WEBVTT 00:00:11.040 --> 00:00:13.080 PROFESSOR: So now, let me make the first 00:00:13.080 --> 00:00:17.040 of our technical asides, which you can skip over if you're not 00:00:17.040 --> 00:00:18.630 interested in the mathematical details 00:00:18.630 --> 00:00:21.570 or for those of you that have a higher-level, 00:00:21.570 --> 00:00:25.500 say, upper-level-undergraduate or even graduate-level 00:00:25.500 --> 00:00:29.160 understanding of transport phenomena and fluid mechanics. 00:00:29.160 --> 00:00:31.350 I'd like to show you some of the equations that 00:00:31.350 --> 00:00:33.000 are behind the results that I've been 00:00:33.000 --> 00:00:35.430 quoting in all the lectures. 00:00:35.430 --> 00:00:38.830 So in particular, let's derive the Wells curve. 00:00:38.830 --> 00:00:45.280 So part of that was a theory of drop settling. 00:00:45.280 --> 00:00:47.950 Here, I will quote a certain result, 00:00:47.950 --> 00:00:50.620 because the derivation would be a lot longer. 00:00:50.620 --> 00:00:55.240 Actually, for that, you could refer to my online class 10.50x, 00:00:55.240 --> 00:00:59.320 which is that if you have a droplet or a particle 00:00:59.320 --> 00:01:06.710 of a radius R and it is settling under gravity-- 00:01:06.710 --> 00:01:10.870 so it has a mass m, and the gravitational force is m g, 00:01:10.870 --> 00:01:13.600 where g is the gravitational acceleration-- 00:01:13.600 --> 00:01:20.310 then there is a flow of fluid around this object. 00:01:20.310 --> 00:01:21.730 And relative to the moving object, 00:01:21.730 --> 00:01:24.280 the flow is going the other way. 00:01:24.280 --> 00:01:26.620 And if you solve for the viscous flow 00:01:26.620 --> 00:01:29.800 around an object being dragged through a fluid, 00:01:29.800 --> 00:01:33.400 then you arrive at the result of Stokes, which 00:01:33.400 --> 00:01:35.390 is the drag on that fluid. 00:01:35.390 --> 00:01:38.200 So if you're falling at a velocity v, 00:01:38.200 --> 00:01:48.259 then the drag force is -- f_d is -- 6*pi 00:01:48.259 --> 00:01:51.590 times the radius of the drop times the viscosity 00:01:51.590 --> 00:01:55.410 of the fluid times the velocity of the drop. 00:01:55.410 --> 00:01:57.140 So we're falling at a velocity v, 00:01:57.140 --> 00:01:59.720 which is the settling velocity. 00:01:59.720 --> 00:02:03.240 And this here is the Stokes drag coefficient, 00:02:03.240 --> 00:02:05.990 which comes from solving the fluid mechanics of viscous flow 00:02:05.990 --> 00:02:10.330 around a sphere translating at a constant speed. 00:02:10.330 --> 00:02:18.500 We can, furthermore, say that the mass of the droplet, of course, 00:02:18.500 --> 00:02:25.990 is 4*pi/3 times the density of the droplet liquid, times 00:02:25.990 --> 00:02:28.060 the radius cubed. 00:02:28.060 --> 00:02:30.180 And so given the mass of the droplet, 00:02:30.180 --> 00:02:33.340 there's a force balance between the gravitational force 00:02:33.340 --> 00:02:37.740 m g and the drag force when the particle reaches a terminal 00:02:37.740 --> 00:02:38.410 velocity. 00:02:38.410 --> 00:02:40.800 So if we think of this v_s as the terminal velocity where 00:02:40.800 --> 00:02:43.510 it's accelerating until there's a balance between the forces 00:02:43.510 --> 00:02:47.040 and is moving at a speed, it's given by this force balance. 00:02:47.040 --> 00:02:48.660 And from that equation, we can solve 00:02:48.660 --> 00:02:54.900 for the settling velocity, which is energy divided by 6*pi*R. 00:02:54.900 --> 00:02:57.110 And to use the same notation as before, 00:02:57.110 --> 00:02:59.460 I'll call this mu_a -- the air -- but generally, it's 00:02:59.460 --> 00:03:02.550 the viscosity of the ambient fluid 00:03:02.550 --> 00:03:05.370 around the particle as it's settling. 00:03:05.370 --> 00:03:15.150 And if we plug in the value for m g, so that's 4/3*pi*rho*g* 00:03:15.150 --> 00:03:19.940 R^3 over 6*pi*R*rho_a. 00:03:19.940 --> 00:03:22.740 And so we simplify that, we end up 00:03:22.740 --> 00:03:32.310 with (2/9)*rho*g*r^2/mu_a. 00:03:32.310 --> 00:03:34.570 So that's the settling speed. 