Hydrogeology: Porosity and Permeability, Aquifer Properties, Darcy’s Law, Groundwater Flow, Well Hydraulics and Exploration

Practice questions (18)

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. A bed of clay and a bed of well-sorted coarse sand are compared. Which statement is correct?

   1. The clay can have the higher total porosity yet a far lower conductivity, because its pores are tiny and hold water tightly
   2. The clay has the higher hydraulic conductivity, because its larger total porosity gives more pore space for water to flow through
   3. The sand has the higher specific retention, because coarse grains hold more water against gravity than fine grains do
   4. The two beds have equal hydraulic conductivity, because conductivity depends only on porosity and not on pore size
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   Answer: A — The clay can have the higher total porosity yet a far lower conductivity, because its pores are tiny and hold water tightly

   Total porosity of clay (often 40 per cent or more) can exceed that of sand, but the pores are tiny and water is bound to the grains, so K is smaller by many orders of magnitude. Conductivity depends on pore size and connection, not on porosity alone, so the claim of higher K for clay is wrong. Fine grains hold more water against gravity, so clay, not sand, has the larger specific retention.

2. A saturated sand has a porosity of 0.38. After free drainage under gravity it still holds water equal to 9 per cent of its total volume. Its specific yield is

   1. 0.09, the volume held by molecular attraction and capillarity
   2. 0.47, the porosity plus the water that stays in the sand
   3. 0.29, the porosity minus the specific retention
   4. 0.38, equal to the total porosity of the sand
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   Answer: C — 0.29, the porosity minus the specific retention

   Porosity splits as n = S_y + S_r. The water that stays after drainage (0.09) is the specific retention, so S_y = 0.38 − 0.09 = 0.29. The value 0.09 is the retention itself, 0.38 would mean that no water is retained at all, and adding the two quantities breaks the relation n = S_y + S_r.

3. When the head in a confined aquifer falls, the water released from storage comes mainly from

   1. gravity drainage of the pores, as in an unconfined aquifer, so that the aquifer progressively desaturates
   2. evaporation of water from the capillary fringe lying above the piezometric surface of the aquifer
   3. downward drainage of the overlying unsaturated zone into the top of the confined aquifer
   4. expansion of the water and compaction of the skeleton as pore pressure falls, the aquifer staying saturated
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   Answer: D — expansion of the water and compaction of the skeleton as pore pressure falls, the aquifer staying saturated

   In a confined aquifer the pores stay full: a head fall lowers the pore pressure, so the water expands slightly and the skeleton compacts, which defines the small storativity S = S_s b. Gravity drainage of pores, with specific yield, belongs to unconfined aquifers. Evaporation from a capillary fringe and drainage from above do not describe release from confined storage.

4. Two piezometers at the same site have their screens at elevations of 90 m and 70 m above a datum. Water stands at 98 m above the datum in the shallow piezometer and at 95 m above the datum in the deep one. The vertical flow between the two screens is

   1. downward, because total head is 98 m at the shallow screen and 95 m at the deep screen, so head falls with depth
   2. upward, because the pressure head is 25 m at the deep screen against 8 m at the shallow one, and water moves towards lower pressure
   3. zero, because both piezometers sit in the same vertical column and so share the same total head
   4. upward, because the deeper screen lies at the lower elevation, and groundwater always moves from low elevation to high elevation
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   Answer: A — downward, because total head is 98 m at the shallow screen and 95 m at the deep screen, so head falls with depth

   Total head is the water level above the datum: 98 m at the shallow screen and 95 m at the deep one. Flow goes from high to low total head, so it is downward, a recharge setting. The pressure heads (8 m and 25 m) are a distractor: pressure is higher at depth, yet flow does not follow pressure. Elevation alone does not set the direction either, and different total heads mean there is flow.

