Soil–Water–Plant Relationships and Irrigation: Crop Water Requirement, Scheduling, Conveyance, Application Methods and Efficiencies
2. Water requirement of crops: evapotranspiration, consumptive use, duty and delta
Evapotranspiration (ET) is the combined evaporation from soil and transpiration from plants; consumptive use is ET plus the small amount of water built into the plant tissue, and in practice the two are used interchangeably. It is estimated in two steps: a reference evapotranspiration ET₀ from weather data — by the FAO-56 Penman–Monteith equation (the standard method), by radiation or temperature methods (Hargreaves, Blaney–Criddle) where data are scarce, or from pan evaporation, ET₀ = K_p × E_pan with a pan coefficient K_p less than 1 — and then the crop evapotranspiration ET_c = K_c × ET₀, where the crop coefficient K_c follows the crop’s growth stages (low at establishment, peaking at mid-season, falling at maturity). With E_pan = 8 mm/day, K_p = 0.7 and K_c = 0.9, ET_c = 5.04 mm/day.
The net irrigation requirement is ET_c less the effective rainfall (and plus any leaching requirement); the gross irrigation requirement divides it by the application efficiency. Over a season, the total depth of water a crop needs is its delta Δ; the area that one cumec of continuous flow can irrigate over the crop’s base period B is its duty D. Since one cumec for one day is 86 400 m³, D = 8.64 B/Δ with D in ha/cumec, B in days and Δ in metres: a crop of Δ = 1.2 m over B = 120 days has a duty of 864 ha/cumec. Duty rises with efficiency, so improving conveyance and application raises the area a canal can serve.
3. Irrigation scheduling; design of irrigation channels and underground pipelines
Scheduling decides when to irrigate and how much. The approaches are soil-based (irrigate when the measured depletion reaches the MAD, or when tensiometers read a set suction), climate-based (a water balance of ET_c against rainfall, or the IW/CPE ratio — a fixed depth of irrigation water applied each time cumulative pan evaporation reaches a set amount), plant-based (canopy temperature, leaf water potential), and irrigating at the critical growth stages when water is scarce (crown-root initiation in wheat, flowering and grain formation in most cereals). The depth applied is the net depth divided by the application efficiency, and the interval is the net depth divided by the daily ET_c.
Irrigation channels on the farm are designed for uniform flow with Manning’s equation, the discharge they must carry, a permissible velocity (high enough to prevent silting, low enough to prevent scour in an unlined channel), side slopes stable for the soil, and freeboard. For a given area the most efficient (best hydraulic) section has the least wetted perimeter: for a rectangle the width is twice the depth (b = 2y) and the hydraulic radius is y/2; for a trapezoid it is half a regular hexagon, again with R = y/2. Lining (concrete, brick, plastic film) cuts seepage — the largest loss in unlined channels — and allows higher velocities and smaller sections. Underground pipelines of concrete or PVC deliver water with almost no seepage or evaporation loss and take no land; they are designed with Hazen–Williams or Darcy–Weisbach for the head available, and fitted with stands, air vents, pressure-relief and alfalfa valves to control and distribute flow.
4. Surface, sprinkler and micro-irrigation: design and evaluation
Surface irrigation spreads water by gravity over the soil: border strips (long strips between low ridges, for close-growing crops on gentle slopes), check basins (level plots enclosed by bunds, for rice and orchards and for leaching), and furrows (small channels between crop rows). Its design is a race between the advance of the water front down the field and the infiltration behind it: the inflow rate, length and slope are chosen so that the far end receives the required depth without excessive deep percolation at the head or runoff at the tail. The opportunity time at each point is the recession time minus the advance time.
Sprinkler irrigation sprays water under pressure through rotating or fixed nozzles. The discharge of a nozzle is q = C_d a √(2gh), and the application rate over a rectangular spacing is I = q/(S_l × S_m) — which must stay below the soil’s basic infiltration rate to avoid ponding and runoff. A sprinkler of 0.5 L/s on a 12 m × 12 m spacing applies 0.5 × 3600/144 = 12.5 mm/h. Uniformity is judged from catch-can tests by Christiansen’s uniformity coefficient, CU = 100(1 − Σ|x − x̄|/(n x̄)): catches of 8, 10, 12 and 10 mm (mean 10, total absolute deviation 4) give CU = 90 per cent. Wind distorts the pattern, so spacings are reduced in windy sites.
