Water Resources: Global Water, Surface Hydrology and Groundwater

The first half of Section 4 of the GATE Environmental Science and Engineering (ES) paper. Section 4 is two subjects under one heading — water resources, which is hydrology, and environmental hydraulics, which is fluid mechanics — so it takes two chapters, and this is the hydrology one. It follows the syllabus’s three sub-headings. Global water resources: the structure and properties of the water molecule that make water behave as it does, where the Earth’s water is, what threatens it and how it is conserved. Surface water: the hydrological cycle and the water balance, precipitation, infiltration, evapotranspiration and runoff, flow hydrographs and the unit hydrograph, the stage–discharge relationship, reservoir capacity, surface-water management and rainwater harvesting. Groundwater: geological formations as aquifers, the vadose and saturated zones, confined and unconfined aquifers and their parameters, Darcy’s law, and steady flow to wells.

1. Global water resources: structure, properties, distribution, threats and conservation

The water molecule is bent (H–O–H about 104.5°) and strongly polar, so neighbouring molecules hydrogen-bond. That one fact explains the properties an engineer relies on: a high specific heat (about 4.18 kJ/kg·K), which lets lakes and oceans buffer climate; a high latent heat of vaporisation, which makes evaporation the dominant energy sink at the land surface; a high surface tension, which drives capillary rise in soils; a high dielectric constant, which makes water the "universal solvent" and therefore easy to pollute; and a density maximum near 4 °C, which is why ice floats and lakes freeze from the top down, with liquid water surviving beneath. The same density–temperature curve sets up the thermal stratification of lakes treated in Section 5.

Of all the water on Earth, about 97% is saline, in the oceans. Only about 2.5–3% is fresh, and roughly two-thirds of that is locked in ice caps and glaciers, most of the rest is groundwater, and the lakes, rivers, soil moisture and atmosphere together hold well under 1% of the fresh water. The resource people actually use is therefore a small, renewable flux — the part of precipitation that becomes runoff and recharge each year — rather than the enormous stock. The threats follow from that: over-abstraction of groundwater faster than recharge (falling water tables, land subsidence, saline intrusion in coastal aquifers); pollution from sewage, industry and agriculture; the silting of reservoirs; the loss of wetlands; and a changing climate that alters when and where rain falls. Water conservation acts on both sides of the balance — on demand, through leak control in distribution, metering and pricing, efficient irrigation (drip and sprinkler), recycling and reuse; and on supply, through rainwater harvesting, artificial recharge and watershed treatment that keeps rain on the land long enough to infiltrate.

🎯 A stock is not a supply
A large aquifer can be pumped for decades at a rate well above its recharge, and the wells keep yielding while the stock falls — which is why groundwater overdraft is noticed so late. The sustainable yield is set by the recharge flux, not by the volume stored; mining the stock is a one-time withdrawal.

2. The hydrological cycle and the water balance: precipitation, infiltration, evapotranspiration and runoff

The hydrological cycle is a closed loop: water evaporates from the oceans and land, is carried as vapour, condenses and falls as precipitation, and returns to the sea as runoff or groundwater flow. For any catchment over any period the water balance is P − R − G − E − T = ΔS: precipitation minus surface runoff, net groundwater outflow, evaporation and transpiration equals the change in storage. Over a long period ΔS ≈ 0 and the balance says that runoff is what precipitation leaves after evapotranspiration. Point rainfall is converted to an areal average by the arithmetic mean (flat country, dense gauges), the Thiessen polygon method (each gauge weighted by the area nearer to it than to any other) or the isohyetal method (areas between lines of equal rainfall, which can use the forecaster’s knowledge of the terrain).

Infiltration is the entry of water into the soil. Its capacity falls during a storm as the surface pores fill and the soil wets, and Horton’s equation describes that decay: f = f_c + (f₀ − f_c)e^(−kt), where f₀ is the initial and f_c the final (steady) infiltration capacity. Integrating gives the cumulative infiltration F(t) = f_c t + (f₀ − f_c)(1 − e^(−kt))/k. For design a single average loss rate is often enough: the φ-index is the constant rate above which all rainfall becomes direct runoff. Evaporation is measured with pans (the Class A pan reads high and is multiplied by a pan coefficient, about 0.7) and estimated by energy balance or mass transfer; transpiration is the water plants draw from the soil and release through their leaves, and together they are evapotranspiration, whose potential value depends only on climate while the actual value is limited by the water available. Runoff is what reaches the stream: surface runoff and quick interflow make direct runoff; slow groundwater discharge is base flow. For a small catchment the peak runoff rate is estimated by the rational method, Q = CiA/3.6 (Q in m³/s, i in mm/h, A in km²), where C is the runoff coefficient and i the intensity of a storm lasting the time of concentration.

