Solar panel and battery for a house (7 h sun)

The solar panel is sized from the daily energy demand divided by the combined efficiency and the sunshine hours; the battery is sized to supply one day's load within an allowed depth of discharge.

Assumptions: all loads are AC through the inverter; one day of autonomy; depth of discharge (DoD) = 0.6.

1. Daily energy demand

2. Solar panel size

$$\begin{aligned} E_{PV} &= \frac{855}{\eta_{inv}\,\eta_{b}} = \frac{855}{0.8 \times 0.8} = 1335.9\ \text{Wh/day} \\ P_{PV} &= \frac{1335.9}{7\ \text{h}} = 190.8\ \text{W}_p \end{aligned}$$

Select 200 Wp (two 100 Wp, 12 V panels in parallel). Array current $\approx 200/12 = 16.7$ A, so use a 12 V, 20 A charge controller.

3. Battery size

Taking both the inverter and battery efficiencies into account:

$$\begin{aligned} E_{b} &= \frac{855}{\eta_{inv}\,\eta_{b}} = \frac{855}{0.8 \times 0.8} = 1335.9\ \text{Wh} \\ \text{Ah (usable)} &= \frac{1335.9}{12} = 111.3\ \text{Ah} \\ C &= \frac{111.3}{0.6} = 185.5\ \text{Ah} \end{aligned}$$

Select a 12 V, 200 Ah deep-cycle battery.

4. Inverter (extra)

Connected load 215 W × 1.25 ≈ 269 W → 12 V, 300 VA inverter.

Answer: Solar panel ≈ 191 Wp → 200 Wp; battery ≈ 186 Ah → 12 V, 200 Ah.