00:03:34.570 --> 00:03:36.190 And this is a pretty important concept, 00:03:36.190 --> 00:03:39.010 so I'll just sketch it here. 00:03:39.010 --> 00:03:40.740 So if we want to know what's the settling 00:03:40.740 --> 00:03:44.870 speed as a function of the radius of the drop, 00:03:44.870 --> 00:03:48.000 then you see it grows like R^2. 00:03:48.000 --> 00:03:48.890 So it's like this. 00:03:52.660 --> 00:03:57.060 And then to put a scale on that, if we have a particle that 00:03:57.060 --> 00:04:02.430 is 3 microns -- so that's really an aerosol particle -- 00:04:02.430 --> 00:04:06.730 then the settling speed is around 1 millimeter 00:04:06.730 --> 00:04:11.430 a second if we use the density of water 00:04:11.430 --> 00:04:15.150 and the viscosity of air for this formula. 00:04:15.150 --> 00:04:17.310 And so that's already a fairly slow settling speed, 00:04:17.310 --> 00:04:18.430 millimeter per second. 00:04:18.430 --> 00:04:19.620 So you can already see the particles that 00:04:19.620 --> 00:04:21.630 are in the micron range will be suspended 00:04:21.630 --> 00:04:23.190 in the air for a long period of time, 00:04:23.190 --> 00:04:25.800 as long as they don't evaporate away. 00:04:25.800 --> 00:04:29.020 And so that is now the second part of the calculation. 00:04:29.020 --> 00:04:31.620 Oh, and I should the finish the first part here. 00:04:31.620 --> 00:04:38.460 What we're left with is that the settling time is L over v_s. 00:04:38.460 --> 00:04:41.500 And that's the formula that we had before, 00:04:41.500 --> 00:04:50.010 which is 9*mu_a*L divided by 2*rho*g*r^2. 00:04:50.010 --> 00:04:53.670 So this is our first part of the Wells curve. 00:04:53.670 --> 00:04:58.930 So if I draw the Wells curve over here 00:04:58.930 --> 00:05:02.030 in the traditional way, where I plot on the horizontal 00:05:02.030 --> 00:05:05.540 axis the size of the particle, and in a downward axis, 00:05:05.540 --> 00:05:13.030 we draw the time, then we have a curve like this for settling. 00:05:13.030 --> 00:05:15.410 And the reason it's drawn down, I guess maybe the feeling 00:05:15.410 --> 00:05:18.190 that as you're sort of falling down a particle a certain size, 00:05:18.190 --> 00:05:20.690 you hit this curve and that's when you've settled a distance 00:05:20.690 --> 00:05:23.480 L and fallen out of the air. 00:05:23.480 --> 00:05:26.640 So now, let's look at evaporation, 00:05:26.640 --> 00:05:29.160 which is our second topic. 00:05:29.160 --> 00:05:30.460 I've lost my blue [marker] -- here it is. 00:05:37.780 --> 00:05:40.240 So these droplets are getting very small 00:05:40.240 --> 00:05:41.200 as they're evaporating, 00:05:41.200 --> 00:05:44.930 and it's happening very quickly, as we shall show in a moment. 00:05:44.930 --> 00:05:49.510 And so a natural assumption is that the process 00:05:49.510 --> 00:05:56.110 is limited by the diffusion of water vapor 00:05:56.110 --> 00:06:01.540 away from the droplet, because essentially we 00:06:01.540 --> 00:06:04.730 have this little droplet here with a certain size 00:06:04.730 --> 00:06:07.580 R, which is now going to be varying with time. 00:06:07.580 --> 00:06:09.850 So it has a radius R(t). 00:06:09.850 --> 00:06:13.660 And it's really close to the surface. 00:06:13.660 --> 00:06:15.700 There is an equilibrium concentration -- 00:06:15.700 --> 00:06:17.950 we'll call it c_w -- 00:06:17.950 --> 00:06:19.750 of water, which depends on the temperature. 00:06:19.750 --> 00:06:22.720 So that's kind of the saturation concentration 00:06:22.720 --> 00:06:26.020 of water vapor in the air. 00:06:26.020 --> 00:06:28.360 But then, if the water is going to evaporate more, 00:06:28.360 --> 00:06:30.760 it would create more concentration, which would then 00:06:30.760 --> 00:06:32.080 re-condense on the particle. 