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Workshop: Simple Solar Math

Words & Images Branden Austen

A typical shopping list for a solar system consists of an 85 Watt solar panel, cabling, a regulator (to prevent the panel from overcharging the battery), a deep cycle battery (100 Ah rated at 20 hours) and a battery box.

An average 85 Watt solar panel will collect an average annual daily high of 5.5 peak sun hours. Camping in summer will increase that average daily collection.

85 Watts x 5.5 hours, divided by the voltage (the nominal output specified by the panel manufacturer), will give you less than 38 Ah per day. This is usually not enough to run all your power needs.

To improve this figure, try “sun tracking”. In other words, place the panel at an angle to maximise sun power, so that it faces North East in the morning, Due North in the middle of the day and North West in the afternoon. In this way, you will collect possibly 7-8 hours per day, or even more.

Follow this approach, and the sum changes: 85 Watts x 7.5 hours, divided by the panel voltage, gives 53 Ah of collected power.

This is closer to meeting your daily needs when camping. Assume that you start with a fully-charged 100 Amp deep cycle battery. On the first day, you use 46.5 Ah, leaving a balance of 53.5 Ah. On the second day, the solar panels replace 50 Amps, bringing the battery back up to full power.

How long will the battery last? This is determined, in the first place, by how large your deep cycle battery is. This figure is expressed in Amps over 20 hours (that is, Ah). To protect your battery, try to maintain a 30% safety buffer: don’t ever discharge your battery totally. Secondly, you need to calculate the draw of the appliances you want to run off the system.


Typical camping power draws

50-litre fridge:

(Runs at 45 Watts per hour, 32% efficiency running at 30⁰C ambient)
45 Watts x R hours, divided by voltage 12V =
30 Amps per day (30 Ah)


2 x 13 Watts fluorescent about 3 hours a day
26 W x 3 ÷ 12V =
6.5 Amps per day (6.5 Ah)

Television via an inverter:

35 Watts x 90% efficiency for inverter,
say 2.5 hours a day
35 W x loss = 39 W x 2.5 ÷ 12V =
8 Amps per day (8 Ah)


10 Watts x 3 hours ÷ 1 2V =
2.5 Amps per day (2.5 Ah)

Total daily load

47 Amps consumed per day =
47 Ah

Divide 70% of the battery rating by the daily load.
A 100 Ah battery will provide approximately 1.5 days of power in this situation.


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