---
title: "Septic tank and drain field, 22 persons"
subject: "Sanitary Engineering"
chapter: 9
key: "semester_6/sanitary_engineering:septic-tank-drain-field-22"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/septic-tank-drain-field-22
site: BCT Engineering
---

# Septic tank and drain field, 22 persons

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

- Asked in: 2075 Bhadra, 8 marks
- Video: [Ultimate Guide: How To Design A Septic Tank For Multiple Users | Components And Operation Explained](https://www.youtube.com/watch?v=GOwxOeKJt_4) — Just Learn Civil
- 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-drain-field-22
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-9

## Question

A household having 22 persons produces 135 liter/person/day of sewage. Design a septic tank and drain field to dispose the sewage in a soil having infiltration rate of 35 liters/m²/day. Assume that the septic tank is cleaned once in three years.

## Short answer

Septic tank internal dimensions 3.3 m × 1.2 m × 1.8 m (including 0.3 m freeboard) with two chambers. Total liquid volume 5.78 m³. Drain field: 5 dispersion trenches each 28.5 m long × 0.6 m wide, providing 84.86 m² bottom infiltration area at 35 L/m²/d rate.

## Answer

**Given:** 22 persons, 135 L/c/d, infiltration 35 L/m²/d, desludging every 3 years. Detention 24 h assumed.

#### Septic tank

- Sewage flow: $Q = 22 \times 135\ \text{L/c/d} = 2970\ \text{L/d} = 2.970\ \text{m}^3\text{/d}$
- Sewage (settling) zone for 24 h detention: $V_1 = 2.970\ \text{m}^3$
- Sludge digestion zone: $0.0425 \times 22 = 0.935\ \text{m}^3$
- Digested sludge storage for 3 years: $(0.085/3) \times 3 \times 22 = 1.870\ \text{m}^3$

$$
V = 2.970 + 0.935 + 1.870 = 5.78\ \text{m}^3
$$

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

$$
\begin{aligned} \text{Plan area} &= 5.78/1.5 = 3.85\ \text{m}^2 \\ B &= \sqrt{3.85/3} \approx 1.2\ \text{m}, \quad L = 3.85/1.2 \approx 3.3\ \text{m} \end{aligned}
$$

Check: $3.3 \times 1.2 \times 1.5 = 5.94\ \text{m}^3 \ge 5.78\ \text{m}^3$ (safe).

**Septic tank:** internal size $3.3\ \text{m} \times 1.2\ \text{m} \times 1.8\ \text{m}$ (including 0.3 m free board), in two chambers: first chamber $2.20$ m long (2/3), second $1.10$ 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.

#### Drain field (dispersion trenches)

- Infiltration area: $A = 2970/35 = 84.86\ \text{m}^2$ (trench bottom area)
- Trench width $w = 0.6$ m, so total length $= 84.86/0.6 = 141.4$ m
- The maximum length per trench is about 30 m, so adopt **5 trenches, each 28.5 m long** (85.5 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 the digested sludge storage for 3 years calculated here?
   A. (0.085/3) x 3 x 22 = 1.87 m^3
   B. 0.0425 x 22 = 0.935 m^3
   C. 0.085 x 22 x 3 = 5.6 m^3
   D. 2.97 x 3 = 8.9 m^3
   Answer: A. Storage is (0.085/3) m^3 per person per year times years times persons. (Key point: Digested sludge storage)

2. Which is the total septic tank volume for 22 persons?
   A. 2.97 m^3
   B. 5.78 m^3
   C. 3.90 m^3
   D. 8.9 m^3
   Answer: B. V = 2.970 + 0.935 + 1.870 = 5.78 m^3. (Key point: Total volume)

3. Infiltration area of the drain field is found from which relation?
   A. Tank volume divided by infiltration rate
   B. Daily flow multiplied by infiltration rate
   C. Daily flow divided by infiltration rate (2970/35)
   D. Tank plan area times 35
   Answer: C. A = Q/rate = 2970/35 = 84.86 m^2 of trench bottom. (Key point: Drain field area)

4. Total trench length is 141.4 m and maximum length per trench is about 30 m. What is adopted?
   A. 2 trenches of 70 m
   B. 3 trenches of 47 m
   C. 10 trenches of 14 m
   D. 5 trenches of 28.5 m
   Answer: D. Length limit of about 30 m means at least 5 trenches, each 28.5 m giving 85.5 m^2. (Key point: Trench layout)

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