Estimating areal rainfall
MethodHow the gauges are weightedBest when
Arithmetic meanEquallyFlat terrain, evenly spread gauges
Thiessen polygonBy the area closer to each gauge than to any otherUneven gauge spacing; the weights are fixed once for a network
IsohyetalBy the area between isohyets, at their mean valueHilly terrain; the most accurate when drawn with skill
⚠️ Horton’s f is a rate; F is a depth
With f₀ = 8 cm/h, f_c = 2 cm/h and k = 1.5 h⁻¹, the rate after 2 h is f = 2 + 6e^(−3) = 2.30 cm/h, but the depth infiltrated in those 2 h is F = 2 × 2 + 6(1 − e^(−3))/1.5 = 4 + 3.80 = 7.80 cm. Multiplying the final rate by the time (4.6 cm) ignores the high early rates; multiplying f₀ by the time (16 cm) ignores the decay.

3. Hydrographs, the unit hydrograph, stage–discharge, reservoirs and rainwater harvesting

A flow hydrograph plots discharge against time at a stream section. It has a rising limb, a crest, and a recession limb, and it sits on base flow; separating the base flow (by a straight line from the start of rise to a point on the recession) leaves the direct runoff hydrograph, whose area is the volume of direct runoff. The unit hydrograph of duration D is the direct runoff hydrograph produced by 1 cm of effective rainfall falling uniformly over the catchment at a constant rate for D hours. It rests on two assumptions — linearity (a storm of 3 cm effective rain gives ordinates three times as large) and time invariance (the same storm gives the same response whenever it falls) — so a complex storm is handled by superposing lagged, scaled unit hydrographs. Because its volume is 1 cm over the catchment, the area under it fixes the catchment area: A = ΣQ·Δt/(0.01 m). A D-hour unit hydrograph is converted to another duration with the S-curve (the response to an infinite series of D-hour unit storms), and the flood hydrograph is then routed through reservoirs or channels.

Discharge is measured occasionally but stage (water level) is recorded continuously, so a stage–discharge relationship (rating curve) is fitted, usually Q = C(G − a)ⁿ, where G is the gauge reading and a the gauge reading of zero flow; taking logarithms makes it a straight line fitted by least squares. Reservoir capacity is sized so that storage carries the flow through dry spells: the mass-curve (Rippl) method draws cumulative inflow and a demand line and reads the largest gap, and the equivalent arithmetic is the sequent-peak algorithm, K_t = max(0, Kt−1 + D_t − Q_t), whose largest K is the required active storage. Beneath the active storage lies dead storage reserved for sediment. Surface-water management covers allocation among users, minimum environmental flows, flood control and the quality of the stored water. Rainwater harvesting collects rain from roofs and paved areas for storage or recharge: the harvestable volume is the catchment area times the rainfall times a runoff coefficient (about 0.8 for a hard roof), V = A·P·C, and the storage tank is sized for the longest dry period at the planned draw.

🧠 The sequent-peak table in one pass
With a constant demand of 7 units and inflows 10, 4, 2, 3, 12, 15: K = 0, then 0 + 7 − 4 = 3, 3 + 7 − 2 = 8, 8 + 7 − 3 = 12, 12 + 7 − 12 = 7, and max(0, 7 + 7 − 15) = 0. The largest K, 12 units, is the storage needed. Note that it is the run of three dry months, not the single driest one, that sets it.

4. Groundwater: aquifers, their parameters, Darcy’s law and steady well hydraulics

Below the ground surface the vadose (unsaturated) zone holds water under suction, with air in the larger pores; beneath it the saturated zone begins at the water table, where the pore pressure is atmospheric. A formation that both stores and transmits water in useful quantities is an aquifer (sands, gravels, fractured or karstic rock); one that stores water but transmits it very slowly is an aquitard (silty clay); one that neither is an aquiclude (massive clay) or aquifuge (unfractured granite). An unconfined aquifer has the water table as its upper boundary; a confined aquifer is sandwiched between confining layers and its water is under pressure, so a well rises to the piezometric surface and flows freely (an artesian well) if that surface is above the ground. A perched water table sits on a local lens of clay above the main one.

