Question

A paper mill uses liquid chlorine, delivered in 90-ton railroad tank cars, as a pulp bleaching agent. One volume of liquid chlorine produces approximately 450 volumes of vapor under normal atmospheric temperature and pressure. The density of liquid chlorine is 103 pounds per cubic foot. In the event of rupture and vapor release of 20% of the tank car contents, how much vapor by volume would be released? If the release were in a closed building with a 30-foot ceiling height without ventilation, how large would the building have to be (in square miles of floor space) to contain the thoroughly mixed vapor-air ratio within the OSHA PEL? The logical conclusion to this exercise is that, with or without ventilation, it is more practical to unload chlorine tank cars outdoors.

Answer 1

One volume of liquid chlorine = 450 volumes of vapor

Volume of liquid chlorine released, volume of liquid chlorine released = 0.2 * 90 = 18 tonnes

In terms of volume, vapor released = 18 * 14.67 ft³ = 264 ft³

Now, total chlorine by weight in the tank car = 0.2 * 90 * 14.67 * 103 = 1,35,990 pounds = 2801 kg

Taking OSHA PEL for chlorine as 1 ppm, or 1 mg per liter, the volume of space needed to contain the chlorine in case of spill within safety limits

= 2801 * 10⁶ liters = 2801 * 10⁶ * .0353 ft³ = 98.88 million ft³

Hence, for a 30 ft high building, we need to area to be ( 98.88 million ft³ / 30 ft )

= 3295 thousand ft²,

= 3.55 sq miles, which is generally impractical for the area of a building

Hence, it is more practical to unload chlorine tank cars outdoors