Dave Keenan, 7 May 1998
Last updated 29 Nov 1999

http://users.bigpond.net.au/d.keenan

I had been wondering for some time how one should decide on the size of a rain water tank. e.g. Given my available collection area (i.e. roof) and my daily average water usage, how big should I make my storage tank so as to supply a reasonable fraction of my requirements from rain? If you know of published info about this I'd appreciate a reference (I haven't looked).

Thanks to Keith Moodie at the Queensland Department of Natural Resources (DNR), I was able to answer this by daily simulation for the past 40 years, specifically for the 5.6 km square containing my house! My house is in Bardon, Brisbane. The daily rainfall data is interpolated from that of surrounding met stations and was supplied by Queensland Department of Natural Resources and Bureau of Meteorology at http://www.dnr.qld.gov.au/silo/datadril.html (The SILO data drill). To use this, you need to type in your latitude and longitude in degrees and minutes. This may be found or estimated via the Bureau of Meteorology's list of Rainfall Reporting Stations at http://www.bom.gov.au/climate/rflstns/ although you will need to convert decimal fractions of a degree to minutes by multiplying them by 60.

I then simulated the roof/tank/usage system using a spreadsheet.

The biggest surprise for me, was not that you needed a rather large roof and about a year's worth of storage to give 100% supply, but that even with a small roof and only 1 day's storage you can supply 15% to 20% of your needs. Just to clarify, if your average daily water usage was 1000 L, a year's storage would mean a 365,250 L tank (pretty much out of the question in the suburbs unless your house is built on top of it, in which case it would need to be about 2 m deep) but 1 day's storage means only a piddling 1000 L tank (e.g. 800 mm dia., 2 m high). 10 to 50 days storage is quite reasonable.

I provide the results here as a pair of charts and below as a set of linearised rules derived from the charts. Strictly speaking these only apply to my specific area but they should also work for nearby areas since they factor out the average rainfall and therefore only assume the particular time distribution statistics of the rainfall. In future I hope to find out just how general they are and how many other parameters are needed, apart from average rainfall, to characterise a location for this purpose.

You first need to know your average daily rainfall (average annual rainfall divided by 365.25) (e.g. 3.2 mm), the area of your roof (e.g. 150 m²), and your daily water usage (e.g. 600 L). Then you work out your collection factor, which is the average amount of water collected by your roof in a given period divided by the average amount you use in that period.

One thing that this tells us immediately (about our example situation) is that no matter how much storage we have, we'll never be able to supply more than 80% of our needs from rain falling on our roof, unless we can reduce our usage.

Assume that we have room for a 27,000 L tank (3.8 m dia., 2.4 m high). We convert this to days of storage by dividing by the daily usage.

Now with these two numbers we can look up the charts to find the rain fraction f, interpolating between the curves if necessary. Either of the two charts will do. They both represent the same 3-dimensional surface, but from different views to allow you to check your interpolations.

The rain fraction tells us what fraction of our usage will be supplied from the rainwater collection and storage system. For the above example, we read a rain fraction of about 0.64 from the charts. This means that 64% of our usage will come from rain and we will have to find the other 36% from some other (typically more expensive) source.

The following approximate rules for calculating the rain fraction f within ±0.15, work for collection factors between 0.5 and 2.0, and storage up to 100 days.

• Half a day's storage gives a rain fraction of 10% irrespective of collection factor.
• For a collection factor of 1.0, every doubling of storage gives a 9.5% increase in the rain fraction. This logarithmic relationship is described as a slope of 9.5% per octave (in music, an octave is a doubling of frequency).
• The percentage in the above rule increases by 3% for each doubling of the collection factor. i.e.

• For collection factors less than 1.0, if a result greater than the collection factor is obtained above then the rain fraction is the collection factor.
• For collection factors greater than 1.0, if a result greater than 1.0 is obtained above then the rain fraction is 1.0.

So for our example, the collection factor of 0.8 gives us a slope of about 8.5% per octave.

Relative to 0.5 days storage, 45 days of storage represents 6.5 doublings, since log₂(45/0.5) ~= 6.5.

So that's 10% for the half days storage, plus 6.5 * 8.5% = 55%, for a total of 65%, i.e. a rain fraction of 0.65. This agrees well with the value of 0.64 obtained from the charts.

The first 0.5 mm of rain falling on any day is diverted away from the tank to avoid pollution. e.g. 100 L for a 200 m² roof.

Half of the usage is for outdoors, e.g. watering gardens, and is therefore less on days when it rains more than average and more on days when it rains less than average. This component was varied linearly so that on days that received 0.256 of average daily rainfall, outdoor usage was normal. On days that received no rain, outdoor usage was 1.256 times normal. On days that received more than 1.256 times average rainfall the outdoor usage was zero. The number 1.256 was found by trial and error to make the average usage come out the same. It may be crudely interpreted as saying that 4 out of every 5 days are dry.

This is an extremely crude and probably unreliable way of modelling outdoor water usage since it takes no account of water storage in the soil or evaporation rates etc. However it is probably better than not modelling it at all and many people don't take these factors into account either, when deciding whether and how much to water.