The aquifer parameters are: porosity n, the void fraction; specific yield S_y, the fraction of the volume that drains by gravity (always less than n, the difference being the specific retention held by surface tension); hydraulic conductivity K (m/d), which depends on both the medium and the fluid, as distinct from the intrinsic permeability k = Kμ/(ρg), a property of the medium alone; transmissivity T = Kb for an aquifer of thickness b, the flow per unit width under unit gradient; and the storage coefficient S, the volume released per unit surface area per unit decline of head — about S_y (0.05–0.3) for an unconfined aquifer, but only 10⁻⁵ to 10⁻³ for a confined one, whose water comes from the compression of the aquifer and the expansion of water rather than from drainage of pores. Darcy’s law states that the flow is proportional to the hydraulic gradient: Q = KiA, or the Darcy flux q = Q/A = Ki. The water actually moves only through the pores, so the seepage (average linear) velocity is v = Ki/n — faster than q, and the one that says how quickly a contaminant arrives. Darcy’s law holds for laminar flow, which in practice means a grain Reynolds number below about 1.

Steady flow to a well that fully penetrates the aquifer follows from Darcy’s law applied to cylinders around it. In a confined aquifer (Thiem): Q = 2πKb(h₂ − h₁)/ln(r₂/r₁) = 2πT(s₁ − s₂)/ln(r₂/r₁), with h the head and s the drawdown at radii r₁ < r₂. In an unconfined aquifer (Dupuit–Thiem) the saturated thickness itself falls towards the well, so the heads enter squared: Q = πK(h₂² − h₁²)/ln(r₂/r₁). The radius of influence R is where the drawdown vanishes, and putting r₂ = R, h₂ = H gives the yield of a well from its drawdown. The cone of depression of two neighbouring wells overlap and their drawdowns add, which is why well spacing matters.

Confined and unconfined aquifers compared
PropertyUnconfinedConfined
Upper boundaryThe water table, at atmospheric pressureA confining layer; water under pressure
Storage coefficient≈ S_y, about 0.05–0.3About 10⁻⁵ to 10⁻³
Source of water pumpedGravity drainage of poresAquifer compression and water expansion
Steady well equationQ = πK(h₂² − h₁²)/ln(r₂/r₁)Q = 2πKb(h₂ − h₁)/ln(r₂/r₁)
Vulnerability to surface pollutionHigh: recharged directly from aboveLower: protected except at its recharge area
⚠️ Darcy flux is not the speed of the water
With K = 50 m/d, i = 0.002 and n = 0.25, the Darcy flux is q = 0.1 m/d, but a tracer travels at v = q/n = 0.4 m/d — four times faster. Using q to estimate the arrival time of a contaminant at a well overestimates the warning time by the factor 1/n.

Key takeaways

  • About 97% of the Earth’s water is saline and most fresh water is ice or groundwater; the usable resource is the annual renewable flux, so sustainable yield is set by recharge, not by storage.
  • Water balance P − R − G − E − T = ΔS; Horton f = f_c + (f₀ − f_c)e^(−kt) with F = f_c t + (f₀ − f_c)(1 − e^(−kt))/k; rational method Q = CiA/3.6 in m³/s, mm/h and km².
  • A D-hour unit hydrograph is the response to 1 cm of effective rain in D hours; linearity and time invariance allow superposition, and its area over 1 cm gives the catchment area.
  • Rating curve Q = C(G − a)ⁿ; reservoir storage by mass curve or sequent peak K_t = max(0, Kt−1 + D_t − Q_t); rainwater harvest V = A·P·C.
  • T = Kb; S ≈ S_y unconfined but 10⁻⁵–10⁻³ confined; Darcy q = Ki and seepage velocity Ki/n; Thiem Q = 2πKb Δh/ln(r₂/r₁), Dupuit Q = πK(h₂² − h₁²)/ln(r₂/r₁).

Practice questions (15)

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.

  1. Lakes in cold climates freeze from the top down and keep liquid water beneath the ice chiefly because:

    1. water has its maximum density near 4 °C, so colder water and ice stay on top
    2. ice has a higher density than liquid water
    3. dissolved salts lower the freezing point at depth
    4. the specific heat of water rises steeply below 4 °C
    Show answer

    Answer: A — water has its maximum density near 4 °C, so colder water and ice stay on top

    Hydrogen bonding gives water a density maximum near 4 °C. As the surface cools below that, the cooler water is lighter and stays on top, then freezes; ice, being less dense still, floats and insulates the water below. Option B has the densities the wrong way round.
  2. Which of the following statements about the Earth’s water are correct?

