2072 Asoj · 4 marks

Sludge volume from sedimentation tank, 10 MLD

Short answer

Dry solids removed: 1500 kg/d. Sludge volume at 95% moisture and sp. gr. 1.02 is 29.4 m³/d.

Determine the volume of sludge produced in a sewage sedimentation tank from the following data: Flow rate = 10 million liters/day; Suspended solids content in raw sewage = 250 mg/l; Sedimentation tank removes 60% of suspended solids; Specific gravity of sludge = 1.02; Moisture content of sludge = 95%.

Volume of sludge $V = \dfrac{W_s}{\rho_w \, G_{sl}\,(1-p)}$

Solids in raw sewage: $10\times10^6\ \text{L/d} \times 250\ \text{mg/L} = 2500$ kg/d
Solids removed (60%): $W_s = 1500$ kg/d
Solids fraction of sludge: $1 - 0.95 = 0.05$

V = \frac{1500}{1000 \times 1.02 \times 0.05} = 29.41\ \text{m}^3\text{/d}

Answer: volume of sludge $\approx 29.4\ \text{m}^3$ per day.

✓ Answer checked by Claude Opus 5.5 (high effort)

Sludge volume and digester, 4 MLD · Sanitary Engineering, chapter 8
Sludge digestion tank, 6.5 MLD · Sanitary Engineering, chapter 8
Yearly sludge volume, 6 MLD · Sanitary Engineering, chapter 8
Sludge digestion tank, 812.5 m³ sed. tank · Sanitary Engineering, chapter 8
Sludge volume from PST and SST · Sanitary Engineering, chapter 8
Standard rate filter, 3 MLD, 175 mg/l, NRC · Sanitary Engineering, chapter 7
Rectangular sedimentation tank for 10 MLD · Water Supply Engineering, chapter 6
Rectangular sedimentation tank, 9 million L peak demand · Water Supply Engineering, chapter 6
Sedimentation tank 6 m x 18 m x 3 m, 4 ML/day · Water Supply Engineering, chapter 6
Rectangular sedimentation tank, 8x10^6 L/day · Water Supply Engineering, chapter 6

All Sanitary Engineering chapter 8 questions →

Sludge volume from sedimentation tank, 10 MLD

Related questions