Sludge volume and digester, 4 MLD

The mass of dry solids stays the same while the water content falls, so $V = \dfrac{W_s}{\rho_w\,G_{sl}\,(1-p)}$.

Fresh sludge (95% moisture)

• Solids in raw sewage: $4\times10^6 \times 220\times10^{-6} = 880$ kg/d
• Removed (55%): $W_s = 484.0$ kg/d

$$V_1 = \frac{484.0}{1000 \times 1.02 \times 0.05} = 9.49\ \text{m}^3\text{/d}$$

Digested sludge (80% moisture)

$$V_2 = \frac{484.0}{1000 \times 1.02 \times 0.20} = 2.37\ \text{m}^3\text{/d}$$

(solids loss during digestion is neglected, and the sludge specific gravity is taken as 1.02 in both cases.)

Digester capacity (T = 80 days)

$$\begin{aligned} V &= \left[V_1 - \tfrac{2}{3}(V_1 - V_2)\right] T \\ &= \left[9.49 - \tfrac{2}{3}(9.49 - 2.37)\right] \times 80 \\ &= 4.75 \times 80 = 380\ \text{m}^3 \end{aligned}$$

Adopt effective depth 6 m (assumed):

$$\frac{\pi}{4} D^2 \times 6 = 380 \Rightarrow D = 8.98\ \text{m} \approx 9.0\ \text{m}$$

Provide a circular digester of 9.0 m diameter and 6 m effective depth, plus 0.5 m free board and a hopper bottom (slope 1 vertical to 1.5 horizontal) with a sludge withdrawal pipe.

Answer: fresh sludge 9.49 m³/d; digested sludge 2.37 m³/d; digester capacity 380 m³ (diameter about 9.0 m).