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18-Env-A6 Solid Waste Engineering and Management · December 2013

Question 7 of 16: Annual Landfill Area for a Population of 50,000

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

Notes on this paper

Reference texts: Tchobanoglous, Theisen & Vigil, Integrated Solid Waste Management: Engineering Principles and Management Issues; Freeze & Cherry, Groundwater (Darcy's Law, vadose zone); CCME, Guidance Document on Landfill Gas Management; EGBC/Engineers Canada 04-Env-A6/18-Env-A6 examination syllabus.

Question 7: Annual Landfill Area for a Population of 50,000 (5 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.

Given. Population $= 50{,}000$; compacted refuse depth $= 4\ \text{m}$ (cover material excluded, as stated).

Find. The annual land area (excluding buffer zone) required to place one year's compacted refuse at this depth.

Approach. Estimate the annual mass of waste generated from an assumed per-capita generation rate, convert to compacted volume using an assumed compacted density, then divide by the stated refuse depth to get area.

Check: neither a generation rate nor a compacted density is given in this question. A per-capita generation rate of 2.0 kg/cap/day is assumed (a typical North American municipal design value, Tchobanoglous). The compacted density is taken as 600 kg/m³, reused from this same paper's own Q15 data ("density of well compacted landfill = 600 kg/m³") for internal consistency rather than inventing a separate figure.
  1. Annual mass of waste generated. $$m = 50{,}000 \times 2.0\ \text{kg/cap/d} \times 365\ \text{d/yr} = 36{,}500{,}000\ \text{kg/yr} = 36{,}500\ \text{t/yr}$$
  2. Annual compacted volume. $$V = \dfrac{m}{\rho_{compacted}} = \dfrac{36{,}500{,}000\ \text{kg/yr}}{600\ \text{kg/m}^3} = 60{,}833\ \text{m}^3/\text{yr}$$
  3. Annual area required (at 4 m refuse depth). $$A = \dfrac{V}{\text{depth}} = \dfrac{60{,}833\ \text{m}^3/\text{yr}}{4\ \text{m}} = 15{,}208\ \text{m}^2/\text{yr} = \boxed{1.52\ \text{ha/yr}}$$
QuantityValue
Annual waste generated36,500 t/yr
Annual compacted volume60,833 m³/yr
Annual area required (excl. buffer, excl. cover)15,208 m²/yr ≈ 1.52 ha/yr