![]() | Irrigation Training Manual: Planning, Design, Operation and Management of Small-Scale Irrigation Systems (Peace Corps, 1994, 151 p.) |
![]() | ![]() | Training session |
![]() | ![]() | Section 4: Developing water sources |
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1. List different types of material that can be used to construct a diversion.
Answer:
earth |
sand bags |
rocks |
gabions |
logs |
concrete |
2. Describe what a gabion is and explain how it can be used in diversion dams.
Answer:
A gabion is a basket made of heavy-duty wire mesh, placed in the stream bed, and filled with rock. It is used as a building block in construction of diversions.
3. Describe the sequential steps in developing a spring source.
Answer:
1. Locate the origin of the flow.
2. Dig large hole.
3.
Measure the flow from the spring.
4. Fill in the hole with gravel or rocks to
protect spring.
5. Dig diversion ditch above spring.
4. Explain the situations in which you would consider a hand-dug well for irrigation; a commercially drilled well.
Answer:
Hand-dug - Shallow water table, low flow rate required, labor is cheap and available.
Drilled - Deep water tables, hard formations such as rock, larger flow rates required, availability of equipment or commercial well drillers.
5. A farmer comes to you and asks for advice on a secondhand, 5 HP, centrifugal pump that he has just bought. He wants to sprinkler irrigate a 0.5 hectare plot but knows nothing about the pump capacity. What factors must be taken into consideration?
Answers:
1. What flow rate is required?
2. What head is
required?
3. Does the pump have adequate flow and head capacity?
6. Describe the equation needed to calculate total dynamic head in a pumping system. (Assume velocity head is negligible.)
Answer:
tdh = hz + hp + hf
hz = elevation head (vertical distance pump must
raise water)
hp = pressure head (pressure required to operate the
sprinkler system, drip system, etc.)
hf = friction head (pressure
losses due to flow through pipe)
7. List the general specifications that you should provide a pump supplier when selecting a pump.
- amount of water to be pumped
- pumping head
- horsepower
requirements
- operating efficiency
- type of prime mover and transmission
device
8. In a communal system, a farmer is given water 6 hrs/day at a flow rate of 100 L/min. a) If the farmer irrigates every 3 days, how large of a storage pond must he construct to store the water? b) If, at peak use, his crops need 5 cm of water per week, which he will apply through a sprinkler system, how much land can he irrigate with the pond? (No water on Saturdays.)
Answer:
a) Volume = Q x T = 100 L/min x 6 hrs/day x 3 days x 1
m3/1000 L x 60 min/hr = 108 m3. The pond can have an
average surface area of 100 m and be 1.1 m in depth. However, a freeboard is
generally needed. Thus, add 0.3 meters of required freeboard. Pond should be 10
m x 10 m x
1.4 m.
b) He will apply 2.5 cm, or 0.025 m twice a week. Area = volume/depth = 108 m3/.025 m = 4320 m2