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{\bf Question}
Suppose that an apartment room contains 40$m^3$ of air and that it
is initially free of carbon monoxide. At midnight a smoker enters
the room and lights a cigarette that produces carbon monoxide at a
rate of $1.2\times 10^{-6} m^3/sec$. A window is open so that
fresh air enters the room at a rate of $3\times10^{-3}m^3/sec$. In
the room the air and pollutants are quickly mixed by a fan and
mixed air then exits through another window at the same rate as it
entered. Determine the concentration of carbon monoxide in the
room as a function of time. Extended exposure to carbon monoxide
at concentrations as low as 0.00012 is harmful to the human body.
At what time is this dangerous concentration reached? If a second
smoker enters the room 15 minutes after the first smoker how long
does it take to reach the danger level?
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{\bf Answer}
Equation for volume of air in room $V(t)$ is:\\ rate of change of
air in room = air flow in - air flow out \\ $dV/dt = 0$\\ Hence
volume of air in room is constant = 40
Equation for the Carbon dioxide in the room (where M(t) = m$^3$ of
carbon dioxide in the room) is:\\ Rate of change of M(t) = rate in
- rate out\\ $\ds \frac{dM}{dt}=1.2\time10^{-6} - 3\times10^{-3}
\frac{M}{40}$\\ where $\ds \frac{M}{40}$ is the concentration of
carbon dioxide in the room. \\ Initial condition is $M(0)=0$\\ The
equation is linear so use integrating factor to give $$
\frac{d}{dt}\left(e^{7.5\times10^{-5} t} M \right) =
1.2\times10^{-6}e^{7.5\times10^{-5} t} $$ Using the initial
conditions then gives\ $\ds
M=1.6\times10^{-2}\left(1-e^{-7.5\times10^{-5} t} \right)$\\ The
concentration in the room becomes 0.00012 when $\ds
0.00012=\frac{M}{40}=4.0\time10^{-4}\left(1-e^{-7.5\times10^{-5}
t} \right)$\\ which is approximately 1.3hrs.
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If the second smoker enters the room after 15minutes then\\ at
t=15mins $\ds M=1.6\times10^{-2}\left(1-e^{-7.5\times10^{-5}
(15\times 60)} \right)$\\ (recall all measurements are in seconds)
After this time the equation for the carbon dioxide must include
the second smoker so that: $\ds \frac{dM}{dt}=2(1.2\time10^{-6}) -
3\times10^{-3} \frac{M}{40}$\\ which has a solution $\ds
M=3.2\times10^{-2}\left(1-Be^{-7.5\times10^{-5} t} \right)$\\
Again consider when $\ds \frac{M}{40}=0.00012$ to give
$t\approx50$minutes.
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