---
title: "Grit chamber for 58 MLD at 25°C"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:grit-58mld-25c"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-58mld-25c
site: BCT Engineering
---

# Grit chamber for 58 MLD at 25°C

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2075 Bhadra, 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-58mld-25c
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

Design a grit chamber to remove grit size of diameter more than 0.2 mm present in 58 MLD of sewage at a temperature of 25°C. Assume specific gravity of grit and organic matters as 2.65 and 1.2 respectively. Adopt k = 0.06 and f = 0.03 to calculate critical velocity.

## Short answer

Settling velocity 0.0400 m/s (Hazen), scour velocity 0.228 m/s; adopt v_h=0.20 m/s. A_c=3.36 m², B=3.4 m, H=1.0 m. L_theo=5.00 m, L=max(1.25×,30 s)=6.5 m. Dimensions: 6.5 m × 3.4 m × 1.0 m water depth, total depth 1.55 m.

## Answer

Grit: $d=0.2$ mm, $G=2.65$; organic matter $G=1.2$; $T=25^\circ$C; $k=0.06$, $f=0.03$.

#### Step 1: Design flow

$$
Q = \frac{58\times 10^6}{1000\times 86400} = 0.6713\ \text{m}^3/\text{s}
$$

#### 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 ($G=1.2$), $v_c=\sqrt{\dfrac{8k}{f}(1.2-1)\,g\,d} = 0.079$ 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

$$
\begin{aligned}
A_c &= \frac{Q}{v_h} = \frac{0.6713}{0.20} = 3.356\ \text{m}^2\\
\text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 3.36\ \text{m (adopt 3.4 m)}\\
L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0400} = 5.00\ \text{m}\\
L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(6.25,\ 6.00) = 6.25\ \text{m (adopt 6.5 m)}
\end{aligned}
$$

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

#### Step 5: Checks

$$
\begin{aligned}
\text{Detention time} &= \frac{L}{v_h} = \frac{6.5}{0.20} = 32\ \text{s}\ (\text{within 30-90 s})\\
\text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.6713\times 86400}{3.4\times 6.5} = 2624\ \text{m}^3/\text{m}^2\text{/day}
\end{aligned}
$$


**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 = 3.4 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 $L\times B\times H = 6.5\ \text{m}\times 3.4\ \text{m}\times 1.0\ \text{m}$ (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 58 MLD converted to m3/s?
   A. 58/86400
   B. 58x10^6 /(1000x86400) = 0.6713
   C. 58 x 1000
   D. 58x10^6/3600
   Answer: B. MLD is million litres per day; divide by 1000 for m3 and 86400 for seconds. (Key point: Flow)

2. What is vs at 25°C for 0.2 mm, G = 2.65?
   A. 0.228 m/s
   B. 0.4 m/s
   C. 0.04 m/s
   D. 0.004 m/s
   Answer: C. 418 x 1.65 x 0.04 x 1.45 = 40 mm/s. (Key point: Settling velocity)

3. Why is v_h = 0.20 m/s adopted?
   A. Just below the scour velocity 0.228 m/s but within 0.2-0.3 m/s to carry organics
   B. It is derived from detention time
   C. Larger than scour velocity
   D. Equal to settling velocity
   Answer: A. It avoids scouring grit and still carries organics. (Key point: Velocity chosen)

4. Which length governs?
   A. 5 m
   B. 30 v_h = 6.0 m
   C. 8 m
   D. 1.25 x 5.0 = 6.25 m, adopted 6.5 m
   Answer: D. 6.25 m exceeds the 6.0 m from minimum detention. (Key point: Length)

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