &= 5.5\sqrt{\frac{100}{1.6} + \frac{150000}{0.9 \times 150}} = 5.5\sqrt{62.5 + 1111.11} \\ &= 5.5 \times 34.258 = 188.42\ \text{kV} \end{aligned}$$

$N_c = 2$:

$$V = 5.5\sqrt{62.5 + 555.56} = 5.5 \times 24.861 = 136.73\ \text{kV}$$

Capability: m.f. at 100 km $= 2.75 - \frac{20}{80}(0.5) = 2.625$.

• 132 kV D/C: economical voltage 136.7 kV is very close to 132 kV; capability 228.7 MW (52% margin); if one circuit trips the other still carries 114 MW (76% of load).
• 220 kV S/C: economical voltage 188.4 kV lies between 132 and 220 kV; capability is more than twice the need and there is no backup circuit.

Answer: 132 kV, double circuit (closest to economical voltage, adequate capability, better reliability).

Given: generation $P = 250$ MW. Assumed $\cos\phi = 0.9$, $Z_c = 400\ \Omega$, standard voltages 66/132/220/400 kV. Options:

• A: 190 km to 220 kV grid bus
• B: 150 km to a grid bus of voltage V (we may choose V)
• C: 160 km to 132 kV grid bus

Economical voltage, $V = 5.5\sqrt{L/1.6 + 1000P/(0.9 \times 150 N_c)}$

Sample (B, 1 ckt): $5.5\sqrt{93.75 + 1851.85} = 5.5\sqrt{1945.60} = 242.60$ kV.

All values point to 220 kV (S/C) or 132–220 kV (D/C).

Capability check (m.f. from Table A-1, interpolated)

(Capability $= \text{m.f.} \times V^2/400 \times N_c$.)

Assessment

• Option C (132 kV bus): even a double circuit carries only 196 MW, and the economical voltage (176–243 kV) is far above 132 kV. It would need 3 circuits or a step-up to 220 kV at the grid end. Rejected.
• Option A (220 kV bus, 190 km): S/C is just short (249.6 MW), so a 220 kV double circuit is needed over the longest route; highest cost and largest losses.
• Option B (150 km): shortest route; choosing V = 220 kV, a single circuit carries 279.8 MW (12% margin), matching the economical voltage of 242.6 kV. Lowest line cost, lowest loss, best voltage regulation.

Selection: Option B, V = 220 kV, single circuit, 150 km. If the plant evacuation requires N-1 security, the same route with a 220 kV double circuit is the next choice; it is still cheaper than option A's double circuit because the route is 40 km shorter.

FALSE. For a 765 kV line the transient stability limit is less than the thermal limit (for practical line lengths).

• Thermal limit depends on conductor area; a 765 kV line uses 4–6 sub-conductor bundles, so its thermal capacity is very large (about 4000–6000 MW, i.e. 2–3 × SIL).
• The SIL of a 765 kV line is about $765^2/260 \approx 2250$ MW. Because EHV lines are long (300–600 km), the power limited by stability/voltage drop is only about 1–1.5 × SIL.
• Hence in EHV/UHV lines power transfer is stability-limited, not thermally limited; only short lines (< 80 km) are thermally limited.

That is why series compensation and FACTS are used on such lines: they raise the stability limit toward the thermal limit.

Given: $P = 160$ MW, $L = 100$ km. Power factor not given, assumed 0.9; $Z_c = 400\ \Omega$; standard voltages 66, 132, 220, 400 kV.

