Question

A process equipment emits 5 kg/h of volatile organic compounds (VOCs). If a hood placed over the process equipment captures 95% of the VOCs, then the fugitive emission in kg/h is

• A. 0.25
• B. 4.75
• C. 2.50
• D. 0.48

Hint:

The fugitive emission is the uncollected portion of the total emissions. It can be calculated by multiplying the total emission by the percentage that is not captured.

Answer 1

The Correct Option is A

We are given that the process equipment emits 5 kg/h of volatile organic compounds (VOCs). A hood placed over the equipment captures 95% of the VOCs. This means the amount of VOCs captured by the hood is: \[ \text{Captured VOCs} = 95% \times 5 \, \text{kg/h} = 0.95 \times 5 = 4.75 \, \text{kg/h}. \] The fugitive emissions refer to the amount of VOCs not captured by the hood, which is the remaining 5%. So, the fugitive emission is: \[ \text{Fugitive Emission} = 5% \times 5 \, \text{kg/h} = 0.05 \times 5 = 0.25 \, \text{kg/h}. \] Thus, the fugitive emission is 0.25 kg/h. The correct answer is option (A).