4.1 Water source
4.2 Distribution of water supplies
The first question a planner may ask when dealing with livestock watering problems on rangeland areas is whether surface or groundwater will be used. In many circumstances there is no choice since only one source of water can physically or economically be developed. But assuming that both ground and surface water can be considered, Table 8 may guide the planner in tentatively selecting the most convenient source of water on which to concentrate the investigation efforts.
Water harvesting techniques, although they are briefly described in this manual, are not mentioned in Table 8 since they can be considered only if neither surface nor groundwater can offer a solution.
Table 8: TENTATIVE SELECTION OF THE MOST SUITABLE WATER SOURCE
|
Presence of shallow aquifer |
Period of use of water supply |
Topography |
Rainfall |
Morphology Soil characteristics |
Suggested water source to investigate |
|
SHALLOW CONTINUOUS AQUIFER |
|
GROUNDWATER (dug wells) |
|||
|
|
Permanent rangeland (throughout the dry season) |
flat to moderate slopes |
|
GROUNDWATER |
|
|
moderate to steep slopes |
200-300 mm |
|
GROUNDWATER |
||
|
300-600 mm |
No suitable sites for surface water mobilization |
GROUNDWATER |
|||
|
|
- Marked valleys |
SURFACE WATER (impounded reservoirs) |
|||
|
NO SHALLOW CONTINUOUS AQUIFER |
temporary rangeland (2-3 months after the rainy season) |
flat slopes |
|
GROUNDWATER |
|
|
moderate slopes |
Sandy soils |
GROUNDWATER |
|||
|
Clayey and sandy soils |
SURFACE WATER (excavated reservoirs) |
||||
|
rocky soils |
SURFACE WATER (impounded reservoirs) |
||||
|
steep slopes |
|
SURFACE WATER (impounded reservoirs) |
|||
It is clear that one of the criteria for location of the water supplies will be the availability of water but it is also a good practice to set up a preliminary layout of the water supply system in order to ensure an even utilisation of the forage resources, and plan the additional investigations necessary for siting the water supplies.
The layout will simply consist of fixing the average distance that should occur from a water supply to another in order to allow all the livestock that can be fed 5 years out of 10 from a given rangeland, to be suitable watered. The following procedure is suggested to determine the distribution of the water supplies over the considered rangeland.
a. First check if a 20 km interval between water supplies is feasible
The distance of 20 km corresponds to twice the average walking capacity of cattle under semi-arid conditions. Over an area of 30 000 ha (circle of 20 km diameter), the number of Tropical Livestock Units (TLU) which should be watered from each supply is equal to 30 000 divided by the carrying capacity (Ccap in ha/TLU) corresponding to the average rainfall year.
Assuming maximum daily water intake of 40 l/TLU, the maximum daily requirement (Vmaxday, in m3/day) to be met by each water supply will be
Vmaxday, should then be compared to the amount of water expected, Qexp in m3/day, that each water supply can deliver in 12-16 hours daily operation. If Vmaxday is less than Qexp, then take 20 km as an average distance from one water supply to another.
b. If the maximum daily water production Qmax of one supply is not sufficient to meet The requirements of the livestock which an be fed by a 30 000 ha rangeland area (circle of 20 km diameter), then determine the optimum interval between water sources according to the following procedure:
Using the same symbols as before, the maximum number of TLU which can be watered from each supply will be:
The rangeland area (in ha) necessary to feed NTLU will be NTLU x Ccap and eventually the maximum interval Dmax between water supplies will be approximately:
in 
Example:
Determine the optimum distance between water supplies on a rangeland with a carrying capacity of 10 ha/TLU and assuming that average daily water production from each supply cannot exceed 20 m3/day.
Vmaxday = 1200/10 = 120m3/day
Since Vmaxday is more than the maximum daily discharge of an average water supply, the distance will be less than 20 km and comes:
Dmax = 0.6 20 x 10 = 8 km
This example is for cattle and would be modified accordingly for other animals.
