5th year peak load from consumer survey

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.