Rainwater harvesting consists of 5 components:

Every living creature and vegetation must have water in order to survive.

Only 3% of all water on Earth is fresh water and 68% of the water is frozen ice on the North and South Poles. All humans and most animals as well as nearly all vegetation can only survive on fresh water free of salt and minerals. Water for irrigation must also be fresh water and applied sparingly by flood, furrow or drip irrigation, otherwise the irrigated soil will turn saline and unproductive for many years to come.

The other 97% of the water on planet Earth is saline seawater in which whales, fish, corals and plankton flourish. Slightly salty water may be used for watering livestock and other animals, although fresh water is healthier.

Nearly half of all deep boreholes are dry or contain saline (brackish) water with minerals harmful to humans, animals and plants. Where fresh water is pumped up from very deep boreholes it is called mining fossil water because the water withdrawn cannot be replaced by rains.

Therefore, there are only two sustainable solutions to the world's increasing demand for water, namely rainwater harvesting and water conservation.

Fresh water sources are replenished in a water cycle through the following activities:

Water can be harvested at several stages in the water cycle, such as:

The quality of water from all these sources can be tested by either a portable testing kit or by a laboratory if the samples can be delivered before deteriorating due to heat over long distances. Contaminated water may be treated by several methods, such as SODIS (Solar disinfection), boiling, water filters, crushed seeds from the Moringa tree, ultra-violet rays (UV) or artificial chemicals.

Although it is known that clouds can be seeded with chemicals to produce rains, the practice is expensive and unsustainable. It should therefore be realized that rainfall varies from region to region and from one year to another beyond peoples' manipulation and interference.

All fresh water comes from rains, including water in deep boreholes which originates from rains infiltrated into the underground thousands or millions of years ago. Fresh water can only be obtained from four main sources; rainwater harvesting, shallow ground water, deep ground water and desalination of which the two latter options are too expensive to be discussed here.

The North-East monsoon coming from India brings rains to East Africa from October to December every year. The South-East monsoon brings rains from March to June. Rain falls where and when the two monsoons meet. The area of convergence is called the Inter-Tropical Convergence Zone. This zone follows the apparent movement of the sun, north and south, bringing rain in its wake. But the pattern is influenced by mountain ranges, Lake Victoria and periodic westerly winds from the Atlantic.

The two monsoons bring an annual average of about 600 mm rainfall to the semi-arid eastern and northern parts of Kenya, while the highland zone at Mount Kenya has a mean average of 1,000 mm and the Lake Victoria zone has a mean average of 1,800 mm. The rainfall pattern of East Africa is also presented as a map with different colours for the various average annual rainfalls.

Although regions with low and erratic rainfalls appear to be unsuitable for rainwater harvesting, it has been proved many times that rainwater harvesting is the most viable water supply system in arid and semi-desert regions. Rainwater harvesting in dry regions is viable when the following aspects are considered and applied:

The following two examples show that rainwater harvesting is viable in regions with little rainfall provided the catchment area is enlarged accordingly:

Example 1
A roof with a catchment area of 100 square metres (m2.) and an annual rainfall of 800 millimetres (mm) can supply 72,000 litres of water (100 m2 roof x 800 mm rain = 80,000 litres minus 10% loss = 72,000 litres).

Example 2
A roof in semi-desert regions with an annual rainfall of 200 mm has to be 4 times larger to supply an equal volume of 72,000 litres of water, because the rainfall is only ¼ of the 800 mm rain shown in Example 1.

To design successful rainwater harvesting systems, it is important to know:

If these two figures are not known, the storage capacity of a water reservoir and its spillways cannot be designed properly. The ruins of such improperly designed water projects can be witnessed in most parts of Africa.

The size of a catchment area is measured in square metres (m2) or in hectares (ha). 1 ha consists of 10,000 m2 which is equal to 2.47 acres. An acre is equal to 4,047 m2.

Farmers measure their acreage of land by walking 70 paces, each pace with a length of 3 feet, equal to 0.915 metre, around the four sides of a square. 70 walking paces x 0.915 metres are equal to 64.05 metres. When two sides of the square are multiplied with each other the result is 4,102 m2 (64.05 x 64.05), this is close to the actual area of 4,047 m2 for 1 acre.

When the size of a catchment area and the volume of rainwater falling on that catchment area have been found, the volume of rainwater that can be harvested is be calculated by multiplying the length with the horizontal width of the roof.

For example: Length 20 m x horizontal width 5 m = 100 square metres

Farmers may admit: "Okay, it has rained but it was only a drizzle". Nevertheless, a drizzle of 10 mm rain can be sufficient to produce the required volume of water, if the catchment area is sufficiently large. The relationship between rainfall and catchment area can be explained by the following example:

A drizzle of 10 mm rain on10,000 m2 (1 hectare = 100 m x 100 m) area of a roof, rock or tarmac road, it produces 100,000 litres of run-off water minus about 20 % loss = 80,000 litres.

