Distribution System Design

Aerial bundled conductor (ABC) is an overhead LT (or HT) line in which the phase conductors, and usually an insulated or bare neutral messenger, are each covered with XLPE (cross-linked polyethylene) insulation and twisted together into a single bundle. The bundle hangs from poles by its messenger (neutral) wire using suspension and dead-end clamps, instead of bare conductors on cross-arms and pin insulators.

   Bare LT line             ABC LT line

  o    o    o    o          (bundle on one clamp)
 ---------------------          _____
 |    cross-arm     |          / R Y \
 |                  |         | B  N |  XLPE
 |      pole        |          \_____/  insulated
                                 |
                               pole

A typical LT ABC is 3 × 95 mm² + 1 × 70 mm² (messenger) + 1 × 16 mm² (street light), aluminium.

Advantages

• Safety: insulated conductors; little risk of electrocution from accidental contact, so safer in crowded urban areas.
• Reduced theft (pilferage): direct hooking ("katiya") on bare lines is very hard, cutting non-technical losses, a major issue in Nepal.
• Higher reliability: no faults from tree branches, birds, kites, wind-driven clashing of conductors; fewer outages.
• Less right-of-way and clearance: no cross-arms; can run close to buildings and through trees; less tree trimming.
• Lower reactance: conductors are close together, so reactance (~0.1 Ω/km vs ~0.3 Ω/km) and voltage drop are lower.
• Easy installation: fewer accessories (no cross-arms, pin insulators), poles can be shorter; quick to string.
• Better appearance and lower maintenance cost.

Disadvantages

• Higher initial cost of conductor (about 1.5–2 times bare conductor).
• Lower current rating for the same size, because insulation reduces heat dissipation.
• Heavier bundle, needing stronger poles or shorter spans.
• Fault location is difficult since conductors are covered; insulation damage may not be visible.
• Tapping of service connections needs special insulation piercing connectors.
• XLPE insulation ages under ultraviolet light and heat if quality is poor.
• Repair and jointing need trained staff and special tools.

ABC is now widely used by NEA for LT distribution in urban and loss-prone areas.

A load centre is the point from which a distribution transformer feeds a group of consumers through LT feeders. Its location and the transformer size decide the LT voltage drop, losses and cost of the scheme. The considerations differ for rural and urban grid-extension schemes.

Selecting the load centre

• Centre of gravity of load: place the transformer at or near the load-weighted centre, $x = \dfrac{\sum P_i x_i}{\sum P_i}$, so that LT feeders are short and balanced.
• LT feeder length limit: keep feeder lengths within the voltage drop limit (about ±5–10% in Nepal; LT reach typically 0.5–1 km urban, up to about 1.5–2 km rural).
• Accessibility: near roads/tracks for transport, installation, maintenance and replacement.
• Short HT tap: close to the 11 kV line route, so the HT tap line is short.
• Safe and clear site: away from schools, playgrounds, flooding, landslide areas; land availability (government or community land).
• Future growth direction of the settlement.

Selecting the distribution transformer

• Design-year peak demand: rating ≥ forecast 5th-year (or 10th-year) peak kVA including losses; choose the next standard size (10, 16, 25, 50, 100, 200, 300, 500 kVA).
• Loading for minimum loss: avoid gross oversizing (high no-load loss) and overloading (high copper loss, ageing).
• Phase: single-phase for scattered small domestic loads; three-phase where motors or larger loads exist.
• Standardisation: few sizes to reduce spare stock.
• Efficiency/loss capitalisation: low-loss (e.g. amorphous core) units where load factor is low.

Rural vs urban

In rural schemes, it is often cheaper to extend the 11 kV line and use several small transformers near each cluster than one large transformer with long LT lines; in urban schemes, transformers are placed so that each covers a compact block within thermal and voltage limits.

TRUE.

Thermal limit in urban areas:

• Urban areas have high load density (many kW per km²), so feeders are short but carry large currents.
• Over a short length, voltage drop ($\Delta V \propto I \cdot l$) stays within limits even at high current. The limiting factor is the current-carrying (thermal) capacity of the conductor or cable: overheating, sag and insulation damage.
• So urban feeders are designed by thermal (ampacity) limit.

Higher losses in rural areas:

• Rural loads are scattered with low load density, so feeders (11 kV and LT) are very long for small loads. They are designed by the voltage drop limit, not thermal limit.
• Power loss $= 3I^2R$, and $R$ is proportional to length. For the same delivered power, long feeders with small conductors give a much higher percentage loss.
• Rural distribution transformers are often lightly loaded most of the day (low load factor), so no-load (core) losses form a large share of the energy supplied.
• Poor maintenance, unbalanced single-phase loading, long service drops and theft add to losses.

Example: a 10 kW load at the end of a 0.5 km LT feeder loses about 4 times less than the same load at 2 km (loss ∝ length for fixed current).

Thus urban feeders reach their thermal limit first, while rural feeders, though lightly loaded, have higher % losses, typically 15–25% in rural Nepal compared with about 8–12% in urban feeders.

The route of an LT (and HT) distribution line is the path along which poles are erected. A good route gives the shortest, safest and cheapest line that serves the consumers well and is easy to maintain.

General criteria

1. Shortest practical length: a route close to a straight line between the transformer and load clusters reduces cost, voltage drop and losses. Avoid unnecessary bends, since each angle point needs a stronger pole with stays.
2. Follow roads and tracks: lines along the road side (within the road right-of-way) give easy access for construction, operation and maintenance, and avoid crossing private land.
3. Proximity to consumers: the line should pass near the houses to keep service drops short (typically within about 30–40 m), reducing service cable cost and losses.
4. Load balance: arrange feeders so that load is shared equally among phases and feeders, and so the transformer remains near the load centre.
5. Right-of-way and land issues: prefer public land; minimise compensation, disputes and tree cutting. Avoid passing over buildings.
6. Safety clearances: maintain statutory clearances from buildings, roads, footpaths, telecom lines and other power lines (as per Nepal Electricity Regulations); avoid schools, playgrounds and fuel stations.
7. Terrain and soil: avoid landslide-prone slopes, river banks, flood plains, marshy and unstable ground; choose firm ground for pole foundations.
8. Crossings: minimise crossings of rivers, highways, railways and other lines; cross at right angles where needed.
9. Environment and aesthetics: avoid forests, heritage sites and protected areas; use ABC or underground cable in sensitive or crowded areas.
10. Future expansion: leave room for new consumers and for future upgrading (e.g. to three-phase or larger conductors).
11. Span and pole positions: keep uniform spans (about 40–50 m for LT) and place poles where they will not obstruct traffic or drainage.