Step 1: Most economical voltage

$$V = 5.5\sqrt{\frac{L}{1.6} + \frac{P \times 1000}{\cos\phi\ N_c \times 150}}$$

$N_c = 1$:

$$\begin{aligned} V &= 5.5\sqrt{62.5 + \frac{160000}{0.9 \times 150}} = 5.5\sqrt{62.5 + 1185.19} \\ &= 5.5\sqrt{1247.69} = 5.5 \times 35.323 = 194.27\ \text{kV} \end{aligned}$$

$N_c = 2$:

$$V = 5.5\sqrt{62.5 + 592.59} = 5.5\sqrt{655.09} = 140.77\ \text{kV}$$

Step 2: Capability check

m.f. at 100 km (interpolating 80 km → 2.75, 160 km → 2.25):

$$\text{m.f.} = 2.75 - \frac{100-80}{80} \times 0.5 = 2.625$$

Step 3: Selection

• $V_{econ}$ for D/C (140.8 kV) is very close to standard 132 kV, and 132 kV D/C carries 228.7 MW (43% margin).
• With one circuit out, the other still carries 114 MW (71% of load), giving good reliability.
• 220 kV S/C is technically adequate but its $V_{econ}$ (194.3 kV) is not close to 220 kV, it has about twice the needed capacity, and has no backup circuit; insulation and substation cost are higher.

Answer: 132 kV, double circuit.

Given: $P = 250$ MW, $L = 160$ km. The appendix is not printed with this paper, so the standard IOE appendix is used: $V_{econ} = 5.5\sqrt{L/1.6 + 1000P/(150N_c\cos\phi)}$, Table A-1 (m.f.), Table A-2 (withstand voltages) and Table A-3 (FOV of 254 × 154 mm discs). Assumed $\cos\phi = 0.9$, $Z_c = 400\ \Omega$.

a) Standard voltage

$N_c = 1$:

$$\begin{aligned} V &= 5.5\sqrt{\frac{160}{1.6} + \frac{250000}{0.9 \times 150}} = 5.5\sqrt{100 + 1851.85} \\ &= 5.5 \times 44.180 = 242.99\ \text{kV} \end{aligned}$$

$N_c = 2$:

$$V = 5.5\sqrt{100 + 925.93} = 5.5 \times 32.030 = 176.17\ \text{kV}$$

Nearest standard voltage: 220 kV (132 kV also checked for D/C).

b) Number of circuits

m.f. at 160 km = 2.25.

220 kV single circuit is adequate (about 9% margin) and closest to $V_{econ}$. Selected: 220 kV, single circuit (a double circuit only if N-1 reliability or future growth demands it).

c) Number of disc insulators for lightning

Insulator design for lightning (from Tables A-2 and A-3), 220 kV system, maximum system voltage 245 kV row:

• Impulse withstand voltage (BIL) required by Table A-2: 900 kV.
• Multiply by flashover withstand ratio, non-atmospheric condition factor and factor of safety:

$$\begin{aligned} V_{FO,req} &= 900 \times 1.15 \times 1.1 \times 1.1 \\ &= 900 \times 1.3915 = 1252.35\ \text{kV} \end{aligned}$$

• From Table A-3, impulse FOV: 13 discs = 1185 kV (< 1252.35), 14 discs = 1265 kV (> 1252.35).

Answer: 220 kV, single circuit, 14 discs per string.

Raising the transmission voltage reduces current, and therefore reduces loss and conductor material, but it increases insulation, tower and equipment cost. This trade-off decides the voltage level.

For a 3-phase line carrying power $P$ at power factor $\cos\phi$:

$$I = \frac{P}{\sqrt{3}\,V\cos\phi}$$

1. Effect on power loss

$$P_{loss} = 3I^2R = 3\left(\frac{P}{\sqrt{3}V\cos\phi}\right)^2 \frac{\rho L}{a} = \frac{P^2 \rho L}{V^2\cos^2\phi\ a}$$

• For the same conductor, power loss $\propto 1/V^2$: doubling the voltage cuts loss to one quarter.
• Percentage loss $= P_{loss}/P \propto P/V^2$, so efficiency improves with voltage.
• Corona loss, however, rises with voltage (Peek's formula: $\propto (V_p - V_d)^2$) and becomes important above about 220 kV; bundled conductors are then needed.

