18-Env-A6 Solid Waste Engineering and Management · May 2015
Question 5 of 12: Clay Liner Thickness for Leachate Seepage Control
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, May 2015 — 04-Env-A6 / 18-Env-A6, Solid Waste Engineering and Management. 3 hours duration, closed book, no calculator beyond an approved Casio/Sharp model, one letter-sized aid sheet permitted. All 12 questions constitute a complete paper (100 marks total).
Reference texts: Tchobanoglous, Theisen & Vigil, Integrated Solid Waste Management: Engineering Principles and Management Issues; Vesilind, Worrell & Reinhart, Solid Waste Engineering; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Freeze & Cherry, Groundwater; CCME, Guidance Document on Landfill Gas Management; Canadian Environmental Protection Act, 1999.
Question 5: Clay Liner Thickness for Leachate Seepage Control (6 marks)
$0.2\ \text{L/d per unit area} = 0.0002\ \text{m/d}$
Leachate ponding head above clay layer, $h$
$1\ \text{m}$ (held constant by pumping)
Boundary condition
Water table at the base of the clay layer
Find. The clay layer thickness $L$ that limits the downward seepage flux to $q = 0.2\ \text{L/day per m}^2$.
Vertical Darcy seepage through the clay liner: leachate ponds at 1 m above the clay, driving downward flow through thickness L to the water table at the liner base.
Approach. Apply the exam-supplied Darcy relation $Q=-KA\,dh/dL$ in flux form ($q=Q/A=K\,dh/dL$), taking the 1 m ponding head as the driving head loss across the layer and solving for the thickness $L$ that reduces the resulting flux to the 0.2 L/day/m² target.
Convert the target flux to consistent units. $q = 0.2\ \text{L/(d}\cdot\text{m}^2) = \dfrac{0.2\times10^{-3}\ \text{m}^3}{\text{d}\cdot\text{m}^2} = 0.0002\ \text{m/d}$.
Write Darcy's Law in flux form and solve for L. $q = K\dfrac{dh}{dL} = K\dfrac{h}{L}$, so $L = \dfrac{K\,h}{q}$.
Substitute the given values. $L = \dfrac{(0.0008\ \text{m/d})(1\ \text{m})}{0.0002\ \text{m/d}} = \boxed{4.0\ \text{m}}$.
Check
A stricter hydrogeologic treatment would add the layer's own thickness to the driving head, since the water table (zero total head, by the stated boundary condition) sits at the BASE of the clay rather than at its top: $q=K(h+L)/L$. Because $K=0.0008$ m/d already exceeds the 0.2 L/day/m² target (0.0002 m/d) — and $q\to K$ as $L\to\infty$ — that stricter form shows no finite thickness of THIS clay can ever reach the target flux by gravity seepage alone; a thicker layer only asymptotically approaches, never beats, $K$ itself. The boxed 4.0 m therefore reflects the direct, single-step application of the exam's own supplied formula (the intended method for a 6-mark, no-calculator question) taking the 1 m ponding as the sole driving head; the companion insight — that meeting a flux target below the material's own $K$ genuinely requires a lower-permeability material or a composite liner, not simply more clay — is worth stating as the engineering takeaway.
