---
title: "Septic tank and dispersion trenches, 20 users"
subject: "Sanitary Engineering"
chapter: 9
key: "semester_6/sanitary_engineering:septic-tank-dispersion-trench-20"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/septic-tank-dispersion-trench-20
site: BCT Engineering
---

# Septic tank and dispersion trenches, 20 users

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

- Asked in: 2072 Magh, 8 marks
- Video: [C09N01: Design of Septic Tank with Effluent Discharged to (Dispersion Trench or  Soak Pit)](https://www.youtube.com/watch?v=yxI1hmQMs2g) — Indra Lal Subedi
- 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-dispersion-trench-20
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-9

## Question

Design the septic tank and dispersion trenches in Nepalese perspectives for 20 users.

## Short answer

Septic tank internal dimensions 2.8 m × 1.0 m × 1.8 m (including 0.3 m freeboard) with two chambers. Total liquid volume 4.15 m³. Dispersion trenches: 2 trenches each 27 m long × 0.6 m wide, providing 32.4 m² bottom infiltration area at 50 L/m²/d rate.

## Answer

**Assumptions (Nepal practice):** water supply 100 L/c/d, sewage 80% of this = 80 L/c/d; detention 24 h; desludging every 3 years; infiltration rate 50 L/m²/d (sandy loam).

#### Septic tank

- Sewage flow: $Q = 20 \times 100\times0.8\ \text{L/c/d} = 1600\ \text{L/d} = 1.600\ \text{m}^3\text{/d}$
- Sewage (settling) zone for 24 h detention: $V_1 = 1.600\ \text{m}^3$
- Sludge digestion zone: $0.0425 \times 20 = 0.850\ \text{m}^3$
- Digested sludge storage for 3 years: $(0.085/3) \times 3 \times 20 = 1.700\ \text{m}^3$

$$
V = 1.600 + 0.850 + 1.700 = 4.15\ \text{m}^3
$$

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

$$
\begin{aligned} \text{Plan area} &= 4.15/1.5 = 2.77\ \text{m}^2 \\ B &= \sqrt{2.77/3} \approx 1.0\ \text{m}, \quad L = 2.77/1.0 \approx 2.8\ \text{m} \end{aligned}
$$

Check: $2.8 \times 1.0 \times 1.5 = 4.20\ \text{m}^3 \ge 4.15\ \text{m}^3$ (safe).

**Septic tank:** internal size $2.8\ \text{m} \times 1.0\ \text{m} \times 1.8\ \text{m}$ (including 0.3 m free board), in two chambers: first chamber $1.90$ m long (2/3), second $0.90$ 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: $A = 1600/50 = 32.00\ \text{m}^2$ (trench bottom area)
- Trench width $w = 0.6$ m, so total length $= 32.00/0.6 = 53.3$ m
- The maximum length per trench is about 30 m, so adopt **2 trenches, each 27.0 m long** (32.4 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. How is sewage flow found when only a water supply rate is assumed?
   A. Sewage = 80% of water supply: 20 x 100 x 0.8 = 1600 L/d
   B. Sewage equals 100% of water supply
   C. Sewage = 50% of supply
   D. Sewage = 20% of supply
   Answer: A. About 80% of water supply becomes sewage, so Q = 1.6 m^3/d. (Key point: Design flow)

2. What is the total septic tank volume?
   A. 3.45 m^3
   B. 4.15 m^3
   C. 1.6 m^3
   D. 5.0 m^3
   Answer: B. V = 1.600 + 0.850 + 1.700 = 4.15 m^3. (Key point: Total volume)

3. Infiltration area is 32 m^2 and trench width is 0.6 m. What total trench length is needed?
   A. 19.2 m
   B. 32 m
   C. 53.3 m
   D. 106 m
   Answer: C. Length = 32/0.6 = 53.3 m, then split into trenches not longer than about 30 m. (Key point: Trench length)

4. Which arrangement describes the trench layers?
   A. Plain soil with a solid pipe
   B. Concrete bed with a sealed pipe
   C. Sand only with no pipe
   D. Gravel layer with perforated pipe, sand/gravel cover, filter layer, then soil
   Answer: D. The perforated pipe sits in gravel, covered with sand/gravel, a geotextile or straw filter and soil. (Key point: Trench construction)

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