2. Effect on energy loss

• Energy loss $= P_{loss,max} \times LLF \times 8760$ kWh/year, where $LLF$ is the loss load factor.
• Since peak loss $\propto 1/V^2$, annual energy loss and its cost also fall as $1/V^2$. Over a 25–30 year life this saving is large.

3. Effect on voltage drop and regulation

• Drop $\approx I(R\cos\phi + X\sin\phi)$; percentage drop $\propto P\,L/V^2$. Higher voltage gives better regulation and allows longer lines.
• Power capability (SIL $= V^2/Z_c$) rises with $V^2$.

4. Effect on conductor economy

• For the same loss (or same efficiency), required area

$$a = \frac{P^2\rho L}{P_{loss}V^2\cos^2\phi} \propto \frac{1}{V^2}$$

• So volume (and cost) of conductor material $\propto 1/V^2$. Also lighter conductors need lighter towers.
• Limit: at EHV the minimum size is set by corona/radio interference, not by current, so further saving is small.

5. Effect on insulator economy

• Number of discs in a string rises about in proportion to voltage (e.g. 66 kV: 5–6, 132 kV: 9–10, 220 kV: 14–15, 400 kV: 23–24).
• Longer strings need longer cross-arms, larger phase spacing, taller towers (ground clearance), wider right-of-way.
• Substation equipment (transformers, breakers, arresters) costs more at higher voltage.
• So insulation-related cost $\propto V$ (roughly), and rises faster at EHV where switching surges govern.

6. Overall economy

Total annual cost $C = A/V + BV$ has a minimum at $V_{econ} = \sqrt{A/B}$. Above this voltage the extra insulation cost exceeds the saving in conductor and loss; below it losses dominate. Hence the economical voltage rises with the power and distance, and the result is rounded to the nearest standard voltage (66, 132, 220, 400 kV).

Surge impedance loading (SIL) is the power delivered when a lossless line is terminated in its surge impedance $Z_c = \sqrt{L/C}$:

$$SIL = \frac{V_L^2}{Z_c}\ \text{MW}$$

At SIL the reactive power generated by line capacitance equals that absorbed by inductance, so the voltage is flat along the line.

Dependency on surge impedance

• SIL $\propto 1/Z_c$. Lowering $Z_c$ (bundled conductors, compact phase spacing, shunt capacitance increase, series compensation) raises SIL and hence the power limit.
• Typical $Z_c$: about 400 Ω for single-conductor lines, 250–300 Ω for bundled EHV lines.

Dependency on length

• Max power $P = \frac{V_sV_r}{Z_c\sin\beta l}$ (lossless line). As length $l$ increases, $\sin\beta l$ increases, so power in multiples of SIL falls.
• Short lines (< 80 km) are limited by thermal rating (about 2.5–3 × SIL); medium lines (80–320 km) by voltage drop (about 1.5–2.5 × SIL); long lines (> 320 km) by stability (about 1 × SIL or less).

So power transfer limit $= \text{m.f.}(L) \times V^2/Z_c$: it increases with $V^2$, decreases with $Z_c$, and decreases with line length.

When the chosen voltage and circuit cannot carry the required power, the designer can either raise the system voltage or add circuits. Their effects are different.

Increasing the system voltage

• Capability rises as $V^2$: $SIL = V^2/Z_c$. Going from 132 kV to 220 kV raises SIL from 43.6 MW to 121 MW (×2.78).
• Current falls as $1/V$, so $I^2R$ loss falls as $1/V^2$ and regulation improves.
• Improves the stability limit ($P_{max} = V_sV_r/X$), so it is the effective remedy for long lines that are stability- or regulation-limited.
• Insulation, tower height, clearances, right-of-way and substation cost rise; corona increases.
• Gives no redundancy: a single circuit outage interrupts the whole power.
• Suitable when: power and distance are large, the line is long, and the grid already has that higher voltage level.