00:06:32.080 --> 00:06:35.800 So in order for it to continue evaporating, 00:06:35.800 --> 00:06:39.230 that water vapor that is produced has to diffuse away. 00:06:39.230 --> 00:06:43.210 So there's going to be a gradient of water vapor going 00:06:43.210 --> 00:06:50.130 outwards from c_w -- [it] is the concentration at position R. 00:06:50.130 --> 00:06:55.170 And then far away, there is a sort of diffusion layer 00:06:55.170 --> 00:06:57.810 thickness, delta. 00:06:57.810 --> 00:07:00.210 And far beyond the diffusion layer thickness, 00:07:00.210 --> 00:07:03.540 the concentration is going to approach the equilibrium 00:07:03.540 --> 00:07:05.640 concentration in the ambient air. 00:07:05.640 --> 00:07:11.150 c at infinity is going to be c_w times the relative humidity. 00:07:11.150 --> 00:07:13.200 So that's the ratio of the concentration of water 00:07:13.200 --> 00:07:16.590 vapor in the air to the saturation concentration, 00:07:16.590 --> 00:07:19.790 c_w, by definition. 00:07:19.790 --> 00:07:21.210 Now, the math problem that we have 00:07:21.210 --> 00:07:24.720 to solve for this diffusion problem -- with a moving boundary 00:07:24.720 --> 00:07:28.110 in this case, though we'll assume it's pseudosteady -- 00:07:28.110 --> 00:07:31.620 is dc/dt is the diffusion coefficient of water 00:07:31.620 --> 00:07:33.300 times the Laplacian of c, so just 00:07:33.300 --> 00:07:36.520 the diffusion equation with these two boundary conditions. 00:07:36.520 --> 00:07:39.600 Now, an interesting aspect a three-dimensional spherical 00:07:39.600 --> 00:07:42.030 diffusion is that at first the diffusion layer grows, 00:07:42.030 --> 00:07:44.620 but it very quickly reaches a steady state. 00:07:44.620 --> 00:07:47.220 And if we assume that that diffusion time 00:07:47.220 --> 00:07:51.100 to reach this distance delta is fast, 00:07:51.100 --> 00:07:52.650 so they reach a steady state, so it's 00:07:52.650 --> 00:07:55.750 a kind of quasisteady or pseudosteady shrinking 00:07:55.750 --> 00:07:58.050 of the droplet with sort of a diffusion layer around it 00:07:58.050 --> 00:08:01.510 that's always kind of at the steady value, 00:08:01.510 --> 00:08:03.900 then it turns out that this diffusion layer 00:08:03.900 --> 00:08:05.400 is on the order of the particle size. 00:08:05.400 --> 00:08:07.950 So as the particle shrinks, the diffusion layer also shrinks. 00:08:07.950 --> 00:08:09.840 But it has a well-defined thickness, 00:08:09.840 --> 00:08:11.480 as opposed to diffusion in one or two 00:08:11.480 --> 00:08:13.230 dimensions, where the diffusion layer just 00:08:13.230 --> 00:08:14.680 keeps growing out to infinity. 00:08:14.680 --> 00:08:16.130 For example, like square root of time -- 00:08:16.130 --> 00:08:19.050 you don't reach a steady state in an infinite domain. 00:08:19.050 --> 00:08:21.310 So the bottom line of this calculation, 00:08:21.310 --> 00:08:23.280 which I will not go through right now, 00:08:23.280 --> 00:08:27.010 is that the flux of water on the surface 00:08:27.010 --> 00:08:31.330 is the area of the surface at a given moment, where the size is 00:08:31.330 --> 00:08:37.600 R, times essentially Fick's law, where 00:08:37.600 --> 00:08:40.120 the driving force, the change in concentration 00:08:40.120 --> 00:08:43.390 from the surface to the bulk, is c_w times 00:08:43.390 --> 00:08:46.900 one minus relative humidity, the diffusivity of water, 00:08:46.900 --> 00:08:49.800 and then divided by delta, the diffusion layer thickness. 00:08:49.800 --> 00:08:52.320 And it turns out that with these coefficients here, 00:08:52.320 --> 00:08:53.750 it turns out to be exactly R. 00:08:53.750 --> 00:08:55.630 So this is not a scaling result, but actually 00:08:55.630 --> 00:08:59.470 an exact result for pseudo steady spherical diffusion 00:08:59.470 --> 00:09:01.400 of water vapor. 00:09:01.400 --> 00:09:03.070 So now, we have the flux on the surface. 00:09:03.070 --> 00:09:04.540 It's uniform on the surface. 00:09:04.540 --> 00:09:07.360 And it's [assumed] to be pseudosteady. 00:09:07.360 --> 00:09:10.060 And so then I can write down that the change 00:09:10.060 --> 00:09:13.120 in the size of the water droplet volume, which 00:09:13.120 --> 00:09:20.170 is (4*pi/3)*R^3 is equal to minus 00:09:20.170 --> 00:09:25.820 the volume of a water molecule times the flux of water. 