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.

table 9.3 Loads in Table 9.2, by Type of Power

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)

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Solar Panel Basics

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