Increasing the number of circuits

• Capability rises only in proportion to $N_c$: two circuits give 2 × capability.
• Each circuit carries less current, so conductor per circuit is smaller; total losses fall about by half for the same conductor.
• Provides reliability (N-1): if one circuit is out, the other still carries a large share.
• Allows staged construction (string one circuit now, second later).
• A double-circuit tower is cheaper than two single-circuit lines and uses one right-of-way.
• Suitable when: the line is short to medium, power is moderate, the grid bus voltage is fixed (e.g. connection to a 132 kV substation), or high reliability is needed (generation evacuation).

Comparison

In practice both $N_c = 1$ and $N_c = 2$ are evaluated with the economic voltage formula, rounded to standard voltages, checked against the capability curve, and the cheapest adequate option is chosen.

Given: $P = 300$ MW, $L = 200$ km. Assumed $\cos\phi = 0.9$, $Z_c = 400\ \Omega$, standard voltages 66, 132, 220, 400 kV.

Most economical voltage

$$V = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{\cos\phi \times N_c \times 150}}$$

$N_c = 1$:

$$\begin{aligned} V &= 5.5\sqrt{\frac{200}{1.6} + \frac{300000}{0.9 \times 150}} = 5.5\sqrt{125 + 2222.22} \\ &= 5.5\sqrt{2347.22} = 5.5 \times 48.448 = 266.46\ \text{kV} \end{aligned}$$

$N_c = 2$:

$$V = 5.5\sqrt{125 + 1111.11} = 5.5\sqrt{1236.11} = 5.5 \times 35.158 = 193.37\ \text{kV}$$

Capability check

m.f. at 200 km (between 160 km: 2.25 and 240 km: 1.75):

$$\text{m.f.} = 2.25 - \frac{40}{80}(0.5) = 2.0$$

Selection

• For a single circuit, $V_{econ} = 266.5$ kV is nearer 220 kV, but 220 kV S/C carries only 242 MW. 400 kV S/C is far above the economic voltage (very costly insulation, towers and substations for 2.7 times the needed capacity).
• For double circuit, $V_{econ} = 193.4$ kV rounds to 220 kV, and 220 kV D/C carries 484 MW (61% margin). With one circuit out, the other still carries 242 MW (81% of load).

Answer: 220 kV, double circuit.

Given: $P = 260$ MW. Assumed $\cos\phi = 0.9$, $Z_c = 400\ \Omega$. Options:

• A: 140 km to a 220 kV grid bus
• B: 250 km to a grid bus of voltage V (chosen by designer)
• C: 200 km to a 132 kV grid bus

a. Voltage level and number of circuits

Economical voltage $V = 5.5\sqrt{L/1.6 + 260000/(0.9 \times 150 \times N_c)}$; the power term is 1925.93 for $N_c = 1$ and 962.96 for $N_c = 2$.

Sample (A, 1 ckt): $5.5\sqrt{87.5 + 1925.93} = 5.5 \times 44.871 = 246.79$ kV.

Capability $= \text{m.f.} \times V^2/400 \times N_c$ (m.f. interpolated from Table A-1):

Comparative assessment

• Option C: the bus is 132 kV, but even a 132 kV double circuit carries only 174 MW; the economical voltage (181–249 kV) is much higher. It would need 3 circuits or a new 220/132 kV substation. Rejected.
• Option B: longest route (250 km). 220 kV S/C is inadequate; a 220 kV D/C or 400 kV S/C is needed. Highest cost, largest loss and right-of-way.
• Option A: shortest route (140 km), grid bus already 220 kV, so no extra transformation. A 220 kV single circuit carries 287.4 MW (10.5% margin) and matches $V_{econ} = 246.8$ kV. Lowest capital cost and loss.