The final route is fixed after a walk-over survey, considering cost, technical limits and the views of the local community.

TRUE.

In rural areas the load is small and scattered over a large area, so the primary feeders are very long. A higher primary distribution voltage (e.g. 33 kV instead of 11 kV) is often the better choice.

Justification:

• For a given power $P$, current is $I = \dfrac{P}{\sqrt{3} V \cos\phi}$. Raising $V$ reduces current proportionally.
• Voltage drop: $\%\Delta V \propto \dfrac{P \cdot l}{V^2}$. Going from 11 kV to 33 kV allows about 9 times the load-distance (kW·km) for the same % drop, so long rural feeders stay within the voltage limit.
• Power loss: $P_{loss} = \dfrac{P^2 R}{V^2 \cos^2\phi}$, i.e. 1/9th at 33 kV compared with 11 kV for the same conductor.
• Smaller conductor: since current is low, small conductors can be used, and fewer intermediate substations are needed.
• Rural feeders are limited by voltage drop, not thermal capacity, so the higher voltage directly solves the main design problem.

Example: in Nepal, NEA uses 33 kV lines to reach remote hill districts, and steps down to 11 kV or directly to 400 V near load clusters.

Limitation: higher voltage needs costlier insulators, transformers, switchgear and clearances, so it is justified only when the load-distance is large. For short feeders with small loads, 11 kV (or single-wire earth return) remains cheaper.

Consider a three-phase feeder of length $l$ with resistance $r$ and reactance $x$ per phase per unit length, supplying a total current $I$ at power factor $\cos\phi$.

Case 1: Load concentrated at the end

The full current $I$ flows through the whole length:

$$\begin{aligned} \Delta V_{end} &= I (r\cos\phi + x\sin\phi)\, l = I Z_{eff} l \\ P_{loss,end} &= 3 I^2 r l \end{aligned}$$

Case 2: Uniformly distributed load

 S                              End
 |-----|-----|-----|-----|-----|
 I    i(x)  load taken uniformly
 <------ x ------><-- l-x -->

The load is taken uniformly along the feeder, $I/l$ per unit length. At distance $x$ from the source, the current still flowing is the load beyond $x$:

$$i(x) = I\left(1 - \frac{x}{l}\right)$$

Voltage drop: drop in element $dx$ is $d(\Delta V) = i(x)(r\cos\phi + x_L\sin\phi)\,dx$ (writing $x_L$ for reactance per unit length):

$$\begin{aligned} \Delta V_{uni} &= (r\cos\phi + x_L\sin\phi)\int_0^l I\left(1 - \frac{x}{l}\right)dx \\ &= (r\cos\phi + x_L\sin\phi)\, I\left[x - \frac{x^2}{2l}\right]_0^l \\ &= (r\cos\phi + x_L\sin\phi)\, I \frac{l}{2} = \frac{1}{2}\Delta V_{end} \end{aligned}$$

Power loss: loss in element $dx$ is $3\,i(x)^2 r\,dx$:

$$\begin{aligned} P_{loss,uni} &= 3r\int_0^l I^2\left(1 - \frac{x}{l}\right)^2 dx \\ &= 3 r I^2 \left[-\frac{l}{3}\left(1 - \frac{x}{l}\right)^3\right]_0^l \\ &= 3 r I^2 \frac{l}{3} = \frac{1}{3} P_{loss,end} \end{aligned}$$

Result

Physical meaning: for voltage drop, a uniformly distributed load acts as if the whole load were placed at the mid-point ($l/2$). For loss, it acts as if the whole load were at one-third of the length ($l/3$) from the source.

These results are used in LT feeder design, where load is roughly uniform along the line: voltage drop $= \sqrt{3}\, I (R\cos\phi + X\sin\phi)/2$ (line value) and loss $= I^2 R$ for the three phases, with $R$, $X$ the total feeder values.

A shunt capacitor supplies reactive power locally, reducing the reactive current in every section upstream of it. The optimum size minimises the total $I^2R$ loss.

Assumptions: all node voltages = 1 p.u. (given), so section current magnitude² = $P^2 + Q^2$ in p.u.; base 1000 kVA; transformer impedances neglected.

Step 1: Loads in p.u.

Step 2: Section flows

Step 3: Loss as a function of capacitor size

With capacitor $Q_c$ at C, the reactive flow in every section reduces by $Q_c$:

$$P_L = \sum R_i \left[P_i^2 + (Q_i - Q_c)^2\right]$$

For maximum loss reduction:

$$\begin{aligned} \frac{dP_L}{dQ_c} &= -2\sum R_i (Q_i - Q_c) = 0 \\ Q_c &= \frac{\sum R_i Q_i}{\sum R_i} \\ &= \frac{1(0.2854) + 2(0.2254) + 3(0.16)}{1 + 2 + 3} \\ &= \frac{1.2162}{6} = 0.2027 \text{ p.u.} \end{aligned}$$ $$Q_c = 0.2027 \times 1000 = 202.7 \text{ kVAR}$$

Step 4: Check of loss reduction (p.u.)

Loss falls from 0.545 p.u. to 0.299 p.u., a reduction of about 45%. (The given impedances are very large, so the absolute p.u. losses are only for comparison.)

Note: the optimum $Q_c$ (0.203 p.u.) is larger than the reactive load at C (0.16 p.u.) because upstream sections also carry the reactive power of A and B; the result is a weighted average of section reactive flows.

Answer: Capacitor size ≈ 203 kVAR at node C (a standard 200 kVAR bank would be used).

The distribution transformer steps 11 kV (or 33 kV) down to 400/230 V for consumers. Its size and location decide the voltage at consumers' terminals, the LT and transformer losses and the cost of the scheme.