Micro-irrigation — drip (trickle), micro-sprinklers and bubblers — applies water slowly and often to the root zone only. An emitter discharge follows q = kHˣ: x = 0.5 for a turbulent-flow orifice or long-path emitter, x near 1 for laminar flow, and x near 0 for a pressure-compensating emitter, which is why those keep uniform discharge along long laterals on slopes. An orifice-type emitter giving 4 L/h at 10 m head gives 4 × √1.44 = 4.8 L/h at 14.4 m. Uniformity is judged by the emission uniformity EU = 100 q_n/q_a, with q_n the mean of the lowest quarter of emitter discharges and q_a the mean of all. The daily water need of a widely spaced crop is ET_c × plant spacing area × a wetted or canopy factor ÷ efficiency: 5 mm/day on a 3 m × 3 m spacing with a factor of 0.6 and 90 per cent efficiency is 30 L per plant per day. Filters (screen, disc, sand media), fertigation and flushing keep emitters from clogging, the main cause of failure.
5. Irrigation efficiencies
| Efficiency | Definition | What lowers it |
|---|---|---|
| Conveyance, η_c | water delivered to the field ÷ water diverted from the source | seepage and evaporation in channels |
| Application, η_a | water stored in the root zone ÷ water delivered to the field | deep percolation and tail-water runoff |
| Storage, η_s | water stored in the root zone ÷ water needed to fill it to field capacity | under-irrigation |
| Distribution, η_d | 100(1 − ȳ/d), ȳ the mean absolute deviation of depth from its mean d | uneven application |
| Overall (project), η_o | η_c × η_a (and any intermediate distribution efficiencies) | every loss in the chain |
The efficiencies chain: if 1000 L/s is diverted, 800 L/s reaches the field and 70 per cent of that is stored in the root zone, η_c = 80 per cent, η_a = 70 per cent and η_o = 56 per cent. Application efficiency and storage efficiency pull against each other — applying too little keeps η_a high and η_s low, applying too much does the opposite — so a good irrigation is judged on both, and on uniformity. Water-use efficiency is a different ratio: crop yield per unit of water (kg/m³), either per unit of ET (crop WUE) or per unit of water applied (field WUE).
Key takeaways
- FC is about 1/3 bar, PWP about 15 bar; available depth d = (FC − PWP)/100 × (ρ_b/ρ_w) × D.
- ET_c = K_c × ET₀, with ET₀ from Penman–Monteith or K_p × E_pan; the interval is net depth ÷ daily ET_c.
- Duty D = 8.64B/Δ (ha/cumec, days, metres).
- Sprinkler rate = q/(S_l × S_m) and must stay below the intake rate; CU = 100(1 − Σ|x − x̄|/(nx̄)); drip q = kHˣ.
- η_c × η_a gives the overall efficiency; application and storage efficiencies trade off against each other.
- The best hydraulic rectangular channel has b = 2y and R = y/2.
Practice questions (13)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A soil has a field capacity of 28 per cent and a permanent wilting point of 12 per cent (both by dry weight) and a bulk density of 1.5 g/cm³. What depth of available water is held in a 0.9 m root zone, in mm?
Numerical answer — type the value.
Show answer
Answer: 216
d = (FC − PWP)/100 × ρ_b × D = 0.16 × 1.5 × 900 mm = 216 mm. Leaving out the bulk density treats a percentage by weight as a percentage by volume and gives 144 mm.For the soil and root zone above (216 mm of available water), irrigation is given when 50 per cent of the available water has been used. If the crop uses 6 mm/day, what is the irrigation interval, in days?
Numerical answer — type the value.
Show answer
Answer: 18
Net depth = 0.5 × 216 = 108 mm; interval = 108/6 = 18 days. Irrigating only when the whole 216 mm is gone (36 days) lets the soil reach wilting point, which scheduling is designed to prevent.A crop has a delta of 1.2 m and a base period of 120 days. What is its duty, in ha/cumec?
Numerical answer — type the value.
Show answer
Answer: 864
D = 8.64B/Δ = 8.64 × 120/1.2 = 864 ha/cumec. Using Δ = 120 cm gives 8.64, a hundred times too small, because the constant 8.64 already assumes Δ in metres.Class A pan evaporation is 8 mm/day, the pan coefficient is 0.7 and the crop coefficient at mid-season is 0.9. What is the crop evapotranspiration, in mm/day (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 5.04
ET₀ = K_p × E_pan = 0.7 × 8 = 5.6 mm/day; ET_c = K_c × ET₀ = 0.9 × 5.6 = 5.04 mm/day. Dividing by the pan coefficient (8/0.7) inflates ET₀, since a pan evaporates more than a crop surface, not less.Of 1000 L/s diverted into a canal, 800 L/s reaches the fields, and 70 per cent of the water delivered to the fields is stored in the root zone. What is the overall irrigation efficiency, in per cent?
