Agricultural Drainage, Salinity Control, Groundwater Hydrology, Wells and Pumps
2. Leaching requirement, salinity control and water quality
Every irrigation adds salt, and unless some water passes below the root zone the salt accumulates. The leaching requirement is the fraction of applied water that must drain through the root zone to hold its salinity at a level the crop tolerates. In the steady-state (USSL) form, LR = EC_iw/EC_dw — the electrical conductivity of the irrigation water over the allowable conductivity of the drainage water. With EC_iw = 2 dS/m and EC_dw = 8 dS/m, LR = 0.25, and a crop whose ET is 600 mm needs a total application of 600/(1 − 0.25) = 800 mm, of which 200 mm leaches. A widely used alternative (FAO) is LR = EC_w/(5EC_e − EC_w), with EC_e the soil-extract salinity the crop tolerates.
| Index or class | Definition | What it warns of |
|---|---|---|
| Electrical conductivity (EC) | conductance of the water in dS/m, proportional to dissolved salts | salinity hazard — osmotic stress on the crop |
| Sodium adsorption ratio (SAR) | Na/√((Ca + Mg)/2), concentrations in meq/L | sodium hazard — dispersion and loss of permeability |
| Residual sodium carbonate (RSC) | (CO₃ + HCO₃) − (Ca + Mg), in meq/L | calcium precipitating, leaving sodium to dominate |
| Saline soil | EC_e > 4 dS/m, ESP < 15, pH usually < 8.5 | reclaimed by leaching with good water and drainage |
| Sodic (alkali) soil | ESP > 15, EC_e < 4 dS/m, pH usually > 8.5 | needs an amendment such as gypsum to replace sodium by calcium, then leaching |
With Na = 20, Ca = 5 and Mg = 3 meq/L, SAR = 20/√4 = 10. Reuse of drainage water — for salt-tolerant crops, by blending with fresh water, or in cyclic use alternating with good water — stretches supply, but the salt it carries must still leave the system somewhere, which is why reuse schemes need an outlet for the concentrated end.
3. Groundwater occurrence, movement and flow in aquifers; exploration and recharge
Below the water table the pores are saturated. An aquifer stores and transmits water in usable quantity; an aquitard transmits slowly, an aquiclude holds water but does not transmit it, and an aquifuge neither holds nor transmits. An unconfined aquifer has the water table as its upper surface; a confined aquifer lies between impermeable layers and is under pressure, so a well in it rises to the piezometric surface (and flows at the surface if that lies above ground — an artesian well); a leaky (semi-confined) aquifer is bounded by an aquitard; a perched aquifer sits on a local lens above the main water table. The water an unconfined aquifer yields by gravity drainage per unit volume is its specific yield S_y; what it retains against gravity is its specific retention; and S_y + S_r = porosity. A confined aquifer releases water by compression and expansion, far less per unit head — its storage coefficient is typically orders of magnitude smaller than S_y.
Groundwater moves by Darcy’s law, v = Ki, where v is the Darcy (apparent) velocity over the whole cross-section; the actual seepage velocity through the pores is v/n_e, larger by the effective porosity: K = 50 m/day and i = 0.002 give v = 0.1 m/day and, with n_e = 0.25, a seepage velocity of 0.4 m/day. The transmissivity of a confined aquifer of thickness b is T = Kb. Steady radial flow to a fully penetrating well (Thiem) gives Q = 2πKb(h₂ − h₁)/ln(r₂/r₁) in a confined aquifer and Q = πK(h₂² − h₁²)/ln(r₂/r₁) in an unconfined one, with h₁ and h₂ the heads at observation radii r₁ and r₂. Unsteady flow is described by the Theis solution, s = (Q/(4πT))W(u) with u = r²S/(4Tt), and for small u (less than about 0.01) by the Cooper–Jacob straight-line approximation s = (2.30Q/(4πT)) log₁₀(2.25Tt/(r²S)), from which a pumping test yields T and S.
Exploration locates aquifers before drilling: surface geophysics — electrical resistivity soundings (Wenner and Schlumberger arrays; fresh-water sands and gravels show higher resistivity than clays or saline zones) and seismic refraction — together with geological and remote-sensing studies, test drilling and borehole logging. Recharge is estimated by the water-table fluctuation method (recharge = S_y × rise + draft over the period), by rainfall-infiltration factors, and by water-balance budgets. Artificial recharge adds water deliberately through spreading basins, percolation tanks, check dams, recharge pits and shafts, and injection wells.
4. Types of wells; design and construction of water wells
Open (dug) wells are large-diameter, shallow and store water in their own volume, suiting low-yield aquifers; dug-cum-bore wells add a bore at the bottom; tube wells — shallow or deep, cased and screened — tap confined or unconfined aquifers at depth; cavity wells draw from a cavity formed under a stable clay layer; filter-point wells are small shallow tube wells driven in sand. Drilling is by the cable-tool (percussion) method in hard or bouldery formations, the direct rotary method with circulating mud for speed in unconsolidated material, the reverse rotary method for large-diameter gravel-packed wells, and jetting or boring for shallow wells.
