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Digester volume for sludge, moisture 92%

Short answer

For 92% moisture, 500 kg/d dry solids (70% volatile), 50% VS destruction, 30-day digestion, 12-day storage: raw sludge volume V1=6.150 m3/d, digested sludge V2=3.153 m3/d. Using parabolic formula V=[V1-2/3(V1-V2)]td+V2ts, digester volume = 162.4 m3.

A sewage treatment plant produces sludge having a moisture content of 92% with a dry solids of 500 kg/d in which 70% solids are volatile. The specific gravity of volatile and fixed solids are 1.02 and 2.65 respectively. Assuming parabolic digestion of 30 days and 12 days of monsoon storage, calculate the volume of conventional rate digester required to produce digested sludge with a moisture content of 90%.

For a conventional-rate digester the capacity with parabolic variation of sludge volume is

V=[V1−23(V1−V2)]td+V2 tsV=\left[V_1-\frac{2}{3}(V_1-V_2)\right]t_d+V_2\,t_s

where V1V_1 = raw sludge volume per day, V2V_2 = digested sludge volume per day, td=30t_d=30 days (digestion), ts=12t_s=12 days (storage during monsoon).

Assumption: 50% of the volatile solids are destroyed in digestion (usual 40-60%).

Raw sludge (500 kg dry solids/day, 70% volatile, moisture 92% i.e. 8% solids):

VS=350 kg,FS=150 kgMass of wet sludge=5000.08=6250 kg, water=5.750 m3Volume of solids=3501020+1502650=0.3997 m3V1=5.750+0.3997=6.150 m3/day\begin{aligned} VS &= 350\ \text{kg},\quad FS=150\ \text{kg}\\ \text{Mass of wet sludge} &= \frac{500}{0.08}=6250\ \text{kg},\ \text{water}=5.750\ \text{m}^3\\ \text{Volume of solids} &= \frac{350}{1020}+\frac{150}{2650}=0.3997\ \text{m}^3\\ V_1 &= 5.750+0.3997=6.150\ \text{m}^3/\text{day} \end{aligned}

Digested sludge (VS left =175=175 kg, FS =150=150 kg, solids =325=325 kg, 90% moisture so 10% solids):

Mass of wet sludge=3250.10=3250 kg, water=2.925 m3Volume of solids=1751020+1502650=0.2282 m3V2=2.925+0.2282=3.153 m3/day\begin{aligned} \text{Mass of wet sludge} &= \frac{325}{0.10}=3250\ \text{kg},\ \text{water}=2.925\ \text{m}^3\\ \text{Volume of solids} &= \frac{175}{1020}+\frac{150}{2650}=0.2282\ \text{m}^3\\ V_2 &= 2.925+0.2282=3.153\ \text{m}^3/\text{day} \end{aligned}

Volume of digester:

V=[6.150−23(6.150−3.153)]×30+3.153×12=162.4 m3V=\left[6.150-\frac{2}{3}(6.150-3.153)\right]\times30+3.153\times12=162.4\ \text{m}^3

Answer: digester volume ≈162.4\approx 162.4 m³ (say 162 m³).

✓ Answer checked by Claude Opus 5.5 (high effort)

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