---
title: "Grit chamber for 190 l/s"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:grit-190-ls"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-190-ls
site: BCT Engineering
---

# Grit chamber for 190 l/s

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2078 Chaitra, 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/grit-190-ls
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a grit chamber for wastewater flow of 190 liter/sec with surface overflow rate = 2 cm/s and detention time = 60 sec. Take specific gravity of organic and inorganic particles are 1.2 and 2.65 respectively. Assume size of both organic and inorganic materials as 0.21 m. Take k = 0.06 and f = 0.03.

## Short answer

Given SOR=0.02 m/s and detention 60 s: A=9.50 m², H=1.20 m. v_h=0.20 m/s (between organic scour 0.081 and grit scour 0.233 m/s). L=v_h t=12.0 m, B=0.8 m. Chamber: 12.0 m × 0.8 m × 1.20 m water depth, total depth 1.75 m, detention 60 s.

## Answer

The flow is $Q=190$ l/s $=0.19$ m³/s. The particle size is read as 0.21 mm (the "0.21 m" in the question is a typing slip). $k=0.06$, $f=0.03$, SOR (surface overflow rate) $=2$ cm/s $=0.02$ m/s, detention time $=60$ s.

#### Step 1: Surface area and depth

$$
\begin{aligned}
Q &= 0.190\ \text{m}^3/\text{s}\\
\text{Surface area } A &= \frac{Q}{SOR} = \frac{0.190}{0.020} = 9.50\ \text{m}^2\\
\text{Volume } V &= Q\,t = 0.190\times 60 = 11.40\ \text{m}^3\\
\text{Depth } H &= \frac{V}{A} = 1.20\ \text{m}
\end{aligned}
$$

#### Step 2: Scour velocity for the two kinds of particles

$$
v_c=\sqrt{\frac{8k}{f}(G-1)\,g\,d}
$$

| Particle | $G$ | $d$ (mm) | $v_c$ (m/s) |
|---|---|---|---|
| Organic | 1.2 | 0.21 | 0.081 |
| Inorganic (grit) | 2.65 | 0.21 | 0.233 |

The horizontal velocity must be **above** the organic scour velocity (organic matter stays suspended and is carried through) and **below** the grit scour velocity (settled grit is not re-suspended). Adopt $v_h = 0.20$ m/s, which satisfies 0.081 < 0.20 < 0.233 m/s. (The flow velocity should also be near 0.2-0.3 m/s in practice.)

#### Step 3: Dimensions

$$
\begin{aligned}
L &= v_h\,t = 0.20\times 60 = 12.00\ \text{m}\\
B &= \frac{A}{L} = \frac{9.50}{12.00} = 0.79\ \text{m}\\
\text{Check: } A_c &= B\,H = 0.79\times 1.20 = 0.950\ \text{m}^2 \approx \frac{Q}{v_h} = 0.950\ \text{m}^2
\end{aligned}
$$

Adopt $L = 12.0$ m, $B = 0.8$ m, water depth $H = 1.20$ m, plus 0.25 m grit storage and 0.30 m free board (total depth about 1.75 m). Provide two units in parallel.

**Answer: grit chamber $L\times B\times H = 12.0\times 0.8\times 1.20$ m, flow velocity 0.20 m/s, detention 60 s, surface overflow rate 2 cm/s.**

## Check yourself

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

1. How is surface area found from SOR?
   A. A = Q/SOR = 0.19/0.02 = 9.5 m2
   B. A = Q x SOR
   C. A = Q t
   D. A = Q/v
   Answer: A. Area is flow over surface overflow rate. (Key point: Surface area)

2. How is depth found?
   A. Q/A
   B. v x t
   C. Q x A
   D. Volume Q t = 11.4 m3 divided by area 9.5 m2 = 1.2 m
   Answer: D. Depth = V/A. (Key point: Depth)

3. What condition must the horizontal velocity satisfy?
   A. Equal to the settling velocity
   B. Above the grit scour velocity
   C. Below the organic scour velocity
   D. Above organic scour velocity (0.081) and below grit scour velocity (0.233 m/s)
   Answer: D. Organics must stay carried along while settled grit must not scour. (Key point: Velocity range)

4. How are L and B found?
   A. L = v_h t = 12 m, B = A/L = 0.8 m
   B. B = Q t
   C. L = A/Q
   D. L = SOR x t
   Answer: A. Length is velocity times detention time and width is area over length. (Key point: Length and width)

## Close variants set in other exams

- [Grit chamber for 200 l/s, SOR 2](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Agrit-200-ls-sor-2) — Sanitary Engineering, chapter 7, asked 2×

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- [Grit chamber for 60 MLD, 0.23 mm](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Agrit-60mld-0p23) — Sanitary Engineering, chapter 7, asked 2×
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- [Rectangular grit chamber for 10 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Agrit-10mld-0p2mm) — Sanitary Engineering, chapter 7

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