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PROFESSOR: So now,
let me make the first

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of our technical asides, which
you can skip over if you're not

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interested in the
mathematical details

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or for those of you that
have a higher-level,

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say, upper-level-undergraduate
or even graduate-level

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understanding of transport
phenomena and fluid mechanics.

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I'd like to show you some
of the equations that

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are behind the
results that I've been

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quoting in all the lectures.

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So in particular, let's
derive the Wells curve.

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So part of that was a
theory of drop settling.

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Here, I will quote
a certain result,

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because the derivation
would be a lot longer.

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Actually, for that, you could
refer to my online class 10.50x,

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which is that if you have
a droplet or a particle

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of a radius R and it is
settling under gravity--

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so it has a mass m, and the
gravitational force is m g,

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where g is the
gravitational acceleration--

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then there is a flow of
fluid around this object.

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And relative to
the moving object,

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the flow is going the other way.

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And if you solve
for the viscous flow

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around an object being
dragged through a fluid,

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then you arrive at the
result of Stokes, which

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is the drag on that fluid.

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So if you're falling
at a velocity v,

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then the drag
force is -- f_d is -- 6*pi

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times the radius of the
drop times the viscosity

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of the fluid times the
velocity of the drop.

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So we're falling
at a velocity v,

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which is the settling velocity.

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And this here is the
Stokes drag coefficient,

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which comes from solving the
fluid mechanics of viscous flow

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around a sphere translating
at a constant speed.

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We can, furthermore, say that the
mass of the droplet, of course,

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is 4*pi/3 times the density
of the droplet liquid, times

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the radius cubed.

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And so given the
mass of the droplet,

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there's a force balance
between the gravitational force

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m g and the drag force when the
particle reaches a terminal

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velocity.

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So if we think of this v_s as
the terminal velocity where

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it's accelerating until there's
a balance between the forces

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and is moving at a speed, it's
given by this force balance.

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And from that
equation, we can solve

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for the settling velocity, which
is energy divided by 6*pi*R.

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And to use the same
notation as before,

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I'll call this mu_a -- the
air -- but generally, it's

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the viscosity of
the ambient fluid

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around the particle
as it's settling.

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And if we plug in the value
for m g, so that's 4/3*pi*rho*g*

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R^3 over 6*pi*R*rho_a.

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And so we simplify
that, we end up

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with (2/9)*rho*g*r^2/mu_a.

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So that's the settling speed.

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And this is a pretty
important concept,

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so I'll just sketch it here.

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So if we want to know
what's the settling

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speed as a function of
the radius of the drop,

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then you see it
grows like R^2.

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So it's like this.

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And then to put a scale on
that, if we have a particle that

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is 3 microns -- so that's
really an aerosol particle --

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then the settling speed
is around 1 millimeter

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a second if we use
the density of water

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and the viscosity of
air for this formula.

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And so that's already a
fairly slow settling speed,

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millimeter per second.

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So you can already
see the particles that

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are in the micron
range will be suspended

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in the air for a
long period of time,

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as long as they
don't evaporate away.

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And so that is now the second
part of the calculation.

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Oh, and I should the
finish the first part here.

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What we're left with is that
the settling time is L over v_s.

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And that's the formula
that we had before,

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which is 9*mu_a*L divided
by 2*rho*g*r^2.

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So this is our first
part of the Wells curve.

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So if I draw the
Wells curve over here

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in the traditional way, where
I plot on the horizontal

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axis the size of the particle,
and in a downward axis,

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we draw the time, then we have
a curve like this for settling.

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And the reason it's drawn
down, I guess maybe the feeling

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that as you're sort of falling
down a particle a certain size,

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you hit this curve and that's
when you've settled a distance

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L and fallen out of the air.

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So now, let's look
at evaporation,

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which is our second topic.

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I've lost my blue [marker] -- here it is.

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So these droplets are
getting very small

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as they're evaporating,

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and it's happening very quickly,
as we shall show in a moment.

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And so a natural assumption
is that the process

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is limited by the
diffusion of water vapor

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away from the droplet,
because essentially we

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have this little droplet
here with a certain size

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R, which is now going
to be varying with time.

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So it has a radius R(t).

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And it's really
close to the surface.

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There is an equilibrium
concentration --

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we'll call it c_w --

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of water, which depends
on the temperature.

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So that's kind of the
saturation concentration

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of water vapor in the air.

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But then, if the water is
going to evaporate more,

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it would create more
concentration, which would then

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re-condense on the particle.

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So in order for it to
continue evaporating,

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that water vapor that is
produced has to diffuse away.

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So there's going to be a
gradient of water vapor going

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outwards from c_w -- [it] is the
concentration at position R.