Numerical answer — type the value.
Show answer
Answer: 56
η_c = 800/1000 = 80 per cent; η_a = 70 per cent; η_o = 0.8 × 0.7 = 56 per cent. Averaging the two (75 per cent) or adding the losses (50 per cent) both misread a chain of efficiencies, which multiply.Before irrigation, 80 mm of water is needed to bring the root zone to field capacity. After irrigation, 60 mm has been stored in the root zone. What is the water storage efficiency, in per cent?
Numerical answer — type the value.
Show answer
Answer: 75
η_s = stored/needed = 60/80 = 75 per cent. It is independent of how much was applied; dividing the stored water by the applied water would give the application efficiency, a different measure.Sprinklers each discharging 0.5 L/s are laid out on a 12 m × 12 m square spacing. What is the application rate, in mm/h?
Numerical answer — type the value.
Show answer
Answer: 12.5
I = q/(S_l × S_m) = 0.5 × 3600 L/h/144 m² = 1800/144 = 12.5 mm/h, since 1 L/m² = 1 mm. Leaving q in L/s gives 0.0035 mm, an hourly rate short by the factor 3600.Four catch cans in a sprinkler test collect 8, 10, 12 and 10 mm. What is Christiansen’s uniformity coefficient, in per cent?
Numerical answer — type the value.
Show answer
Answer: 90
Mean x̄ = 10 mm; absolute deviations 2, 0, 2, 0 sum to 4; CU = 100(1 − 4/(4 × 10)) = 90 per cent. Summing the signed deviations gives zero and a false CU of 100 per cent.An orifice-type drip emitter (discharge exponent 0.5) delivers 4 L/h at an operating head of 10 m. What does it deliver at 14.4 m, in L/h?
Numerical answer — type the value.
Show answer
Answer: 4.8
q = kH^0.5, so q₂ = q₁√(H₂/H₁) = 4 × √1.44 = 4 × 1.2 = 4.8 L/h. Scaling linearly with head (5.76 L/h) assumes laminar flow; a pressure-compensating emitter would stay at 4 L/h.Mango trees are planted at 3 m × 3 m. The crop evapotranspiration is 5 mm/day, the wetted-area factor is 0.6 and the drip system efficiency is 90 per cent. What volume of water must be applied per plant per day, in litres?
Numerical answer — type the value.
Show answer
Answer: 30
V = ET_c × area × factor/efficiency = 5 mm × 9 m² × 0.6/0.9 = 45 × 0.6/0.9 = 30 L/day (1 mm over 1 m² is 1 L). Multiplying by the efficiency instead of dividing gives 24.3 L, too little to meet the crop’s need after losses.The permanent wilting point of a soil corresponds approximately to a soil-water suction of:
Show answer
Answer: A — 15 bar (1500 kPa)
The conventional wilting point is 15 bar, where most crops can no longer extract water fast enough to recover turgor. About 1/3 bar is field capacity (0.1 bar is sometimes used for sandy soils); 31 bar lies near the hygroscopic coefficient, drier still.Which of the following statements about soil-moisture measurement are correct?
Show answer
Answer: A — A tensiometer works only in the wet range, up to a suction of roughly 0.8 bar; B — A neutron probe relies on fast neutrons being slowed mainly by hydrogen in soil water; C — The gravimetric method, oven-drying at 105 °C, is the reference against which other methods are calibrated
Beyond about 0.8 bar air enters a tensiometer’s cup and it breaks tension; the neutron probe counts neutrons slowed by hydrogen; oven-drying is the reference method. TDR measures the travel time of a pulse set by the soil’s dielectric constant — the gypsum block is the resistance method, so the fourth statement is false.For a lined rectangular irrigation channel of the most economical (best hydraulic) section, the bed width b and the flow depth y are related by:
Show answer
Answer: A — b = 2y
For a fixed area A = by, the wetted perimeter P = b + 2y = A/y + 2y is least when dP/dy = 0, i.e. A = 2y², so b = 2y and R = A/P = y/2. b = y is a square section, which has a larger perimeter for the same area.