5. Which statement about the Darcy flux q and the seepage velocity v_s in a saturated sand is correct?

   1. v_s = q·n_e, so the pore water moves more slowly than the Darcy flux because the grains obstruct it
   2. v_s = q/n_e, so pore water moves faster than q suggests, as flow uses only part of the cross-section
   3. v_s equals q, because the Darcy flux is the average speed at which the water moves through the pores
   4. q = v_s/n_e, so the Darcy flux is larger than the actual speed of the water in the pores
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   Answer: B — v_s = q/n_e, so pore water moves faster than q suggests, as flow uses only part of the cross-section

   The Darcy flux is the discharge divided by the whole cross-sectional area, but water can pass only through the effective pores, so the true average speed is v_s = q/n_e, which exceeds q because n_e < 1. Travel times and tracer speeds therefore use v_s. The flux is not the pore speed, and the relations that multiply by n_e invert the geometry of the flow.

6. Darcy’s linear law is least likely to hold in which of the following situations?

   1. Slow laminar flow through fine sand under a small natural hydraulic gradient of about 0.001
   2. Fast flow through coarse gravel or an open fracture near a pumped well, at Reynolds number above 10
   3. Slow seepage of water through a thick silt layer under a small difference of head across it
   4. Regional laminar flow in a sandstone aquifer at a speed of only a few centimetres per day
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   Answer: B — Fast flow through coarse gravel or an open fracture near a pumped well, at Reynolds number above 10

   Darcy’s law assumes laminar, viscous-dominated flow, valid up to a Reynolds number of roughly 1 to 10. Near a pumped well in gravel or an open fracture, velocities are high enough for inertial and turbulent losses, so the relation between flux and gradient becomes non-linear. Slow flow in fine sand, silt or a regional sandstone aquifer is the standard domain of the law.

7. In a pumping test on a confined aquifer, drawdown plotted against log time first follows one straight line and later a second straight line with twice the slope. This is best explained by

   1. a perennial river acting as a recharge boundary, equivalent to an image injection well
   2. a single impermeable boundary, equivalent to an image well pumping at the same rate
   3. a leaky overlying aquitard that begins to supply water to the aquifer
   4. the aquifer becoming unconfined at late times, so that the storativity rises towards the specific yield
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   Answer: B — a single impermeable boundary, equivalent to an image well pumping at the same rate

   An impermeable boundary reflects the cone of depression, which is equivalent to a second well of the same sign pumping at the same rate, so after its effect arrives the drawdown rate doubles and the semi-log slope doubles. A recharge boundary or a leaky aquitard supplies water and flattens the curve, and a delayed-yield effect flattens the middle part rather than doubling the final slope.

8. A fully penetrating well pumps 1500 m³/d from a confined aquifer of transmissivity 300 m²/d. Assume steady radial flow. The difference in drawdown between observation wells at 20 m and 200 m from the well is closest to

   1. 0.80 m, from using log₁₀(200/20) = 1 in place of the natural logarithm
   2. 0.92 m, from using 4π in place of 2π in the denominator
   3. 1.83 m, from [Q/(2πT)] ln(200/20)
   4. 11.5 m, from Q/T multiplied by ln(200/20), omitting 2π
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   Answer: C — 1.83 m, from [Q/(2πT)] ln(200/20)

   Thiem’s equation gives s₁ − s₂ = [Q/(2πT)] ln(r₂/r₁) = [1500/(2π × 300)] × ln 10 = 0.7958 × 2.3026 = 1.83 m. The value 0.80 uses a base-10 logarithm, which is wrong in this form of the equation, 0.92 uses the Theis denominator 4π, and 11.5 forgets the 2π.

9. A vertical electrical sounding in an alluvial plain shows a layer of low resistivity between 25 m and 60 m depth. Which interpretation is most defensible before drilling?

   1. It is a fresh-water sand aquifer, because water always lowers resistivity, so fresh-water sand has the lowest resistivity of any layer
   2. It is dry sand, because air-filled pores conduct electric current better than water-filled pores
   3. It is unweathered crystalline bedrock, because such rock has lower resistivity than saturated sediments
   4. It may be clay or sand saturated with saline water, so logs, lithology and water-quality data are needed to tell them apart
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   Answer: D — It may be clay or sand saturated with saline water, so logs, lithology and water-quality data are needed to tell them apart

   Both clay and saline-water-saturated sand have low resistivity, and only one of them is a useful fresh-water aquifer, so the sounding cannot decide by itself. Fresh-water sand has moderate to high resistivity, not the lowest. Dry sand is highly resistive, not conductive, and fresh unweathered crystalline rock is far more resistive than saturated sediments.

