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Septic tank and soak pit, 10 users, 85 l/m²/d

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 1.79 m³. Soak pit: one circular pit of 1.0 m diameter and 3 m effective depth, providing 9.41 m² infiltration area at 85 L/m²/d rate.

Design a septic tank and soak pit for 10 number of users of a house at a place where infiltration capacity of soil is 85 liters per square meter per day. The peak sewage flow rate 80 lpcd and the septic tank is cleaned at an interval of 2 years. The ground water table lies 6 m below the ground level.

Given: 10 users, 80 L/c/d (peak), desludging every 2 years, infiltration 85 L/m²/d, ground-water 6 m deep. Detention 24 h assumed; soak pit depth 3 m keeps 3 m clearance above ground-water.

Septic tank

  • Sewage flow: Q=10×80 L/c/d=800 L/d=0.800 m3/dQ = 10 \times 80\ \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: 0.0425×10=0.425 m30.0425 \times 10 = 0.425\ \text{m}^3
  • Digested sludge storage for 2 years: (0.085/3)×2×10=0.567 m3(0.085/3) \times 2 \times 10 = 0.567\ \text{m}^3
V=0.800+0.425+0.567=1.79 m3V = 0.800 + 0.425 + 0.567 = 1.79\ \text{m}^3

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

Plan area=1.79/1.5=1.19 m2B=1.19/3≈0.7 m,L=1.19/0.7≈1.8 m\begin{aligned} \text{Plan area} &= 1.79/1.5 = 1.19\ \text{m}^2 \\ B &= \sqrt{1.19/3} \approx 0.7\ \text{m}, \quad L = 1.19/0.7 \approx 1.8\ \text{m} \end{aligned}

Check: 1.8×0.7×1.5=1.89 m3≥1.79 m31.8 \times 0.7 \times 1.5 = 1.89\ \text{m}^3 \ge 1.79\ \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=800/85=9.41 m2A = Q/\text{rate} = 800/85 = 9.41\ \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=9.41⇒D=9.41π×3=1.00≈1.0 m\pi D h = 9.41 \Rightarrow D = \frac{9.41}{\pi \times 3} = 1.00 \approx 1.0\ \text{m}

Check: π×1.0×3=9.42 m2≥9.41 m2\pi \times 1.0 \times 3 = 9.42\ \text{m}^2 \ge 9.41\ \text{m}^2.

Soak pit: one circular pit, 1.0 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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