---
title: "Sludge volume from PST and SST"
subject: "Sanitary Engineering"
chapter: 8
key: "semester_6/sanitary_engineering:sludge-volume-pst-sst-20mld"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-volume-pst-sst-20mld
site: BCT Engineering
---

# Sludge volume from PST and SST

Sanitary Engineering · Chapter 8: Sludge Treatment and Disposal

- Asked in: 2074 Bhadra, 8 marks
- Video: [Volume of sludge produced by PST & SST | Trickling filter Numerical | 2019 PU | TU,PU,KU,POU](https://www.youtube.com/watch?v=Pwnl8Vkzbjs) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sludge-volume-pst-sst-20mld
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-8

## Question

The biological process occurs in trickling filter. PST removes 60% suspended solids and 30% of BOD. Determine the volume of sludge produced by PST as well as SST with the following data. Sp. gr. of inorganic solids = 2.65, Sp. gr. of organic solid = 1.02; Flow of sewage = 20*10⁶ l/d; BOD5 of sewage = 220 mg/l; Suspended solids in the sewage = 280 mg/l; Water content of the sludge = 95%.

## Short answer

PST removes 3360 kg/d dry solids, giving 66.5 m³/d sludge at 95% moisture. Trickling filter removes 2618 kg/d BOD, producing 1047 kg/d dry humus solids, giving 20.7 m³/d SST sludge (also 95% moisture).

## Answer

Sludge volume $V = \dfrac{W_s}{\rho_w \, G_{sl}\,(1-p)}$, where $W_s$ is the dry solids mass, $G_{sl}$ the specific gravity of wet sludge and $p$ the water fraction.

**Assumptions** (not given in the data): suspended solids are 70% organic and 30% inorganic; trickling filter removes 85% of the BOD reaching it; humus (SST sludge) produced is 0.4 kg dry solids per kg BOD removed; SST sludge also has 95% water.

#### Specific gravity of sludge

$$
\frac{1}{G_s} = \frac{0.7}{1.02} + \frac{0.3}{2.65} \Rightarrow G_s = 1.251
$$

$$
\frac{1}{G_{sl}} = \frac{0.05}{1.251} + \frac{0.95}{1} \Rightarrow G_{sl} = 1.0101
$$

#### Sludge from PST

- Suspended solids in raw sewage: $20\times10^6 \times 280\times10^{-6} = 5,600$ kg/d
- Removed in PST (60%): $W_s = 3,360$ kg/d

$$
V_{PST} = \frac{3,360}{1000 \times 1.0101 \times 0.05} = 66.5\ \text{m}^3\text{/d}
$$

#### Sludge from SST (humus)

- BOD in raw sewage: $20\times10^6 \times 220\times10^{-6} = 4,400$ kg/d
- BOD after PST (30% removed): $3,080$ kg/d
- BOD removed in trickling filter (85%): $2,618$ kg/d
- Dry humus solids: $0.4 \times 2,618 = 1,047$ kg/d

$$
V_{SST} = \frac{1,047}{1000 \times 1.0101 \times 0.05} = 20.7\ \text{m}^3\text{/d}
$$

**Answer: PST sludge $\approx 66.5\ \text{m}^3$/d; SST sludge $\approx 20.7\ \text{m}^3$/d** (total about 87.3 m³/d).

## Check yourself

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

1. Why is the sludge specific gravity found from 1/Gs = 0.7/1.02 + 0.3/2.65 and then 1/Gsl = 0.05/Gs + 0.95/1?
   A. Gsl is the average of 1.02 and 2.65
   B. Water is ignored
   C. Gsl equals 2.65 since it is the largest
   D. Solids G is combined from organic and inorganic parts, then combined with water
   Answer: D. Weight fractions are combined harmonically, first for the solids (Gs = 1.251), then with the water (Gsl = 1.0101). (Key point: Sludge G)

2. Dry solids removed in PST are
   A. 3360 kg/d (60% of 5600 kg/d)
   B. 5600 kg/d
   C. 2240 kg/d
   D. 1047 kg/d
   Answer: A. Raw SS = 20 x 10^6 x 280 x 10^-6 = 5600 kg/d; 60% is removed. (Key point: PST solids)

3. BOD removed in the trickling filter is found as
   A. 4400 x 0.85 = 3740 kg/d
   B. (4400 x 0.7) = 3080 kg/d reaching it, times 0.85 = 2618 kg/d
   C. 4400 x 0.3 = 1320 kg/d
   D. 3080 x 0.15 = 462 kg/d
   Answer: B. PST first removes 30% of BOD; the filter then removes 85% of what reaches it. (Key point: BOD to trickling filter)

4. The sludge volumes from PST and SST are about
   A. 20.7 m3/d and 66.5 m3/d
   B. 87.3 m3/d each
   C. 66.5 m3/d and 20.7 m3/d
   D. 3360 m3/d and 1047 m3/d
   Answer: C. V = Ws/(1000 x 1.0101 x 0.05): 3360 gives 66.5; humus 0.4 x 2618 = 1047 kg gives 20.7 m3/d. (Key point: Sludge volumes)

## Related in Sanitary Engineering

- [Sludge volume and digester, 4 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-4mld-digester) — Sanitary Engineering, chapter 8
- [Sludge volume from sedimentation tank, 10 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-10mld-250) — Sanitary Engineering, chapter 8
- [HR trickling filter, BOD 280 mg/l](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-hr-5mld-280) — Sanitary Engineering, chapter 7
- [Effluent BOD of two-stage trickling filter](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atwo-stage-trickling-effluent-bod) — Sanitary Engineering, chapter 7
- [Yearly sludge volume, 6 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Asludge-volume-yearly-6mld) — Sanitary Engineering, chapter 8
- [Standard rate trickling filter, 4 MLD, NRC](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Atrickling-standard-4mld-180-nrc) — Sanitary Engineering, chapter 7

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