A tube well’s screen is placed against the best water-bearing layers, with a slot size chosen from the aquifer’s grain-size curve so that most of the sand is held back; where the aquifer is fine or uniform, a graded gravel pack is placed around the screen. The well is then developed — by surging, over-pumping or backwashing — to remove fines and the drilling mud and to raise the permeability next to the screen. Its performance is summarised by the specific capacity, the yield per unit drawdown: 20 L/s at 4 m drawdown is 5 L/s per metre. The well loss caused by flow through the screen and casing adds to the aquifer loss, so specific capacity falls as the pumping rate rises.
5. Classification, characteristics, selection and installation of pumps
Pumps are positive-displacement — reciprocating (piston, plunger) and rotary (gear, screw) — which deliver a nearly fixed volume per stroke or revolution whatever the head, or rotodynamic, which add energy by a spinning impeller: centrifugal (radial-flow, volute or diffuser casing, for moderate flows at high head), mixed-flow, and axial-flow (propeller, for large flows at low head). Vertical turbine and submersible pumps are multistage centrifugal pumps for deep wells; the jet pump and air-lift pump have no moving parts in the well. The type follows from the specific speed N_s = N√Q/H^(3/4): low for centrifugal, intermediate for mixed-flow, high for axial-flow.
A centrifugal pump’s characteristic curves at constant speed plot head, power and efficiency against discharge: the head falls as the discharge rises, the efficiency peaks at the best efficiency point, and the power drawn rises with discharge. The operating point is where the pump curve crosses the system curve (static lift plus friction losses rising as Q²). The affinity laws for one pump at changed speed are Q ∝ N, H ∝ N² and P ∝ N³: doubling the speed from 1450 to 2900 rev/min takes a pump from 20 L/s at 25 m to 40 L/s at 100 m and multiplies its power by eight. The power required is P = ρgQH/η — 0.02 m³/s against 30 m at 60 per cent efficiency needs 9.81 kW. Pumps in series add their heads at the same discharge; in parallel they add their discharges at the same head.
Selection and installation. A pump is chosen so that its best efficiency point sits at the required duty; a centrifugal pump above the water surface must be primed (its casing and suction pipe filled) before starting, and fitted with a foot valve and strainer. The suction lift is limited because the pressure at the impeller eye must stay above the vapour pressure: atmospheric pressure at sea level supports a water column of only 10.33 m, and friction, velocity head and the pump’s required NPSH (net positive suction head) cut the practical lift well below that. Too great a suction lift causes cavitation — vapour bubbles forming and collapsing on the impeller, with noise, loss of head and pitting — which is why deep wells use submersible or turbine pumps set below the water level.
Key takeaways
- Q = DC × A; Hooghoudt q = (8K_b d_e h + 4K_a h²)/L², reducing to L² = 4Kh²/q for drains on the barrier.
- LR = EC_iw/EC_dw, and total water = ET/(1 − LR); SAR = Na/√((Ca + Mg)/2) in meq/L.
- Saline: EC_e > 4, ESP < 15, leach it; sodic: ESP > 15, add gypsum first.
- Thiem: Q = 2πKb Δh/ln(r₂/r₁) confined, πK(h₂² − h₁²)/ln(r₂/r₁) unconfined; S_y + S_r = porosity; seepage velocity = v/n_e.
- Affinity laws Q ∝ N, H ∝ N², P ∝ N³; pump power = ρgQH/η; series adds head, parallel adds discharge.
Practice questions (14)
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 subsurface drainage system for 50 ha is designed for a drainage coefficient of 1 cm/day. What is the design discharge, in L/s (to one decimal place)?
Numerical answer — type the value.
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Answer: 57.9
Volume per day = 0.01 m × 50 × 10⁴ m² = 5000 m³; Q = 5000/86 400 = 0.0579 m³/s = 57.9 L/s. Dividing by 3600 instead of 86 400 treats the coefficient as an hourly depth and gives 1389 L/s.Parallel drains are to hold the water table 0.8 m above drain level midway between them, with a drainage coefficient of 2 mm/day. The soil has K = 1 m/day above and below the drains, and the equivalent depth to the impermeable layer is 3 m. What drain spacing does Hooghoudt’s equation give, in m (to one decimal place)?
Numerical answer — type the value.
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Answer: 104.3
L² = (8Kd_e h + 4Kh²)/q = (8 × 1 × 3 × 0.8 + 4 × 1 × 0.64)/0.002 = (19.2 + 2.56)/0.002 = 10 880, so L = 104.3 m. Leaving out the flow below the drains (the 8Kd_e h term) gives only 35.8 m; forgetting to convert q from mm to m gives 3.3 m.Drains resting on an impermeable layer must keep the water table 1.0 m above them midway, in a soil with K = 0.5 m/day, for a drainage coefficient of 5 mm/day. What is the drain spacing, in m?
Numerical answer — type the value.
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Answer: 20
With d = 0, L² = 4Kh²/q = 4 × 0.5 × 1²/0.005 = 400, so L = 20 m. Using 8K instead of 4K (borrowing the below-drain term) gives 28.3 m.An irrigation water contains 20 meq/L of Na⁺, 5 meq/L of Ca²⁺ and 3 meq/L of Mg²⁺. What is its sodium adsorption ratio?
