## Info

Time

Example 9.5

A remote cottage has the loads listed in Table 9.2. Find the average load and peak power to be satisfied by a 12 V PV system with an inverter.

Table 9.2 Loads for Cottage in Example 9.5

Appliance type

Description

Power type

Period of operation

Lights

3, 25 W compact fluorescent bulbs, daily

DC

Nighttime 5 h each

Light

11 W compact fluorescent bulb, daily

AC

Nighttime 5 h

Water pump

50 W (6 A start current), daily

DC

Daytime 2 h

Oven

500 W, 3 times a week

AC

Daytime 1.5 h

Steam iron

800 W, once a week

AC

In Table 9.3, the loads for this application are separated according to type of power. Because no information is given about the time schedule of the loads, these are assumed to occur simultaneously.

 Appliance Power Power (W) Run time Energy/day Energy/ type type (h) (Wh) week (Wh) Lights DC 3 X 25 = 75 W 5 375 2625 Lights AC 11 W 5 55 385 Water pump DC 50 W 2 100 700 Oven AC 500 W 1.5 — 2250 Steam iron AC 800 W 1.5 — 1200

From Table 9.3, the following can be determined:

Average AC load = (385 + 2250 + 1200)/7 = 547.9Wh/d

Peak DC load = 6 X 12 + 75 = 147W (the maximum occurs when the pump starts, 6 X 12 > 50W)