A room has a characteristic time (tau) of 1.3 hrs. If the concentration of a con

Question

A room has a characteristic time (tau) of 1.3 hrs. If the concentration of a contaminant is 100 ppm at closing time, 5:00 PM, what will be the concentration when A opens again at 8:00 AM? We first must ask how long the dilution ventilation the to remove the contaminant. The time from 5:00 PM one evening to 8:00 AM the next morning is 15 hours. We can assume that while the shop is closed the generator next mo therefore the equilibrium concentration will also be zero. The concentration after 15 hours is given by. C(t = 15 hrs) = C_o e^-t/t = 100ppm Times e^-15 hr/1.3 hr = 9.7 times 10^-4 ppm = 0.97 ppb A room measures 50' Times 30' Times 20'. The dilution volume flow rate is 3,000 CFM. At 8:00 am the concentration of a contaminant is 50 ppm. Assuming no additional contaminant is added to the workplace air, what will be the concentration of the contaminant at 8:00 pm?

Explanation / Answer

fluid residence time(characteristic time) = room volume/dilution volume flow rate   

= 30000 feet3/3000 feet3/min=10 min=0.1 hrs

using the above equation= C(T=12 hrs)=C0 exp(-12/0.1)=50exp(-120)=3.88 * 10^(-51) very very small

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