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Septic tank and soak pits, 15 users, Terai

Short answer

Compute volumes: settling 0.675 m³, digestion 0.638 m³, storage 1.700 m³, total 3.01 m³. Tank 2.3 m × 0.9 m × 1.5 m liquid depth + 0.3 m freeboard, two chambers. Soak pit: infiltration area 13.5 m², one circular pit diameter 2.2 m, depth 2.0 m, bottom >1 m above groundwater.

Design a septic tank and numbers of soak pit for 15 numbers of average users. Rate of sewage discharge is 45 lpcd. Cleaning period of septic tank is 4 years interval. Assume other necessary data if required. Location of construction is in Terai area.

Assumptions: Terai has a high ground-water table (about 3 m) and flat land, so soak pits are kept shallow (effective depth 2.0 m) and infiltration rate for sandy soil is taken as 50 L/m²/d; detention 24 h; pit diameter at most 3 m.

Septic tank

  • Sewage flow: Q=15×45 L/c/d=675 L/d=0.675 m3/dQ = 15 \times 45\ \text{L/c/d} = 675\ \text{L/d} = 0.675\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=0.675 m3V_1 = 0.675\ \text{m}^3
  • Sludge digestion zone: 0.0425×15=0.638 m30.0425 \times 15 = 0.638\ \text{m}^3
  • Digested sludge storage for 4 years: (0.085/3)×4×15=1.700 m3(0.085/3) \times 4 \times 15 = 1.700\ \text{m}^3
V=0.675+0.638+1.700=3.01 m3V = 0.675 + 0.638 + 1.700 = 3.01\ \text{m}^3

Adopt liquid depth d=1.5d = 1.5 m and L:B≈3:1L:B \approx 3:1.

Plan area=3.01/1.5=2.01 m2B=2.01/3≈0.9 m,L=2.01/0.9≈2.3 m\begin{aligned} \text{Plan area} &= 3.01/1.5 = 2.01\ \text{m}^2 \\ B &= \sqrt{2.01/3} \approx 0.9\ \text{m}, \quad L = 2.01/0.9 \approx 2.3\ \text{m} \end{aligned}

Check: 2.3×0.9×1.5=3.11 m3≥3.01 m32.3 \times 0.9 \times 1.5 = 3.11\ \text{m}^3 \ge 3.01\ \text{m}^3 (safe).

Septic tank: internal size 2.3 m×0.9 m×1.8 m2.3\ \text{m} \times 0.9\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.551.55 m long (2/3), second 0.750.75 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 pits

  • Required infiltration area: A=Q/rate=675/50=13.50 m2A = Q/\text{rate} = 675/50 = 13.50\ \text{m}^2 (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
  • Adopt effective depth h=2.0h = 2.0 m
πDh=13.50⇒D=13.50π×2.0=2.15≈2.2 m\pi D h = 13.50 \Rightarrow D = \frac{13.50}{\pi \times 2.0} = 2.15 \approx 2.2\ \text{m}

Check: π×2.2×2.0=13.82 m2≥13.50 m2\pi \times 2.2 \times 2.0 = 13.82\ \text{m}^2 \ge 13.50\ \text{m}^2.

Soak pit: one circular pit, 2.2 m internal diameter and 2.0 m deep, lined with dry brick/stone masonry with open joints (honeycomb), filled with brick bats/gravel, and covered with a slab.

Because of the high water table, keep the pit bottom at least 1 m above the highest ground-water level; if this cannot be done use dispersion trenches or an evapotranspiration mound instead.

✓ Answer checked by Claude Opus 5.5 (high effort)

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