When determining what size collector you need, you must consider two key factors: insolation level and energy requirements. Energy requirement will usually take into consideration the volume of water and rise in temperature required. Once you know these factors you can determine the size collector you require. The bigger the collector you have, the more hot water, but you should make an economically sound decision. Generally it is wise to select a size which will provide you with 90% of your hot water needs in the summer.

Although it may seem strange to use a value of only 90% for summer solar contribution, it is for good reason. It is normal to size based on 100% of your summer hot water energy needs, with a percentage provided throughout other months, lowest obviously in winter. That is based on normal water usage, but often, and particularly in the summer, water usage patterns may not be that normal, with cooler than normal showers taken in hot weather, and greater possibility of the house being vacant for one or two days each week (weekends). As such, using a target value of 90% will probably actually result in a system that is able to supply more than 100% of your hot water needs in the summer, without excessive heat production, which can lead to water loss via pressure release and a waste of energy.

If you are looking for general "rule of thumb" information for sizing our world class solar thermal systems you are in the right place. Although not full of technical jargon this will give the average consumer a good idea of what to expect in a quality Apricus installation.

There are various methods for calculating hot water usage based on household occupancy. Apricus recommends the following sizing guidelines for most domestic households. Please note this is for a household with optimal (~8L/min / 2.1Gpm) tap flow rates. 100L / 26Gallons per adult 50L / 13 Gallons per child Add additional usage for a washing machine (if it uses hot water) and dishwasher.

If the household has high flow shower-heads installed, don’t even consider installing solar until that has been remedied!

Therefore for the average family of 4, about 300L / 80Gallons of hot water is used per day. Please note this is based on actual tap flow volume at the target hot water temperature and therefore does not represent the usage of hot water from the tank; that value will be 30-40% lower, depending on the thermostat setting and cold water temperature. Please use Apricus document AS-1.6.6 to calculate cold-hot water ratio.

As a general rule, meeting 100% of hot water needs will provide annual contribution of: Cold Region = 50-60% Mild Region = 60-70% Hot Region = 70-80%

Trying to achieve greater than these levels is only viable if there is a means of using or dissipating the additional heat created in the summer. For example, if the above household installed 2 x 30tube collectors, they could probably achieve an annual average contribution of around 85-90% (100% is impossible), but would have twice as much heat as they need in the summer. There are various methods for dealing with excess heat, namely:

- Using a tank cooling feature on controller, to cool tank at night.
- Install a heat dissipator - Use excess heat for pool/spa/hot-tub heating
- Install collectors at high angle to minimize summer output

(Metric Units)

> Household using 300L of water per day @ 45⁰C.
> Summer cold water temperature of 18⁰C
> Average summer insolation level of 6kWh/m²/day

Step 1. Determine temperature rise
→ 45 – 18 = 27⁰C temp rise
Step 2. Determine energy requirement
→ 300L x 27⁰C = 8,100kcal (1kcal raises 1L of water by 1⁰C)
→ 8100 / 859.8 = 9.42kWh (1kWh = 859.8kcal)
Step 3. Determine solar collector output/tube (using average conversion of 70%)
→ 6kWh * 70% conversion = 4.2kWh per m² of collector absorber area
→ 4.2kWh * 0.08m² absorber area = 0.336kWh/tube/day
Step 4. Determine tube requirements
→ 9.42kWh ÷ 0.336 = 28 tubes

(Imperial Units)

> Household using 80Gallons of water per day @ 114⁰F.
> Summer cold water temperature of 65⁰F
> Average summer insolation level of 1,902Btu/ft2/day

Step 1. Determine temperature rise
→ 114 – 65 = 49⁰F temp rise
Step 2. Determine energy requirement
→ 80Gallons = 667.2lb (1 Gallon = 8.34lb of water)
→ 667.2lb x 49⁰F = 32,692Btu (1Btu raises 1lb of water by 1⁰F)
Step 3. Determine solar collector output/tube
→ 1902Btu * 70% conversion = 1331Btu per ft² of collector absorber area
→ 1331 * 0.86ft² absorber area = 1144Btu/tube/day
Step 4. Determine tube requirements
→ 32,692 ÷ 1144 = 28.6 tubes

So from this simple calculation, it is shown that a 30 tube collector would suit this household perfectly, providing 100% of their hot water needs in the summer, and a percentage throughout the year dependant on their hot water usage patterns and solar radiation levels.

When considering all factors, the following average solar conversion values can be used for Apricus solar collectors:

Cold Weather Day = 60%
Mild Weather Day = 65%
Hot Weather Day = 70%

If you want to calculate the heat output for a summer’s day, then use the Hot Weather conversion value of 70%. If the daily solar radiation level is 6kWh/m2/day, then the calculation is very straightforward.

→ 6 x 0.70 = 4.2kWh/m2/day (per m2 of solar collector absorber area)
→ 4.2 x 2.4m2 absorber area (30tubes) = 10kWh

So the total day output for the collector can be expected to reach around 10kWh on a clear sunny day in mid summer, if solar radiation levels are 6kWh/m2/day.

When considering all factors, the following average solar conversion values can be used for Apricus solar collectors:

Cold Region = 60%
Mild Region = 65%
Hot Region = 70%

Using the average annual solar insolation level, a simple calculation can be completed. For a Mild region, the average value is around 4.5kWh/m2/day.


→ 4.5 x 0.70 = 3.15kWh/m2/day (per m2 of solar collector absorber area)
→ 3.15 x 2.4m2 absorber area (30tubes) = 7.56kWh

So the average daily heat output from a 30tube collector is 7.56kWh, which is enough to heat 250L / 66Gallons of water by 26oC / 46oF.