Considerations for size (rating)

1. Design-year peak demand: rating is based on the forecast peak (usually 5th year, sometimes 10th) in kVA, including diversity, contribution factors, power factor and loss allowance: $S = P_{peak}/\cos\phi$.
2. Standard ratings: choose the next standard size (e.g. 10, 16, 25, 50, 100, 200, 300, 500 kVA) to keep spares and stock limited.
3. Load growth: allow for future growth, but avoid gross oversizing. A unit loaded at about 50–80% of rating at peak is usual.
4. Losses: an oversized unit has high no-load (core) losses throughout the year, which matters when load factor is low (rural). An undersized unit has high copper loss and overheating. Loss capitalisation (total owning cost) decides the economic size.
5. Overload capability: short-time overloads allowed by IEC 60076-7 loading guides, considering ambient temperature.
6. Type of load: motor starting, single- vs three-phase load, harmonics.
7. Number of units: one large unit vs several small units; small units near loads reduce LT length.
8. Transport and mounting: pole-mounted units are limited to about 200–300 kVA; larger ones need plinth mounting.

Considerations for location

1. Load centre: at or near the centre of gravity of the load, $\bar{x} = \sum P_i x_i/\sum P_i$, to minimise LT length, voltage drop and loss.
2. Voltage drop limit: all consumers should be within the allowable LT feeder length.
3. Proximity to HT line: short 11 kV tap line.
4. Accessibility: near roads for installation, maintenance and replacement.
5. Safety: away from schools, playgrounds and crowded places; proper clearances, fencing and earthing.
6. Site conditions: firm ground, free from flooding, landslide and waterlogging; space for earthing pits.
7. Land availability and public acceptance.
8. Future expansion of the settlement.

   11 kV line
 ======+=========================
       | HT tap (short)
      [T] at load centre
   ____|____________
  |    |       |    |   LT feeders
  o o  o o   o o  o o   consumers

The final choice balances capital cost against losses and voltage quality.

Distribution transformers may be single-phase (11 kV/230 V, or SWER/phase-to-phase primary) or three-phase (11/0.4 kV). Each suits a different type of load and area.

Single-phase distribution transformer

Suitable where:

• Loads are small, scattered and mainly domestic (lighting, fans, TV): rural villages, hilly settlements.
• Load per cluster is small (5–25 kVA), so a three-phase unit would be lightly loaded with high no-load loss.
• The HT line can be a single-phase tap (two wires or SWER), much cheaper than three-phase.

Advantages: low cost, light (easy pole mounting and transport in hills), lower no-load loss, can be placed very close to small clusters, reducing LT length.

Disadvantages: cannot supply three-phase motors; may unbalance the three-phase feeder unless taps are distributed across phases; lower capacity.

Three-phase distribution transformer

Suitable where:

• Load density is high (urban, semi-urban areas, market centres).
• There are three-phase loads: motors, mills, pumps, workshops, industries, irrigation.
• Larger capacities (25–500 kVA) are needed.

Advantages: supplies both 3-phase (400 V) and 1-phase (230 V) loads; about 15–20% less material and cost per kVA than three single-phase units; balanced loading of the HT system; better efficiency at large ratings.

Disadvantages: higher cost for small loads, heavier, needs a three-phase HT line, high no-load loss when lightly loaded.

In practice, single-phase units are used for rural electrification of small, dispersed settlements, and three-phase units for urban, commercial and industrial areas.

An area (distribution) substation steps down sub-transmission voltage (e.g. 66 or 33 kV) to primary distribution voltage (11 kV) and feeds a number of feeders. Its site is chosen considering:

1. Load centre: as near as possible to the centre of the present and future load, so 11 kV feeders are short, with low voltage drop and loss.
2. Supply source: close to the incoming sub-transmission line for short, cheap connection.
3. Feeder exits: enough space and routes for outgoing feeders in all directions.
4. Land: enough area for present needs and future expansion (more transformers and bays); reasonable cost; clear ownership.
5. Access: road access for heavy transformers and maintenance vehicles.
6. Site conditions: level, firm ground with good drainage; free from flooding, landslides and pollution; good soil resistivity for earthing.
7. Environment and safety: away from residential crowding, schools and hospitals; noise and visual impact acceptable; fire safety.
8. Communication and water facilities for operation.
9. Cost: total cost of land, site preparation, lines and losses should be minimum.
10. Regulations and public acceptance: comply with local by-laws and get community consent.

Consider a three-phase feeder of length $l$ (km) with resistance $r$ and reactance $x$ (Ω per phase per km). The load is uniformly increasing along the feeder: the load density is zero at the sending end and rises linearly to its maximum at the far end. The total sending-end current is $I$ at power factor $\cos\phi$.

 load density
  ^                    /|
  |                 /   |
  |              /      |
  |           /         |
  |        /            |
  |     /               |
  +--------------------------> x
  S (x=0)            far end (x=l)

Current at distance x

Load density (current per km) $= k x$. Total current:

$$I = \int_0^l kx\,dx = \frac{k l^2}{2} \Rightarrow k = \frac{2I}{l^2}$$

The current flowing at distance $x$ is the load beyond $x$:

$$i(x) = \int_x^l k s\,ds = \frac{k}{2}(l^2 - x^2) = I\left(1 - \frac{x^2}{l^2}\right)$$

Voltage drop (per phase)

$$\begin{aligned} \Delta V &= \int_0^l i(x)(r\cos\phi + x_L \sin\phi)\,dx \\ &= (r\cos\phi + x_L\sin\phi)\, I\left[x - \frac{x^3}{3l^2}\right]_0^l \\ &= \frac{2}{3}\, I\,(r\cos\phi + x_L\sin\phi)\,l \end{aligned}$$

where $x_L$ is the reactance per km. The line voltage drop is $\sqrt{3}$ times this.

Power loss (three phases)

$$\begin{aligned} P_{loss} &= 3\int_0^l i(x)^2 r\,dx = 3 I^2 r \int_0^l \left(1 - \frac{x^2}{l^2}\right)^2 dx \\ &= 3 I^2 r \left[x - \frac{2x^3}{3l^2} + \frac{x^5}{5l^4}\right]_0^l \\ &= 3 I^2 r\, l\left(1 - \frac{2}{3} + \frac{1}{5}\right) = \frac{8}{15}\left(3 I^2 r l\right) \end{aligned}$$

Comparison

So with uniformly increasing load, the voltage drop is $\frac{2}{3}$ and the power loss is $\frac{8}{15}$ of the end-loaded values $I(r\cos\phi + x_L\sin\phi)l$ and $3I^2 r l$. Equivalently, the load may be treated as lumped at $\frac{2}{3}l$ for voltage drop and at $\frac{8}{15}l$ for loss.