Therefore, a drizzle of 10 mm rains may be sufficient for harvesting a required volume of water, if the catchment area is large enough. Huge volumes of water can thus be harvested from e.g. roads because they have large and hard surface catchment areas.

For example: a drizzle of 10 mm rain on a road 6 metres wide and 1 km (1,000 metres) long road can supply the following volume of water:
A 6 m x 1,000 m road x 10 mm minus 20% loss = 48,000 litres = 48 m3 of water.

When designing a reservoir to hold water from a road catchment it is important that spillways can discharge safely the overflowing water during storms.

For example: A rain storm of 75 mm falling on a 2 km long road produces about:
75 mm rain x 6 m x 2,000 m road minus 20% = 720,000 litres = 720 m3 of water.

If the storage capacity, such as a pond is 500 m3, then the spillways must be capable of discharging 220 m3 of water (720 minus 500) in a few minutes or the pond will be damaged or perhaps washed away by the flood of incoming water.

Catchment of rainwater from roads is potentially the cheapest water source in dryland where there are no sandy riverbeds (luggahs or wadis). Tarmacked roads produce more run-off water than dirt roads but the water may contain harmful tar components for people and livestock and should therefore be used for irrigation only.

There are two main types of storage, namely:

Almost every type of storage reservoir loses some of its stored water to evaporation, seepage or leakage.

Open water reservoirs, such as; tanks without roofs, ponds, earth dams and rock catchment dams, lose water due to evaporation. In hot and windy climates the evaporation rate may be over 3 mm per day which is equal to losing a depth of 0.9 metre of water in a month. Water tanks and rock catchments should therefore be roofed but that is not feasible for ponds and earth dams.

Rainwater stored in the voids between the sand particles of riverbeds is the most economical water storage because up to 350 litres (35%) of water can be extracted from 1 cubic metre of coarse sand while only about 10% of water can be extracted from fine structured sand because its voids are smaller.

In hot regions without sandy riverbeds, rainwater can be stored in situ between the voids between soil particles, e.g. downstream of ponds and earth dams or in terraced land or in seasonal macro and micro irrigation structures as described in "Water Storage".

In addition to evaporation loses, water stored in ponds and earth dams also lose water to seepage through the floor of these reservoirs. Fortunately, some of the seepage loses from ponds and dams can be utilized by sinking a shallow well into a seepage area. However, the combined losses of seepage and evaporation during long dry seasons usually result in the ponds and earth dams drying up except for very large earth dams.

The seepage loses through the floor of ponds and earth dams can be partly sealed by either:

Loss of water from leaking water tanks, water pipes, water taps should be prevented by using either bitumen paste or cement mortar.

Many rainwater harvesting installations do not perform as well as expected because of unsatisfactory gutters. It is therefore important to give careful attention to the materials used, the way the gutters are fabricated and the way they are installed. Ways of fabricating and installing low-cost gutters are described below.

The best known gutter is semi-circular and made either of galvanised iron sheet or PVC. Gutters are laid in gutter brackets nailed onto facia-boards or in V-shaped tree branches nailed to the rafters with a gradient sloping towards the water tank. Bamboos and Sisal poles can also be used as gutters when split in two halves.

Cut galvanized plain iron sheets of gauge 26 or 28 into three strips, each being 200 cm long and 33.3 cm wide by marking the sheets with a thick wire, about 40 cm long, with each end having a sharp bend and a pointed end to scratch a line. The distance between the two bends must be 33.3 cm in order to make equal width of the cut sheets.

The shape of the metal strips depends on whether they shall be splash-guards or gutters with one of the shapes shown below: namely the V-shaped gutter, the square gutter and the semi-circular gutter.

Gutters should be installed with a gradient of 10 cm depth for every 10 m length of a roof. This gradient of 1:100 will facilitate rain water running off the gutter with high velocity and no water will be wasted due to overflow. The high velocity of the water will transport leaves and debris to the inlet sieve without blocking the gutter.

A gradient of 1:100 is distance is found by dividing the length of a roof with a factor of 100.
For example; if a roof is 20 m long, the hanger at the water tank must be 20 cm lower than the other hanger to get the desired gradient of 1:100.

Two hangers are attached to the first gutter with one hanger at the middle and the other hanger at the end of the gutter. Bitumen is smeared on the inner end of first gutter before a second gutter is laid into it and so on until the whole length of gutter is installed.

The required storage capacity of a water reservoir depends on:

While the number of days in a dry season can be estimated fairly easy, such as 180 days without rain in a semi-arid region, the volume of water required for each of the 180 days can be calculated using the following guidelines on daily requirements of water for a rural homestead:

This example shows that a rural homestead in a dry area could use 3 types of water sources: