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

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?

• 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. Jul 15 '20 at 19:41
• Related: covid-19 from the perspective of fluid mechanics: doi.org/10.1017/jfm.2020.330 Jul 15 '20 at 19:50
• @BarocliniCplusplus That make sense. I forgot that the PPM refers to molecules. Jul 15 '20 at 21:12

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.

• 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. Jul 15 '20 at 21:43
• 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. Jul 15 '20 at 21:47
• 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. Jul 16 '20 at 3:42
• 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. Jul 16 '20 at 15:20
• Delightful answer that sets out assumptions, reasoning, qualifications and uncertainties. Exemplary. Aug 25 '20 at 10:29