---
title: "Septic tank and soak pits, 15 users, Terai"
subject: "Sanitary Engineering"
chapter: 9
key: "semester_6/sanitary_engineering:septic-tank-15-users-terai"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/septic-tank-15-users-terai
site: BCT Engineering
---

# Septic tank and soak pits, 15 users, Terai

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

- Asked in: 2068 Magh (old course), 10 marks
- 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-15-users-terai
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-9

## Question

Design a septic tank and numbers of soak pit for 15 numbers of average users. Rate of sewage discharge is 45 lpcd. Cleaning period of septic tank is 4 years interval. Assume other necessary data if required. Location of construction is in Terai area.

## Short answer

Compute volumes: settling 0.675 m³, digestion 0.638 m³, storage 1.700 m³, total 3.01 m³. Tank 2.3 m × 0.9 m × 1.5 m liquid depth + 0.3 m freeboard, two chambers. Soak pit: infiltration area 13.5 m², one circular pit diameter 2.2 m, depth 2.0 m, bottom >1 m above groundwater.

## Answer

**Assumptions:** Terai has a high ground-water table (about 3 m) and flat land, so soak pits are kept shallow (effective depth 2.0 m) and infiltration rate for sandy soil is taken as 50 L/m²/d; detention 24 h; pit diameter at most 3 m.

#### Septic tank

- Sewage flow: $Q = 15 \times 45\ \text{L/c/d} = 675\ \text{L/d} = 0.675\ \text{m}^3\text{/d}$
- Sewage (settling) zone for 24 h detention: $V_1 = 0.675\ \text{m}^3$
- Sludge digestion zone: $0.0425 \times 15 = 0.638\ \text{m}^3$
- Digested sludge storage for 4 years: $(0.085/3) \times 4 \times 15 = 1.700\ \text{m}^3$

$$
V = 0.675 + 0.638 + 1.700 = 3.01\ \text{m}^3
$$

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

$$
\begin{aligned} \text{Plan area} &= 3.01/1.5 = 2.01\ \text{m}^2 \\ B &= \sqrt{2.01/3} \approx 0.9\ \text{m}, \quad L = 2.01/0.9 \approx 2.3\ \text{m} \end{aligned}
$$

Check: $2.3 \times 0.9 \times 1.5 = 3.11\ \text{m}^3 \ge 3.01\ \text{m}^3$ (safe).

**Septic tank:** internal size $2.3\ \text{m} \times 0.9\ \text{m} \times 1.8\ \text{m}$ (including 0.3 m free board), in two chambers: first chamber $1.55$ m long (2/3), second $0.75$ 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: $A = Q/\text{rate} = 675/50 = 13.50\ \text{m}^2$ (only the side wall below the inlet is counted; the bottom is ignored as it clogs)
- Adopt effective depth $h = 2.0$ m

$$
\pi D h = 13.50 \Rightarrow D = \frac{13.50}{\pi \times 2.0} = 2.15 \approx 2.2\ \text{m}
$$

Check: $\pi \times 2.2 \times 2.0 = 13.82\ \text{m}^2 \ge 13.50\ \text{m}^2$.

**Soak pit:** one circular pit, 2.2 m internal diameter and 2.0 m deep, lined with dry brick/stone masonry with open joints (honeycomb), filled with brick bats/gravel, and covered with a slab.

Because of the high water table, keep the pit bottom at least 1 m above the highest ground-water level; if this cannot be done use dispersion trenches or an evapotranspiration mound instead.

## Check yourself

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

1. Why is the soak pit kept shallow (2 m) in Terai?
   A. High ground-water table
   B. Soil is rocky
   C. Low rainfall
   D. To save slab cost only
   Answer: A. Terai has water table near 3 m, and the pit bottom must remain at least 1 m above it. (Key point: Terai soak pit)

2. Digested sludge storage for 4 years and 15 users?
   A. 1.275 m^3
   B. 1.700 m^3
   C. 0.638 m^3
   D. 2.55 m^3
   Answer: B. (0.085/3) x 4 x 15 = 1.700 m^3. (Key point: Storage)

3. Total tank volume for Q = 0.675 m^3/d?
   A. 2.0 m^3
   B. 4.2 m^3
   C. 3.01 m^3
   D. 1.7 m^3
   Answer: C. V = 0.675 + 0.638 + 1.700 = 3.01 m^3. (Key point: Total volume)

4. Area 13.5 m^2, h = 2.0 m: which diameter?
   A. 1.1 m
   B. 4.3 m
   C. 6.75 m
   D. 2.2 m
   Answer: D. D = 13.5/(pi x 2) = 2.15, adopted 2.2 m. (Key point: Pit diameter)

5. If the pit bottom cannot stay 1 m above ground water, what is suggested?
   A. Dispersion trenches or an evapotranspiration mound
   B. Deeper soak pit
   C. Direct discharge to a stream
   D. Omit the soak pit
   Answer: A. Shallow trenches or an ET mound avoid the high water table. (Key point: Alternative)

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