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How far downwind of a covid-19 emitter has a 1000-fold reduction in PPM?

Assuming that covid-19 (0.12μm diameter) particles have the aerodynamic properties of a gas:

Particles < 20 μm behave same as gas – Low settling velocity http://home.engineering.iastate.edu/~leeuwen/CE%20524/Presentations/Dispersion_Handout.pdf

How far would someone have to be downwind of a covid-19 particle emitter for an at least 1000-fold reduction in PPM?

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    $\begingroup$ One thing I want to note is a 'unit miscommunication': Viruses aren't measured in PPM (parts per million or molecules per million molecules). PPM is reserved for gases. Aerosols are usually expressed in units of mass per unit volume. Viruses, I can't exactly speak about. $\endgroup$ – BarocliniCplusplus Jul 15 at 19:41
  • $\begingroup$ Related: covid-19 from the perspective of fluid mechanics: doi.org/10.1017/jfm.2020.330 $\endgroup$ – Jean-Marie Prival Jul 15 at 19:50
  • $\begingroup$ @BarocliniCplusplus That make sense. I forgot that the PPM refers to molecules. $\endgroup$ – polcott Jul 15 at 21:12
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As I said in my comment, viruses aren't my area (concerning units). But the answer is actually unitless. The exact answer you seek is dependent on a couple of different variables:

  • The speed of the exhalation (cough or breathing)+speed of wind
  • The atmospheric stability

Of course, the viral load of the exhalation is also important, but since you're asking for a ratio, I won't list that.

The Gaussian Plume model can be used to describe pollutants. The Gaussian Plume model is: $$\frac{C}{Q}=\frac{1}{2\pi U \sigma_y \sigma_z}\exp\left(-\frac{1}{2}\left[\frac{y}{\sigma_y}\right]^2-\frac{1}{2}\left[\frac{z-H}{\sigma_z}\right]^2\right)$$,

where $C$ is the concentration, $Q$ is the emission rate, $U$ is the wind speed (or worst case-scenario is cough speed), $x$ is the downwind distance, $z$ is the distance from the ground, and $\sigma_x, \sigma_z$ are the horizontal and vertical diffusion constants.

$\sigma_y, \sigma_z$ are functions of stability and wind speed. The actual formulas are... iffy. Theoretically they can be described as $$\sigma_{y,z}=\sqrt{\frac{2K_{y,z}x}{U}}$$, where $K$ is the eddy viscosity. There are empirical formulas based on stability classifications too.

Lets assume that the receiver is directly downwind and at the same height (z=H and y=0). Therefore the equation collapses to: $$\frac{C}{Q}=\frac{1}{2\pi U \sigma_y \sigma_z}$$. Since the distance downwind is expressed only by $\sigma_{y,z}$, then to answer your question, we need to isolate $\sigma_{y,z}$. $$\sigma_y \sigma_z=\frac{1000}{2\pi U}$$

Let's just say, for the sake of argument, consider two different scenarios:

  1. Empirical $\sigma_{y,z}$ (refer to this for values of constants)

$$(ax^b)\times (cx^d+f)=acx^{b+d}+afx^b=\frac{1000}{2\pi U}$$

And this is complicated to solve. So I'll leave solving for $x$ as an exercise to the reader.

  1. Constant $K_{y,z}$

$$\sqrt{\frac{2K_{z}x}{U}}\sqrt{\frac{2K_{y}x}{U}}=\frac{1000}{2\pi U}$$ $$\frac{4K_{z}K_{y}x^2}{U^2}=\left(\frac{1000}{2\pi U}\right)^2=\frac{10^6}{4\pi^2U^2}$$ $$x=\sqrt{\frac{10^6 K_{z}K_{y}}{16\pi^2}}$$ $$x=\frac{10^3}{4\pi}\sqrt{K_{z}K_{y}}$$

If we assume that $K_z=K_y=K$, then the equation simplifies further. $$x=\frac{K10^3}{4\pi}$$ There are many ways on how to calculate $K$. One simple formula for $K$ can be used if mixing length theory is invoked

$$K=\frac{ku_*z}{\phi(\frac{z}{L})}$$, where $k$ is the von-Karman constant, $u_*$ is the friction velocity, $\phi_M(\frac{z}{L})$ is the Businger-Dyer function for momentum, and $L$ is the Monin-Obukhov length. If we assume neutral atmospheric stability,$\phi_M(\frac{z}{L})=1$ and the resultant equation can be written as:

$$x=\frac{ku_*z10^3}{4\pi}$$

More assumptions! Let's assume z=1.5 m and $u_*=1.11$ m s$^{-1}$ (corresponds to a 5 kt wind speed at 10 m with a roughness length of 0.1 m under neutral stability). Plugging those numbers in gets

$x=35$ m

Of course, there were a LOT of assumptions to get to this point. But this is what I have to offer.

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  • $\begingroup$ Side note: In way 2, I just assumed that OP is requesting $\frac{C}{Q}=\frac{1}{1000}$ even though they aren't even the same unit (emission rate vs concentration). But it reduces the dimensionality of the problem. I couldn't get to that point in way 1. I also invoked the Log-wind profile in way 2. $\endgroup$ – BarocliniCplusplus Jul 15 at 21:43
  • $\begingroup$ That is very helpful. I would assume that the most turbulent flow would travel the least distance and the most laminar flow would travel the most distance. What I am trying to find out is a safe distribution of large crowds outdoors, erring on the safe side. $\endgroup$ – polcott Jul 15 at 21:47
  • $\begingroup$ Before I mark that as an accepted answer I want to know your best estimate of ± error variance. It looks like a really great answer. It turns out to be exactly the same as my 100 feet wild guess. $\endgroup$ – polcott Jul 16 at 3:42
  • $\begingroup$ Since it's an equation, and the estimated value of 35 m just used typical values, an error variance isn't really a thing. But if you plug in some more extreme values, I get $(0.6, 318)$ m. When I tried the empirical approach (instead of solving for x, I just graphed the solution) I found answers ranging from 7 m to 199 m, depending on the stability classification. Certainly those have errors too, but I am not entirely sure the uncertainty of the model itself. $\endgroup$ – BarocliniCplusplus Jul 16 at 15:20
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    $\begingroup$ Delightful answer that sets out assumptions, reasoning, qualifications and uncertainties. Exemplary. $\endgroup$ – Anton Aug 25 at 10:29

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