    1. About 97% of it is saline
    2. Most of the fresh water is held in ice caps and glaciers
    3. Rivers and lakes hold most of the fresh water that is not ice
    4. Over-abstraction of a coastal aquifer can draw sea water into it
    Show answer

    Answer: A — About 97% of it is saline; B — Most of the fresh water is held in ice caps and glaciers; D — Over-abstraction of a coastal aquifer can draw sea water into it

    (a) and (b) are the standard budget: about 97% saline, and roughly two-thirds of the fresh water frozen. (d) is saline intrusion, which follows when pumping lowers the fresh-water head below what holds the salt-water wedge back. (c) is false: groundwater, not surface water, holds most of the unfrozen fresh water; lakes and rivers hold well under 1% of it.
  3. A 2 km² urban catchment has a runoff coefficient of 0.5. By the rational method, what is the peak runoff, in m³/s, for a rainfall intensity of 60 mm/h lasting the time of concentration? Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 16.67

    Q = CiA/3.6 = 0.5 × 60 × 2/3.6 = 60/3.6 = 16.67 m³/s. The factor 3.6 converts mm/h × km² into m³/s (1 mm/h over 1 km² is 1000 m³/h = 0.278 m³/s). Forgetting it gives 60, which is in no consistent unit.
  4. Infiltration capacity follows Horton’s equation with f₀ = 8 cm/h, f_c = 2 cm/h and k = 1.5 h⁻¹. If rainfall intensity exceeds the capacity throughout, what total depth of water infiltrates in the first 2 hours, in cm? Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 7.8

    F = f_c t + (f₀ − f_c)(1 − e^(−kt))/k = 2 × 2 + (8 − 2)(1 − e^(−3))/1.5 = 4 + 6 × 0.9502/1.5 = 4 + 3.80 = 7.80 cm. Using only the final rate (2 × 2 = 4 cm) misses the high early capacity; using f at t = 2 h (2.30 cm/h) times 2 h gives 4.60 cm.
  5. Which of the following are assumptions of unit hydrograph theory?

    1. The effective rainfall is uniform over the catchment
    2. The ordinates of direct runoff are proportional to the depth of effective rainfall
    3. A storm of given duration produces the same response whenever it falls
    4. Base flow is included in the unit hydrograph ordinates
    Show answer

    Answer: A — The effective rainfall is uniform over the catchment; B — The ordinates of direct runoff are proportional to the depth of effective rainfall; C — A storm of given duration produces the same response whenever it falls

    (a) uniform effective rain, (b) linearity and (c) time invariance are the assumptions that allow superposition. (d) is false: the unit hydrograph is a direct runoff hydrograph, obtained after the base flow has been separated; base flow is added back only when a flood hydrograph is built.
  6. The ordinates of a 4-hour unit hydrograph, read at 4-hour intervals, sum to 300 m³/s. What is the area of the catchment, in km²?

    Numerical answer — type the value.

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    Answer: 432

    The runoff volume is ΣQ·Δt = 300 × 4 × 3600 = 4.32 × 10⁶ m³, and it equals 1 cm = 0.01 m over the catchment, so A = 4.32 × 10⁶/0.01 = 4.32 × 10⁸ m² = 432 km². Forgetting that the depth is 1 cm, not 1 m, gives 4.32 km².
  7. The rating curve at a gauging site is Q = 20(G − 1)^1.5, with Q in m³/s and the gauge reading G in m. What is the discharge when the gauge reads 5 m, in m³/s?

    Numerical answer — type the value.

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    Answer: 160

    Q = 20(5 − 1)^1.5 = 20 × 4^1.5 = 20 × 8 = 160 m³/s. The constant a = 1 m is the gauge reading at zero flow; forgetting to subtract it gives 20 × 5^1.5 = 223.6 m³/s.
  8. A reservoir must supply a constant 7 Mm³ per month. The inflows in six successive months are 10, 4, 2, 3, 12 and 15 Mm³. Using the sequent-peak method with the reservoir full at the start, what active storage is required, in Mm³?

    Numerical answer — type the value.

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    Answer: 12

    K_t = max(0, Kt−1 + 7 − Q_t): 0, 3, 8, 12, 7, 0. The largest cumulative deficit, 12 Mm³, is the storage needed. The single worst month (deficit 5) understates it, because the deficits of months 2 to 4 accumulate before the wet months refill the reservoir.
  9. A house has a roof area of 150 m² and the annual rainfall is 800 mm. Taking a runoff coefficient of 0.8, how much rainwater can be harvested in a year, in m³?