10. In a granite-gneiss terrain, which site is the best first target for a bore well?

    1. A hill crest where fresh massive granite is exposed and no joints or weathering are visible
    2. A flat area with fresh massive gneiss at the surface and no lineament mappable on the imagery
    3. A steep scarp with thin soil over massive rock, where runoff is high and infiltration is poor
    4. A topographic low at a lineament intersection, over thick weathering on fractured rock
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    Answer: D — A topographic low at a lineament intersection, over thick weathering on fractured rock

    In crystalline rock the water is held in the weathered mantle and in fractures below it, so the best sites are lows with thick weathering and intersecting lineaments, which concentrate recharge and fracturing. Fresh massive rock, a hill crest or a steep scarp with thin soil offers little storage, little infiltration and few open fractures.

11. Which of the following statements about aquifer storage properties are correct?

    1. For the same head decline an unconfined aquifer releases far more water per unit area than a confined one, because S_y is much larger than confined storativity
    2. Specific retention is the volume of water that drains freely from a unit volume of the formation under gravity
    3. Transmissivity T = Kb has the dimensions of length squared per unit time
    4. Storativity of a confined aquifer is expressed in units of inverse metres
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    Answer: A — For the same head decline an unconfined aquifer releases far more water per unit area than a confined one, because S_y is much larger than confined storativity; C — Transmissivity T = Kb has the dimensions of length squared per unit time

    S_y (0.01 to 0.3) is orders of magnitude larger than confined S (about 10⁻⁵ to 10⁻³), and T = K × b has units m²/d. Specific retention is the water held against gravity, which is the opposite of what drains freely, and storativity S = S_s b is dimensionless; the quantity with units of inverse length is the specific storage S_s.

12. Which of the following statements about groundwater flow are correct?

    1. Groundwater always flows from a point of higher pressure to a point of lower pressure, whatever their elevations
    2. For flow parallel to layers the effective conductivity is the thickness-weighted arithmetic mean, and for flow across the layers it is the harmonic mean
    3. At a boundary between two media, flow lines refract so that tan θ₁/tan θ₂ = K₁/K₂, with the angles measured from the normal
    4. The seepage velocity is always smaller than the Darcy flux, because the pore water is slowed by the grains
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    Answer: B — For flow parallel to layers the effective conductivity is the thickness-weighted arithmetic mean, and for flow across the layers it is the harmonic mean; C — At a boundary between two media, flow lines refract so that tan θ₁/tan θ₂ = K₁/K₂, with the angles measured from the normal

    The thickness-weighted arithmetic mean (parallel flow) and harmonic mean (flow across layers) are the standard layered-medium results, and the tangent law of refraction is tan θ₁/tan θ₂ = K₁/K₂. Water moves down total head, so a deeper point may have the higher pressure and still receive no flow from the shallower one. Seepage velocity v_s = q/n_e is larger than the Darcy flux, not smaller.

13. Which of the following statements about well hydraulics are correct?

    1. The Theis solution assumes a confined, homogeneous, isotropic aquifer of infinite extent and a fully penetrating well of negligible diameter
    2. The Cooper-Jacob straight-line method is valid when u is large, at the earliest times after pumping begins
    3. Well loss is the part of the drawdown proportional to Q, whereas aquifer loss is proportional to Q²
    4. A recharge boundary such as a perennial river acts like an image injection well and makes the drawdown flatten
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    Answer: A — The Theis solution assumes a confined, homogeneous, isotropic aquifer of infinite extent and a fully penetrating well of negligible diameter; D — A recharge boundary such as a perennial river acts like an image injection well and makes the drawdown flatten

    The listed Theis assumptions are the standard ones, and a recharge boundary behaves as an image injection well so drawdown flattens. The Cooper-Jacob line is valid for small u (later times), not large u. In s_w = BQ + CQ², the linear term BQ is the aquifer loss and the quadratic term CQ² is the well loss, the reverse of the statement given.

