Question

Before entering an underground utility vault to do repairs, a work crew analyzed the gas in the vault and found that it contained 29 mg/m3of hydrogen sulfide. Because the allowable exposure level is 14 mg/m3, the work crew began ventilating the vault with a blower. If the volume of the vault is 160 m3and the flow rate of contaminant-free air is 10 m3/min, how long will it take to lower the hydrogen sulfide level to a level that will allow the work crew to enter? Assume the manhole behaves as a CMFR and that hydrogen sulfide is nonreactive in the time period considered.

Answer 1

Answer:

11.65 minutes.

Explanation:

See the attached picture for detailed explanation.

Answer 2

Answer:

11.65 min

Explanation:

Hydrogen sulfide is very poisonous and as a result, it is essential to reduce the concentration of the gas to the lowest possible value to minimize its effects. The time taken to reduce the amount of hydrogen sulfide in the system to the allowable limit can be estimated as shown below:

t = (V/Q)*ln(Ci/Co) = (160/10)*ln(14/29) = 16*0.728 = 11.65  min