Question

Determine the TWA of the following chlorine exposure. Is the worker overexposed to the ACGIH TLV updated in 2017?

2 hours at None detected

1 hour at 0.2 ppm

3 hours at 0.1 ppm

2 hours at 0.15 ppm

Answer 1

TWA or the time weighted average is the average exposure to a harmful substance for a specific period.

The TWA can be calculated by:

.

TWA= Citi + C2t2 + ... + Cntn ti + t2 + ... + tn

where
C1, C2, C3,
are the concentration of the substance and
t1, t2, t3,
are the time of exposure to the harmful substance.

In the given problem, the employee spends 2 hours where there is no chlorine detected

So $C_1=0$ and .

t1 = 2

The employee also spends 1 hour at 0.2 ppm

So $C_2=0.2$ .  .$t_2=1$

He also spends another 3 hours at 0.1 ppm

i.e. $C_3=0.1$ .  .$t_3=3$

and also spends another 2 hours at 0.15 ppm

$C_4=0.15$ .  .$t_4=2$

Therefore the employee’s eight-hour TWA is:

$TWA=\frac{\left (0\times 2 \right )+\left (0.2\times 1 \right )+\left (0.1\times 3 \right )+\left (0.15\times 2 \right )}{8}$

i.e.  .$TWA=\frac{0+0.2+0.3+0.3}{8}$

$TWA=\frac{0.8}{8}=0.1$

Therefore employee's 8 hour TWA of chlorine exoposure is 0.1 ppm.

Now, the ACGIH TLV for Chlorine is 0.5 ppm. So, we can say that the worker's TWA (0.1 ppm) is less than that of the TLV value for chlorine (0.5 ppm).

Therefore the worker is not overexposed to chlorine.