    Numerical answer — type the value.

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    Answer: 96

    V = A × P × C = 150 × 0.8 × 0.8 = 96 m³. Leaving out the runoff coefficient gives 120 m³, which ignores the losses to wetting, splash and first-flush diversion.
  10. A confined aquifer and an unconfined aquifer of similar material show the same decline in head over the same area. Compared with the unconfined aquifer, the confined one has released:

    1. far less water, because its storage coefficient is orders of magnitude smaller
    2. the same volume, because the porosity is the same
    3. more water, because it is under pressure
    4. no water at all until the head falls below the confining layer
    Show answer

    Answer: A — far less water, because its storage coefficient is orders of magnitude smaller

    An unconfined aquifer drains its pores, S ≈ S_y ≈ 0.05–0.3; a confined aquifer stays saturated and yields water only by compression and expansion, S ≈ 10⁻⁵–10⁻³. The same head decline therefore releases hundreds to thousands of times less water. Option D is wrong because elastic storage releases water as soon as the head falls.
  11. An aquifer has hydraulic conductivity 50 m/d and porosity 0.25, and the hydraulic gradient is 0.002. What is the average seepage velocity of the groundwater, in m/d?

    Numerical answer — type the value.

    Show answer

    Answer: 0.4

    Darcy flux q = Ki = 50 × 0.002 = 0.1 m/d; the water moves only through the pores, so v = q/n = 0.1/0.25 = 0.4 m/d. Quoting 0.1 m/d gives the flux per unit total area, not the speed of the water.
  12. A well fully penetrates a confined aquifer 20 m thick with K = 40 m/d. Under steady pumping the drawdowns at observation wells 10 m and 100 m away differ by 2 m. What is the pumping rate, in m³/d? Give the answer to the nearest whole number.

    Numerical answer — type the value.

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    Answer: 4366

    Thiem: Q = 2πKb(s₁ − s₂)/ln(r₂/r₁) = 2π × 40 × 20 × 2/ln 10 = 10 053/2.3026 = 4366 m³/d. Using log₁₀ 10 = 1 instead of ln 10 gives 10 053 m³/d, too large by 2.303; T = Kb = 800 m²/d is the product that matters.
  13. A well pumps an unconfined aquifer with K = 30 m/d at steady state. The saturated thicknesses at observation wells 10 m and 100 m from it are 36 m and 40 m. What is the discharge, in m³/d? Give the answer to the nearest whole number.

    Numerical answer — type the value.

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    Answer: 12443

    Dupuit–Thiem: Q = πK(h₂² − h₁²)/ln(r₂/r₁) = π × 30 × (1600 − 1296)/ln 10 = π × 30 × 304/2.3026 = 28 651/2.3026 = 12 443 m³/d. Applying the confined formula with an average thickness of 38 m gives 2π × 30 × 38 × 4/2.3026 = 12 443 as well, because h₂² − h₁² = (h₂ + h₁)(h₂ − h₁) — the two agree exactly when the average thickness is used. Enter it as 12443.
  14. The water table in an unconfined aquifer of specific yield 0.25 falls by 2 m over an area of 1 km². What volume of water has been removed from storage, in m³?

    Numerical answer — type the value.

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    Answer: 500000

    ΔV = S_y × A × Δh = 0.25 × 10⁶ × 2 = 5 × 10⁵ m³. Using the porosity instead of the specific yield overstates it, because the specific retention stays in the pores against gravity. Enter it as 500000.
  15. Which of the following statements about aquifers and Darcy’s law are correct?

    1. Transmissivity is the hydraulic conductivity multiplied by the saturated thickness
    2. Specific yield is always less than porosity
    3. Darcy’s law holds for turbulent flow through coarse gravel
    4. An aquitard stores water but transmits it only slowly
    Show answer

    Answer: A — Transmissivity is the hydraulic conductivity multiplied by the saturated thickness; B — Specific yield is always less than porosity; D — An aquitard stores water but transmits it only slowly

    (a) T = Kb. (b) S_y = n − specific retention, and the retention is never zero. (d) is the definition of an aquitard. (c) is false: Darcy’s law is a laminar-flow law, valid for a grain Reynolds number below about 1; in coarse gravel near wells the flow can become non-Darcian.