14. Two observation wells 300 m apart in a sand aquifer have heads of 52.4 m and 51.2 m. The hydraulic conductivity is 15 m/d and the effective porosity is 0.20. The time taken by groundwater to travel 600 m along the flow line at the seepage velocity is ____ days.

    Numerical answer — type the value.

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    Answer: 2000

    Gradient i = (52.4 − 51.2)/300 = 0.004. Darcy flux q = K i = 15 × 0.004 = 0.06 m/d. Seepage velocity v_s = q/n_e = 0.06/0.20 = 0.30 m/d. Time = 600/0.30 = 2000 days. Using q instead of v_s would give 10 000 days.

15. An unconfined aquifer of area 25 km² has specific yield 0.08. Between the pre-monsoon and post-monsoon surveys the water table rose by 2.4 m. This rise is net of a seasonal pumping draft of 1.3 million m³ taken from the aquifer in the same period; other inflows and outflows are negligible. By the water-table fluctuation method, gross recharge (net storage gain plus draft) is ____ million m³, to one decimal place.

    Numerical answer — type the value.

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    Answer: 6.1

    Net storage gain = S_y × Δh × A = 0.08 × 2.4 × 25 × 10⁶ = 4.8 × 10⁶ m³ = 4.8 million m³. Because that rise is after the draft was withdrawn, gross recharge = 4.8 + 1.3 = 6.1 million m³. Reporting 4.8 would understate recharge by the amount pumped out.

16. Three horizontal layers lie one above another: 4 m thick with K = 12 m/d, 2 m thick with K = 0.3 m/d and 4 m thick with K = 6 m/d. The effective vertical hydraulic conductivity of the sequence, to two decimal places, is ____ m/d.

    Numerical answer — type the value.

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    Answer: 1.30

    For flow across layers K_v = Σb_i/Σ(b_i/K_i). Total thickness = 10 m. Σ(b_i/K_i) = 4/12 + 2/0.3 + 4/6 = 0.3333 + 6.6667 + 0.6667 = 7.6667 d. K_v = 10/7.6667 = 1.30 m/d. The arithmetic mean, 7.26 m/d, is the horizontal conductivity and is wrong for vertical flow, where the thin 0.3 m/d layer controls.

17. Cooper-Jacob analysis of a pumping test at Q = 2400 m³/d gives a drawdown of Δs = 0.8 m per log₁₀ cycle of time. The straight line cuts zero drawdown at t₀ = 0.002 d in an observation well at r = 30 m. Use T = 2.303Q/(4πΔs) and S = 2.25Tt₀/r². The storativity S, expressed as a multiple of 10⁻³ and to two decimal places, is ____ × 10⁻³.

    Numerical answer — type the value.

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    Answer: 2.75

    T = 2.303 × 2400/(4π × 0.8) = 5527.2/10.053 = 549.8 m²/d. S = 2.25 × 549.8 × 0.002/30² = 2.474/900 = 2.749 × 10⁻³, i.e. 2.75 × 10⁻³. The slope gives T and only then does the intercept give S; using r² = 30 instead of 900 would give a value 30 times too large.

18. A clean saturated sand has porosity 0.20 and pore-water resistivity 15 Ωm. Use Archie’s law ρ₀ = ρ_w a φ^(−m) with a = 1 and m = 2. The bulk resistivity of the sand is ____ Ωm.

    Numerical answer — type the value.

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    Answer: 375

    Formation factor F = a φ^(−m) = 0.20^(−2) = 25. Bulk resistivity ρ₀ = F × ρ_w = 25 × 15 = 375 Ωm. Using φ^m instead of φ^(−m) would give 0.6 Ωm, which is far below the pore-water resistivity and therefore impossible for a rock that also contains insulating grains.