Electrical Load Characteristics and Load Forecast

Load factor (LF) is the ratio of average load to peak load. Loss load factor (LLF) is the ratio of average power loss to the loss at peak load. Because loss is proportional to $I^2$, LLF always lies between $LF^2$ and $LF$. The usual relation is

$$LLF = k\,LF + (1-k)\,LF^2, \qquad k \approx 0.2\text{–}0.3$$

Definitions

For a period $T$ with peak load $P_m$, and assuming constant voltage and power factor:

$$LF = \frac{\frac{1}{T}\int_0^T P\,dt}{P_m}, \qquad LLF = \frac{\frac{1}{T}\int_0^T P^2\,dt}{P_m^2}$$

since loss $\propto I^2 \propto P^2$.

Suitable load curve

Take a two-step daily load curve. The load is at peak $P_m$ for time $t$, and at a lower load $P_1$ for the rest of the time $(T - t)$.

 P
 Pm |-----+
    |     |
 P1 |     +--------------------+
    |                          |
    +-----+--------------------+--> time
    0     t                    T

Then

$$LF = \frac{P_mt + P_1(T-t)}{P_mT}, \qquad LLF = \frac{P_m^2t + P_1^2(T-t)}{P_m^2T}$$

Limiting cases

Case 1: the off-peak load is zero ($P_1 \to 0$, a short sharp peak).

$$LF = \frac{t}{T}, \qquad LLF = \frac{t}{T} \;\Rightarrow\; LLF = LF$$

Case 2: the peak lasts a very short time ($t \to 0$, a nearly flat load at $P_1$).

$$LF = \frac{P_1}{P_m}, \qquad LLF = \frac{P_1^2}{P_m^2} \;\Rightarrow\; LLF = LF^2$$

Every real load curve lies between these two extremes, so

$$LF^2 \le LLF \le LF$$

General relation

LLF is expressed as a weighted combination of the two limits:

$$LLF = k\,LF + (1-k)\,LF^2$$

Here $k$ is found from measured load curves of the system. For typical distribution feeders $k = 0.3$ (Buller and Woodrow), giving $LLF = 0.3LF + 0.7LF^2$. Some utilities use $k = 0.2$ for urban feeders, giving $LLF = 0.2LF + 0.8LF^2$.

Example: for $LF = 0.5$, $LLF = 0.3(0.5) + 0.7(0.25) = 0.325$. The energy loss per year is then $P_{loss,peak}\times0.325\times8760$ kWh.

The 5th-year peak load of the load center is the sum of each consumer class's own peak, multiplied by its contribution (coincidence) factor to the system peak.

Method

For each class:

1. Consumers electrified by the 5th year: $N_5 = \text{coverage}\times N_{potential} = 0.7N$.
2. Monthly consumption per consumer in the 5th year, with 5% growth per year compounded over 5 years:

$$E_5 = E_0(1+0.05)^5 = 1.27628\,E_0$$

3. Daily energy per consumer:

$$E_d = \frac{E_5}{\text{effective days per month}}$$

4. Peak demand per consumer, using the load factor:

$$P_{peak} = \frac{\text{average demand}}{LF} = \frac{E_d}{24\times LF}$$

5. Class peak $= N_5\times P_{peak}$. Its contribution to the load-center peak = class peak × contribution factor.

Domestic class (sample)

$$\begin{aligned} N_5 &= 0.7\times100 = 70\\ E_5 &= 30\times1.27628 = 38.288\ \text{kWh/month}\\ E_d &= 38.288/30 = 1.2763\ \text{kWh/day}\\ P_{peak} &= \frac{1.2763}{24\times0.25} = 0.2127\ \text{kW}\\ \text{Class peak} &= 70\times0.2127 = 14.890\ \text{kW}\\ \text{Contribution} &= 1.0\times14.890 = 14.890\ \text{kW} \end{aligned}$$

All classes

5th-year peak load

$$P_{peak,5} = 14.890 + 1.489 + 0.744 = 17.12\ \text{kW}$$

Answer: the 5th-year peak load of the load center is about 17.1 kW. If the survey year is counted as year 1, so that growth applies for only 4 years, the result is $17.12/1.05 = 16.31$ kW.

TRUE.

Small area load forecasting predicts the location, amount and timing of load growth in small zones (a few hectares to a few km², or a feeder/distribution transformer service area). The simulation (land-use) method is better suited for this than trending.

Justification:

• In a small area the load does not grow smoothly. It stays almost zero for years, then grows very fast when the land is developed, and then saturates (S-shaped growth curve). Trending of past load data cannot predict when this sudden growth will start.
• Many small areas have little or no load history (vacant land, new housing plots), so there is nothing to extrapolate.
• The simulation method uses land-use data (residential, commercial, industrial, vacant), zoning plans, road and infrastructure development, distance to urban centres and per-hectare load densities of each land-use class. It models how and where land will be developed and converts it into load.
• It can therefore forecast load in areas where development has not yet started, and it gives the spatial (map-wise) distribution of load needed to site substations, transformers and feeders.
• It also allows "what if" studies (e.g. a new highway or industrial estate).

