# Pressure change due to fan removing air from a non-airtight room

The following problem occurred to me today:

Suppose a $100\mathrm{cfm}$ fan is pushing air out of a large room which is airtight except for a $10 \mathrm{cm}^2$ hole. The air pressure outside the room is $101.3\mathrm{kPa}$, and the atmosphere consists of $80\%$ nitrogen and $20\%$ oxygen. All temperatures are $300\mathrm{K}$. What is the equilibrium pressure inside the room?

I believe I have given enough information to determine the resulting pressure. My inclination is to take a naïve statistical mechanics approach to the problem. This leads me down the following train of thought:

• Particles are distributed evenly inside and outside the room, with a cubic meter containing $N_i=\frac{P_i}{k_BT}$ and $N_o=\frac{P_o}{k_BT}$ molecules respectively, where $P_i$ is the pressure inside and $P_o$ the pressure outside.
• Velocities are distributed symmetrically and speeds satisfy a Boltzmann distribution, so for a fixed type of molecule $M$ the distribution of $v_z$, the $z$-coordinate of velocity squared, is $$D_M(v_z)=\int_{S^2}\mathrm{exp}\left(\frac{-m_Mv_z^2}{2k_BT\cos\theta^2}\right)\mathrm{d}\theta\mathrm{d}\phi$$ where $m_M$ is the mass of the molecule $M$.
• The number of molecules of $M$ that strike an area $A$ over time $t$ from inside is $$\int_0^\infty Ax_MN_i\int_{d/t}^{\infty} D_M(v_z)\mathrm{d}v_z\mathrm{d}d$$ where $d$ denotes the distance along the $z$-axis of the molecule from the area and $x_M$ denotes the fraction of all molecules which are $M$. Similarly, the number of molecules of $M$ that strike an area $A$ over time $t$ from outside is $$\int_0^\infty Ax_MN_o\int_{d/t}^{\infty} D_M(v_z)\mathrm{d}v_z\mathrm{d}d.$$
• The difference between these two, summed for oxygen and nitrogen, must be equal to the number of molecules of $M$ removed by the fan, which is $.0472\cdot N_it$.

Thus we have $$.0472\cdot N_it=A(N_i-N_o)\int_0^\infty \int_{d/t}^{\infty} \left(x_{N^2}D_{N^2}+x_{O^2} D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d$$ so solving for $N_i$ we get \begin{align} N_i &=\frac{AN_o\int_0^\infty \int_{d/t}^{\infty} \left(x_{N^2}D_{N^2}+x_{O^2} D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d}{A\int_0^\infty \int_{d/t}^{\infty} \left(x_{N^2}D_{N^2}+x_{O^2} D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d-.0472t}\\ &=\frac{.1\mathrm{m}^2\cdot\frac{101.3\mathrm{kPa}\cdot \mathrm{m}^3}{1.38e-23 \mathrm{J}/\mathrm{K}\cdot 300\mathrm{K}}\int_0^\infty \int_{d/1\mathrm{s}}^{\infty} \left(.8D_{N^2}+.2 D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d}{.1\mathrm{m}^2\int_0^\infty \int_{d/1\mathrm{s}}^{\infty} \left(.8D_{N^2}+.2 D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d-.0472\mathrm{s}}\\ &=\frac{2.45e24 \mathrm{m}^{2}\int_0^\infty \int_{d/1\mathrm{s}}^{\infty} \left(.8D_{N^2}+.2 D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d}{.1\mathrm{m}^2\int_0^\infty \int_{d/1\mathrm{s}}^{\infty} \left(.8D_{N^2}+.2 D_{O^2}(v_z)\right)\mathrm{d}v_z\mathrm{d}d-.0472\mathrm{s}}\\ \end{align} which I could probably evaluate using Mathematica if my desktop were working, but sadly it is not. From there it would be trivial to calculate $P_i$.

My question is whether my reasoning up to this point is correct, and whether there are any factors I have left out. Additionally, is there an easier way to calculate $P_i$?

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Wow, I'm just a chemical engineer but statistical mechanics certainly isn't how I would have solved this problem. Where exactly is the fan located? If it is freestanding in the large room then the fan will just recycle the air in circles and the pressure in the room will equilibrate with the outside pressure. If the fan is located in the hole then the equilibrium pressure in the room will depend on the head curve of the fan and will be outside pressure minus the maximum head developed by the fan at zero flow. –  Jason Waldrop Aug 29 '12 at 17:06
@JasonWaldrop I guess I was unclear. The fan is blowing air out of another hole, at a rate of 100cfm at 101.3kPa (and the inside pressure should be close enough to this for the difference not to matter). –  Alex Becker Aug 29 '12 at 19:21