The volume is reduced to 40% of its original value, so V2 =

360/(1000 x 1.02 x 0.2) = 1.76 m3/d

2069 Bhadra · 8 marks

Digester diameter from 500 m³ sed. tank

Short answer

Flow 3000 m³/d, dry solids removed 360 kg/d. Fresh sludge 17.65 m³/d, digested sludge 7.06 m³/d (40% of original). Digester volume 318 m³, diameter 8.2 m for 6 m depth.

Sewage with a suspended solid content of 200 mg/l flows continuously in a sedimentation tank of 500 m³ capacity. Sewage is detained in the sedimentation tank for 4 hours. Sixty percentages of solids are removed in the sedimentation tank during its detention. The sludge produced in the sedimentation tank has moisture content of 98% and specific gravity of 1.02. The sludge from sedimentation tank is fed to digester for its digestion. The volume of sludge is reduced to 40% of its original volume during digestion. Calculate the diameter of digester if its effective depth is 6 m. Assume detention period in the digester is 30 days.

Flow

Q = \frac{\text{Volume}}{\text{detention time}} = \frac{500}{4} = 125\ \text{m}^3\text{/h} = 3000\ \text{m}^3\text{/d}

Sludge from the sedimentation tank

Solids in sewage: $3000\times1000 \times 200\times10^{-6} = 600$ kg/d
Removed (60%): $W_s = 360$ kg/d

V_1 = \frac{360}{1000 \times 1.02 \times (1-0.98)} = 17.65\ \text{m}^3\text{/d}

Digested sludge

V_2 = 0.40 \times 17.65 = 7.06\ \text{m}^3\text{/d}

Digester volume (T = 30 days)

\begin{aligned} V &= \left[V_1 - \tfrac{2}{3}(V_1 - V_2)\right] T \\ &= \left[17.65 - \tfrac{2}{3}(17.65 - 7.06)\right] \times 30 \\ &= 10.59 \times 30 = 317.6\ \text{m}^3 \end{aligned}

Diameter (effective depth 6 m)

\frac{\pi}{4}D^2 \times 6 = 317.6 \Rightarrow D = \sqrt{\frac{4 \times 317.6}{\pi \times 6}} = 8.21\ \text{m}

Answer: diameter of digester $\approx 8.2$ m (say 8.3 m), digester volume 318 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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