00:09:25.820 --> 00:09:31.520 So that's basically my volume or mass balance of water. 00:09:31.520 --> 00:09:38.690 So if I plug this in here, then I get dR/dt is equal to -- 00:09:38.690 --> 00:09:40.750 let's see, collecting all the terms here -- 00:09:40.750 --> 00:09:44.590 so derivative of R^3 is 3*R^2. 00:09:44.590 --> 00:09:46.490 So the 3's cancel. 00:09:46.490 --> 00:09:48.380 And then I have a 4*pi*R^2, 00:09:48.380 --> 00:09:50.370 which cancels this 4*pi*R^2. 00:09:50.370 --> 00:10:00.930 So I just have dR/dt is -v_w*D_w*c_w*(1-RH)/R. 00:10:00.930 --> 00:10:06.200 If I put this R on the other side here, then I have -- 00:10:06.200 --> 00:10:10.510 I'll just continue the derivation here -- 00:10:10.510 --> 00:10:17.320 I have R dR/dt is equal to all this stuff. 00:10:17.320 --> 00:10:25.010 So -v_w*D_w*c_w*(1-RH). 00:10:25.010 --> 00:10:26.960 And then this expression here can 00:10:26.960 --> 00:10:34.800 be written as 1/2 the derivative of R^2. 00:10:34.800 --> 00:10:39.890 So what we find is that R^2 is linear in time. 00:10:39.890 --> 00:10:44.000 And then using the boundary condition that we start out 00:10:44.000 --> 00:10:47.840 at a certain initial value, R_0, then I'm 00:10:47.840 --> 00:10:54.910 going to get the R(t) is the initial value R_0 times 00:10:54.910 --> 00:10:59.560 the square root of 1 minus t over a certain evaporation 00:10:59.560 --> 00:11:01.240 time. 00:11:01.240 --> 00:11:07.160 And that evaporation time is given here 00:11:07.160 --> 00:11:17.500 by (R_0)^2 times basically all these coefficients here, 00:11:17.500 --> 00:11:21.340 where I'll separate out the effective humidity, 00:11:21.340 --> 00:11:23.720 and then a bunch of other coefficients, which you can see 00:11:23.720 --> 00:11:26.630 have units of length squared over time, 00:11:26.630 --> 00:11:28.180 because R_0 is a length squared. 00:11:28.180 --> 00:11:30.920 So it's effectively some kind of diffusivity. 00:11:30.920 --> 00:11:33.080 And what we get from this calculation 00:11:33.080 --> 00:11:34.910 is that this effective diffusivity 00:11:34.910 --> 00:11:37.700 that goes into this expression is -- there's 00:11:37.700 --> 00:11:39.050 a factor 2 from this guy -- 00:11:39.050 --> 00:11:44.380 there's a 2*v_w*D_w*c_w. 00:11:44.380 --> 00:11:47.240 And if you plug in values for water vapor, 00:11:47.240 --> 00:11:49.670 for the saturation pressure, and the diffusivity, 00:11:49.670 --> 00:11:53.030 and the volume of water in air, then this coefficient 00:11:53.030 --> 00:11:58.190 turns out to be 1.2e-9 meters squared 00:11:58.190 --> 00:12:02.200 for second for pure water. 00:12:02.200 --> 00:12:05.420 And that's where you get now the second part of the Wells 00:12:05.420 --> 00:12:08.000 theory, which is the evaporation, which gives you 00:12:08.000 --> 00:12:10.230 a curve looking like this. 00:12:15.590 --> 00:12:18.270 So that there's this sort of in this theory 00:12:18.270 --> 00:12:25.820 a natural crossover between large drops, which in this case 00:12:25.820 --> 00:12:28.070 here are ones that are large enough to settle out 00:12:28.070 --> 00:12:30.260 of the fluid before they evaporate, 00:12:30.260 --> 00:12:38.400 and small drops, which evaporate. 00:12:38.400 --> 00:12:41.070 On the other hand, for true biological fluids 00:12:41.070 --> 00:12:44.420 that appear in respiratory droplets, 00:12:44.420 --> 00:12:49.170 the evaporation is limited by solutes and salts, which 00:12:49.170 --> 00:12:51.240 stop the evaporation and, in fact, 00:12:51.240 --> 00:12:53.740 can attract even more water in some cases. 00:12:53.740 --> 00:12:57.060 So that the evaporation part of it is not as accurate, 00:12:57.060 --> 00:13:00.030 and we tend to see that the settling part is more 00:13:00.030 --> 00:13:03.510 important to consider, given an equilibrium distribution 00:13:03.510 --> 00:13:05.100 of droplets that has been measured 00:13:05.100 --> 00:13:10.520 and is understood to come from different types of respiration.