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18-Env-B2 Water Resources · May 2016

Question 5 of 6: Groundwater Zones and Darcy Flow in a Sand Aquifer

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2016 — 04-Env-B2 / Water Resources. 3 hours duration; closed book; Casio or Sharp approved calculator only. Six Problems are printed; any five constitute a complete paper (the first five answered are marked). Each Problem is worth 20 marks. All six are solved below for completeness.

Reference texts. Linsley, Kohler & Paulhus, Hydrology for Engineers (3rd ed.); Chow, Open-Channel Hydraulics; Freeze & Cherry, Groundwater; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Ontario Ministry of the Environment, Stormwater Management Planning and Design Manual (2003); Fisheries Act, Ontario Water Resources Act, Clean Water Act, 2006 (Ontario).

Problem 5: Groundwater Zones and Darcy Flow in a Sand Aquifer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Groundwater Zone Definitions

Vertical zonation of subsurface water, ground surface downward
ZoneDefinition
i. Soil water zoneThe near-surface, unsaturated belt extending from ground level to the base of the root zone, where water is held against gravity by capillary and adsorptive forces and is available for plant uptake and evapotranspiration; moisture content fluctuates directly with infiltration and ET.
ii. Intermediate (vadose) zoneThe unsaturated zone below the root/soil-water zone and above the capillary fringe, through which infiltrating "gravitational" water percolates downward as a film around soil grains on its way to recharge; it may be thin or absent where the water table is shallow.
iii. Capillary zone (capillary fringe)The zone immediately above the water table where pore water is drawn upward by surface-tension (capillary) forces and pore pressure remains below atmospheric; it can be nearly saturated even though it sits above the free water table, with its thickness controlled by pore/grain size (thicker in fine-grained soils).
iv. Saturated zone (zone of saturation)The zone below the water table where all interconnected pore spaces are completely filled with water at pressure greater than atmospheric; it is the zone tapped by wells and is what is normally meant by "the aquifer."

(b) Darcy Flow Through the Sand Aquifer

Given. Steady flow through a confined sand aquifer:

Given data
QuantitySymbolValue
Piezometric head gradient$i$0.01
Hydraulic conductivity$K$2 m/d
Effective porosity$n_e$0.3
Aquifer thickness—15 m
Aquifer width—1 km = 1000 m
Travel distance$L$100 m
Check: the source prints the conductivity as "2 mid" — read as 2 m/d (metres per day), the standard unit for hydraulic conductivity in this class of problem ("mid" is evidently a misprint of "m/d").

Find. (a) specific discharge $q$ and seepage velocity $v_s$; (b) volumetric flow rate $Q$; (c) travel time $t$ for groundwater to move 100 m.

Approach. Apply Darcy's Law to get the specific discharge, divide by effective porosity for the true (seepage) pore velocity, multiply by the aquifer's cross-sectional area for the volumetric flow, and divide the travel distance by the seepage velocity for the travel time.

  1. (a) Specific discharge (Darcy flux).$$q=Ki=(2)(0.01)=\boxed{0.02\ \text{m/d}}.$$Seepage (average linear pore) velocity:$$v_s=\frac{q}{n_e}=\frac{0.02}{0.3}=\boxed{0.0667\ \text{m/d}}.$$
  2. (b) Volumetric flow rate. Cross-sectional area of the aquifer, $A_{aq}=15\times1000=15{,}000$ m²:$$Q=qA_{aq}=(0.02)(15{,}000)=\boxed{300\ \text{m}^3/\text{d}}.$$
  3. (c) Travel time over 100 m. Using the seepage velocity (the true average pore-water velocity, not the specific discharge):$$t=\frac{L}{v_s}=\frac{100}{0.0667}=1500\ \text{d}=\boxed{4.11\ \text{yr}}.$$
QuantityValue
Specific discharge, $q$0.02 m/d
Seepage velocity, $v_s$0.0667 m/d
Volumetric flow rate, $Q$300 m³/d
Travel time (100 m)1500 d ≈ 4.11 yr