Modelling gas concentration in a room with exhaust fan Asumme a ceramic kiln is placed inside a room of volume $V \ m^3$. There is an exhaust fan installed in the room to extract all the kiln fumes to the outside of the room (and building) with a constant flow of $Q_o \ m^3/h$. As the kiln rises in temperature, clay releases gas compounds. For a mass $m$ of clay we have a certain Loss On Ignition of $L$. So the total gas released from the firing is $m · L\  kg$. We also know the total firing time $T$ during which gas is being released. We can assume that the gas inflow is constant in time: $I_f = \frac{m · L}{T} \ kg/h$.
I want to approximate how much clay can be introduced in the kiln so that the gas concentration in the room doesn't exceed a certain amount.
My first thoughts were to model the following differential equations to represent $y(t)$ as the gas mass in the room at a time t in $kg$:
$$y(0) = 0$$
$$y'(t) = I_f - \frac{Q_o}{V} y(t)$$
With solution:
$$y(t) = \frac{I_f}{Q_o} V \left(1-e^{-\frac{Q_o}{V}t}\right)$$
However, I feel this equation is assuming the "outflow" isn't constant with time and it depends on the concentration at any given time $y(t)/V$, and that's why we obtain an exponential decay.
I'm missing something else, for example, if the fan flow is $Q_o \ m^3/h$, does it mean it would take $\frac{V}{Q_o}$ hours to evacuate all the gas? I'm not a physicist, so any suggestion would be appreciated.
 A: Given the assumptions you have listed, your solution is correct. The differential equation implicitly also assumes that the time for the released gas compounds to completely mix with the rest of the air in the room is negligible.
If you plot the solution, you will see the typical behaviour of some saturation process.


I feel this equation is assuming the "outflow" isn't constant with time and it depends on the concentration at any given time $y(t)$, and that's why we obtain an exponential decay.

This is exactly right. If you start your exhaust fan without any contaminant in the room, it cannot possibly extract any. When you start putting a little contaminant in it, it can only extract a little. So by increasing the concentration of the pollutant, you also increase the ability of the fan to extract it, simply because there is more of it contained in the volume of air that is removed.
This game continues until the ability of the exhaust fan to expel the contaminant matches the rate at which the contaminant is released into the room. Then the concentration just stays constant.
Of course, once you turn off the kiln, this balance is no longer maintained, and the concentration will decrease exponentially for the same reasons given above. You can write down an equivalent ODE with an appropriate initial condition.
Taking the expression of the time-dependent mass $y(t)$, you can immediately see that the highest mass in equilibrium is
$$m_\mathrm{E}= \frac{I_f V}{Q_0} [\mathrm{kg}]$$
which gives a maximum concentration of
$$c_\mathrm{E}= \frac{I_f}{Q_0} \left[\frac{\mathrm{kg}}{\mathrm{m^3}}\right]$$

If we want to respect the fact that the saturation concentration may not be reached during the time $T$, we can explicitly calculate the concentration at its highest point, which is at $t=T$.
$$c_\mathrm{max} = \frac{I_f}{Q_0}\left(1 - e^{-\frac{Q_0 T}{V}}\right) $$
A: Well, typically exhaust fans are not rated to be vaccum pumps, so I think it is reasonable to say that while the fan removes some volume of air at the rate $Q_0$, an equal volume of air keeps leaking in from the outside. 
This on the surface makes no difference, but presuming that you don't have an air pollution crisis on at the minute, the outer air is fresh, so has none of the harmful gases the air inside had.
BTW let us abbreviate harmful gasses as Hg to save space. Mercury is toxic anyway :).
So the air is removed and replaced at the rate $Q_0$, and that removes harmful gases from the room. But how much? Well we may take the air to evenly mixed, so
$$
  \begin{aligned}
    \text{Concentration of Hg in the room} =& \text{Concentration of Hg in the air the fan removed} \\
    =& \frac{\text{Amount of Hg removed}}{\text{Amount of air removed}} \\ 
    \implies \text{Amount of Hg removed} =& \text{Amount of Air removed} \cdot \text{ Concentration of Hg in the room} \\
    =& \text{Amount of Hg in the room} \cdot \text{Fraction of Air removed}  \\
    =& y \left( t\right) \cdot \frac{\text{Amount of air removed}}{\text{Amount of air in the room}} = y \left( t\right) \cdot \frac{Q_0 t}{V} \\
    \implies \text{Rate of removal of Hg} =& y \left( t\right) \cdot \frac{Q_0}{V}
  \end{aligned}
$$
As we can see, while the outflow of air is constant, but the outflow of Hg isn't, since we cannot selectively remove Hg only, but the also the air it is in. As the concentration of Hg increases, the amount we can throw out with the air also increases.
This mean that your upwards-exponential-decay is completely correct, as long as we don't hit time $T$ when the production stops. After that it should start exponetially decaying (downwards) at the same rate from whatever concentration it had at time $T$.
A: As mentioned in the other answers, the exponential solution assumes that the gas produced from the kiln mixes completely and instantaneously with the air in the room. If the kiln gases are hot, we might expect them to rise towards to the ceiling due to buoyancy rather than mix. The layer of buoyant air at the ceiling would contain all the kiln gases.
Assuming the ventilation is exhausting gases from the buoyant layer, and the ventilation inflow is at low level and not causing mixing between the buoyant layer and the rest of the air in the room, then all the kiln gases will be removed from the room in a finite time, rather than an exponential decay.
Edit:
To estimate the concentration of the kiln gas at steady state we need to estimate the height of the layer of buoyant air at the top of the room. The kiln will give rise to a plume of hot air rising off it and it is that plume that feeds the buoyant layer. Conservation of volume for the layer gives $Q_0=Q_p$, the flow rate from the plume must match the ventilation rate.
Treating the kiln as a point source of buoyancy of strength $B$ (a bit of a stretch but fine for a first go), the flow rate in the plume at the base of the layer is
$$
Q_p = \frac{6\alpha}{5}\left(\frac{9\alpha B}{10}\right)^{1/3}(H-h)^{5/3},
$$
where $H$ is the height of the room, $h$ is the height of the layer and $\alpha\approx0.1$ is the plume entrainment coefficient (taken from Morton, Taylor, Turner 1956).
Matching the flow rates and rearranging, we find the depth of the buoyant layer to be
$$
h=H-\left(\frac{5Q_0}{6\alpha}\right)^{3/5}\left(\frac{10}{9\alpha B}\right)^{1/5}.
$$
Defining the room floor area as $S$, the steady state gas concentration is therefore
$$
y_\infty = \frac{I_f S}{Q_0}\left(H-\left(\frac{5Q_0}{6\alpha}\right)^{3/5}\left(\frac{10}{9\alpha B}\right)^{1/5}\right).
$$
We see that the steady state mass is reduced if the kiln gas is confined to a layer rather than mixed throughout the room.