Selection: Option A, 220 kV. A single circuit is the most economical. Because the line evacuates a whole 260 MW plant, a 220 kV double-circuit tower (or D/C tower with one circuit strung first) is advisable for N-1 security: each circuit alone then carries 287 MW, i.e. full output.

b. Number of standard discs per string (220 kV line)

Factor $k = \text{FWR} \times \text{NACF} \times \text{FOS} = 1.15 \times 1.1 \times 1.1 = 1.3915$. For the 220 kV class the Table A-2 row gives dry 435 kV, wet 395 kV, impulse 900 kV. Highest system voltage taken as 245 kV (standard for 220 kV).

(i) Lightning (external) over-voltage

$$V_{req} = 900 \times 1.3915 = 1252.35\ \text{kV}$$

Impulse FOV: 13 discs = 1185 kV, 14 discs = 1265 kV → 14 discs.

(ii) Power-frequency over-voltage (wet condition governs)

$$V_{req,wet} = 395 \times 1.3915 = 549.64\ \text{kV}$$

Wet FOV: 13 discs = 520 kV, 14 discs = 555 kV → 14 discs. (Dry: $435 \times 1.3915 = 605.30$ kV → 11 discs; less severe.)

(iii) Switching (internal) over-voltage, assuming switching surge ratio SSR = 2.8 for 220 kV:

$$\begin{aligned} V_{ss} &= SSR \times \frac{\sqrt{2}\,V_{max}}{\sqrt{3}} = 2.8 \times \frac{\sqrt{2} \times 245}{\sqrt{3}} = 560.12\ \text{kV (peak)} \\ V_{req} &= 560.12 \times 1.3915 = 779.40\ \text{kV (peak)} = \frac{779.40}{\sqrt{2}} = 551.12\ \text{kV (rms)} \end{aligned}$$

Compared with wet FOV: 14 discs = 555 kV → 14 discs.

Answer: 14 standard discs (254 × 154 mm) per suspension string; one extra disc (15) is commonly used in tension strings.

FALSE (as a general statement).

• Power capability rises with the square of voltage, $SIL = V^2/Z_c$, but only linearly with the number of circuits. Raising 132 kV to 220 kV increases capability by $(220/132)^2 = 2.78$ times, while adding a second 132 kV circuit gives only 2 times.
• Losses fall as $1/V^2$ with higher voltage but only as $1/N_c$ with more circuits.
• For long lines, the stability limit $P_{max} = V_sV_r/X$ improves strongly with voltage; adding a parallel circuit only halves $X$.
• Each extra circuit adds conductors, insulators, wider/heavier towers and extra substation bays.

So to meet a technical requirement (capability, regulation, stability) for large power over long distance, raising the voltage is more effective. Adding circuits is preferred only when the grid voltage is fixed, the line is short/medium, or when reliability (N-1) is the main need.

Given: $P = 180$ MW. Assumed $\cos\phi = 0.9$, $Z_c = 400\ \Omega$. The distance to the 220 kV bus is illegible ("2_0 km"); it is taken as 200 km, and the check below shows the conclusion holds for any value from 200 to 290 km.

Options:

• A: ≈200 km to a 220 kV grid bus
• B: 300 km to a grid bus of voltage V (designer's choice)
• C: 180 km to a 132 kV grid bus

Economical voltage

$V = 5.5\sqrt{L/1.6 + 180000/(0.9 \times 150 \times N_c)}$, power term = 1333.33 ($N_c = 1$) or 666.67 ($N_c = 2$).

Sample (C, 2 ckt): $5.5\sqrt{112.5 + 666.67} = 5.5 \times 27.914 = 153.52$ kV.

Capability (m.f. × $V^2/400$ × $N_c$)

Comparative assessment

• B is the longest and needs a double circuit (220 kV S/C fails); most expensive.
• C is shortest but 132 kV D/C is at its limit, current is 67% higher (more $I^2R$ loss), and there is no margin for growth.
• A: 220 kV single circuit matches $V_{econ}$ (210 kV), connects directly to a 220 kV bus, has 34% spare capability and lower losses. The 220 kV S/C remains adequate up to a length where m.f. $= 180/121 = 1.488$, i.e. about 292 km, so the choice is valid whatever the illegible distance (200–290 km).