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And then far away, there is
a sort of diffusion layer

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thickness, delta.

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And far beyond the
diffusion layer thickness,

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the concentration is going
to approach the equilibrium

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concentration in
the ambient air.

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c at infinity is going to be
c_w times the relative humidity.

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So that's the ratio of
the concentration of water

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vapor in the air to the
saturation concentration,

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c_w, by definition.

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Now, the math
problem that we have

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to solve for this diffusion
problem -- with a moving boundary

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in this case, though we'll
assume it's pseudosteady --

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is dc/dt is the diffusion
coefficient of water

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times the Laplacian
of c, so just

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the diffusion equation with
these two boundary conditions.

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Now, an interesting aspect a
three-dimensional spherical

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diffusion is that at first
the diffusion layer grows,

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but it very quickly
reaches a steady state.

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And if we assume that
that diffusion time

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to reach this distance
delta is fast,

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so they reach a
steady state, so it's

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a kind of quasisteady or
pseudosteady shrinking

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of the droplet with sort of
a diffusion layer around it

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that's always kind of
at the steady value,

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then it turns out that
this diffusion layer

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is on the order of the
particle size.

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So as the particle shrinks, the
diffusion layer also shrinks.

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But it has a
well-defined thickness,

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as opposed to
diffusion in one or two

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dimensions, where the
diffusion layer just

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keeps growing out to infinity.

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For example, like
square root of time --

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you don't reach a steady
state in an infinite domain.

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So the bottom line
of this calculation,

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which I will not go
through right now,

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is that the flux of
water on the surface

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is the area of the surface at a
given moment, where the size is

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R, times essentially
Fick's law, where

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the driving force, the
change in concentration

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from the surface to
the bulk, is c_w times

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one minus relative humidity,
the diffusivity of water,

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and then divided by delta,
the diffusion layer thickness.

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And it turns out that with
these coefficients here,

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it turns out to be exactly R.

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So this is not a scaling
result, but actually

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an exact result for pseudo
steady spherical diffusion

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of water vapor.

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So now, we have the
flux on the surface.

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It's uniform on the surface.

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And it's [assumed]
to be pseudosteady.

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And so then I can write
down that the change

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in the size of the water
droplet volume, which

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is (4*pi/3)*R^3
is equal to minus

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the volume of a water molecule
times the flux of water.

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So that's basically my volume
or mass balance of water.

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So if I plug this in here,
then I get dR/dt is equal to --

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let's see, collecting
all the terms here --

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so derivative of R^3
is 3*R^2.

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So the 3's cancel.

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And then I have
a 4*pi*R^2,

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which cancels this
4*pi*R^2.

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So I just have dR/dt is
-v_w*D_w*c_w*(1-RH)/R.

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If I put this R on the other
side here, then I have --

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I'll just continue
the derivation here --

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I have R dR/dt is equal
to all this stuff.

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So -v_w*D_w*c_w*(1-RH).

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And then this
expression here can

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be written as 1/2 the
derivative of R^2.

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So what we find is that R^2
is linear in time.

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And then using the boundary
condition that we start out

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at a certain initial
value, R_0, then I'm

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going to get the R(t) is
the initial value R_0 times

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the square root of 1 minus
t over a certain evaporation

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time.

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And that evaporation
time is given here

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by (R_0)^2 times basically
all these coefficients here,

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where I'll separate out
the effective humidity,

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and then a bunch of other
coefficients, which you can see

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have units of length
squared over time,

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because R_0 is a length squared.

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So it's effectively some
kind of diffusivity.

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And what we get from
this calculation

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is that this
effective diffusivity

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that goes into this
expression is -- there's

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a factor 2 from this guy --

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there's a 2*v_w*D_w*c_w.

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And if you plug in
values for water vapor,

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for the saturation pressure,
and the diffusivity,

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and the volume of water in
air, then this coefficient

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turns out to be 1.2e-9
meters squared

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for second for pure water.

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And that's where you get now
the second part of the Wells

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theory, which is the
evaporation, which gives you

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a curve looking like this.

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So that there's this
sort of in this theory

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a natural crossover between
large drops, which in this case

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here are ones that are
large enough to settle out

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of the fluid before
they evaporate,

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and small drops,
which evaporate.

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On the other hand, for
true biological fluids

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that appear in
respiratory droplets,

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the evaporation is limited
by solutes and salts, which

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stop the evaporation
and, in fact,

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can attract even more
water in some cases.

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So that the evaporation part
of it is not as accurate,

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and we tend to see that
the settling part is more

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important to consider, given
an equilibrium distribution

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of droplets that
has been measured

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and is understood to come from
different types of respiration.