Septic tank and soak pit, household of 80

Given: 80 persons, 120 L/c/d, desludging every 2 years, infiltration 80 L/m²/d. Assume 24 h detention and soak pit effective depth 3 m (ground-water table deep).

Septic tank

• Sewage flow: $Q = 80 \times 120\ \text{L/c/d} = 9600\ \text{L/d} = 9.600\ \text{m}^3\text{/d}$
• Sewage (settling) zone for 24 h detention: $V_1 = 9.600\ \text{m}^3$
• Sludge digestion zone: $0.0425 \times 80 = 3.400\ \text{m}^3$
• Digested sludge storage for 2 years: $(0.085/3) \times 2 \times 80 = 4.533\ \text{m}^3$

$$V = 9.600 + 3.400 + 4.533 = 17.53\ \text{m}^3$$

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

$$\begin{aligned} \text{Plan area} &= 17.53/2.0 = 8.77\ \text{m}^2 \\ B &= \sqrt{8.77/3} \approx 1.8\ \text{m}, \quad L = 8.77/1.8 \approx 4.9\ \text{m} \end{aligned}$$

Check: $4.9 \times 1.8 \times 2.0 = 17.64\ \text{m}^3 \ge 17.53\ \text{m}^3$ (safe).

Septic tank: internal size $4.9\ \text{m} \times 1.8\ \text{m} \times 2.3\ \text{m}$ (including 0.3 m free board), in two chambers: first chamber $3.30$ m long (2/3), second $1.60$ 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: $A = Q/\text{rate} = 9600/80 = 120.00\ \text{m}^2$ (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
• Adopt effective depth $h = 3$ m

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

$$N = \frac{120.00}{28.27} = 4.24 \Rightarrow 5\ \text{pits}$$

Soak pits: 5 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.