Peak load, load factor and energy from load pattern

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.