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Yearly sludge volume, 6 MLD

Short answer

Dry solids removed: 1170 kg/d, sludge volume 22.9 m³/d (assuming sp. gr. 1.02, 95% moisture). Yearly volume = 8374 m³.

A sedimentation tank treats 6 mld of sewage containing 300 mg/l of suspended solids. The tank removes 65% of the suspended solids. Compute the volume of the sludge produced yearly if the moisture content of the sludge is 95%.

The specific gravity of sludge is not given, so take Gsl=1.02G_{sl} = 1.02 (usual value for sewage sludge).

  • Solids in raw sewage: 6×106×300×10−6=18006\times10^6 \times 300\times10^{-6} = 1800 kg/d
  • Solids removed (65%): Ws=1170W_s = 1170 kg/d
V=Wsρw Gsl (1−p)=11701000×1.02×0.05=22.94 m3/dV = \frac{W_s}{\rho_w\,G_{sl}\,(1-p)} = \frac{1170}{1000 \times 1.02 \times 0.05} = 22.94\ \text{m}^3\text{/d} Vyear=22.94×365=8,374 m3V_{year} = 22.94 \times 365 = 8,374\ \text{m}^3

Answer: about 22.9 m³/day, i.e. about 8,374 m³ of sludge per year.

✓ Answer checked by Claude Opus 5.5 (high effort)

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