---
title: "High-volume sampler and concentration calculation"
subject: "Environment and Pollution Control"
chapter: 1
key: "semester_7/environment_and_pollution_control:high-volume-sampler-concentration"
source: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/high-volume-sampler-concentration
site: BCT Engineering
---

# High-volume sampler and concentration calculation

Environment and Pollution Control · Chapter 1: Air pollution

- Asked in: year not given, 4+4 marks
- Video: [Ambient air sampling by using High Volume Air Sampler (HVAS) #AirPollution #AirQualityMonitoring](https://www.youtube.com/watch?v=ePmqxkL36Xk) — Dr. Neel Talati
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/high-volume-sampler-concentration
- Whole chapter: https://www.bctengineer.com/md/semester_7/environment_and_pollution_control/chapter-1

## Question

(a) Describe the principle and working of a high-volume (hi-vol) sampler used for measuring suspended particulate matter. (b) A hi-vol sampler runs for 24 hours at a mean flow rate of 1.5 m3/min. The filter weighs 3.215 g before and 3.412 g after sampling. Calculate the TSP concentration in micrograms per cubic metre. Also convert an SO2 reading of 0.05 ppm to micrograms per cubic metre at 25 degrees C and 1 atm (molecular weight of SO2 = 64.06; molar volume = 24.45 L/mol).

## Short answer

The mass of particulates collected (0.197 g) divided by the sampled air volume (2160 m3) gives TSP concentration of 91.2 μg/m3. SO2 concentration is converted using C = (ppm × MW × 1000) / molar volume, yielding 131 μg/m3.

## Answer

#### (a) High-volume sampler
Principle: a large volume of air is drawn through a pre-weighed filter. Particles are retained on the filter, and the mass gained divided by the air volume gives the concentration.

```text
        rain-proof roof
       ________________
      /  air inlet gap \
     /__________________\
        | filter paper |  (20 x 25 cm glass fibre)
        |______________|
        |   support    |
        |______________|
        |  blower +    |--> flow meter / recorder
        |  motor       |
        |______________|
```
Working:
1. A conditioned glass-fibre filter is weighed ($W_1$) and placed in the holder.
2. A motor-driven blower draws air at about 1.1-1.7 $\text{m}^3$/min through the filter. The roof keeps out rain and large particles (about > 100 $\mu$m).
3. The flow rate is recorded at the start and end (or continuously) and the mean value used.
4. After 24 h the filter is conditioned and re-weighed ($W_2$).

Concentration $= \dfrac{(W_2 - W_1)\times 10^6}{V}\ \mu\text{g/m}^3$. It measures total suspended particulates (TSP); a PM10 inlet can be fitted for PM10.

#### (b) Numerical
Volume of air sampled:
$$
V = 1.5\ \text{m}^3/\text{min} \times 24 \times 60\ \text{min} = 2160\ \text{m}^3
$$
Mass of particulates:
$$
m = 3.412 - 3.215 = 0.197\ \text{g} = 197\,000\ \mu\text{g}
$$
$$
C = \frac{197\,000}{2160} = 91.2\ \mu\text{g/m}^3
$$

**SO$_2$ conversion:**
$$
C = \frac{0.05 \times 64.06 \times 1000}{24.45} = 131.0\ \mu\text{g/m}^3
$$
(0.05 ppm = $0.05\times10^{-6}$ mol per mol of air; 1 ppm of a gas = $\text{MW}\times 1000/24.45\ \mu\text{g/m}^3$.)

**Answer: TSP = 91.2 $\mu$g/m$^3$; 0.05 ppm SO$_2$ = 131 $\mu$g/m$^3$.**

## Check yourself

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

1. How is TSP concentration obtained with a hi-vol sampler?
   A. Mass gained by the filter divided by volume of air drawn
   B. Volume of air divided by mass of filter
   C. Flow rate multiplied by time only
   D. Pressure drop across the filter
   Answer: A. A pre-weighed filter retains particles; mass gain divided by air volume gives concentration. (Key point: Principle of hi-vol sampler)

2. Air volume for 1.5 m3/min over 24 h is:
   A. 36 m3
   B. 90 m3
   C. 2160 m3
   D. 129 600 m3
   Answer: C. 1.5 x 24 x 60 = 2160 m3. (Key point: Air volume sampled)

3. The filter gains 0.197 g over 2160 m3. The TSP concentration is about:
   A. 9.12 ug/m3
   B. 91.2 ug/m3
   C. 912 ug/m3
   D. 0.0912 ug/m3
   Answer: B. 197 000 ug / 2160 m3 = 91.2 ug/m3. (Key point: TSP calculation)

4. Which expression converts ppm of SO2 to ug/m3 at 25 °C and 1 atm?
   A. ppm x 24.45 / (MW x 1000)
   B. ppm x MW x 1000 / 24.45
   C. ppm x MW / 24.45 x 10^6
   D. ppm x 24.45 x MW
   Answer: B. ug/m3 = ppm x MW x 1000 / 24.45, giving 131 ug/m3 for 0.05 ppm SO2. (Key point: ppm to ug/m3)

## Related in Environment and Pollution Control

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