Selection: Option A, 220 kV, single circuit (a D/C tower with one circuit strung may be used for future reliability).

FALSE.

• Raising voltage reduces current, so conductor cost and $I^2R$ loss fall ($\propto 1/V^2$).
• But insulators, cross-arms, tower height, clearances, right-of-way, corona loss and substation equipment cost all rise with voltage.
• Total cost $C = A/V + BV$ has a minimum at $V_{econ} = \sqrt{A/B}$. Below $V_{econ}$ a higher voltage is cheaper; above it, a higher voltage is costlier.

So for a given $P$ and $L$ there is one most economical voltage (e.g. about 219 kV for 200 MW over 160 km, so 220 kV, not 400 kV). Choosing a voltage higher than this increases total cost.

FALSE (strictly). The capability (in multiples of SIL) decreases with line length, but not in inverse proportion.

• If it were inversely proportional, m.f. × length would be constant. From the capability curve: $80 \times 2.75 = 220$, $160 \times 2.25 = 360$, $320 \times 1.35 = 432$, $640 \times 0.75 = 480$; the product is not constant.
• The curve has three regions: short lines (thermal limit, about 3 × SIL, nearly independent of length), medium lines (voltage-drop limit), long lines (stability limit, $P \propto 1/\sin\beta l$).

So the correct statement is: power transfer capability falls non-linearly as length increases.

Given: $P = 200$ MW, $L = 120$ km. Assumed $\cos\phi = 0.9$, $Z_c = 400\ \Omega$, factors FWR = 1.15, NACF = 1.1, FOS = 1.1 (usual design values), highest system voltage of 220 kV class = 245 kV, disc spacing 154 mm.

a. Economical voltage, capability, standard voltage and circuits

$$V = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{\cos\phi\,N_c \times 150}}$$

$N_c = 1$: $V = 5.5\sqrt{75 + 1481.48} = 5.5 \times 39.452 = 216.99$ kV $N_c = 2$: $V = 5.5\sqrt{75 + 740.74} = 5.5 \times 28.561 = 157.09$ kV

m.f. at 120 km $= 2.75 - \frac{40}{80}(0.5) = 2.50$.

$V_{econ}$ for S/C (217 kV) almost equals 220 kV. Selected: 220 kV, single circuit (vertical/triangular formation, one earth wire).

b. Number of insulator discs

$k = 1.15 \times 1.1 \times 1.1 = 1.3915$

Switching: $V_{ss} = 2.8 \times \sqrt{2} \times 245/\sqrt{3} = 560.12$ kV peak; $\times k = 779.40$ kV peak $= 551.12$ kV rms.

Number of discs = 14 per suspension string.

c. Air clearances

Minimum air clearance (6.5 in per 10 kV of maximum phase rms voltage + 8 in):

$$\begin{aligned} V_{ph} &= \frac{245}{\sqrt{3}} = 141.45\ \text{kV} \\ a &= 6.5 \times \frac{141.45}{10} + 8 = 99.94\ \text{in} = 2.54\ \text{m} \end{aligned}$$

ii. Insulator string length (14 discs × 154 mm, fittings neglected):

$$l = 14 \times 0.154 = 2.156\ \text{m} \approx 2.16\ \text{m}$$

i. Cross-arm length. With 45° string swing the conductor moves $l\sin45°$ toward the tower and must still be $a$ away from the tower body:

$$CA = l\sin45° + a = 2.156 \times 0.7071 + 2.54 = 1.52 + 2.54 = 4.06\ \text{m}$$

iii. Separation of conductors

• Vertical separation between cross-arms: the conductor hangs $l$ below its cross-arm and must be $a$ above the cross-arm below:

