5th year peak load, 4% growth, 60% coverage

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.