(If the load density instead decreases linearly from the source to zero at the far end, the same method gives $\Delta V = \frac{1}{3}I Z l$ and $P_{loss} = \frac{1}{5}(3I^2 r l)$.)

Losses in an existing distribution network are technical ($I^2R$ in lines and transformers, core loss) and non-technical (theft, metering and billing errors). The main options, with their limitations, are:

Notes on choice:

• Capacitor placement and load balancing are cheapest and usually done first.
• Reconductoring and voltage upgrading give large savings but need cost–benefit study (energy saved × tariff vs capital cost).
• In Nepal, where non-technical loss has been large, ABC, meter replacement and anti-theft drives (NEA loss reduction campaign) have given the largest gains.

Every option should be justified by comparing the present worth of energy saved with its investment.

FALSE.

Justification:

A transformer has two kinds of loss:

• No-load (core) loss $P_i$: constant whenever the transformer is energised, regardless of load.
• Load (copper) loss: $P_{cu} = x^2 P_{cu,FL}$, where $x$ is the fraction of rated load.

Efficiency is maximum when copper loss equals core loss, i.e. at load fraction

$$x_{\eta,max} = \sqrt{\frac{P_i}{P_{cu,FL}}}$$

which is typically 40–60% of rating for distribution transformers.

When the transformer is significantly under-loaded:

• Copper loss becomes very small, but the core loss continues for all 8760 hours.
• The energy delivered is small, so the percentage energy loss rises.

Example: 100 kVA unit, $P_i = 200$ W, $P_{cu,FL} = 1100$ W, LLF taken as 0.3 for illustration.

The 20 kVA case carries one-third of the load but has two-thirds of the losses, so loss per kWh delivered is about twice as high.

Therefore gross under-loading (oversized transformers) wastes energy through core loss, as well as capital. Moderate under-loading near the maximum-efficiency point is beneficial; very light loading is not. This is why transformers should be sized for the realistic design-year demand and why low-loss cores are used where load factor is low.

Percentage energy loss = (core energy loss + copper energy loss) / energy sold × 100. Copper energy loss uses the loss of load factor.

Assumption: month = 30 days = 720 h.

Step 1: Load factor

$$LF = \frac{\text{Energy}}{P_{max} \times T} = \frac{4800}{20 \times 720} = 0.3333$$

Step 2: Loss of load factor

$$\begin{aligned} LLF &= 0.3\,LF + 0.7\,LF^2 \\ &= 0.3(0.3333) + 0.7(0.3333)^2 \\ &= 0.1000 + 0.0778 = 0.17778 \end{aligned}$$

Step 3: Copper loss at peak load

$$\begin{aligned} S_{peak} &= \frac{20}{0.85} = 23.529 \text{ kVA} \\ P_{cu,peak} &= 300 \times \left(\frac{23.529}{25}\right)^2 = 300 \times 0.8858 = 265.74 \text{ W} \end{aligned}$$

Step 4: Monthly energy losses

$$\begin{aligned} E_{cu} &= 0.26574 \times 0.17778 \times 720 = 34.02 \text{ kWh} \\ E_{core} &= 0.072 \times 720 = 51.84 \text{ kWh} \\ E_{loss} &= 34.02 + 51.84 = 85.86 \text{ kWh} \end{aligned}$$

Step 5: Percentage energy loss

$$\%E_{loss} = \frac{85.86}{4800} \times 100 = 1.79\%$$

(Based on energy input $4800 + 85.86 = 4885.86$ kWh, it is $85.86/4885.86 = 1.76\%$.)

Answer: Percentage energy loss ≈ 1.79% of energy sold (monthly loss ≈ 85.9 kWh: 51.84 kWh core loss + 34.02 kWh copper loss).

Note that core loss is the larger part, because the load factor is low (0.33): the transformer is energised all month but carries peak load only briefly.

Choosing a distribution transformer's location, voltage levels and LT layout sets the voltage quality, losses and cost for all consumers it serves.

Location

• Load centre: at the centre of gravity of the load ($\bar{x} = \sum P_i x_i / \sum P_i$) so LT feeders are short and equally loaded.
• Voltage drop: farthest consumer within allowed drop (about 5–10%).
• HT access: near the 11 kV line for a short tap.
• Accessibility and safety: near roads; away from schools and crowds; firm, flood-free ground; space for earthing.
• Future growth and land availability.

Primary voltage level

• Depends on the available supply and distance: 11 kV is standard in Nepal; 33 kV for long rural feeders, where loss ∝ $1/V^2$ and permissible load-distance ∝ $V^2$.
• Higher voltage costs more in insulation, transformers and switchgear, so it is used only where load × distance is large.
• Standardisation with the existing network and spare stock.

Secondary voltage level

• 400 V three-phase, 230 V single-phase (Nepal standard) for domestic and small commercial loads.
• Single-phase 230 V only for small, scattered domestic loads; three-phase where motors exist.
• Large consumers may be supplied directly at 11 kV with their own transformers.

Layout

• Radial LT feeders (2–4 per transformer) in different directions, so each feeder is short and loads are balanced on phases.
• Feeder length decided by voltage drop (rural) or thermal limit (urban).
• Follow roads; keep service drops short.
• Use ABC in theft-prone or congested areas, underground cable in dense city centres.
• Provide LT fuses/MCCBs per feeder, lightning arresters and drop-out fuses on HT side, proper earthing.
• Leave room for splitting the area with an extra transformer as load grows.

        11 kV
   ======+======
         |
        [T] 11/0.4 kV
   ______|______
   |     |     |
  F1    F2    F3   LT radial feeders

With the same load density (kW per km) and the same conductor, doubling the length of a uniformly loaded line doubles both its total load and its resistance.

Original line (length $l$, uniform load density $\lambda$ kW/km, sending-end current $I \propto \lambda l$, resistance $R = rl$):

$$P_{loss,1} = 3 \times \frac{I^2 R}{3} = I^2 r l$$

(for a uniformly distributed load the loss is one-third of the end-loaded value).

Doubled line (length $2l$): total load $= 2\lambda l$, so current $I_2 = 2I$; resistance $R_2 = 2rl$:

$$\begin{aligned} P_{loss,2} &= (2I)^2 \times r(2l) = 8 I^2 r l \\ \frac{P_{loss,2}}{P_{loss,1}} &= 8 \end{aligned}$$

Result:

• Power loss increases 8 times ($\propto l^3$).
• The load supplied only doubles, so the percentage power loss increases 4 times ($\propto l^2$).
• Similarly, voltage drop ($\propto I \times l$) increases 4 times.

