Solar panel and battery for a house (6 h sun)

A stand-alone solar home system is sized by finding the daily energy demand, adding inverter and battery losses to get the energy the panel must produce, and sizing the battery 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; no extra wiring loss beyond the given efficiencies.

1. Daily energy demand

2. Solar panel size

Energy passes through the battery and the inverter, each 80% efficient:

$$\begin{aligned} E_{PV} &= \frac{695}{0.8 \times 0.8} = 1085.9\ \text{Wh/day} \\ P_{PV} &= \frac{1085.9}{6\ \text{h}} = 181.0\ \text{W}_p \end{aligned}$$

Select 200 Wp (two 100 Wp, 12 V panels in parallel). Panel 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:

$$\begin{aligned} E_{b} &= \frac{695}{\eta_{inv}\,\eta_{b}} = \frac{695}{0.8 \times 0.8} = 1085.9\ \text{Wh} \\ \text{Ah (usable)} &= \frac{1085.9}{12} = 90.5\ \text{Ah} \\ C &= \frac{90.5}{DoD} = \frac{90.5}{0.6} = 150.8\ \text{Ah} \end{aligned}$$

A 150 Ah battery would be just short with no margin, so select a 12 V, 200 Ah deep-cycle (tubular lead-acid) battery, which also gives some margin for cloudy days.

4. Inverter (extra)

Connected load 175 W × 1.25 ≈ 219 W → 12 V, 250–300 VA inverter.

Answer: Solar panel ≈ 181 Wp → 200 Wp; battery ≈ 151 Ah → 12 V, 200 Ah.