Trending works well for large areas where growth is smooth and history is long; for small areas the simulation method gives more accurate spatial forecasts, though it needs more data and effort.

The loss of load factor (LLF) is the ratio of the average power loss to the power loss at peak load over a period. Since loss is proportional to the square of load (at constant voltage and power factor):

$$LLF = \frac{\sum P_i^2 t_i}{P_{max}^2 \, T}, \qquad LF = \frac{\sum P_i t_i}{P_{max} \, T}$$

The empirical relation is $LLF = K_1 \, LF + K_2 \, LF^2$ with $K_1 + K_2 = 1$ (so that LLF = 1 when LF = 1). Two equations are thus available to find $K_1$ and $K_2$.

Step 1: Tabulate the load pattern

Peak demand $P_{max} = 10$ MW, $T = 24$ h.

Step 2: Load factor

$$\begin{aligned} LF &= \frac{74}{10 \times 24} = 0.30833 \\ LF^2 &= 0.09507 \end{aligned}$$

Step 3: Loss of load factor

$$LLF = \frac{444}{10^2 \times 24} = \frac{444}{2400} = 0.185$$

Step 4: Solve for K1 and K2

Put $K_2 = 1 - K_1$:

$$\begin{aligned} LLF &= K_1 LF + (1 - K_1) LF^2 \\ K_1 &= \frac{LLF - LF^2}{LF - LF^2} \\ &= \frac{0.185 - 0.09507}{0.30833 - 0.09507} = \frac{0.08993}{0.21327} \\ &= 0.4217 \\ K_2 &= 1 - 0.4217 = 0.5783 \end{aligned}$$

Check: $0.4217 \times 0.30833 + 0.5783 \times 0.09507 = 0.1300 + 0.0550 = 0.185$ ✓

Answer: $K_1 \approx 0.42$, $K_2 \approx 0.58$, i.e. $LLF = 0.42\,LF + 0.58\,LF^2$ for this system.

The commonly used Buller–Woodrow values are $K_1 = 0.3$, $K_2 = 0.7$; this load pattern (a sharp evening peak with zero load at night) gives a somewhat larger $K_1$. The value of LLF (0.185) lies between $LF^2$ (0.095) and $LF$ (0.308), as it must.

Loss of load factor (LLF), also called loss factor, is the ratio of the average power loss in a system element over a period to the power loss at peak load during that period:

$$LLF = \frac{\text{Average power loss}}{\text{Power loss at peak load}} = \frac{\text{Energy loss in time } T}{P_{loss,peak} \times T}$$

Since copper loss varies as the square of the current, $LLF = \frac{1}{T}\int_0^T \left(\frac{P}{P_{max}}\right)^2 dt$. It always lies between $LF^2$ and $LF$, and an empirical relation is used in practice:

$$LLF = 0.3\,LF + 0.7\,LF^2$$

Significance in distribution design:

• Energy loss in feeders and transformer windings is found simply as $E_{loss} = P_{loss,peak} \times LLF \times 8760$ kWh/year, without a full hourly load curve.
• It is used to cost the losses over the life of the system, which decides the economic conductor size and transformer rating.
• Because LLF < LF, the % energy loss is lower than the % peak power loss ($\%E_{loss} = \%P_{loss,peak} \times LLF/LF$); designers use this to set acceptable peak loss limits.
• It shows the benefit of improving load factor (flatter load means lower loss per kWh sold).

Coincidence factor is the ratio of the maximum demand of a group of consumers taken together to the sum of the individual maximum demands of those consumers:

$$\text{Coincidence factor} = \frac{\text{Maximum demand of the group}}{\sum \text{Individual maximum demands}} = \frac{1}{\text{Diversity factor}} \le 1$$

It is less than 1 because consumers do not all reach their own peaks at the same time. For example, if 50 houses each have a peak of 2 kW but the group peak is 60 kW, the coincidence factor is $60/100 = 0.6$.

Significance in distribution design:

• The peak load on a distribution transformer, LT feeder or service main is found as (number of consumers) × (individual peak) × (coincidence factor). Without it, equipment would be badly oversized.
• It decreases as the number of consumers increases, so a transformer serving many consumers needs less capacity per consumer than a service drop serving one.
• It lets the designer choose smaller, cheaper transformers and conductors while still meeting the actual peak, and gives realistic voltage drop and loss calculations.
• Wrong (too high) values give idle capacity and high no-load losses; too low values cause overloading.

