--- title: "Landfill volume and area calculation" subject: "Environment and Pollution Control" chapter: 5 key: "semester_7/environment_and_pollution_control:landfill-area-volume-numerical" source: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/landfill-area-volume-numerical site: BCT Engineering --- # Landfill volume and area calculation Environment and Pollution Control · Chapter 5: Solid waste - Asked in: year not given, 6 marks - Video: [ENV E 320 - Lecture 6: Example Sizing a Landfill](https://www.youtube.com/watch?v=NvW3K_RmvW4) — Matthew Verbyla - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/landfill-area-volume-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/environment_and_pollution_control/chapter-5 ## Question A town of 150 000 people produces 0.4 kg of solid waste per person per day. The waste is compacted in a sanitary landfill to a density of 600 kg/m3. Daily and intermediate cover soil adds 25% to the compacted waste volume. The landfill depth is 8 m. Estimate (a) the annual waste in tonnes, (b) the landfill volume needed per year including cover, (c) the area needed for a design life of 20 years, assuming no population growth. ## Short answer Annual waste: 150,000 × 0.4 × 365 ÷ 1000 = 21,900 t. Compacted volume = 21,900×1000÷600 = 36,500 m³/yr; with 25% cover soil = 45,625 m³/yr. For 20 years at 8 m depth, area = 45,625×20÷8 ≈ 114,000 m² (11.4 ha). ## Answer Data: population $= 150\,000$, per capita rate $= 0.4$ kg/day, compacted density $= 600$ kg/m$^3$, cover $= 25\%$, depth $= 8$ m. #### (a) Annual waste $$ \text{Waste/day} = 150\,000\times0.4 = 60\,000\ \text{kg} = 60\ \text{t/day} $$ $$ \text{Waste/year} = 60\times365 = 21\,900\ \text{t/year} $$ #### (b) Volume per year Compacted waste volume: $$ V_w = \frac{21\,900\times1000}{600} = 36\,500\ \text{m}^3/\text{year} $$ Including cover soil: $$ V = 1.25\times36\,500 = 45\,625\ \text{m}^3/\text{year} $$ #### (c) Area for 20 years Total volume $= 45\,625\times20 = 912\,500\ \text{m}^3$. $$ A = \frac{912\,500}{8} = 114\,063\ \text{m}^2 \approx 11.4\ \text{ha} $$ Extra land (about 20-30%) would be needed for roads, buffer zone, leachate treatment and offices. **Answer: 21 900 t/year; 45 625 m$^3$/year; area $\approx$ 11.4 ha (114 000 m$^2$) for 20 years.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 150 000 people at 0.4 kg/person/day, what is the annual waste? A. 60 t/year B. 21 900 t/year C. 219 000 t/year D. 8 000 t/year Answer: B. 150 000 x 0.4 = 60 000 kg/day = 60 t/day; x 365 = 21 900 t/year. (Key point: Annual waste) 2. How is the compacted waste volume per year found? A. Mass divided by density: 21 900 x 1000 / 600 = 36 500 m3 B. Mass multiplied by density C. Mass divided by landfill depth D. Mass divided by 1.25 Answer: A. Volume = mass / density = 21 900 000 kg / 600 kg/m3 = 36 500 m3/year. (Key point: Compacted volume) 3. How is the cover soil allowance of 25% applied? A. Subtract 25% from the waste volume B. Multiply the compacted waste volume by 1.25 to get 45 625 m3/year C. Add 25% to the density D. Multiply the area by 0.25 Answer: B. Cover adds 25% to the compacted waste volume, so V = 1.25 x 36 500 = 45 625 m3/year. (Key point: Effect of cover soil) 4. What is the area needed for 20 years at 8 m depth? A. About 114 000 m2 (11.4 ha) B. About 5 700 m2 C. About 912 500 m2 D. About 1.14 ha Answer: A. Total volume 45 625 x 20 = 912 500 m3; area = 912 500 / 8 = 114 063 m2, about 11.4 ha, before extra land for roads and buffer. 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