$$d_v = l + a = 2.156 + 2.54 = 4.69\ \text{m}$$

• Horizontal separation between conductors on opposite sides of the tower (tower body width neglected):

$$d_h = 2 \times CA = 2 \times 4.063 = 8.13\ \text{m}$$

iv. Height of earth wire above top conductor. With a single earth wire on the tower axis and shielding angle 30°:

$$h = \frac{CA}{\tan30°} = 4.063 \times 1.732 = 7.04\ \text{m}$$

          o  earth wire
          |\
          | \ 30 deg
     7.04 m  \
          |   \
   -------+----o  top phase
   4.06 m |    | string 2.16 m
          |
   4.69 m |
   o------+------o  lower phases
     8.13 m apart
          |
       tower

(Also: minimum ground clearance $= (245-33)/33 + 17 = 23.42$ ft $= 7.14$ m.)

Summary: 220 kV S/C; 14 discs; string length 2.16 m; cross-arm 4.06 m; vertical spacing 4.69 m; horizontal spacing 8.13 m; earth wire 7.04 m above the top conductor.

For bulk power over long distance, both the technical limits and the annual cost push the design toward a higher voltage and/or more circuits.

1. Economical voltage rises with P and L

$$V_{econ} = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{\cos\phi \times N_c \times 150}}$$

Both terms under the root grow with $L$ and $P$. Examples ($\cos\phi = 0.9$, single circuit):

The underlying reason: total cost $C = A/V + BV$, where $A$ (conductor + loss cost) grows with current, i.e. with $P$, and with $L$. The optimum $V = \sqrt{A/B}$ therefore rises.

2. Loss and conductor economy

• Current $I = P/(\sqrt{3}V\cos\phi N_c)$. For large $P$ at low voltage, current becomes very high.
• Loss $\propto P^2L/(V^2 N_c a)$ grows with $P^2$ and $L$; to keep loss at an economic level (typically 2–5%) either $V$ or $N_c$ must increase. For a fixed percentage loss, conductor volume $\propto PL^2/V^2$, so raising $V$ saves material sharply.

3. Capability (technical) criterion

• Power limit $= \text{m.f.}(L) \times N_c \times V^2/Z_c$.
• m.f. falls with length (2.75 at 80 km, 1.0 at 480 km), so long lines carry fewer multiples of SIL. To carry large $P$ over large $L$, $V^2 N_c$ must increase.
• Example: 300 MW over 200 km (m.f. = 2.0): 220 kV S/C gives only 242 MW; 220 kV D/C gives 484 MW.

4. Voltage regulation and stability

• % voltage drop $\propto PL/V^2$; long heavy lines at low voltage give unacceptable regulation.
• Stability limit $P_{max} = V_sV_r/(xL)$: higher $V$ or parallel circuits (halving $X$) raise it.

5. Reliability criterion

• Large blocks of power should not depend on one circuit: a double circuit keeps supply on N-1 outage. This is especially important for generation evacuation lines.

6. Cost per MW-km

• Cost of a line rises less than proportionally with voltage, while capability rises as $V^2$. So cost per MW transmitted falls at higher voltage when the line is well loaded. Similarly a double-circuit tower costs about 1.5–1.6 times a single-circuit tower but carries twice the power.
• Right-of-way per MW is much smaller for high-voltage or double-circuit lines.

7. Choice between higher V and more circuits

• Raise voltage first for long lines (stability/regulation limited, capability $\propto V^2$).
• Add circuits when reliability is needed or grid voltage is fixed.
• Often both: e.g. 400 kV D/C for 1000+ MW over 300 km.

Hence, from the economic criterion (minimum of capital + loss cost per MW) and supported by technical checks, a higher voltage and/or more circuits is advisable for larger power over longer distances. The final standard value is selected by checking capability, regulation and cost of nearby standard options.