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Sludge treatment aims; digester volume (94%)

Short answer

Aims: reduce volume, stabilize organics, destroy pathogens, improve dewatering, recover biogas and manure, protect environment. For 94% moisture, 500 kg/d dry solids (70% volatile), 50% VS destruction, 30-day digestion, 12-day storage: raw sludge volume V1=8.233 m3/d, digested sludge V2=3.153 m3/d. Using parabolic formula V=[V1-2/3(V1-V2)]td+V2ts, digester volume = 183.2 m3.

Write down the aims of sludge treatment. A sewage treatment plant produces sludge having a moisture content of 94% with a dry solid of 500 kg/d in which 70% solids are volatile. The specific gravity of volatile and fixed solids are 1.02 and 2.65. 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%.

Aims of sludge treatment

  1. To reduce the volume (water content) so handling, transport and disposal become cheaper.
  2. To stabilise the organic matter so that it does not putrefy and cause odour.
  3. To destroy pathogens and make the sludge safe to handle and reuse.
  4. To make the sludge easy to dewater and dry.
  5. To recover resources: biogas (methane) for energy and the dried sludge as manure/soil conditioner.
  6. To protect the environment and public health from nuisance, flies and water pollution.

Digester volume

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, tdt_d = digestion period, tst_s = storage period.

Assumption: 50% of the volatile solids are destroyed in the digester (usual 40-60% for conventional rate).

Raw sludge (500 kg dry solids/day, 70% volatile, moisture 94% i.e. 6% solids):

VS=350 kg,FS=150 kgMass of wet sludge=5000.06=8333 kg, water=7833 kg=7.833 m3Volume of solids=3501.02×1000+1502.65×1000=0.3997 m3V1=7.833+0.3997=8.233 m3/day\begin{aligned} VS &= 350\ \text{kg},\quad FS=150\ \text{kg}\\ \text{Mass of wet sludge} &= \frac{500}{0.06}=8333\ \text{kg},\ \text{water}=7833\ \text{kg}=7.833\ \text{m}^3\\ \text{Volume of solids} &= \frac{350}{1.02\times1000}+\frac{150}{2.65\times1000}=0.3997\ \text{m}^3\\ V_1 &= 7.833+0.3997=8.233\ \text{m}^3/\text{day} \end{aligned}

Digested sludge (VS destroyed =0.5×350=175=0.5\times350=175 kg; remaining VS =175=175 kg, FS =150=150 kg; total solids =325=325 kg at 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 (td=30t_d=30 days, ts=12t_s=12 days):

V=[8.233−23(8.233−3.153)]×30+3.153×12=183.2 m3V=\left[8.233-\frac{2}{3}(8.233-3.153)\right]\times30+3.153\times12=183.2\ \text{m}^3

Answer: digester volume ≈183.2\approx 183.2 m³ (say 183 m³).

✓ Answer checked by Claude Opus 5.5 (high effort)

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