Responsibility factor (also called contribution factor) of a consumer or consumer class is the ratio of its load at the time of the system (or group) peak to its own maximum demand:

$$\text{Responsibility factor } c_i = \frac{\text{Load of consumer } i \text{ at time of system peak}}{\text{Maximum demand of consumer } i} \le 1$$

For example, a commercial class with a 10 kW peak at noon that draws only 5 kW at the evening system peak has a responsibility factor of 0.5.

The system peak is then $P_{system} = \sum c_i P_{max,i}$, and the coincidence factor of the group is $\sum c_i P_{max,i} / \sum P_{max,i}$.

Significance in distribution design:

• At a load centre with several consumer classes (domestic, commercial, industrial), the class peaks occur at different hours. The load centre peak is the sum of each class peak weighted by its responsibility factor, not their plain sum.
• It is the key input for sizing distribution transformers and feeders in load forecasting (e.g. "5th year contribution factor to peak").
• It identifies which class is responsible for the peak; tariffs (time-of-day pricing) and demand-side measures can then target that class.
• Using correct factors avoids oversizing equipment and reduces capital cost and no-load losses.

FALSE.

The ratio of average demand to peak demand of a feeder is its load factor:

$$LF = \frac{\text{Average demand}}{\text{Peak demand}} = \frac{\text{Energy in period } T}{P_{peak} \times T}$$

If average consumption were almost equal to the peak, the load factor would be close to 1, which happens only for a flat, round-the-clock load (e.g. a continuous-process industry).

Justification:

• A distribution feeder mainly supplies domestic and commercial consumers whose use varies strongly through the day: low at night, moderate in the day, and a sharp evening peak (lighting, cooking, TV).
• Typical feeder load factors are about 0.3–0.5 for residential/rural feeders and 0.5–0.7 for mixed urban feeders. So the average demand is usually only 30–60% of the peak.
• Example: a feeder with a 1 MW peak and 9,600 kWh/day has an average of 400 kW, i.e. LF = 0.4.
• This is why the feeder and transformer must be sized for the peak, not the average, and why energy loss is computed using the loss of load factor ($LLF = 0.3LF + 0.7LF^2$) rather than assuming peak loss all day.

A high load factor is desirable (better use of equipment, lower loss per kWh), and utilities try to raise it by demand-side management, but in practice the average is well below the peak.

Demand forecasting at a load centre (a village, town ward or the service area of one distribution transformer) estimates the peak demand and energy at a design year (usually the 5th year after commissioning, sometimes 10th/15th) so that the transformer, LT feeders and HT tap line can be sized. The method used in Nepal (NEA and IOE practice) is a consumer-class based end-use method.

Data required

These come from field survey, census data, NEA billing records of similar electrified areas and socio-economic studies.

Steps

1. Survey and classify consumers: domestic, commercial, non-commercial (institutions, schools, offices), industrial, irrigation, street lighting.
2. Base year consumption: fix average monthly kWh per consumer of each class from similar electrified areas.
3. Project consumer numbers to the design year $n$: $N_n = N_0 (1+p)^n \times \text{coverage factor}$, where $p$ is population growth rate. Include the implementation period in $n$.
4. Project consumption per consumer: $E_n = E_0 (1+g)^n$, where $g$ is consumption growth rate.
5. Daily energy per consumer: $E_{day} = E_n / \text{effective days per month}$.
6. Individual peak from load factor: $P_{ind} = \dfrac{E_{day}}{24 \times LF}$.
7. Class peak: $P_{class} = N_n \times P_{ind}$ (times a coincidence factor if given).
8. Load centre peak: $P_{LC} = \sum (\text{contribution factor} \times P_{class})$.
9. Add losses and convert to kVA: $S = P_{LC}(1 + \text{loss allowance}) / \cos\phi$; then choose the next standard transformer size (25, 50, 100, 200 kVA…).
10. Annual energy: $\sum N_n E_n \times 12$ for revenue and loss studies.

The result gives transformer rating, LT feeder loading, and the demand that the 11 kV feeder must carry.

Load forecasting is the prediction of future peak demand (kW) and energy (kWh) of a system or area, needed for planning generation, transmission and distribution. Methods differ by time horizon and approach.

By time horizon

Main methods

1. Trend analysis (extrapolation): past load data are fitted with a curve (linear $P = a + bt$, exponential $P = a e^{bt}$, polynomial, Gompertz S-curve) and extended into the future. Simple and cheap; suitable for large areas with smooth growth. It ignores causes of growth and fails when conditions change.

2. Econometric (regression) method: load is expressed as a function of factors such as GDP, population, per-capita income, electricity price and number of consumers, e.g. $E = a + b\,(GDP) + c\,(Pop)$. Coefficients come from regression on past data. Explains causes but needs forecasts of the economic variables.

3. End-use method: total demand = Σ (number of appliances/consumers × use per appliance × hours). Built up from consumer classes (domestic, commercial, industrial). Good for new areas and demand-side studies; the load-centre method used by NEA (consumer number × consumption × growth × coverage) is an end-use type method. Needs detailed survey data.

