Peak demand of load center from load pattern

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.