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Septic tank and soak pit, 8 persons

Short answer

Septic tank internal dimensions 1.8 m × 0.7 m × 1.8 m (including 0.3 m freeboard) with two chambers. Total liquid volume 1.82 m³. Soak pit: one circular pit of 1.7 m diameter and 3 m effective depth, providing 16 m² infiltration area at 50 L/m²/d rate.

Design a septic tank and soak pit to dispose the sewage generated from a household of 8 persons. The sewage is generated at the rate of 100 liters/person/day. Assume that septic tank is cleaned once in 3 years and infiltration rate of soil is 50 liters/m²/day.

Given: 8 persons, 100 L/c/d, desludging every 3 years, infiltration 50 L/m²/d. Detention 24 h assumed; soak pit depth 3 m.

Septic tank

  • Sewage flow: Q=8×100 L/c/d=800 L/d=0.800 m3/dQ = 8 \times 100\ \text{L/c/d} = 800\ \text{L/d} = 0.800\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=0.800 m3V_1 = 0.800\ \text{m}^3
  • Sludge digestion zone: 0.0425×8=0.340 m30.0425 \times 8 = 0.340\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×8=0.680 m3(0.085/3) \times 3 \times 8 = 0.680\ \text{m}^3
V=0.800+0.340+0.680=1.82 m3V = 0.800 + 0.340 + 0.680 = 1.82\ \text{m}^3

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

Plan area=1.82/1.5=1.21 m2B=1.21/3≈0.7 m,L=1.21/0.7≈1.8 m\begin{aligned} \text{Plan area} &= 1.82/1.5 = 1.21\ \text{m}^2 \\ B &= \sqrt{1.21/3} \approx 0.7\ \text{m}, \quad L = 1.21/0.7 \approx 1.8\ \text{m} \end{aligned}

Check: 1.8×0.7×1.5=1.89 m3≥1.82 m31.8 \times 0.7 \times 1.5 = 1.89\ \text{m}^3 \ge 1.82\ \text{m}^3 (safe).

Septic tank: internal size 1.8 m×0.7 m×1.8 m1.8\ \text{m} \times 0.7\ \text{m} \times 1.8\ \text{m} (including 0.3 m free board), in two chambers: first chamber 1.201.20 m long (2/3), second 0.600.60 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=800/50=16.00 m2A = Q/\text{rate} = 800/50 = 16.00\ \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
πDh=16.00⇒D=16.00π×3=1.70≈1.7 m\pi D h = 16.00 \Rightarrow D = \frac{16.00}{\pi \times 3} = 1.70 \approx 1.7\ \text{m}

Check: π×1.7×3=16.02 m2≥16.00 m2\pi \times 1.7 \times 3 = 16.02\ \text{m}^2 \ge 16.00\ \text{m}^2.

Soak pit: one circular pit, 1.7 m internal diameter and 3 m deep, lined with dry brick/stone masonry with open joints (honeycomb), filled with brick bats/gravel, and covered with a slab.

✓ Answer checked by Claude Opus 5.5 (high effort)

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