A corporate boardroom contains 4500ft^3 of air initially free of carbon monoxide

Question

A corporate boardroom contains 4500ft^3 of air initially free of carbon monoxide. Each morning (time t=0) the executives arrive and cigarette smoke containing 4% carbon monoxide by volume is introduced into the room at a rate of .3ft^3/min by the smokers. A ceiling fan keeps the air well circulated, and the air and smoke mixture leaves the room at a rate of 10ft^3 /min. After exiting the room the carbon monoxide is removed from the air and smoke mixture via a catalytic converter system before being returned as clean air to the room at the same rate by the building's air system.

a.) explain in detail how to find a linear differential equation for the number of cubic feet, A(t), of carbon monoxide in the room at t minutes. Using the initial condition A(0)=0, solve this equation by hand to determine a formula for A(t).

d.) It is known that a CO concentration of .16% is lethal. Can such life-threatening concentrations appear in the conference room? It is also known that exposure to concentrations above .02% are associated with loss of judgement on the part of the subjects. Since the occupants of the room are senior executives of the company, do you have any recommendations for savvy investors holding stock in the company?


Any help with this would be much appreciated. Thanks!

Explanation / Answer

dA/dt = Input - Output
Input = .3*.04 = .012 ft^3/min
Output = A/4500*10 ft^3/min

dA/dt = .012 - A/450
dA/(.012-A/450) = dt
450dA/(5.4-A) = dt
450ln(5.4-A) = t+C
ln(5.4-A) = (t+C)/450
t is neagive because CO is increasing
5.4-A = Ce^(-t/450)
When t=0, A=0
So 5.4 = C
5.4-A = 5.4e^(-t/450)
A = 5.4 - 5.4e^(-t/450)

As t increases, say after one hour the concentration will be A/4500*100% = [5.4-5.4e^(-1/450)]/4500*100% = .000266%
After say a 10 hour day, A/4500*100% = [5.4-5.4e^(-10/450)]/4500*100% = .00264%
To get to .02% would take about 70 hours....not a likely scenario.

I've worked the numbers, what do you think about the recommendation? Increase ventilation?
To tell you the truth, I would be more concerned about the buildup of CO2 with a ventilation rate of only 10ft^3/min. Exhaled air has a CO2 concentration of around 4% of all exhaled air, a far greater input volume than the amount of CO and discomfort is experienced at room levels of 0.5% CO2.

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