4. Time-series methods: autoregressive (AR), moving average (ARMA/ARIMA) and exponential smoothing models of hourly/daily load. Mainly for short-term forecasting.

5. Simulation (land-use) method: for small-area/spatial forecasting. Land is divided into small cells; land-use change (residential, commercial, industrial) is simulated and converted to load using load density per land-use class. Gives where and when load will appear.

6. Judgmental / Delphi method: opinions of experts are collected and refined in rounds. Used when data are lacking.

7. Artificial intelligence methods: artificial neural networks, fuzzy logic and machine learning models trained on historical load and weather data; mostly for short-term forecasting.

In practice, a combination is used: trending/econometric for the national system, end-use and simulation for distribution areas.

The load centre peak is found by projecting the number of consumers and their consumption to the 5th year, converting energy to peak using the load factor, and combining the class peaks with their contribution factors.

Assumptions: The 5th year is counted after project completion, so with a 2-year implementation period the growth period from the survey is $n = 5 + 2 = 7$ years for both population and consumption.

Formulae

$$\begin{aligned} N_5 &= N_0 (1 + p)^n \times \text{coverage} \\ E_5 &= E_0 (1 + g)^n \\ P_{ind} &= \frac{E_5 / \text{effective days}}{24 \times LF} \\ P_{LC} &= \sum \text{CF} \times N_5 \times P_{ind} \end{aligned}$$

Growth multipliers

$$\begin{aligned} (1.018)^7 &= 1.13301 \\ (1.05)^7 &= 1.40710 \end{aligned}$$

Number of consumers in 5th year

Consumption and individual peak

Example (domestic): $P_{ind} = \dfrac{1.4071}{24 \times 0.25} = 0.23452$ kW.

Class peak and contribution to load centre peak

Answer: 5th year peak load of the load centre ≈ 21.4 kW.

(If the implementation period were ignored, i.e. $n = 5$, the result would be lower; the 2-year period is included because consumers keep growing while the project is being built.)

FALSE.

The peak demand of a load centre is the coincident (simultaneous) maximum of the combined load, not the arithmetic sum of the class peaks.

Justification:

• Different consumer classes reach their maximum demand at different hours. Domestic load peaks in the evening (lighting, cooking), commercial load around midday/afternoon, and offices, schools and institutions during working hours.
• At the hour of the load centre peak, each class draws only a fraction of its own peak. This fraction is the contribution (responsibility) factor $c_i$.
• Therefore

$$P_{LC} = \sum c_i \, P_{class,i} \le \sum P_{class,i}$$

Example: domestic peak 20 kW (evening, $c = 1$), commercial peak 6 kW ($c = 0.5$ in the evening), institutional peak 4 kW ($c = 0.1$):

$$P_{LC} = 20 + 3 + 0.4 = 23.4 \text{ kW, not } 30 \text{ kW}$$

Adding the class peaks directly would oversize the distribution transformer and feeders, raising capital cost and no-load losses. The plain sum is correct only if all classes peak at the same time (all contribution factors equal to 1), which rarely happens.

The load pattern (Table 2) gives the load of each class in per unit of its peak. From it we find the equivalent full-load hours per day, then the peak from the daily energy.

Method:

$$\begin{aligned} h_{eq} &= \sum (\text{p.u. load} \times \text{hours}) \\ \text{Daily energy per consumer} &= \frac{\text{Monthly kWh}}{\text{Effective days}} \\ P_{ind} &= \frac{\text{Daily energy}}{h_{eq}}, \quad P_{class} = N \times P_{ind} \times \text{coincidence factor} \\ LF_{daily} &= \frac{h_{eq}}{24}, \quad LF_{annual} = \frac{\text{Annual energy}}{P_{ind} \times 8760} \end{aligned}$$

Equivalent hours from load pattern

• Class A: $0.1(6) + 0.4(4) + 0.2(4) + 0.2(4) + 1(3) + 0.4(3) = 8.0$ h
• Class B: $0.1(6) + 0.1(4) + 0.5(4) + 1(4) + 0.5(3) + 0.1(3) = 8.8$ h
• Class C: $0.1(6) + 0.1(4) + 1(4) + 1(4) + 0.1(3) + 0.1(3) = 9.6$ h

For each consumer class

The annual LF is lower than the daily LF for B and C because they consume only on effective days.

Transformer load curve (kW) = class peak × p.u. pattern

The transformer peak occurs at 18:00–21:00.

