---
title: "Grit chamber for 55×10⁶ l/d at 27°C"
subject: "Sanitary Engineering"
chapter: 7
key: "semester_6/sanitary_engineering:grit-55e6-ld-27c"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/grit-55e6-ld-27c
site: BCT Engineering
---

# Grit chamber for 55×10⁶ l/d at 27°C

Sanitary Engineering · Chapter 7: Wastewater Treatment

- Asked in: 2070 Magh, 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-55e6-ld-27c
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7

## Question

A town discharges sewage at the 55×10⁶ l/d. The specific gravity of grit particles in that sewage is found from an experiment as 2.65 and the temperature as 27°C. Design grit chamber for removal of grit particles of 0.21 mm. Use: K = 0.06 and f = 0.03.

## Short answer

For55 MLD, designflow0.6366 m\u00b3/s, settlingvelocity0.0459 m\/ s, horizontalvelocity0.2 m\/ s. Cross-section3.18 m\u00b2, depth1 m, width3.2 m. Length from v_h H\/ v_s with25% extra and30 s min is6.0 m. Chamber: L=6.0 m, B=3.2 m, water depth1.0 m (total 1.55 m), velocity0.20 m\/ s; two units.

## Answer

Grit: $d=0.21$ mm, $G=2.65$, $T=27^\circ$C; $k=0.06$, $f=0.03$.

#### Step 1: Design flow

$$
Q = \frac{55\times 10^6}{1000\times 86400} = 0.6366\ \text{m}^3/\text{s}
$$

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

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

$$
v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.21^2\times\frac{3\times 27+70}{100} = 45.93\ \text{mm/s} = 0.0459\ \text{m/s}
$$

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

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

$$
v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.21\times 10^{-3}} = 0.233\ \text{m/s}
$$

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

$$
\begin{aligned}
A_c &= \frac{Q}{v_h} = \frac{0.6366}{0.20} = 3.183\ \text{m}^2\\
\text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 3.18\ \text{m (adopt 3.2 m)}\\
L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0459} = 4.35\ \text{m}\\
L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(5.44,\ 6.00) = 6.00\ \text{m (adopt 6.0 m)}
\end{aligned}
$$

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

#### Step 5: Checks

$$
\begin{aligned}
\text{Detention time} &= \frac{L}{v_h} = \frac{6.0}{0.20} = 30\ \text{s}\ (\text{within 30-90 s})\\
\text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.6366\times 86400}{3.2\times 6.0} = 2865\ \text{m}^3/\text{m}^2\text{/day}
\end{aligned}
$$


**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 = 3.2 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 $L\times B\times H = 6.0\ \text{m}\times 3.2\ \text{m}\times 1.0\ \text{m}$ (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 55 x 10^6 l/d converted into Q in m^3/s?
   A. Divide by 1000 only
   B. Multiply by 1000 and divide by 86400
   C. Divide by 1000 and by 86400 giving 0.6366 m^3/s
   D. Divide by 24 and 1000 giving 2292 m^3/s
   Answer: C. Litres to m^3 is /1000, days to seconds is /86400. (Key point: Design flow conversion)

2. What does the Camp-Shields scour velocity v_c = sqrt(8k(G-1)g d / f) decide in the design?
   A. The horizontal flow velocity that must not re-suspend settled grit
   B. The settling velocity of the grit
   C. The depth of the grit storage
   D. The surface loading
   Answer: A. v_c (0.233 m/s) is the velocity at which settled grit starts to scour, so v_h is taken at or below it (0.20 m/s). (Key point: Scour velocity formula)

3. How is the theoretical length found from settling velocity v_s = 0.0459 m/s?
   A. L = v_s H / v_h
   B. L = v_h H / v_s = 0.20 x 1.0 / 0.0459 = 4.35 m
   C. L = Q / (v_s B)
   D. L = H / v_h
   Answer: B. A particle must settle depth H while travelling L, so L/v_h = H/v_s. (Key point: Theoretical length)

4. Why is the adopted length 6.0 m rather than 1.25 x 4.35 = 5.44 m?
   A. Because 25% is subtracted for turbulence
   B. Because the length is always doubled
   C. Because depth was increased
   D. Because length must give at least 30 s detention, 30 x v_h = 6.0 m governs
   Answer: D. L is the larger of 1.25 L_theo and 30 v_h, and 30 v_h = 6.0 m is larger. (Key point: Final length rule)

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