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Sludge volume from PST and SST

Short answer

PST removes 3360 kg/d dry solids, giving 66.5 m³/d sludge at 95% moisture. Trickling filter removes 2618 kg/d BOD, producing 1047 kg/d dry humus solids, giving 20.7 m³/d SST sludge (also 95% moisture).

The biological process occurs in trickling filter. PST removes 60% suspended solids and 30% of BOD. Determine the volume of sludge produced by PST as well as SST with the following data. Sp. gr. of inorganic solids = 2.65, Sp. gr. of organic solid = 1.02; Flow of sewage = 20*10⁶ l/d; BOD5 of sewage = 220 mg/l; Suspended solids in the sewage = 280 mg/l; Water content of the sludge = 95%.

Sludge volume V=Wsρw Gsl (1−p)V = \dfrac{W_s}{\rho_w \, G_{sl}\,(1-p)}, where WsW_s is the dry solids mass, GslG_{sl} the specific gravity of wet sludge and pp the water fraction.

Assumptions (not given in the data): suspended solids are 70% organic and 30% inorganic; trickling filter removes 85% of the BOD reaching it; humus (SST sludge) produced is 0.4 kg dry solids per kg BOD removed; SST sludge also has 95% water.

Specific gravity of sludge

1Gs=0.71.02+0.32.65⇒Gs=1.251\frac{1}{G_s} = \frac{0.7}{1.02} + \frac{0.3}{2.65} \Rightarrow G_s = 1.251 1Gsl=0.051.251+0.951⇒Gsl=1.0101\frac{1}{G_{sl}} = \frac{0.05}{1.251} + \frac{0.95}{1} \Rightarrow G_{sl} = 1.0101

Sludge from PST

  • Suspended solids in raw sewage: 20×106×280×10−6=5,60020\times10^6 \times 280\times10^{-6} = 5,600 kg/d
  • Removed in PST (60%): Ws=3,360W_s = 3,360 kg/d
VPST=3,3601000×1.0101×0.05=66.5 m3/dV_{PST} = \frac{3,360}{1000 \times 1.0101 \times 0.05} = 66.5\ \text{m}^3\text{/d}

Sludge from SST (humus)

  • BOD in raw sewage: 20×106×220×10−6=4,40020\times10^6 \times 220\times10^{-6} = 4,400 kg/d
  • BOD after PST (30% removed): 3,0803,080 kg/d
  • BOD removed in trickling filter (85%): 2,6182,618 kg/d
  • Dry humus solids: 0.4×2,618=1,0470.4 \times 2,618 = 1,047 kg/d
VSST=1,0471000×1.0101×0.05=20.7 m3/dV_{SST} = \frac{1,047}{1000 \times 1.0101 \times 0.05} = 20.7\ \text{m}^3\text{/d}

Answer: PST sludge ≈66.5 m3\approx 66.5\ \text{m}^3/d; SST sludge ≈20.7 m3\approx 20.7\ \text{m}^3/d (total about 87.3 m³/d).

✓ Answer checked by Claude Opus 5.5 (high effort)

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