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Septic tank and soak pit, hostel of 400

Short answer

Septic tank internal dimensions 11.7 m × 3.9 m × 2.3 m (including 0.3 m freeboard) with two chambers. Total liquid volume 91 m³. Soak pits: 29 circular pits of 3 m diameter and 3 m effective depth, providing 800 m² infiltration area at 50 L/m²/d rate.

If you are asked to design septic tank and soak pit for a hostel with 400 users, calculate sizes. For septic tank, consider sludge digestion rate of 0.0425 m³/person and volume required for storage of digested sludge 0.085 m³/person for 3 years cleaning period.

Assumptions (flow and soil data not given): sewage 100 L/person/day, detention 24 h, infiltration 50 L/m²/d (sandy loam), digestion 0.0425 m³/person and storage 0.085 m³/person for 3 years as given. Soak pit effective depth 3 m, pit diameter 3 m.

Septic tank

  • Sewage flow: Q=400×100 L/c/d=40000 L/d=40.000 m3/dQ = 400 \times 100\ \text{L/c/d} = 40000\ \text{L/d} = 40.000\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=40.000 m3V_1 = 40.000\ \text{m}^3
  • Sludge digestion zone: 0.0425×400=17.000 m30.0425 \times 400 = 17.000\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×400=34.000 m3(0.085/3) \times 3 \times 400 = 34.000\ \text{m}^3
V=40.000+17.000+34.000=91.00 m3V = 40.000 + 17.000 + 34.000 = 91.00\ \text{m}^3

Adopt liquid depth d=2.0d = 2.0 m and L:B≈3:1L:B \approx 3:1.

Plan area=91.00/2.0=45.50 m2B=45.50/3≈3.9 m,L=45.50/3.9≈11.7 m\begin{aligned} \text{Plan area} &= 91.00/2.0 = 45.50\ \text{m}^2 \\ B &= \sqrt{45.50/3} \approx 3.9\ \text{m}, \quad L = 45.50/3.9 \approx 11.7\ \text{m} \end{aligned}

Check: 11.7×3.9×2.0=91.26 m3≥91.00 m311.7 \times 3.9 \times 2.0 = 91.26\ \text{m}^3 \ge 91.00\ \text{m}^3 (safe).

Septic tank: internal size 11.7 m×3.9 m×2.3 m11.7\ \text{m} \times 3.9\ \text{m} \times 2.3\ \text{m} (including 0.3 m free board), in two chambers: first chamber 7.807.80 m long (2/3), second 3.903.90 m (1/3), separated by a partition wall with an opening near mid-depth. Inlet and outlet through T-pipes dipping 0.3 m below water level.

Soak pit

  • Required infiltration area: A=Q/rate=40000/50=800.00 m2A = Q/\text{rate} = 40000/50 = 800.00\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=3h = 3 m

A single pit would be too large, so adopt pits of D=3.0D = 3.0 m (area of each = π×3.0×3=28.27 m2\pi \times 3.0 \times 3 = 28.27\ \text{m}^2).

N=800.0028.27=28.29⇒29 pitsN = \frac{800.00}{28.27} = 28.29 \Rightarrow 29\ \text{pits}

Soak pits: 29 circular pits, each 3.0 m diameter and 3 m deep, honeycomb-lined and filled with brick bats/gravel, spaced at least 3 m (about 2 times the diameter) apart and connected through a distribution box. Soak-pit bottoms must stay at least 1 m above the highest ground-water level.

Such a large number of soak pits is impractical on a small site; in practice a hostel of this size uses several septic tanks in parallel or a series of dispersion trenches / a small package treatment plant followed by disposal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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