This is why LT feeders must be kept short; in a large area it is better to use more transformers placed near load centres than to extend long LT lines.

Energy loss = core loss for all hours + copper loss at peak × loss of load factor × hours.

Assumptions: "additional 10 kVA for subsequent two years" means the load is 85 kVA in year 2 and 95 kVA in year 3. The given copper loss (1100 W) is at rated load (100 kVA). The year has 8760 h.

Step 1: Third-year peak load

$$S_3 = 75 + 10 + 10 = 95 \text{ kVA}$$

Step 2: Loss of load factor

$$\begin{aligned} LLF &= 0.3\,LF + 0.7\,LF^2 = 0.3(0.8) + 0.7(0.8)^2 \\ &= 0.24 + 0.448 = 0.688 \end{aligned}$$

Step 3: Copper loss at 95 kVA

$$P_{cu} = 1100 \times \left(\frac{95}{100}\right)^2 = 1100 \times 0.9025 = 992.75 \text{ W}$$

Step 4: Annual energy losses

$$\begin{aligned} E_{cu} &= 0.99275 \times 0.688 \times 8760 = 5983.19 \text{ kWh} \\ E_{core} &= 0.200 \times 8760 = 1752.00 \text{ kWh} \\ E_{total} &= 5983.19 + 1752.00 = 7735.19 \text{ kWh} \end{aligned}$$

Answer: Energy loss in the third year ≈ 7735 kWh (copper 5983 kWh + core 1752 kWh).

Because the load factor is high (0.8) and the transformer is loaded at 95%, copper loss is the dominant part of the energy loss.

Assumptions:

• The 40 kW peak is shared equally by the four feeders (10 kW each) and the load is uniformly distributed along each feeder.
• The reactance is 0.25 Ω/km per phase.
• The transformer iron loss (200 W) and copper loss (480 W) are the losses at this peak load (the transformer rating is not given).
• % distribution loss is taken as peak power loss as a percentage of peak demand.

Step 1: Current in each feeder

$$I = \frac{P}{\sqrt{3} V \cos\phi} = \frac{10{,}000}{\sqrt{3} \times 380 \times 0.85} = 17.875 \text{ A}$$

Step 2: Feeder impedance

$$\begin{aligned} R &= 0.9 \times 1.5 = 1.35\ \Omega, \quad X = 0.25 \times 1.5 = 0.375\ \Omega \\ \sin\phi &= \sqrt{1 - 0.85^2} = 0.5268 \end{aligned}$$

Step 3: LT line loss

For uniformly distributed load, loss = 1/3 of the end-loaded value:

$$\begin{aligned} P_{loss,feeder} &= \frac{3 I^2 R}{3} = I^2 R = 17.875^2 \times 1.35 = 431.33 \text{ W} \\ P_{loss,LT} &= 4 \times 431.33 = 1725.31 \text{ W} \end{aligned}$$

Step 4: Total distribution loss

$$\begin{aligned} P_{loss} &= 1725.31 + 200 + 480 = 2405.31 \text{ W} \\ \%\text{ loss} &= \frac{2405.31}{40{,}000} \times 100 = 6.01\% \end{aligned}$$

(LT line alone: 4.31%; transformer: 1.70%. On the basis of input power, $2405.31/42{,}405.31 = 5.67\%$.)

Step 5: Voltage drop in LT line

For uniformly distributed load, drop = 1/2 of the end-loaded value:

$$\begin{aligned} \Delta V_L &= \frac{\sqrt{3}\, I (R\cos\phi + X\sin\phi)}{2} \\ &= \frac{\sqrt{3} \times 17.875 \times (1.35 \times 0.85 + 0.375 \times 0.5268)}{2} \\ &= \frac{\sqrt{3} \times 17.875 \times 1.3450}{2} = 20.82 \text{ V} \\ \%\Delta V &= \frac{20.82}{380} \times 100 = 5.48\% \end{aligned}$$

Answer:

• (i) % distribution loss ≈ 6.01% of peak demand (2.41 kW)
• (ii) % voltage drop in each LT feeder ≈ 5.48% (20.82 V)

The voltage drop is within the usual ±10% LT limit (and close to a 5% limit), and losses are acceptable.

Assumptions: load is shared equally among the feeders of each option and uniformly distributed along each feeder; LT line voltage = 400 V (transformer secondary); $\cos\phi = 0.8$, $\sin\phi = 0.6$; $z = 1.1 + j0.25$ Ω/km per phase.

Formulae (uniformly distributed load)

$$\begin{aligned} I &= \frac{P_f}{\sqrt{3} \times 400 \times 0.8} \\ P_{loss} &= n \times I^2 R \quad (R = 1.1\,l) \\ \Delta V_L &= \frac{\sqrt{3}\, I (R\cos\phi + X\sin\phi)}{2} \end{aligned}$$

where $n$ = number of feeders, $P_f = 50/n$ kW per feeder, $l$ = feeder length.

Option I: 4 feeders × 3 km

$$\begin{aligned} P_f &= 12.5 \text{ kW}, \quad I = \frac{12{,}500}{\sqrt{3}(400)(0.8)} = 22.553 \text{ A} \\ R &= 3.3\ \Omega, \quad X = 0.75\ \Omega, \quad R\cos\phi + X\sin\phi = 3.09\ \Omega \\ P_{loss} &= 4 \times 22.553^2 \times 3.3 = 6713.9 \text{ W} \Rightarrow 13.43\% \\ \Delta V &= \frac{\sqrt{3}(22.553)(3.09)}{2} = 60.35 \text{ V} \Rightarrow 15.09\% \end{aligned}$$

Option II: 2 feeders × 4 km

$$\begin{aligned} P_f &= 25 \text{ kW}, \quad I = 45.105 \text{ A} \\ R &= 4.4\ \Omega, \quad X = 1.0\ \Omega, \quad R\cos\phi + X\sin\phi = 4.12\ \Omega \\ P_{loss} &= 2 \times 45.105^2 \times 4.4 = 17{,}903.6 \text{ W} \Rightarrow 35.81\% \\ \Delta V &= \frac{\sqrt{3}(45.105)(4.12)}{2} = 160.94 \text{ V} \Rightarrow 40.23\% \end{aligned}$$

