The other day one of our teams was attending a client going into an empty tank. Our team wanted to take a number of precautions whereas the client did not. The team was concerned that the client was taking unnecessary risks. The client, on the other hand, thought we were going overboard. This conflict was created because of how each party got to their point of view. The point of this blog is to explain how to measure hazards better. Having a proper understanding of how hazards are measured prevents these misunderstandings and helps everyone working with hazards be on the same page.
For my scenario, the tank to be entered is 2 metres in diameter and approximately 3 metres in height. There was X (you can pick any substance you want) in the tank up to the morning. However, the tank was emptied and washed prior to being opened for the job. It has not been ventilated and test results show oxygen at 20.9%, 0 ppm CO and 0 ppm H2S. The question then is, how bad is it?
The first step believe it or not, is to determine the volume of the tank. Tank Volume = ∏r2 x h. Doing the calculation Vol. = 3.14(1)2 x 3 or Tank Volume = 9.42m3 or 9,420 L. Step #2 (the one we always forget) is to remember that 10,000 ppm equals 1%. This is a key piece of data if you want to make any decision regarding risk. The third step is to remember we only need to make the tank dangerous to the height of the workers nose. Now the math is fairly simple. If we need to have 1000 ppm of a substance in this tank to make it dangerous, we need:
(Tank Volume) x (ppm/10,000) = Amount of Substance.
Based on the tank we are using, (3.14(1)2 x 3) x (1,000/10,000)= Litres of a substance
3768 x.1%= 377 L
So if the tank is empty, how do we get 377 litres of the substance by you. So in this scenario you need not worry about a toxic gas.
However lets say the tank has a sludge in it. The sludge contains water, organic material and other solids. The sludge is approximately .2 metres deep. If we take the above tank, how risky is it? To calculate the Amount of Substance, use:
Tank Volume to .2m or (3.14(1)2 x .2) x 1000 (to changes metres to Litres) = 628 L
For you that are about to panic, please remember the first exercise needed 377 litres of a gas, but this calculation gives us the amount of liquid. We now have to change the liquid we have in the tank into a gas. So now we need to get Evaporation data of the substance. For traditional chemicals you would find the info in the MSDS. For unique mixtures, scientists would take the sludge and boil it. They could then find out how fast the material evaporated and the volume the vapour occupies when it boils. Since you won’t have this ability in the field, what’s the next best option?
Well for water, my research (looking at Class A pan evaporation rates from Environment Canada and the National Weather Service (US)) indicates a tank at normal room temperature may lose approximately 3 mm of water per day or 1 mm of water per shift. This evaporation rate is based on temperature, wind, sun, and humidity averages. It is not completely accurate, but a great approximation. With the tank example above, we would gain approximately 10 L of water vapour (a gas) per shift from water evaporation. The calculation is (3.14(1)2 x .001) x 1000 = 9.86 L per shift. So we are a fair ways away from creating a toxic atmosphere in 1 shift. So this again would not make the tank project that risky. If the material inside is more volatile, (e.g paint thinners), the evaporation rate would change, which also changes the hazard.
So hazard evaluation is a science not a guess. Resource material such as MSDS’s, the web, or the library (if you’re old school) are a must. In future blogs I’ll delve into the science of hazard measuring even more.
