---
title: "Rectangular grit chamber for 10 MLD"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:grit-10mld-0p2mm"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-10mld-0p2mm
site: BCT Engineering
---

# Rectangular grit chamber for 10 MLD

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2077 Chaitra, 8 marks
- Video: [Grit Chamber in Sewage Treatment Plant | Grit Chamber | Grit Removal by Easy Civil](https://www.youtube.com/watch?v=b9zjbNyPAOI) — EASY CIVIL
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-10mld-0p2mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a rectangular grit chamber for maximum wastewater flow for 10 MLD to remove particles up to 0.2 mm diameter having sp. gr. 2.65. Settling velocity of grits is found to be 0.02 m/s in average and maintain a flow velocity of 0.3 m/s constant through a flow weir.

## Short answer

Given v_s=0.02 m/s, v_h=0.30 m/s. A_c=0.386 m², H=0.6 m, B=0.7 m. L_theo=9.0 m, L=max(1.25×,30 s)=11.5 m. Grit chamber: 11.5 m × 0.7 m × 0.6 m water depth, total depth 1.15 m, flow velocity 0.30 m/s.

## Answer

Grit particles of $d=0.2$ mm and $G=2.65$ are to be removed from the maximum flow of 10 MLD. The settling velocity ($0.02$ m/s) and the flow velocity ($0.3$ m/s, maintained constant by a proportional weir at the outlet) are given. $k=0.06$ and $f=0.03$ are assumed only for the scour check.

#### Step 1: Design flow

$$
Q = \frac{10\times 10^6}{1000\times 86400} = 0.1157\ \text{m}^3/\text{s}
$$

#### Step 2: Settling velocity of the grit

The settling velocity is given: $v_s = 0.02$ m/s.

#### Step 3: Horizontal (flow-through) velocity

The velocity must not scour settled grit. Scour (critical) velocity (Camp-Shields):

$$
v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.2\times 10^{-3}} = 0.228\ \text{m/s}
$$

The flow velocity is given: $v_h = 0.3$ m/s. (It is slightly above the Camp-Shields scour value for 0.2 mm grit, but a proportional weir holds it constant and the usual 0.3 m/s limit applies.)

#### Step 4: Cross-section and dimensions

$$
\begin{aligned}
A_c &= \frac{Q}{v_h} = \frac{0.1157}{0.30} = 0.386\ \text{m}^2\\
\text{Adopt depth } H &= 0.6\ \text{m} \Rightarrow B = \frac{A_c}{H} = 0.64\ \text{m (adopt 0.7 m)}\\
L_{theo} &= \frac{v_h H}{v_s} = \frac{0.30\times 0.6}{0.0200} = 9.00\ \text{m}\\
L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(11.25,\ 9.00) = 11.25\ \text{m (adopt 11.5 m)}
\end{aligned}
$$

(25% is added for inlet and outlet turbulence, and the length is not allowed to give less than 30 s detention.)

#### Step 5: Checks

$$
\begin{aligned}
\text{Detention time} &= \frac{L}{v_h} = \frac{11.5}{0.30} = 38\ \text{s}\ (\text{within 30-90 s})\\
\text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.1157\times 86400}{0.7\times 11.5} = 1242\ \text{m}^3/\text{m}^2\text{/day}
\end{aligned}
$$


**Total depth** $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.15$ m.

```text
 Plan:   inlet ->[ <-- L = 11.5 m --> ]-> outlet (weir)
                 |  B = 0.7 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 0.6 m
          grit storage 0.25 m  ______________
```
**Answer: provide a grit chamber (at least two units in parallel, each for the full flow so that one can be cleaned) of size $L\times B\times H = 11.5\ \text{m}\times 0.7\ \text{m}\times 0.6\ \text{m}$ (water depth), total depth about 1.15 m, with flow velocity 0.30 m/s.**

## Check yourself

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

1. Why is the settling velocity not computed with Hazen's formula here?
   A. Grit is too small
   B. It is given as 0.02 m/s
   C. Flow is too large
   D. Hazen is invalid at 20°C
   Answer: B. The question provides vs directly. (Key point: Given data)

2. What is the purpose of the flow weir in this problem?
   A. Maintain a constant flow velocity of 0.3 m/s as discharge varies
   B. Measure BOD
   C. Increase settling velocity
   D. Remove organics
   Answer: A. A proportional weir keeps velocity constant. (Key point: Weir)

3. What is the theoretical length?
   A. 0.02 x 0.6 = 0.012 m
   B. v_h H / vs = 0.3 x 0.6/0.02 = 9 m
   C. 0.3 x 0.02 = 0.006 m
   D. 0.6/0.3 = 2 m
   Answer: B. Theoretical length is horizontal velocity times depth over settling velocity. (Key point: Length)

4. Which length was adopted?
   A. 9 m
   B. 6 m
   C. 30 x 0.3 = 9 m
   D. 1.25 x 9 = 11.25 m, adopted as 11.5 m
   Answer: D. 25% allowance for turbulence exceeds the 30 s detention minimum. (Key point: Final length)

## Related in Sanitary Engineering

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