Question

In: Chemistry

Tobacco smoke, containing at least 50 known carcinogens, is often considered the most harmful and widespread...

Tobacco smoke, containing at least 50 known carcinogens, is often considered the most harmful and widespread contaminant of indoor air. Many airports have a designated smoking area to prevent these contaminants from spreading to the rest of the airport. Consider one such smoking area with a volume of 15 m×15 m×4 m that has 35 persons smoking two cigarettes every hour. One of the gases coming out of a cigarette is formaldehyde and each cigarette may be assumed to emit 1.35 mg of formaldehyde. Conversion of formaldehyde to carbon dioxide can be assumed to be first order and the rate of reaction is 0.40 per hour. Outside (fresh) air enters the enclosed space at the rate of 800 m3 per hour. Assume the smoke becomes completely mixed with air and this smoke-mixed air leaves the enclosed space at the same rate that the fresh air enters. Initially the formaldehyde value is zero. (a) What is the concentration of formaldehyde in the room air at steady state? (b) If the threshold for eye irritation due to formaldehyde is 0.05 ppm (parts per million), what minimum flow rate of fresh air needs to be maintained for the enclosed space? [At the assumed temperature of 25°C and pressure of 1 atm for the enclosed space, convert the formaldehyde concentration to ppm using the formula ppm =mg/m3×24.45/M, where M is 30, the molecular weight of formaldehyde]

Solutions

Expert Solution

Amount of Formaldehyde produced /hr = input rate = 35 x 2 x 1,35 mg = 94.5 mg/h of Formaldehyde is produced

Volume of the room = 900 m3

output rate = 800 m3/h x C mg/m3 = 800Cmg/h

Decay rate = KCV = 0.4 h-1 x Cmg/m3 x 900 m3 = 360 Cmg/h

input rate = output rate + decay rate

94.5 = 800Cmg/h + 360 Cmg/h

94.5 mg/h = 1160 Cmg/h

C = 94.5/1160 = 0.0814 mg/m3

so concentration of formaldehyde at steady state is 0.0814 mg/m3

ppm =mg/m3×24.45/M where M=30

=0.0814 x 24.45/30 = 0.0663 ppm

So the concentration of formaldehyde in ppm is 0.0663 ppm

if concentration in ppm has to be 0.05 ppm

then we will need the the concentration of formaldehyde in the room to be 0.0613 mg/m3

input rate = output rate + decay rate

Decay rate is fixed = KCV = 0.4/h x 0.0613 mg/m3 x 900 m3 = 22.068 mg/h

input rate is 94.5mg/h

input rate = output rate + decay rate

94.5 = Q m3/h x 0.0613 mg/m3 + 22.068 mg/h

(94.5 - 22.068)/0.0613 = Q m3/h

Flow of fresh air has to be 1181 m3/h

queen_honey_blossom answered 1 hour ago
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