Option III: 3 feeders × 2 km

$$\begin{aligned} P_f &= 16.667 \text{ kW}, \quad I = 30.070 \text{ A} \\ R &= 2.2\ \Omega, \quad X = 0.5\ \Omega, \quad R\cos\phi + X\sin\phi = 2.06\ \Omega \\ P_{loss} &= 3 \times 30.070^2 \times 2.2 = 5967.9 \text{ W} \Rightarrow 11.94\% \\ \Delta V &= \frac{\sqrt{3}(30.070)(2.06)}{2} = 53.65 \text{ V} \Rightarrow 13.41\% \end{aligned}$$

Summary

Answer: Option III (3 feeders × 2 km) gives the lowest peak loss (11.94%) and voltage drop (13.41%); option II is the worst. All options exceed the usual LT voltage drop limit (about 5–10%), so in practice the area should be split with an additional transformer, a larger conductor, or a higher voltage (11 kV) brought closer to the loads.

Method (optimum single capacitor)

With all node voltages taken as 1 p.u., the reactive current in a section equals its reactive power flow $Q_i$ (p.u.). A shunt capacitor $Q_c$ at the end node reduces the reactive flow in every section between the source and that node from $Q_i$ to $(Q_i - Q_c)$. The loss reduction is

$$\begin{aligned} \Delta P_L &= \sum_i R_i\left[Q_i^2-(Q_i-Q_c)^2\right] = \sum_i R_i\left(2Q_iQ_c-Q_c^2\right) \end{aligned}$$

For maximum loss reduction, $\dfrac{d(\Delta P_L)}{dQ_c} = \sum_i R_i(2Q_i-2Q_c)=0$, so

$$Q_{c,opt} = \frac{\sum_i R_iQ_i}{\sum_i R_i}$$

i.e. the optimum capacitor equals the resistance-weighted average of the reactive flows in the sections it relieves. The ratio does not depend on whether $R$ is in ohms or p.u., so kVAR can be used directly.

 S --(1+j1)-- A --(1+j1)-- B --(1+j2)-- C
              |            |            |
          100 kVA       50 kVA      100 kVA
          0.8 lag      0.85 lag     0.8 lag   [Qc]

Reactive load at each node

Reactive flow in each section

Optimum capacitor

$$Q_c = \frac{146.34+86.34+60}{1+1+1} = \frac{292.68}{3} = 97.56\ \text{kVAR}$$

In p.u. on 1000 kVA: $Q_c = 0.0976$ p.u.

Check of the loss reduction (p.u. on 1000 kVA):

$$\begin{aligned} \Delta P_L &= \sum R_i(2Q_iQ_c - Q_c^2)\\ &= 2(0.09756)(0.29268) - 3(0.09756)^2 = 0.02855\ \text{p.u.} \approx 28.6\ \text{kW} \end{aligned}$$

The capacitor is larger than the 60 kVAR load at C because it also relieves the heavier flows in S–A and A–B. In practice the nearest standard bank (about 100 kVAR) would be used.

Answer: Optimum capacitor at node C ≈ 97.6 kVAR (≈ 100 kVAR standard).

FALSE.

Justification

In urban areas the load density is high and the load is spread almost uniformly along streets. The design does not start from a fixed transformer location; it starts from the LT conductor and transformer rating, and the transformer locations follow from them.

Usual urban design steps:

1. Estimate the load density (kW/km or kVA/km²) of the area from consumer surveys and load forecast.
2. Select a standard LT conductor size (for example ACSR Rabbit/Weasel or ABC cable used by NEA), considering current capacity, economy and uniformity of stores.
3. With this conductor and the permitted LT voltage drop (about 5–8 %) and loss limit, find the maximum length of LT feeder that one transformer can supply.
4. From that service area and the load density, choose the standard transformer size (e.g. 100, 200, 300 kVA).
5. Finally place the transformers at the load centres of the service areas, so that each serves an area of the computed size.

Because the load is everywhere, a transformer can be placed in many points; the controlling constraint is how far the selected conductor can carry power within the voltage-drop limit. So the order in the statement (location first, then conductor) is reversed.

(By contrast, in rural areas the loads are clustered in villages, so the load centre is fixed first by the village location.)

Assumption: "uniformly varying load" is taken as a uniformly increasing load, i.e. the load density rises linearly from zero at the sending end to a maximum at the far end (Gonen's case). No power factor is given, so the maximum possible drop is found using $|z|$ (the drop $I(R\cos\phi+X\sin\phi)$ is largest when it equals $I|Z|$).

Current distribution

Let load density $= kx$ (A/km). The current at distance $x$ from the source is the load beyond $x$:

$$\begin{aligned} I(x) &= \int_x^{l} k\,u\,du = \frac{k}{2}(l^2-x^2),\quad I_s = I(0) = \frac{kl^2}{2}\\ I(x) &= I_s\left(1-\frac{x^2}{l^2}\right) \end{aligned}$$

Voltage drop (per phase)

$$V_D = \int_0^l z\,I_s\left(1-\frac{x^2}{l^2}\right)dx = z I_s\left(l-\frac{l}{3}\right) = \frac{2}{3}I_s z l$$

Power loss (three-phase)

$$P_L = 3\int_0^l r I_s^2\left(1-\frac{x^2}{l^2}\right)^2dx = 3 r I_s^2 l\left(1-\frac{2}{3}+\frac{1}{5}\right) = 3\cdot\frac{8}{15}I_s^2 r l$$

Substitution

$I_s = 100$ A, $l = 16$ km, $R = rl = 0.6\times16 = 9.6\ \Omega$, $|z| = \sqrt{0.6^2+0.3^2} = 0.6708\ \Omega$/km.

$$\begin{aligned} P_L &= 3\times\frac{8}{15}\times100^2\times9.6 = 153{,}600\ \text{W} = 153.6\ \text{kW}\\ V_{D,ph} &= \frac{2}{3}\times100\times0.6708\times16 = 715.5\ \text{V}\\ V_{D,L} &= \sqrt3\times715.5 = 1239.4\ \text{V}\\ \%V_D &= \frac{1239.4}{11000}\times100 = 11.27\ \% \end{aligned}$$

Comparison with other load patterns

Answer: Power loss = 153.6 kW; maximum voltage drop ≈ 715.5 V per phase = 1239 V line (≈ 11.3 % of 11 kV).

