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Landfill volume and area calculation

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).

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.

Data: population =150 000= 150\,000, per capita rate =0.4= 0.4 kg/day, compacted density =600= 600 kg/m3^3, cover =25%= 25\%, depth =8= 8 m.

(a) Annual waste

Waste/day=150 000×0.4=60 000 kg=60 t/day\text{Waste/day} = 150\,000\times0.4 = 60\,000\ \text{kg} = 60\ \text{t/day} Waste/year=60×365=21 900 t/year\text{Waste/year} = 60\times365 = 21\,900\ \text{t/year}

(b) Volume per year

Compacted waste volume:

Vw=21 900×1000600=36 500 m3/yearV_w = \frac{21\,900\times1000}{600} = 36\,500\ \text{m}^3/\text{year}

Including cover soil:

V=1.25×36 500=45 625 m3/yearV = 1.25\times36\,500 = 45\,625\ \text{m}^3/\text{year}

(c) Area for 20 years

Total volume =45 625×20=912 500 m3= 45\,625\times20 = 912\,500\ \text{m}^3.

A=912 5008=114 063 m2≈11.4 haA = \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 m3^3/year; area ≈\approx 11.4 ha (114 000 m2^2) for 20 years.

✓ Answer checked by Claude Opus 5.5 (high effort)

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