Isolated PV sizing: lamps, fan and TV

An isolated PV system is sized by first finding the daily energy demand, then adding the losses to get the PV array rating, the battery capacity for the required autonomy, and an inverter that can carry the connected load.

Assumptions: all loads are AC (through inverter); wiring losses are taken as 20% extra energy; peak sun hours in Nepal = 5 h/day; panel derating (temperature, dust) = 0.9.

1. Daily load energy

2. Energy to be supplied by the battery (DC side)

$$\begin{aligned} E_{inv,in} &= \frac{500}{0.95} = 526.3\ \text{Wh} \\ E_{DC} &= 1.2 \times 526.3 = 631.6\ \text{Wh/day} \end{aligned}$$

3. Battery size

$$\begin{aligned} C &= \frac{E_{DC} \times \text{days of autonomy}}{DoD \times \eta_{b} \times V} \\ &= \frac{631.6 \times 3}{0.6 \times 0.9 \times 12} = 292.4\ \text{Ah} \end{aligned}$$

Select 12 V, 300 Ah (e.g. two 12 V, 150 Ah batteries in parallel).

4. PV panel size

Energy the array must give (battery efficiency 90%):

$$\begin{aligned} E_{PV} &= \frac{631.6}{0.9} = 701.8\ \text{Wh/day} \\ P_{PV} &= \frac{701.8}{5 \times 0.9} = 155.9\ \text{W}_p \end{aligned}$$

Select 160 Wp (two 80 Wp, 12 V panels in parallel).

5. Inverter size

Connected load = 155 W. With a 25% margin for surge (fan motor, TV) and future load:

$$P_{inv} = 1.25 \times 155 = 193.75\ \text{W}$$

Select 12 V DC / 230 V AC, 250 VA pure sine wave inverter.

6. Charge controller (extra)

$I = 1.25 \times 160/12 = 16.7$ A, so a 12 V, 20 A charge controller.

Answer: PV array ≈ 156 Wp → 160 Wp; battery ≈ 292 Ah → 12 V, 300 Ah; inverter ≈ 194 W → 250 VA.