(iii) Contribution factor of each class

$$c = \frac{\text{Class load at transformer peak}}{\text{Class peak}}$$

• Class A: $10/10 = 1.0$
• Class B: $1.406/2.8125 = 0.5$
• Class C: $0.381/3.806 = 0.1$

For the distribution transformer

1. Peak load: $P = 10 + 1.406 + 0.381 = 11.79$ kW. With reactive power $Q = 10\tan(\cos^{-1}0.9) + 1.406\tan(\cos^{-1}0.8) + 0.381\tan(\cos^{-1}0.85) = 6.13$ kVAR, $S = \sqrt{11.79^2 + 6.13^2} = 13.29$ kVA.
2. Annual energy sales (LT losses neglected): $28{,}800 + 7{,}920 + 12{,}000 = 48{,}720$ kWh.
3. Load factor:

$$LF = \frac{48{,}720}{11.787 \times 8760} = 0.472$$

Answer: Transformer peak ≈ 11.79 kW (13.29 kVA), annual LF ≈ 0.47, annual energy sales = 48,720 kWh. Class contribution factors are 1.0, 0.5 and 0.1.

Small area load forecasting predicts the load of small zones (a feeder area, a grid cell of a few hectares) so that the location and timing of new substations, transformers and feeders can be planned.

The load of a small area follows an S-shaped growth curve (Gompertz-type), unlike the smooth growth of a large area:

 Load
  ^                       ___________  saturation
  |                    _/
  |                  _/
  |                 /    rapid growth
  |               _/
  |             _/
  |  __________/   dormant
  +-----------------------------------> Years
     |   I    |    II     |    III

1. Dormant period (I): the land is vacant or farmland; load is almost zero for many years.
2. Rapid growth (II): once development starts (new roads, housing, shops), load rises very fast, often over 5–10 years. The timing of this start is hard to predict.
3. Saturation (III): the area is fully developed; load grows slowly, only through higher consumption per consumer.

Forecasting using the curve:

• Each small area is located on its S-curve. Areas in the growth phase are forecast by fitting a Gompertz curve, $P(t) = a\,e^{-b e^{-ct}}$, to recent data, with the saturation level $a$ taken from land area × load density of the planned land use.
• Areas in the dormant phase cannot be trended; their start of growth is predicted by the simulation (land-use) method using zoning plans and infrastructure development.
• The sum of small-area forecasts is checked against the large-area (system) forecast.

The curve shows why trending alone fails for small areas and why spatial land-use information is required.

Energy loss in a line depends on how the load varies with time, not just on the peak. Since copper loss is proportional to the square of current, the energy loss is linked to the load factor (LF) through the loss of load factor (LLF).

Definitions

For a line of resistance $R$ (per phase) carrying load $P(t)$ at voltage $V$ and power factor $\cos\phi$:

$$\begin{aligned} P_{loss}(t) &= 3 I^2 R = \frac{P(t)^2 R}{V^2 \cos^2\phi} = k\,P(t)^2 \\ E_{loss} &= \int_0^T k P(t)^2 dt = k P_{max}^2 \, T \cdot \underbrace{\frac{1}{T}\int_0^T \left(\frac{P}{P_{max}}\right)^2 dt}_{LLF} \end{aligned}$$

So $E_{loss} = P_{loss,peak} \times LLF \times T$, and $LF = \dfrac{1}{T}\int_0^T \dfrac{P}{P_{max}} dt$.

Limits of LLF in terms of LF

Consider a two-step load: $P_{max}$ for time $t$ and $P_{min}$ for the rest $(T - t)$.

• Case 1, $P_{min} = 0$: $LF = t/T$ and $LLF = t/T$, so $LLF = LF$ (upper limit).
• Case 2, very short peak ($t \to 0$), off-peak $P_{min}$ for nearly all the time: $LF \approx P_{min}/P_{max}$ and $LLF \approx (P_{min}/P_{max})^2$, so $LLF = LF^2$ (lower limit).

Hence

$$LF^2 \le LLF \le LF$$

Real load curves lie between these, giving the empirical relation (Buller–Woodrow):

$$LLF = 0.3\,LF + 0.7\,LF^2$$

Dependency of % energy loss on LF

$$\%E_{loss} = \frac{P_{loss,peak} \times LLF \times T}{P_{max} \times LF \times T} \times 100 = \%P_{loss,peak} \times \frac{LLF}{LF}$$

With the empirical formula, $\dfrac{LLF}{LF} = 0.3 + 0.7\,LF$.

Example: peak loss 8% of peak demand.

Conclusions:

• For a given peak, a lower LF means fewer kWh lost but also fewer kWh sold; the % energy loss is always ≤ % peak power loss.
• For a given energy delivered, a higher LF lowers the peak, and since loss ∝ peak², the absolute energy loss falls.
• Energy loss can never be found by multiplying peak loss by hours; LLF must be used.
• Constant losses (transformer core loss, corona) do not depend on load, so for them LLF = 1.

Small area forecasting predicts the amount, location and timing of load growth in small zones. The two main methods are trending and simulation (land-use). The simulation method is explained below.

Simulation (land-use) method

The service area is divided into small cells (e.g. a grid of 1–10 hectare squares). Instead of extrapolating past load, the method simulates how land use will change and then converts land use into load.