Numerical answer — type the value.
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Answer: 10
SAR = Na/√((Ca + Mg)/2) = 20/√(8/2) = 20/2 = 10. Forgetting to halve the sum gives 20/√8 = 7.07, and forgetting the square root gives 5.Irrigation water has an EC of 2 dS/m and the drainage water may not exceed 8 dS/m. If the crop’s evapotranspiration for the season is 600 mm, what total depth of water must be applied (ignoring other losses), in mm?
Numerical answer — type the value.
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Answer: 800
LR = EC_iw/EC_dw = 2/8 = 0.25, so the applied depth D satisfies D − 0.25D = 600 and D = 600/0.75 = 800 mm, of which 200 mm drains below the root zone. Adding 25 per cent of the ET (750 mm) under-supplies, because the leaching fraction is of the applied water, not of the ET.A well fully penetrates a confined aquifer 20 m thick with K = 30 m/day. At steady state the heads in observation wells 10 m and 100 m from it differ by 2 m. What is the pumping rate, in m³/day (to the nearest whole number)?
Numerical answer — type the value.
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Answer: 3275
T = Kb = 600 m²/day; Q = 2πT(h₂ − h₁)/ln(r₂/r₁) = 2π × 600 × 2/ln 10 = 7539.8/2.3026 = 3274.5 ≈ 3275 m³/day. Using log₁₀(10) = 1 in place of ln 10 gives 7540 m³/day, 2.3 times too much.An aquifer has K = 50 m/day and an effective porosity of 0.25, and the hydraulic gradient is 0.002. What is the average seepage (pore) velocity of the groundwater, in m/day?
Numerical answer — type the value.
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Answer: 0.4
Darcy velocity v = Ki = 50 × 0.002 = 0.1 m/day over the whole cross-section; the water actually moves through the pores only, so v_s = v/n_e = 0.1/0.25 = 0.4 m/day. Reporting 0.1 m/day gives the Darcy flux, not the speed a tracer would travel.A centrifugal pump delivers 20 L/s against a head of 25 m at 1450 rev/min. What head does it develop at 2900 rev/min, in m?
Numerical answer — type the value.
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Answer: 100
H ∝ N², so H₂ = 25 × (2900/1450)² = 25 × 4 = 100 m; the discharge doubles to 40 L/s and the power rises eightfold. Scaling the head linearly gives 50 m.A pump lifts 0.02 m³/s of water through a total head of 30 m with an overall efficiency of 60 per cent. What input power does it need, in kW (to two decimal places)? Take g = 9.81 m/s².
Numerical answer — type the value.
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Answer: 9.81
P = ρgQH/η = 1000 × 9.81 × 0.02 × 30/0.6 = 5886/0.6 = 9810 W = 9.81 kW. Multiplying by the efficiency instead of dividing gives 3.53 kW, the water power times η — less than the water itself receives.A tube well yields 20 L/s with a drawdown of 4 m. What is its specific capacity, in L/s per metre?
Numerical answer — type the value.
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Answer: 5
Specific capacity = Q/s = 20/4 = 5 L/s per metre of drawdown. Multiplying yield by drawdown (80) gives a meaningless product; the index is yield per unit drawdown.For an unconfined aquifer material, the sum of the specific yield and the specific retention equals:
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Answer: A — the porosity
Of the water that fills the pores, part drains by gravity (specific yield) and the rest is held by capillarity and adhesion (specific retention); together they fill the pore space, so their sum is the porosity. The void ratio is V_v/V_s, not V_v/V, and the storage coefficient describes a confined aquifer.Which of the following statements about salt-affected soils are correct?
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Answer: A — A saline soil has EC_e above 4 dS/m and ESP below 15; B — A sodic soil has an exchangeable sodium percentage above 15; C — Gypsum is applied to reclaim a sodic soil
These are the USSL criteria: saline soils have high EC and low ESP, sodic soils high ESP. Gypsum supplies calcium that displaces exchangeable sodium, which is then leached. Leaching alone lowers the salt and lets the sodic clay disperse further, so the fourth statement is false.Which of the following statements about pumps are correct?
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Answer: A — Two identical pumps in series deliver the same discharge at about twice the head; C — Axial-flow pumps have a high specific speed and suit large flows at low head; D — The risk of cavitation rises as the suction lift increases
Series pumps add heads; axial-flow machines sit at the high end of specific speed; a greater suction lift lowers the pressure at the impeller eye towards the vapour pressure. Atmospheric pressure at sea level supports only 10.33 m of water, and the practical lift is less, so a 12 m suction lift is impossible.The Cooper–Jacob straight-line method of analysing a pumping test is valid only when:
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Answer: A — u = r²S/(4Tt) is small, about 0.01 or less — i.e. at large times or near the well
Cooper–Jacob keeps only the first two terms of the series for the well function W(u), which is accurate when u is small — late in the test or close to the well. Steady state is Thiem’s case, and a drawdown that large in an unconfined aquifer violates the assumptions of both methods.