Percentage voltage regulation (≈ percentage voltage drop) of a feeder is the drop divided by the rated voltage. For a given power delivered over a given distance, it varies as $1/V_L^2$.

Derivation

Consider a three-phase feeder of per-phase impedance $R + jX$ supplying $P$ (W) at power factor $\cos\phi$ lagging, line voltage $V_L$.

Line current:

$$I = \frac{P}{\sqrt3\,V_L\cos\phi}$$

Approximate per-phase voltage drop (phasor angle small):

$$V_D = I(R\cos\phi + X\sin\phi)$$

Per-unit (or %) regulation is drop over phase voltage $V_L/\sqrt3$:

$$\begin{aligned} \%VR &= \frac{I(R\cos\phi+X\sin\phi)}{V_L/\sqrt3}\times100\\ &= \frac{\sqrt3}{V_L}\cdot\frac{P}{\sqrt3 V_L\cos\phi}(R\cos\phi+X\sin\phi)\times100\\ &= \frac{P(R + X\tan\phi)}{V_L^2}\times100 \end{aligned}$$

For a given load $P$, power factor and feeder (fixed $R$, $X$):

$$\%VR \propto \frac{1}{V_L^2}$$

Interpretation and consequences

• Doubling the primary voltage (e.g. 11 kV to 22 kV) cuts the % drop to one-quarter for the same load and length.
• Equivalently, for the same % regulation, the load-distance product ($P\times l$) a feeder can carry rises as $V_L^2$:

$$\frac{(P l)_2}{(P l)_1} = \left(\frac{V_{L2}}{V_{L1}}\right)^2$$

• The same reasoning gives % power loss:

$$\%P_L = \frac{3I^2R}{P}\times100 = \frac{P R}{V_L^2\cos^2\phi}\times100 \propto \frac{1}{V_L^2}$$

Example

A feeder with $R = 2\ \Omega$, $X = 2\ \Omega$ carries 1 MW at 0.8 p.f. ($\tan\phi = 0.75$):

Tripling the voltage reduces regulation by a factor of 9, confirming the inverse-square law. This is why long rural feeders and higher loads call for higher primary voltages (33 kV instead of 11 kV).

Assumptions: the 80 kVA peak is shared equally by the four lines, load on each line is uniformly distributed, lines are 3-phase at 400 V, and a month = 30 days = 720 h.

Step 1: Current per line

Peak load per line $= 80/4 = 20$ kVA.

$$I = \frac{20\,000}{\sqrt3\times400} = 28.87\ \text{A}$$

Step 2: Peak power loss per line (UDL)

For a uniformly distributed load, 3-phase loss $= 3\cdot\frac13 I^2 R = I^2 R$, where $R = 1\times1.5 = 1.5\ \Omega$ per phase.

$$P_{L,line} = 28.87^2\times1.5 = 1250\ \text{W}$$

Total peak loss in 4 lines:

$$P_{L,peak} = 4\times1250 = 5000\ \text{W} = 5\ \text{kW}$$

Step 3: Loss load factor

$$\begin{aligned} LLF &= 0.3LF + 0.7LF^2\\ &= 0.3(0.4) + 0.7(0.4)^2 = 0.12 + 0.112 = 0.232 \end{aligned}$$

Step 4: Monthly energy loss

$$\begin{aligned} E_{loss} &= P_{L,peak}\times LLF\times720\\ &= 5\times0.232\times720 = 835.2\ \text{kWh} \end{aligned}$$

Summary

Note: if the whole line load were assumed lumped at the far end, the loss would be three times larger (2505.6 kWh); the uniform-load assumption is the usual one for LT distribution lines serving houses along the route.

Answer: Monthly energy loss in the lines ≈ 835 kWh.

FALSE.

Justification

In rural areas the loads are not uniform; they appear as clusters (villages, settlements, market centres) separated by empty land. The design must therefore start from the load centre, not from the transformer size.

Correct sequence for rural distribution design:

1. Survey the area and identify the settlements (load clusters) and the number and type of consumers.
2. Estimate the present and forecast demand of each cluster (e.g. 0.3–0.5 kW per household with diversity, plus small industries, irrigation pumps).
3. Locate the load centre (centre of gravity of the loads) of each cluster:

$$x = \frac{\sum P_ix_i}{\sum P_i},\quad y = \frac{\sum P_iy_i}{\sum P_i}$$

4. Then select the transformer size (standard 25, 50, 100 kVA) to match the forecast demand of that load centre, keeping reserve for growth.
5. Choose LT conductor and length so that voltage drop and loss at the farthest consumer stay within limits; extend the 11 kV line to the transformer.

So the load centre is chosen first and the transformer is sized to it. Selecting a transformer size first and then "sizing the load centre" reverses cause and effect.

A transformer has two losses: the core (no-load) loss, present all 720 h of the month, and the copper loss, which varies with the square of the load and is averaged using the loss load factor (LLF). Month taken as 30 days = 720 h.

Step 1: Core loss energy

$$E_{core} = 120\times720 = 86\,400\ \text{Wh} = 86.4\ \text{kWh}$$

Step 2: Copper loss at peak

Copper loss varies as $(S/S_{rated})^2$:

$$P_{cu,peak} = 200\times\left(\frac{30}{50}\right)^2 = 200\times0.36 = 72\ \text{W}$$

Step 3: Loss load factor

$$LLF = 0.3(0.3) + 0.7(0.3)^2 = 0.09 + 0.063 = 0.153$$

Step 4: Copper loss energy

$$E_{cu} = 72\times0.153\times720 = 7931.5\ \text{Wh} = 7.93\ \text{kWh}$$

Step 5: Energy delivered in the month

Peak demand in kW $= 30\times0.9 = 27$ kW.

$$E_{out} = 27\times0.3\times720 = 5832\ \text{kWh}$$

Step 6: Percentage loss

$$\begin{aligned} E_{loss} &= 86.4 + 7.93 = 94.33\ \text{kWh}\\ \%\,\text{loss} &= \frac{94.33}{5832}\times100 = 1.62\ \% \end{aligned}$$

(If expressed on input energy, $94.33/(5832+94.33) = 1.59\ \%$.)

Observation: at low load factor the core loss dominates (over 90 % of the loss), so for rural transformers with low LF, low no-load-loss (e.g. amorphous core) designs are preferred.