 Base land-use map + zoning plan
           |
           v
 Global (system) forecast of growth
           |
           v
 Allocate growth to cells
 (preference / suitability scores)
           |
           v
 Future land use in each cell
           |
           v
 x load density (kW/ha) per class
           |
           v
 Cell load map --> substations, feeders

Steps:

1. Data collection: present land use of each cell (residential, commercial, industrial, agricultural, vacant, restricted), roads, railways, rivers, zoning by-laws, population and planned projects.
2. Global forecast: total growth of the whole region (customers or kW) is found from econometric/trending studies.
3. Suitability (preference) scoring: each vacant cell is scored for each land-use class from factors such as distance to roads and town centre, nearby land use, slope, and planned infrastructure.
4. Allocation: the global growth is allocated to the highest-scoring cells year by year, so that land use changes in a realistic spatial pattern.
5. Load conversion: future load of each cell = Σ (area of each land use × load density of that class), with end-use load curves giving the coincident peak.
6. Calibration: results are checked against present loads and the global forecast.

Merits: forecasts load in undeveloped areas, gives a load map useful for siting substations and routing feeders, and allows "what-if" scenarios.

Demerits: needs large data and computing effort.

(The trending method, by contrast, fits an S-curve (Gompertz) to each area's past load; it is simpler but fails for areas without load history.)

The load pattern gives each class's load in per unit of its own peak. The class peak is found from its daily energy and the equivalent full-load hours of the pattern; the load centre peak is then the maximum of the summed class load curves.

Step 1: Equivalent full-load hours per day

$$h_{eq} = \sum(\text{p.u. load} \times \text{duration})$$

• Class A: $0.1(6)+0.4(4)+0.2(4)+0.2(4)+1(3)+0.4(3) = 8.0$ h
• Class B: $0.1(6)+0.1(4)+0.5(4)+1(4)+0.5(3)+0.1(3) = 8.8$ h
• Class C: $0.1(6)+0.1(4)+1(4)+1(4)+0.1(3)+0.1(3) = 9.6$ h

Step 2: Class peak demand

$$P_{class} = N \times \frac{\text{Monthly kWh} / \text{Effective days}}{h_{eq}} \times \text{Coincidence factor}$$

Step 3: Combined load curve (kW)

The maximum occurs during 18:00–21:00.

Step 4: Peak in kVA

Reactive power at peak, $Q = P\tan\phi$:

$$\begin{aligned} Q &= 10(0.4843) + 1.406(0.75) + 0.381(0.6197) \\ &= 4.843 + 1.055 + 0.236 = 6.134 \text{ kVAR} \\ S &= \sqrt{11.787^2 + 6.134^2} = 13.29 \text{ kVA} \end{aligned}$$

The contribution factors at peak are A = 1.0, B = 0.5, C = 0.1.

Answer: Peak demand of the load centre ≈ 11.79 kW (≈ 13.29 kVA at about 0.89 p.f.), occurring between 18:00 and 21:00.

TRUE.

Loss of load factor is defined as

$$LLF = \frac{\text{Average power loss over period}}{\text{Peak power loss}}$$

If the power loss is constant throughout the period, the average loss equals the peak loss, so $LLF = 1$.

Justification:

• Constant (no-load) losses such as transformer core (iron) loss, dielectric loss in cables, corona loss in fair weather and losses in meter potential coils depend on voltage, not on load current. Since voltage is nearly constant, these losses are the same every hour.
• Their energy loss is therefore $E = P_{const} \times 8760$ kWh/year, i.e. the loss acts at its full value for all hours (LLF = 1).
• Example: a 100 kVA transformer with 200 W core loss loses $0.2 \times 8760 = 1752$ kWh/year whether it is lightly or fully loaded.
• Only variable (copper) losses, which vary as $I^2$, need $LLF < 1$, e.g. $LLF = 0.3LF + 0.7LF^2$.

The statement also follows from the formula: a constant load gives $LF = 1$, and $0.3(1) + 0.7(1)^2 = 1$.

The 5th year peak is found by projecting consumer numbers and consumption, converting daily energy to individual peak with the load factor, and combining class peaks using contribution factors.

Growth multipliers

Consumption growth: 5% in year 1, increasing by 1% each year (6%, 7%, 8%, 9%):

$$\begin{aligned} G_c &= 1.05 \times 1.06 \times 1.07 \times 1.08 \times 1.09 = 1.40194 \\ G_p &= (1.02)^5 = 1.10408 \end{aligned}$$

Formulae

$$\begin{aligned} N_5 &= N_0 \times G_p \times \text{coverage} \\ E_5 &= E_0 \times G_c \\ P_{ind} &= \frac{E_5 / \text{effective days}}{24 \times LF} \\ P_{LC} &= \sum CF \times N_5 \times P_{ind} \end{aligned}$$

Number of consumers (5th year)

Consumption and individual peak

Example (commercial): $P_{ind} = \dfrac{1.4019}{24 \times 0.30} = 0.19471$ kW.

