---
title: "Septic tank and soak pits, hotel of 80"
subject: "Sanitary Engineering"
chapter: 9
key: "semester_6/sanitary_engineering:septic-tank-hotel-80-users"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/septic-tank-hotel-80-users
site: BCT Engineering
---

# Septic tank and soak pits, hotel of 80

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

- Asked in: 2074 Bhadra, 8 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-hotel-80-users
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-9

## Question

What would be the internal dimension of a septic tank and numbers of soak pits for an isolated hotel situated at mid-southern zone of Nepal having average 80 numbers of average users? Rate of sewage discharge is 210 lpcd. Cleaning period of septic tank is 3 years. Assume other necessary data if required.

## Short answer

Septic tank internal dimensions 6.2 m × 2.2 m × 2.3 m (including 0.3 m freeboard) with two chambers. Total liquid volume 27 m³. Soak pits: 12 circular pits of 3 m diameter and 3 m effective depth, providing 336 m² infiltration area at 50 L/m²/d rate.

## Answer

**Assumptions:** mid-southern zone (Terai/inner Terai) with sandy loam soil, infiltration rate 50 L/m²/d, ground-water at least 5 m below the surface; detention 24 h, soak pit depth 3 m and diameter 3 m.

#### Septic tank

- Sewage flow: $Q = 80 \times 210\ \text{L/c/d} = 16800\ \text{L/d} = 16.800\ \text{m}^3\text{/d}$
- Sewage (settling) zone for 24 h detention: $V_1 = 16.800\ \text{m}^3$
- Sludge digestion zone: $0.0425 \times 80 = 3.400\ \text{m}^3$
- Digested sludge storage for 3 years: $(0.085/3) \times 3 \times 80 = 6.800\ \text{m}^3$

$$
V = 16.800 + 3.400 + 6.800 = 27.00\ \text{m}^3
$$

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

$$
\begin{aligned} \text{Plan area} &= 27.00/2.0 = 13.50\ \text{m}^2 \\ B &= \sqrt{13.50/3} \approx 2.2\ \text{m}, \quad L = 13.50/2.2 \approx 6.2\ \text{m} \end{aligned}
$$

Check: $6.2 \times 2.2 \times 2.0 = 27.28\ \text{m}^3 \ge 27.00\ \text{m}^3$ (safe).

**Septic tank:** internal size $6.2\ \text{m} \times 2.2\ \text{m} \times 2.3\ \text{m}$ (including 0.3 m free board), in two chambers: first chamber $4.15$ m long (2/3), second $2.05$ 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: $A = Q/\text{rate} = 16800/50 = 336.00\ \text{m}^2$ (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 = $\pi \times 3.0 \times 3 = 28.27\ \text{m}^2$).

$$
N = \frac{336.00}{28.27} = 11.88 \Rightarrow 12\ \text{pits}
$$

**Soak pits:** 12 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.

The large number of pits can be replaced by dispersion trenches or by two tanks in parallel if space is limited.

## Check yourself

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

1. For 80 users at 210 lpcd, what is the daily sewage flow?
   A. 16.8 m^3/d
   B. 8.0 m^3/d
   C. 1.68 m^3/d
   D. 21.0 m^3/d
   Answer: A. Q = 80 x 210 = 16800 L/d = 16.8 m^3/d. (Key point: Flow)

2. What is the total septic tank volume for 3-year cleaning?
   A. 20.2 m^3
   B. 27.0 m^3
   C. 16.8 m^3
   D. 33.6 m^3
   Answer: B. V = 16.8 + 3.4 + 6.8 = 27.0 m^3. (Key point: Total volume)

3. With depth 2.0 m and L:B = 3:1, how are the plan dimensions obtained?
   A. Plan area = 27 x 2, then L = B
   B. Plan area = 16.8/2 then B = 3L
   C. Plan area = 27/2 = 13.5 m^2, B = sqrt(13.5/3), L = 3B
   D. B = 2 m by assumption and L = 27/2
   Answer: C. Plan area = V/d, and with L = 3B, B = sqrt(area/3), giving about 2.2 m by 6.2 m. (Key point: Tank dimensions)

4. Area needed is 336 m^2 and each pit gives 28.27 m^2. How many pits?
   A. 10
   B. 11
   C. 13
   D. 12
   Answer: D. 336/28.27 = 11.88, rounded up to 12 pits. (Key point: Soak pit count)

## Related in Sanitary Engineering

- [Septic tank and soak pits, 15 users, Terai](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aseptic-tank-15-users-terai) — Sanitary Engineering, chapter 9
- [Septic tank and soak pit, 10 users, 85 l/m²/d](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aseptic-tank-10-users-85) — Sanitary Engineering, chapter 9, asked 2×
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- [Septic tank and soak pit, household of 80](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aseptic-tank-soak-pit-80-persons) — Sanitary Engineering, chapter 9, asked 2×
- [Septic tank and soak pit design in Nepal](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aseptic-tank-design-considerations-nepal) — Sanitary Engineering, chapter 9, asked 2×
- [VIP latrine; septic tank for 15 users](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Avip-septic-tank-15-users-80lpcd) — Sanitary Engineering, chapter 9

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