---
title: "Grit chamber for 5 MLD"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:grit-5mld-0p2mm"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-5mld-0p2mm
site: BCT Engineering
---

# Grit chamber for 5 MLD

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2072 Magh, 8 marks
- Video: [Design of Grit Chamber | Grit removal](https://www.youtube.com/watch?v=FhMny8Geo_o) — SavEnviron
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-5mld-0p2mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a grit chamber for the following data: Discharge = 5 MLD; Size of the grit particles = 0.2 mm; Sp. gravity of grit particles = 2.65 at temperature 20°C.

## Short answer

Design flow 0.0579 m\u00b3/s, settling velocity0.0359 m\/ s, horizontal velocity 0.2 m\/ s. Cross-sectional area 0.289 m\u00b2, depth 0.5 m, width 0.6 m. Length from v_h H\/ v_s with 25% extra and 30 s min detention is 6.0 m. Chamber: L=6.0 m, B=0.6 m, water depth 0.5 m (total 1.05 m), velocity 0.2 m\/ s; two units.

## Answer

Grit: $d=0.2$ mm, $G=2.65$, $T=20^\circ$C. $k=0.06$ and $f=0.03$ are assumed (standard values) for the scour check.

#### Step 1: Design flow

$$
Q = \frac{5\times 10^6}{1000\times 86400} = 0.0579\ \text{m}^3/\text{s}
$$

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

Hazen's formula (with temperature correction), $d$ in mm:

$$
v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.2^2\times\frac{3\times 20+70}{100} = 35.86\ \text{mm/s} = 0.0359\ \text{m/s}
$$

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

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

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

Adopt $v_h = 0.20$ m/s (the scour velocity rounded down; the usual range is 0.2-0.3 m/s). This is high enough to keep the light organic matter moving, so it passes on instead of settling.

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

$$
\begin{aligned}
A_c &= \frac{Q}{v_h} = \frac{0.0579}{0.20} = 0.289\ \text{m}^2\\
\text{Adopt depth } H &= 0.5\ \text{m} \Rightarrow B = \frac{A_c}{H} = 0.58\ \text{m (adopt 0.6 m)}\\
L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 0.5}{0.0359} = 2.79\ \text{m}\\
L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(3.49,\ 6.00) = 6.00\ \text{m (adopt 6.0 m)}
\end{aligned}
$$

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

#### Step 5: Checks

$$
\begin{aligned}
\text{Detention time} &= \frac{L}{v_h} = \frac{6.0}{0.20} = 30\ \text{s}\ (\text{within 30-90 s})\\
\text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.0579\times 86400}{0.6\times 6.0} = 1389\ \text{m}^3/\text{m}^2\text{/day}
\end{aligned}
$$


**Total depth** $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.05$ m.

```text
 Plan:   inlet ->[ <-- L = 6.0 m --> ]-> outlet (weir)
                 |  B = 0.6 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 0.5 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 $L\times B\times H = 6.0\ \text{m}\times 0.6\ \text{m}\times 0.5\ \text{m}$ (water depth), total depth about 1.05 m, with flow velocity 0.20 m/s.**

## Check yourself

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

1. What is vs at 20°C for 0.2 mm, G = 2.65?
   A. 0.359 m/s
   B. 0.00359 m/s
   C. 0.2 m/s
   D. 0.0359 m/s
   Answer: D. 418 x 1.65 x 0.04 x 1.3 = 35.9 mm/s. (Key point: Settling velocity)

2. Why is L = 6.0 m rather than 1.25 x 2.79 = 3.49 m?
   A. 3.49 m is too long
   B. Minimum 30 s detention at 0.20 m/s requires 6.0 m
   C. Depth requires it
   D. Width limits it
   Answer: B. Because the flow is small, the 30 s detention rule governs. (Key point: Governing length)

3. Which values were assumed since not given?
   A. Grit size 0.2 mm
   B. k = 0.06 and f = 0.03 for the scour check
   C. G = 2.65
   D. Temperature 20°C
   Answer: B. The question gave G, d and T but not k and f. (Key point: Assumed data)

4. What width follows with H = 0.5 m?
   A. B = 2.9 m
   B. B = 0.289/0.5 = 0.58, adopted 0.6 m
   C. B = 6 m
   D. B = 0.06 m
   Answer: B. Area 0.289 m2 divided by depth 0.5 m. (Key point: Width)

## Related in Sanitary Engineering

- [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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## Related in other subjects

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- [Rectangular sedimentation tank for 10 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-s71) — Water Supply Engineering, chapter 6