---
title: "Grit chamber for 50 MLD"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:grit-50mld-0p2mm"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-50mld-0p2mm
site: BCT Engineering
---

# Grit chamber for 50 MLD

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2073 Bhadra, 8 marks
- Video: [Design a Rectangular Grit Chamber | Numerical | Waste Water Treatment | Sanitary Engineering](https://www.youtube.com/watch?v=IBXJZgc7nY8) — SBK Concept
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-50mld-0p2mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Propose the dimensions of grit chamber for a sewage treatment plant with 50 MLD of sewage flow at 25°C to remove 0.2 mm size of grit having specific gravity of 2.65. The specific gravity of organic matter is 1.02. Assume k = 0.06 and f = 0.03.

## Short answer

Design flow 0.5787 m³/s, settling velocity 0.04 m\/ s, horizontal velocity 0.2 m\/ s. Cross-sectional area 2.9 m², depth 1 m, width 2.9 m. Length from v_h H\/ v_s with 25% extra and min 30 s detention gives 6.5 m. Grit chamber: L=6.5 m, B=2.9 m, water depth 1.0 m (total 1.55 m), flow velocity 0.2 m\/ s; provide two parallel units.

## Answer

Grit: $d=0.2$ mm, $G=2.65$ at $25^\circ$C; organic matter $G=1.02$; $k=0.06$, $f=0.03$.

#### Step 1: Design flow

$$
Q = \frac{50\times 10^6}{1000\times 86400} = 0.5787\ \text{m}^3/\text{s}
$$

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

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

$$
v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.2^2\times\frac{3\times 25+70}{100} = 40.00\ \text{mm/s} = 0.0400\ \text{m/s}
$$

#### 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). For the organic matter ($G=1.02$), $v_c=\sqrt{\dfrac{8k}{f}(1.02-1)\,g\,d} = 0.025$ m/s, so $v_h=0.20$ m/s is well above it: the light organic matter stays in suspension and passes on instead of settling.

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

$$
\begin{aligned}
A_c &= \frac{Q}{v_h} = \frac{0.5787}{0.20} = 2.894\ \text{m}^2\\
\text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 2.89\ \text{m (adopt 2.9 m)}\\
L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0400} = 5.00\ \text{m}\\
L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(6.25,\ 6.00) = 6.25\ \text{m (adopt 6.5 m)}
\end{aligned}
$$

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

#### Step 5: Checks

$$
\begin{aligned}
\text{Detention time} &= \frac{L}{v_h} = \frac{6.5}{0.20} = 32\ \text{s}\ (\text{within 30-90 s})\\
\text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.5787\times 86400}{2.9\times 6.5} = 2653\ \text{m}^3/\text{m}^2\text{/day}
\end{aligned}
$$


**Total depth** $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.55$ m.

```text
 Plan:   inlet ->[ <-- L = 6.5 m --> ]-> outlet (weir)
                 |  B = 2.9 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 1.0 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 $L\times B\times H = 6.5\ \text{m}\times 2.9\ \text{m}\times 1.0\ \text{m}$ (water depth), total depth about 1.55 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 area results for Q = 0.5787 m3/s and v_h = 0.2?
   A. 0.116 m2
   B. 28.9 m2
   C. 2.894 m2
   D. 0.5787 m2
   Answer: C. A = Q/v_h. (Key point: Cross-section)

2. What length is adopted?
   A. 5.0 m
   B. 6.5 m from 1.25 x 5.0 = 6.25 m
   C. 12 m
   D. 3 m
   Answer: B. L_theo = 0.2x1.0/0.04 = 5 m, plus 25%. (Key point: Length)

3. What is the detention time, and is it acceptable?
   A. 65 s, too long
   B. 6.5/0.2 = 32 s, within 30-90 s
   C. 6.5 s, too short
   D. 3.2 s
   Answer: B. Detention = L/v_h. (Key point: Detention check)

## Close variants set in other exams

- [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×

## Related in Sanitary Engineering

- [Grit chamber for 58 MLD at 25°C](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Agrit-58mld-25c) — Sanitary Engineering, chapter 7
- [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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- [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 5 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Agrit-5mld-0p2mm) — Sanitary Engineering, chapter 7

## Related in other subjects

- [Settling tank 30 m x 3 m, 0.03 mm at 25°C; 0.025 mm at 20°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-k72) — Water Supply Engineering, chapter 6, asked 3×
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- [Settling tank 30 m x 3 m, 0.02 mm, SG 2.50](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-bk80) — Water Supply Engineering, chapter 6