Skip to main content

Septic tank and soak pits, hotel of 80

Short answer

Septic tank internal dimensions 6.2 m × 2.2 m × 2.3 m (including 0.3 m freeboard) with two chambers. Total liquid volume 27 m³. Soak pits: 12 circular pits of 3 m diameter and 3 m effective depth, providing 336 m² infiltration area at 50 L/m²/d rate.

What would be the internal dimension of a septic tank and numbers of soak pits for an isolated hotel situated at mid-southern zone of Nepal having average 80 numbers of average users? Rate of sewage discharge is 210 lpcd. Cleaning period of septic tank is 3 years. Assume other necessary data if required.

Assumptions: mid-southern zone (Terai/inner Terai) with sandy loam soil, infiltration rate 50 L/m²/d, ground-water at least 5 m below the surface; detention 24 h, soak pit depth 3 m and diameter 3 m.

Septic tank

  • Sewage flow: Q=80×210 L/c/d=16800 L/d=16.800 m3/dQ = 80 \times 210\ \text{L/c/d} = 16800\ \text{L/d} = 16.800\ \text{m}^3\text{/d}
  • Sewage (settling) zone for 24 h detention: V1=16.800 m3V_1 = 16.800\ \text{m}^3
  • Sludge digestion zone: 0.0425×80=3.400 m30.0425 \times 80 = 3.400\ \text{m}^3
  • Digested sludge storage for 3 years: (0.085/3)×3×80=6.800 m3(0.085/3) \times 3 \times 80 = 6.800\ \text{m}^3
V=16.800+3.400+6.800=27.00 m3V = 16.800 + 3.400 + 6.800 = 27.00\ \text{m}^3

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

Plan area=27.00/2.0=13.50 m2B=13.50/3≈2.2 m,L=13.50/2.2≈6.2 m\begin{aligned} \text{Plan area} &= 27.00/2.0 = 13.50\ \text{m}^2 \\ B &= \sqrt{13.50/3} \approx 2.2\ \text{m}, \quad L = 13.50/2.2 \approx 6.2\ \text{m} \end{aligned}

Check: 6.2×2.2×2.0=27.28 m3≥27.00 m36.2 \times 2.2 \times 2.0 = 27.28\ \text{m}^3 \ge 27.00\ \text{m}^3 (safe).

Septic tank: internal size 6.2 m×2.2 m×2.3 m6.2\ \text{m} \times 2.2\ \text{m} \times 2.3\ \text{m} (including 0.3 m free board), in two chambers: first chamber 4.154.15 m long (2/3), second 2.052.05 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=16800/50=336.00 m2A = Q/\text{rate} = 16800/50 = 336.00\ \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=336.0028.27=11.88⇒12 pitsN = \frac{336.00}{28.27} = 11.88 \Rightarrow 12\ \text{pits}

Soak pits: 12 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.

The large number of pits can be replaced by dispersion trenches or by two tanks in parallel if space is limited.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Sanitary Engineering chapter 9 questions →