Modelling dispersion of
micro-organisms by air.
Arno Swart – RIVM
June 2010
Part 0 - Introduction
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<- University of Utrecht
Master in mathematics
PhD in modelling fluid dynamics
RIVM ->
Centre for Infectious Disease Control
Lab. for Zoonoses and Environmental Microbiology
- Risk Assessment (Q fever, Salmonella in Pigs)
- Modeling (dispersion, heat transfer)
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Contents
General Aerobiology
Air transport models
Box models
Plume models
Exposure
Exercise
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Part I – General Aerobiology
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Aerobiology
Study of airborne microorganisms
Bacteria
Fungi
Parasites
Virusses
Algae
Toxins
‘Bioaerosol’
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airborne biological particles/droplets containing
biological material, dead or alive.
Usually 0.5-30 micrometer
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Sources of bioaerosols
Surface water – splash
Soil & Plants – particles are 'raft' for organisms
Human/animal – Sneeze, cough
Agriculture – mechanical disturbance of soil
Wastewater treatment - splash
Transport - exhaust
Industry - exhaust
Bioterrorism - weapons
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Micro-organisms in the air
L. pneumonia (water droplets)
M. tubercolosis (sneeze)
B. anthracis (anthrax, bioterrorism)
Staphylococcus spp (indoors, from clothes, hair,
nose)
Endotoxin (lipopolysaccharide from cell wall of
gram neg. bacteria. E.g. found in
slaughterhouses)
Enteric virus (e.g. sewage treatment)
Fungi & their mycotoxins (resperatory infection,
allergy)
Amoebae (aerosolized from soil/water, health
effects poorly known)
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Viability in the air
Dependent on e.g.
Relative humidity
Water content
Oxygen
Radiation (UV!)
Temperature (dessication)
Chemicals (ozone, polutants)
Variability between micro-organisms
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Typical viability
Bacteria
Much variability
Gram negative susceptible to oxygen
Spores are very resistant
Viruses
Little variability
Resistant to oxygen
Lipid is a determining factor
Fungi
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Resistant to dissication
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Viability Model
A fraction k of organisms is inactived per time
unit
The result is ‘viable’
V(t+∆t) = V(t) – k ∆t V(t) and V(0)=V0
So, (V(t+∆t) - V(t))/ ∆t = -k, gives dV/dt = -k
This yields an exponential
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V(t) =V0 e -k t
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Part II.5 - Outbreaks
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Sverdlovsk Anthrax outbreak of 1979
Spores released at
microbiology lab.
96 cases and 64 deaths
Officially: Foodborne..
[Meselson, 1994, Science
266]
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Netherlands – Q fever
Caused by airborne Coxiella burnetii
Emitted from diary goat stables
World record of largest epidemic
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Netherlands – Q fever
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Several other examples
Foot and Mouth disease
(e.g. UK 2001: 2000 cases)
Legionella
Named after gathering
of veterans from
‘American Legion’
in 1976
Bacteria found in cooling tower of the
hotel’s air conditioning system: it spread it
through the entire building ..
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Part II – Air transport models
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Eulerian vs Lagrangian
Eulerian – Look at how the field changes through
time at one spot
Lagriangian – Follow the path of a particle while it
flows
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Eulerian models (1/3)
Full CFD (Computational Fluid Dynamics)
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Based on full equations of fluid motion (e.g.
Navier-Stokes)
Solved using e.g. Finite Elements, Finite
Differences
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Eulerian models (2/3)
Box Models
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Simplification of full equations