Contribution to load centre peak

Answer: 5th year peak load of the load centre ≈ 20.5 kW (20.47 kW).

The statement is correct.

The trending method fits a curve (linear, exponential or polynomial) to past load data and extends it into the future. It suits large areas (a whole utility, region or country) because:

• The load of a large area is the sum of many small areas at different stages of growth. Their random, step-like changes average out, so the total grows smoothly and steadily, which a simple curve can follow.
• Large areas have long, reliable load records (years of peak demand and energy sales) to fit the curve.
• Only the total amount of load is needed for generation and transmission planning, not its exact location.
• It is quick, cheap and needs no detailed land-use or end-use data.

For small areas, load stays near zero and then jumps (S-curve), and history is short, so trending gives poor results there; simulation (land-use) methods are used instead.

Load factor is the ratio of average demand to peak demand over a period:

$$LF = \frac{E}{P_{max} \times T}$$

A higher LF means a flatter load curve. Its advantages, for a given energy $E$ to be supplied:

1. Lower peak demand, so smaller plant and lower capital cost

$$P_{max} = \frac{E}{LF \times 8760}$$

Peak demand, and therefore the required kVA of generators, transformers and conductor size, varies inversely with LF. Example: for 876,000 kWh/year, LF = 0.4 needs 250 kW capacity, while LF = 0.8 needs only 125 kW.

2. Lower cost per unit generated

Annual cost = fixed cost (∝ $P_{max}$) + running cost (∝ $E$):

$$\text{Cost per kWh} = \frac{a P_{max} + bE}{E} = \frac{a}{8760\,LF} + b$$

The fixed-charge part per kWh falls as LF rises, so the tariff can be reduced.

3. Lower energy loss

Energy loss in lines $= k P_{max}^2 \times LLF \times 8760$ with $LLF = 0.3LF + 0.7LF^2$. Substituting $P_{max}$:

$$E_{loss} = \frac{k E^2}{8760} \left(\frac{0.3}{LF} + 0.7\right)$$

So raising LF from 0.3 to 0.6 cuts energy loss by about 29% for the same energy delivered.

4. Other advantages

• Better use of installed equipment (less idle capacity).
• Smaller voltage swings between peak and off-peak, easing voltage regulation.
• Generating units run near their best efficiency, with fewer start-ups.
• Lower peak-hour import costs for the utility.

Utilities raise LF by time-of-day tariffs, encouraging night-time industrial use, and other demand-side measures.

FALSE.

The percentage power loss is measured at peak load, while the percentage energy loss is averaged over time. Because losses vary as the square of the load, they are not equal.

Justification:

$$\begin{aligned} \%P_{loss} &= \frac{P_{loss,peak}}{P_{max}} \times 100 \\ \%E_{loss} &= \frac{P_{loss,peak} \times LLF \times T}{P_{max} \times LF \times T} \times 100 = \%P_{loss} \times \frac{LLF}{LF} \end{aligned}$$

Since $LF^2 \le LLF \le LF$, we get $LLF/LF \le 1$, so % energy loss is less than % peak power loss.

Example: % peak power loss = 10%, LF = 0.4. Using $LLF = 0.3LF + 0.7LF^2$:

$$\begin{aligned} LLF &= 0.3(0.4) + 0.7(0.16) = 0.232 \\ \%E_{loss} &= 10 \times \frac{0.232}{0.4} = 5.8\% \end{aligned}$$

So 10% power loss corresponds to only 5.8% energy loss.

The two are equal only when $LF = LLF = 1$, i.e. a perfectly constant load, which never occurs in practical lines. (If constant losses such as corona or transformer core loss are included, those parts have LLF = 1, but the overall figures still differ.)

Coverage factor (electrification coverage factor) is the fraction or percentage of the potential consumers of a load centre who are expected to actually take an electricity connection by the design year:

$$\text{Coverage factor} = \frac{\text{Number of consumers actually connected}}{\text{Number of potential consumers}}$$

For example, if a village has 200 households and 140 are expected to be connected by the 5th year, the coverage factor is 70%.

It is less than 100% because some households cannot afford connection or wiring costs, are far from the LT line, use other sources (solar home systems, micro-hydro), or join gradually after the line is built.

Significance in distribution design:

• Effective consumers in the design year $= N_0 (1+p)^n \times \text{coverage factor}$; this directly sets the peak demand used to size the distribution transformer and LT feeders.
• Too high a value oversizes transformers, increasing cost and no-load losses; too low a value causes early overloading and voltage problems.
• It is used to estimate revenue and the economic viability of rural electrification projects.
• Coverage usually rises over time, so the design must allow for future additions (spare transformer capacity or space for upgrading).

FALSE (for percentage power loss); load factor affects the percentage energy loss.

