Sludge digestion tank, 812.5 m³ sed. tank

Flow from the sedimentation tank

The tank capacity is 812.5 m³ with detention time of 3 h, so

$$Q=\frac{812.5}{3}\times24=6500.0\ \text{m}^3/\text{day}$$

Data and assumptions

Flow $Q=6500$ m³/day; suspended solids (SS) $=250$ mg/l; removal in the primary tank $=55\%$; fresh sludge moisture $=95\%$; digested sludge moisture $=80\%$; specific gravity of sludge $=1.02$; digestion period $=60$ days (2 months). The digested sludge is taken to hold the same dry solids as the fresh sludge (solids loss in digestion is ignored, which is conservative); no extra storage period is added.

Step 1: Dry solids removed per day

$$W_s=Q\times SS\times\text{removal}=6500\times250\times0.55\times10^{-3}=893.8\ \text{kg/day}$$

Step 2: Volume of fresh sludge

$$V_1=\frac{W_s}{(1-0.95)\times S\times1000}=\frac{893.8}{0.05\times1.02\times1000}=17.52\ \text{m}^3/\text{day}$$

Step 3: Volume of digested sludge (volume-moisture relation)

$$V_2=V_1\times\frac{100-95}{100-80}=17.52\times\frac{5}{20}=4.38\ \text{m}^3/\text{day}$$

Step 4: Capacity of the digester (parabolic digestion)

$$V=\left[V_1-\frac{2}{3}(V_1-V_2)\right]t=\left[17.52-\frac{2}{3}(17.52-4.38)\right]\times60=526\ \text{m}^3$$

Step 5: Dimensions

Adopt a circular digester with side water depth 6 m (usual 6-12 m) plus 0.5 m free board and a hopper bottom with 1 : 1 slope:

$$A=\frac{526}{6}=87.6\ \text{m}^2,\qquad D=\sqrt{\frac{4A}{\pi}}=10.56\ \text{m}\ (\text{adopt 11.0 m})$$

Answer: digester capacity $\approx 526$ m³, circular, diameter 11.0 m, side water depth 6 m (plus free board, hopper and gas dome).