Solar panel and inverter for a remote home

The panel is sized from the daily energy demand plus all losses, divided by the peak sun hours; the inverter is sized from the total connected load with a safety margin.

Assumptions: all loads run on AC through an inverter; inverter efficiency 90%; wiring/other losses 10%; battery efficiency 85%; peak sun hours 5 h/day (typical for Nepal); panel derating factor 0.9; system voltage 24 V (because the daily energy is above 1 kWh).

1. Daily energy demand

2. Energy required from the PV array

$$\begin{aligned} E_{inv,in} &= \frac{1150}{0.9} = 1277.8\ \text{Wh} \\ E_{DC} &= 1.1 \times 1277.8 = 1405.6\ \text{Wh} \\ E_{PV} &= \frac{1405.6}{0.85} = 1653.6\ \text{Wh/day} \end{aligned}$$

3. Solar panel size

$$P_{PV} = \frac{E_{PV}}{\text{PSH} \times \text{derating}} = \frac{1653.6}{5 \times 0.9} = 367.5\ \text{W}_p$$

Select a 400 Wp array, e.g. four 100 Wp, 12 V panels (2 in series × 2 in parallel for 24 V).

4. Inverter size

Total connected load = 180 W. With a 25% margin for starting surge and future load:

$$P_{inv} = 1.25 \times 180 = 225\ \text{W}$$

Select a 24 V DC / 230 V AC, 300 VA pure sine wave inverter.

5. Battery (for completeness)

With 2 days of autonomy, DoD 0.6 and battery efficiency 0.85:

$$C = \frac{1405.6 \times 2}{0.6 \times 0.85 \times 24} = \frac{2811.2}{12.24} = 229.7\ \text{Ah}$$

so a 24 V, 250 Ah bank (two 12 V, 250 Ah batteries in series).

Answer: Solar panel ≈ 368 Wp → 400 Wp; inverter ≈ 225 W → 300 VA; (battery ≈ 230 Ah → 24 V, 250 Ah).