Reasons:

• Power loss at any instant is $P_{loss} = 3I^2R$. The percentage power loss at peak is

$$\%P_{loss} = \frac{3 I_{max}^2 R}{P_{max}} \times 100 = \frac{P_{max} R}{V^2 \cos^2\phi} \times 100$$

It depends only on the peak demand, voltage, power factor and resistance (conductor size and length). Load factor does not appear, so two load centres with the same peak but different load factors have the same % power loss.

• Load factor describes how the load varies with time. It therefore affects energy quantities:

$$\%E_{loss} = \%P_{loss} \times \frac{LLF}{LF}, \quad LLF = 0.3LF + 0.7LF^2$$

so $\%E_{loss} = \%P_{loss}(0.3 + 0.7LF)$.

Example: % peak power loss = 8%.

The % power loss stays 8% for both, while the % energy loss changes with LF.

• Indirectly, if the energy consumption is fixed, a higher LF lowers the peak demand and hence the peak power loss. In that sense the planner sees load factor influencing losses, but for a load centre of given peak demand the percentage power loss is independent of load factor.

The 5th year peak of the load centre is found by growing each class's consumption to the 5th year, applying the coverage factor to the consumer number, converting daily energy to individual peak using the 5th year load factor, and adding class peaks weighted by their contribution factors.

Assumption: no population growth is given, so the number of potential consumers stays the same; growth is applied for $n = 5$ years.

Formulae

$$\begin{aligned} N_5 &= N_0 \times \text{coverage factor} \\ E_5 &= E_0 (1 + g)^5 \\ E_{day} &= \frac{E_5}{\text{effective days per month}} \\ P_{ind} &= \frac{E_{day}}{24 \times LF_5} \\ P_{class} &= N_5 \times P_{ind} \\ P_{LC} &= \sum CF_i \times P_{class,i} \end{aligned}$$

Step 1: Consumption growth multiplier

$$(1 + 0.05)^5 = 1.27628$$

Step 2: Effective number of consumers

Step 3: Consumption and individual peak demand

Sample calculation (domestic):

$$P_{ind} = \frac{1.2763}{24 \times 0.25} = \frac{1.2763}{6} = 0.21271 \text{ kW}$$

Step 4: Class peak and contribution to load centre peak

Step 5: Load centre peak

$$P_{LC} = 14.890 + 1.241 + 0.745 = 16.875 \text{ kW}$$

Note that the plain sum of class peaks (24.82 kW) would overestimate the demand by about 47%, because commercial and non-commercial loads peak at other hours.

Answer: 5th year peak load of the load centre ≈ 16.88 kW.

For transformer selection, this would be converted to kVA (e.g. at 0.85 p.f., $16.88/0.85 \approx 19.9$ kVA), plus a loss allowance, so a standard 25 kVA transformer would be suitable.

The method: project each class's consumption to the 5th year, apply the coverage factor to the consumer numbers, convert daily energy to individual peak demand using the load factor, then combine the class peaks using their contribution factors.

Assumption: no population growth is given, so potential consumer numbers stay constant; consumption grows for $n = 5$ years.

Formulae

$$\begin{aligned} N_5 &= N_0 \times \text{coverage} \\ E_5 &= E_0 (1+g)^5 \\ P_{ind} &= \frac{E_5/\text{effective days}}{24 \times LF_5} \\ P_{LC} &= \sum CF_i \times N_{5,i} \times P_{ind,i} \end{aligned}$$

Step 1: Growth multiplier

$$(1.04)^5 = 1.21665$$

Step 2: Connected consumers in 5th year (60% coverage)

Step 3: Consumption and individual peak

Sample (non-commercial):

$$P_{ind} = \frac{0.9733}{24 \times 0.25} = 0.16222 \text{ kW}$$

Step 4: Class peaks and contribution

Step 5: Load centre peak

$$P_{LC} = 24.333 + 2.535 + 0.389 = 27.257 \text{ kW}$$

The non-coincident sum (33.30 kW) is higher because commercial and non-commercial classes do not peak with the domestic evening peak.

Answer: 5th year peak load of the load centre ≈ 27.26 kW.

At about 0.85 p.f. this is roughly 32 kVA; allowing for losses and later growth, a 50 kVA standard transformer would be selected.

Consumption growth factor is the annual rate at which the average electricity consumption (kWh per month or per year) of a consumer of a given class increases, as consumers buy more appliances and use more energy with rising income and living standard.

If $E_0$ is the present average consumption per consumer and $g$ is the consumption growth factor (per year), the consumption after $n$ years is

$$E_n = E_0 (1 + g)^n$$

Example: 30 kWh/month with $g = 5\%$ becomes $30 \times 1.05^5 = 38.3$ kWh/month in the 5th year.

It is different for each consumer class (domestic, commercial, industrial) and is separate from the population (consumer number) growth factor. It is used in load forecasting to find the design-year peak demand of a load centre for sizing transformers and feeders.