Answer: Monthly loss ≈ 94.3 kWh, i.e. about 1.62 % of the energy delivered.

Formulas (uniformly distributed load, 3-phase)

For each feeder of length $l$ carrying sending-end current $I$:

$$\begin{aligned} P_L &= 3\cdot\frac13 I^2 r l = I^2 r l\\ V_{D,ph} &= \frac12 I(r\cos\phi + x\sin\phi)\,l\\ \%V_D &= \frac{V_{D,ph}}{V_L/\sqrt3}\times100 \end{aligned}$$

The peak load is shared equally between the feeders.

Given: $P = 50$ kW, p.f. 0.8 ($\sin\phi = 0.6$), $V_L = 400$ V (secondary of 11/0.4 kV), $r = 1.1\ \Omega$/km, $x = 0.25\ \Omega$/km. $r\cos\phi + x\sin\phi = 1.1(0.8) + 0.25(0.6) = 1.03\ \Omega$/km.

Option I: 4 feeders × 2 km

$$\begin{aligned} P_{feeder} &= 50/4 = 12.5\ \text{kW},\quad I = \frac{12\,500}{\sqrt3\times400\times0.8} = 22.55\ \text{A}\\ P_L &= 4\times(22.55^2\times1.1\times2) = 4\times1118.98 = 4475.9\ \text{W}\\ \%P_L &= \frac{4475.9}{50\,000}\times100 = 8.95\ \%\\ V_{D,ph} &= \frac12\times22.55\times1.03\times2 = 23.23\ \text{V}\\ \%V_D &= \frac{23.23}{230.94}\times100 = 10.06\ \% \end{aligned}$$

Option II: 2 feeders × 4 km

$$\begin{aligned} P_{feeder} &= 25\ \text{kW},\quad I = \frac{25\,000}{\sqrt3\times400\times0.8} = 45.11\ \text{A}\\ P_L &= 2\times(45.11^2\times1.1\times4) = 2\times8951.8 = 17\,903.6\ \text{W}\\ \%P_L &= \frac{17\,903.6}{50\,000}\times100 = 35.81\ \%\\ V_{D,ph} &= \frac12\times45.11\times1.03\times4 = 92.92\ \text{V}\\ \%V_D &= \frac{92.92}{230.94}\times100 = 40.23\ \% \end{aligned}$$

Comparison

Halving the number of feeders doubles both current and length, so loss rises 4 times (∝ $I^2 l$) and drop 4 times (∝ $I l$). Option I is clearly better; even it exceeds the usual LT drop limit (about 5–8 %), suggesting a bigger conductor or more feeders/transformers.

Answer: Option I: 8.95 % loss, 10.06 % drop; Option II: 35.81 % loss, 40.23 % drop.

Formulas (uniformly distributed load, 3-phase)

For each feeder of length $l$ carrying sending-end current $I$:

$$\begin{aligned} P_L &= 3\cdot\frac13 I^2 r l = I^2 r l\\ V_{D,ph} &= \frac12 I(r\cos\phi + x\sin\phi)\,l\\ \%V_D &= \frac{V_{D,ph}}{V_L/\sqrt3}\times100 \end{aligned}$$

The peak load is shared equally between the feeders.

Given: $P = 40$ kW, p.f. 0.8 ($\sin\phi = 0.6$), 4 feeders, $V_L = 380$ V, $l = 2$ km, $r = 1.1$, $x = 0.25\ \Omega$/km.

Current per feeder

$$P_{feeder} = \frac{40}{4} = 10\ \text{kW},\quad I = \frac{10\,000}{\sqrt3\times380\times0.8} = 18.99\ \text{A}$$

i) Percentage peak power loss

$$\begin{aligned} P_{L,feeder} &= I^2 r l = 18.99^2\times1.1\times2 = 793.5\ \text{W}\\ P_{L,total} &= 4\times793.5 = 3174.1\ \text{W}\\ \%P_L &= \frac{3174.1}{40\,000}\times100 = 7.94\ \% \end{aligned}$$

ii) Maximum percentage voltage drop

The maximum drop occurs at the far end of each feeder.

$$\begin{aligned} r\cos\phi + x\sin\phi &= 1.1(0.8)+0.25(0.6) = 1.03\ \Omega/\text{km}\\ V_{D,ph} &= \frac12\times18.99\times1.03\times2 = 19.56\ \text{V}\\ V_{ph} &= 380/\sqrt3 = 219.39\ \text{V}\\ \%V_D &= \frac{19.56}{219.39}\times100 = 8.92\ \% \end{aligned}$$

(Line value of drop $= \sqrt3\times19.56 = 33.88$ V.)

The drop is above the common LT limit of about 5–8 %, so a larger conductor or a shorter LT reach would be needed in practice.

Answer: i) 7.94 % ii) 8.92 %.

The distribution transformer should be placed at the load centre (centre of gravity of the loads), because this minimises the load-moment ($\sum P_i d_i$), and so the LT voltage drop and loss.

Since the villages are 0.6 km from each other, they form an equilateral triangle of side 0.6 km.

              C (30 kW)
             /\
        0.6 /  \ 0.6
           / T  \
          /______\
   A (20 kW) 0.6  B (50 kW)

Coordinates (A at origin, AB along x-axis)

Load centre

$$\begin{aligned} \bar x &= \frac{\sum P_ix_i}{\sum P_i} = \frac{20(0)+50(0.6)+30(0.3)}{100} = \frac{39}{100} = 0.39\ \text{km}\\ \bar y &= \frac{\sum P_iy_i}{\sum P_i} = \frac{30(0.5196)}{100} = 0.156\ \text{km} \end{aligned}$$

Distance from A:

$$d = \sqrt{0.39^2+0.156^2} = 0.42\ \text{km},\quad \theta = \tan^{-1}\frac{0.156}{0.39} = 21.8^\circ$$

Engineering judgement

• The transformer is located about 0.42 km from A, at 21.8° from line AB towards C, i.e. pulled towards B, the largest load.
• In practice it is placed at the nearest accessible point (road side, public land) to this centre, and a 3-phase transformer of about 100–125 kVA standard size (100 kW at ≈0.8–0.9 p.f. plus growth) would be chosen.

Answer: Transformer at (0.39 km, 0.156 km) from A, i.e. 0.42 km from A at 21.8° to AB towards C.