High-volume sampler and concentration calculation

(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.

        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$^3$.