---
title: "Septic tank and dispersion trench, 12 persons"
subject: "Sanitary Engineering"
chapter: 9
key: "semester_6/sanitary_engineering:septic-tank-12-persons-70lpcd"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/septic-tank-12-persons-70lpcd
site: BCT Engineering
---

# Septic tank and dispersion trench, 12 persons

Sanitary Engineering · Chapter 9: Disposal of Sewage from Isolated Buildings

- Asked in: 2080 Chaitra, 5+3 marks
- Video: [Septic Tank Design Numerical-1|Septic tank|Solved Numerical|Civil Engineering](https://www.youtube.com/watch?v=h0r7YFBjg3o) — ISHA MISHRA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/septic-tank-12-persons-70lpcd
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-9

## Question

Assuming 80% contribution of water supply for the sewage, design septic tank and dispersion trenches for a house having 12 persons with an average water supply rate of 70 lpcd. The desludging period of septic tank is 3 years. The infiltration capacity of soil is 30 l/m²/d.

## Short answer

Sewage flow 0.672 m³/d; septic tank volume 2.20 m³ (settling 0.672 + digestion 0.510 + storage 1.020), size 2.1×0.7×1.5 m. Infiltration area 672/30 = 22.4 m²; trench width 0.6 m, total length 37.3 m, so two 19.0 m long trenches, 0.6 m wide, gravel bed and perforated pipe.

## Answer

**Given:** 12 persons, water supply 70 L/c/d, sewage = 80% of supply, desludging every 3 years, infiltration 30 L/m²/d.

#### Septic tank

- Sewage flow: $Q = 12 \times 70\times0.8\ \text{L/c/d} = 672\ \text{L/d} = 0.672\ \text{m}^3\text{/d}$
- Sewage (settling) zone for 24 h detention: $V_1 = 0.672\ \text{m}^3$
- Sludge digestion zone: $0.0425 \times 12 = 0.510\ \text{m}^3$
- Digested sludge storage for 3 years: $(0.085/3) \times 3 \times 12 = 1.020\ \text{m}^3$

$$
V = 0.672 + 0.510 + 1.020 = 2.20\ \text{m}^3
$$

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

$$
\begin{aligned} \text{Plan area} &= 2.20/1.5 = 1.47\ \text{m}^2 \\ B &= \sqrt{1.47/3} \approx 0.7\ \text{m}, \quad L = 1.47/0.7 \approx 2.1\ \text{m} \end{aligned}
$$

Check: $2.1 \times 0.7 \times 1.5 = 2.21\ \text{m}^3 \ge 2.20\ \text{m}^3$ (safe).

**Septic tank:** internal size $2.1\ \text{m} \times 0.7\ \text{m} \times 1.8\ \text{m}$ (including 0.3 m free board), in two chambers: first chamber $1.40$ m long (2/3), second $0.70$ 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.

#### Dispersion trenches

- Infiltration area: $A = 672/30 = 22.40\ \text{m}^2$ (trench bottom area)
- Trench width $w = 0.6$ m, so total length $= 22.40/0.6 = 37.3$ m
- The maximum length per trench is about 30 m, so adopt **2 trenches, each 19.0 m long** (22.8 m² provided), 0.6 m wide and 0.6-1.0 m deep, laid at a grade of 1 in 200 to 1 in 300, at 2 m centre-to-centre spacing.

Each trench has a 0.3 m deep layer of gravel (20-40 mm) with a 100 mm perforated (or open-jointed) distribution pipe laid in it, covered with 50 mm of coarse sand/gravel, a filter layer such as geotextile or straw, and then soil. Effluent is fed from a distribution box.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The sewage flow is 80% of supply, so for 12 persons at 70 lpcd Q =
   A. 840 L/d
   B. 672 L/d
   C. 560 L/d
   D. 1120 L/d
   Answer: B. 12 x 70 x 0.8 = 672 L/d = 0.672 m3/d. (Key point: Sewage flow)

2. Digested sludge storage for 12 persons, 3 years, is
   A. 0.510 m3
   B. 0.340 m3
   C. 1.020 m3
   D. 2.04 m3
   Answer: C. (0.085/3) x 3 x 12 = 1.020 m3 (0.085 m3 per person for 3 years). (Key point: Sludge storage)

3. The infiltration area of dispersion trenches is
   A. 672 x 30
   B. 672/12
   C. 30/672
   D. 672/30 = 22.4 m2
   Answer: D. Area = flow divided by infiltration capacity of soil. (Key point: Trench area)

4. Why are 2 trenches of 19 m provided?
   A. Total length 22.4/0.6 = 37.3 m exceeds the maximum about 30 m per trench
   B. Each trench must be exactly 30 m
   C. Trench width must be 2 m
   D. Only one trench is allowed
   Answer: A. A single trench would be 37.3 m long, over the usual maximum of 30 m, so it is split. (Key point: Trench length)

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