---
title: "Grit chamber for 60 MLD, 0.23 mm"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:grit-60mld-0p23"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-60mld-0p23
site: BCT Engineering
---

# Grit chamber for 60 MLD, 0.23 mm

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2081 Chaitra, 8 marks
- Video: [Unit:7 | Waste Water Treatment | Grit Chamber | Numerical | Sanitary Engineering | Prashant YT |](https://www.youtube.com/watch?v=xxEGHxqk5Gw) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-60mld-0p23
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a grit chamber for a sewage treatment with 60 MLD of sewage flow at 25°C to remove 0.23 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

Use Hazen’s formula: settling velocity 0.0529 m/s; Camp-Shields scour velocity 0.244 m/s. Adopt horizontal velocity 0.20 m/s. Cross-section area 3.47 m² yields width 3.5 m, water depth 1.0 m. Length from 1.25× theoretical and 30 s check is 6.0 m. Grit chamber: 6.0 m × 3.5 m × 1.0 m water depth, total depth 1.55 m.

## Answer

A grit chamber is designed to settle grit of $d=0.23$ mm and $G=2.65$ at 25°C while organic matter ($G=1.02$) is carried forward. The scour velocity is checked with $k=0.06$, $f=0.03$.

#### Step 1: Design flow

$$
Q = \frac{60\times 10^6}{1000\times 86400} = 0.6944\ \text{m}^3/\text{s}
$$

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

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

$$
v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.23^2\times\frac{3\times 25+70}{100} = 52.90\ \text{mm/s} = 0.0529\ \text{m/s}
$$

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

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

$$
v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.23\times 10^{-3}} = 0.244\ \text{m/s}
$$

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 ($G=1.02$), $v_c=\sqrt{\dfrac{8k}{f}(1.02-1)\,g\,d} = 0.027$ 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

$$
\begin{aligned}
A_c &= \frac{Q}{v_h} = \frac{0.6944}{0.20} = 3.472\ \text{m}^2\\
\text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 3.47\ \text{m (adopt 3.5 m)}\\
L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0529} = 3.78\ \text{m}\\
L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(4.73,\ 6.00) = 6.00\ \text{m (adopt 6.0 m)}
\end{aligned}
$$

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

#### Step 5: Checks

$$
\begin{aligned}
\text{Detention time} &= \frac{L}{v_h} = \frac{6.0}{0.20} = 30\ \text{s}\ (\text{within 30-90 s})\\
\text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.6944\times 86400}{3.5\times 6.0} = 2857\ \text{m}^3/\text{m}^2\text{/day}
\end{aligned}
$$


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

```text
 Plan:   inlet ->[ <-- L = 6.0 m --> ]-> outlet (weir)
                 |  B = 3.5 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 $L\times B\times H = 6.0\ \text{m}\times 3.5\ \text{m}\times 1.0\ \text{m}$ (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. How is settling velocity vs found at 25°C?
   A. vs = v x L
   B. vs = Q/A
   C. vs = sqrt(8k/f...)
   D. Hazen: vs = 418(G-1)d^2(3T+70)/100, giving 0.0529 m/s
   Answer: D. Hazen's formula includes a temperature correction. (Key point: Settling velocity)

2. What is the role of the Camp-Shields velocity vc = sqrt(8k(G-1)gd/f) = 0.244 m/s?
   A. It gives the settling speed
   B. It limits horizontal velocity so settled grit is not re-suspended
   C. It gives detention time
   D. It is the velocity of organics settling
   Answer: B. The flow velocity must stay below the scour velocity; 0.20 m/s was adopted. (Key point: Scour velocity)

3. Why was L = 6.0 m adopted instead of 1.25 x 3.78 = 4.73 m?
   A. 30 v_h = 6.0 m is larger, ensuring at least 30 s detention
   B. Because the depth was 0.6 m
   C. Because 4.73 is too long
   D. Because the width was small
   Answer: A. The length is the larger of 1.25 L_theo and 30 v_h. (Key point: Length governed by)

4. How is the width found?
   A. A = Q/v_h = 3.472 m2, B = A/H = 3.5 m for H = 1 m
   B. B = Q x vs
   C. B = v_h / vs
   D. B = L / H
   Answer: A. Area comes from flow over velocity; width is area over chosen depth. (Key point: Cross-section)

## Close variants set in other exams

- [Grit chamber for 50 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Agrit-50mld-0p2mm) — Sanitary Engineering, chapter 7, asked 2×

## Related in Sanitary Engineering

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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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