Irrigation water: turn a supplied depth into a volume
A water depth tells you how much water is spread over each unit of area. It does not tell you the whole volume until the area is known. A second distinction matters when some supplied water does not become useful water in the defined root-zone target: net volume and gross supply are different quantities.
All targets, efficiencies, tank sizes and method constraints in this lesson are fictional teaching inputs. They are not crop recommendations or measured performance claims. The task is to interpret the supplied quantities, calculate consistently and explain what the answer covers.
1. Why one millimetre can mean one litre
Imagine a uniform water layer 1 mm deep over 1 m². Convert the depth first: 1 mm = 0.001 m. Volume = area × depth = 1 m² × 0.001 m = 0.001 m³. Because 1 m³ = 1,000 litres, that volume is 1 litre.
This gives a convenient identity: volume in litres = area in square metres × depth in millimetres. Alternatively, convert depth to metres and calculate cubic metres. Do not multiply square metres by an unconverted millimetre value and then label the result cubic metres.
Text alternative: a rectangular footprint represents 1 m × 1 m, an area of 1 m². A shallow layer above it is labelled 1 mm, or 0.001 m. Multiplying gives 0.001 m³, equivalent to 1 litre. The picture represents a depth-to-volume calculation rather than an instruction to leave standing water on soil.
2. Name the boundary of the efficiency
Net volume is the useful water volume required by the question’s defined target. Gross volume is the supply needed at the stated starting point. Here efficiency is defined as net volume divided by gross volume. Thus net = gross × efficiency, and gross = net ÷ efficiency. Write 75% as 0.75 before dividing.
The boundary must stay consistent. If an exercise gives overall efficiency from tank outlet to the target root zone, use it once for that complete route. Do not apply the same loss adjustment again at each stage. In real systems different efficiency definitions may refer to different boundaries; a percentage without that context is incomplete information.
Worked case 1: area and an ideal depth
Plot A is a rectangle 25 m long and 10 m wide. The supplied net target is 6 mm. Its area is 25 × 10 = 250 m². Net volume = 250 × 6 = 1,500 litres. In cubic metres, 250 × 0.006 = 1.5 m³, which confirms the litre calculation.
For this first case alone, the model assumes no losses, so gross equals net. Doubling the area with the same depth would double the volume. Doubling both area and depth would multiply volume by four. These relationships are useful checks even before doing detailed arithmetic.
Worked case 2: account for a supplied loss fraction
The same 250 m² plot still needs the exercise’s net 1,500 litres, but overall efficiency from supply point to target is now given as 75%. Gross supply = 1,500 ÷ 0.75 = 2,000 litres. Check forward: 2,000 × 0.75 = 1,500 litres useful at the stated target.
The remaining 500 litres is outside that useful target in the model. It is 25% of the gross 2,000 litres. It is not correct to add 25% of the net volume: 1,500 × 1.25 = 1,875 litres would deliver only 1,406.25 useful litres at the given efficiency. The percentage belongs to the gross supply, which is why division is needed.
Worked case 3: decide what one tank can cover
A tank has a usable supply of 1,800 litres. Keep the 75% efficiency and the 6 mm net target. Useful volume = 1,800 × 0.75 = 1,350 litres. Supported area = 1,350 ÷ 6 = 225 m². The tank therefore falls short of the full 250 m² plot under the stated model.
The gross shortfall is 2,000 − 1,800 = 200 litres. The net shortfall is 1,500 − 1,350 = 150 litres. These are consistent because 200 × 0.75 = 150. If the useful 1,350 litres were averaged over the whole plot, the net depth would be 1,350 ÷ 250 = 5.4 mm, below the exercise’s 6 mm target.
3. Compare methods using the supplied constraints
Volume arithmetic cannot by itself choose an irrigation method. The following cards specify the constraints for a fictional comparison. Treat them as facts of the exercise, not universal ratings of every installation.
| Method | Supplied constraints | Interpretation |
|---|---|---|
| Drip | Available equipment requires clean, filtered water; filtration is currently unavailable | The water-quality requirement is unresolved, even if a low gross volume is attractive |
| Sprinkler | Site is exposed to frequent strong wind; the supplied system requires a pressure source not yet available | Both distribution conditions and the missing resource weaken this option |
| Surface | Only a surveyed, suitably graded level plot is accepted by this case; the fictional site has uneven ground | The required site condition has not been demonstrated |
For these cards, none is ready to select confidently. “More information or a resolved constraint is needed” is a valid answer. It is better than choosing a fashionable method while ignoring a mandatory condition. A different set of supplied site and resource facts could change the decision.
4. Three practice tasks with explained answers
Task 1: work in hectares
A fictional area of 0.4 ha has a supplied net target of 12 mm. Find its net volume in cubic metres.
Answer: 0.4 ha = 4,000 m²; 12 mm = 0.012 m. Volume = 4,000 × 0.012 = 48 m³, or 48,000 litres. Another check is that 1 mm over 1 ha equals 10 m³, so 0.4 × 12 × 10 also gives 48 m³.
Task 2: reverse the calculation
A model plot of 120 m² receives a gross 900 litres with 80% overall efficiency. What net depth does that represent?
Answer: Net volume = 900 × 0.80 = 720 litres. Net depth = 720 ÷ 120 = 6 mm. Dividing the gross 900 directly by the area would give 7.5 mm at the supply-volume equivalent, not the useful net depth requested.
Task 3: correct a percentage error
A learner writes “gross = 960 × 0.60 = 576 litres” for a net requirement of 960 litres and efficiency of 60%. Repair it.
Answer: Gross = 960 ÷ 0.60 = 1,600 litres. The forward check is 1,600 × 0.60 = 960. With efficiency below 100%, the necessary gross supply must exceed the net requirement.
5. A final unit and meaning check
Mark every number as area, depth, volume or efficiency. Keep the same plot and time period throughout. Check whether the answer is net or gross, and whether efficiency covers the whole stated route. Finally, remember that a mathematically correct volume does not validate the supplied depth for a real crop; that would require agricultural information not present in this exercise.
Frequently asked questions
How many litres are in one cubic metre?
There are 1,000 litres in 1 m³.
What does one millimetre over one hectare equal?
It equals 10 m³, or 10,000 litres, because a hectare contains 10,000 m².
Why divide by efficiency?
Efficiency is net divided by gross. Rearranging that relationship gives gross equal to net divided by efficiency.
Can net exceed gross in this model?
No. The model’s efficiency is at most 100%, so useful target volume cannot exceed its supplying volume.
Is 75% a recommended efficiency?
No. It is a supplied classroom value used to practise the relationship, not a claim about a particular system.
Does a tank’s labelled size prove usable supply?
No. Use the usable volume supplied by the question. A nominal capacity and available water need not be the same.
Can I choose a method from volume alone?
No. Site conditions, water quality, distribution requirements and available resources also affect the decision.
Technical references: FAO water-depth and volume calculations and FAO distinction between net and gross irrigation need.
WhatsApp Us Now