Suitable for indoors modelling
Subdivide room into ‘cells’
Models obstacles, inlets, outlets and timedependence
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Eulerian models (3/3)
Plume models
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Again a simplification of the full equations
Suitable for outdoor use
Assumes constant emission and steady state
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Part III – Box Models
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Derivation of Equation
Some notation
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C(x,y,z,t) is concentration in [1/m3]
(x,y,z) is Cartesian frame of reference [m]
∆ is change in value, e.g. ∆C [1/m3]
(u,v,w) are components of ‘wind’ in [m/s]
Q is a source in [m3/s]
M is a source in [1/s]
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The Volume Element
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Convection Diffusion
∆V = ∆x ∆y ∆z (volume element)
Rate of change of mass: ∆V∆C/∆t
Equals what flows in (flux): ∆y∆z C u1
Minus what flows out (flux): ∆y∆z C u0
Minus diffusion (Fick’s law): ∆y∆z (D ∆C/∆x)
Plus source terms: Q
Do not consider gravitational force
∆V∆C/∆t = ∆y∆z[ C( u1 – u0 ) - D ∆C/∆x + Q]
∆C/∆t = C ∆u/∆x - D ∆C/∆x2 + M
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Convection Diffusion
Do the same for wind v and w
Use math notation ∂C/∂t, etc.
∂C
∂C
∂C
∂C
∂ 2C ∂ 2 C ∂ 2C
=u
+v
+w
− D( 2 + 2 + 2 ) + M
∂t
∂x
∂y
∂z
∂x
∂y
∂z
This reads “change in time” = “what flows
in/out” – “what diffuses out” + “source terms”
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Convection-Diffusion
Basis for all air transport phenomena
But very hard to solve
Specialize for specific needs:
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Indoor dispersion
Outdoor dispersion
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Box Models
For modelling contained dispersion
Divide the room into ‘boxes’ or ‘cells’ and
calculate what happens within a box, and
between boxes.
This account based on Nicas, 2001, Am.
Ind. Hyg. J. 62
Note: get value of ‘D’ from algorithm in
Drivas et al. 1996, Indoor Air 6
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Box model - dispersion
Inside a box we consider diffusion:
∂C
∂ 2 C ∂ 2C ∂ 2 C
= − D( 2 + 2 + 2 ) + M
∂t
∂x
∂y
∂z
Diffusion in space due to M can be solved:
C (r , t ) =
M
(4πDt )
3
e
r2
−
4 Dt
This is in spherical coordinates (i.e. r2=x2+y2+z2
is a radius)
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Box model - Stochastic
Consider r as a random variable R
The diffusion equation is actually the pdf of
a Normal Distribution with
E[R(t)]=0
Var[R(t)]=2Dt
Moreover, the distribution can be split like
NR(0,2Dt) = NX(0,2Dt) Ny(0,2Dt) Nz(0,2Dt)
The ‘Drivas’ model
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Box Models – Random Walk
We can use a ‘Random Walk’ to model the
distribution (actually kind of ‘Lagrangian’)
In the x-direction the particle may
Go left (-∆x) with probability p
Go right (+ ∆x) with probability p
Hold, with probability h = 1-2p
Note that
E[X(∆t)] = -∆x p + ∆x p + 0*h = 0
But.. what should p be?
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Box Models – Jumps
We find the right p by checking the needed
variance
Lets take n steps of time ∆t, so t=n∆t
Due to independence of steps
Var[X(t)]=Var[X(n∆t)] = nVar[X (∆t)]
Var[X(∆t)] = E[(X(∆t)-E[X(∆t)])2] = E[X(∆t)2]
=p(-∆x)2+p(∆x)2+h(0-0) = 2p(∆x)2
Compare with Var[X(∆t)] = 2Dt = 2Dn∆t
2Dn∆t = 2np(∆x)2 and p=D∆t/(∆x)2
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Box Models – Walls
Walls are easy, we adjust the jump
probabilities. E.g. for a wall to the left
In the x-direction the particle may
Go left (-∆x) with probability 0
Go right (+ ∆x) with probability p
Hold, with probability h = 1-p
This actually models total reflection
Another option is deposition
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Diffusion as a Markov Chain
Make probability
vectors Px, Py and Pz
Calculate probability
of moving from
(x,y,z)0 to (x,y,z)1
In t= n∆t seconds
0
0 
p
1 − p
 p 1− 2 p

0
p

Px = 
 0
1− 2 p
p
p 


0
1 − p
p
 0
M
P(0 → 1) =
( Pxn ) x1 , x0 ( Pyn ) y1 , y0 ( Pzn ) z1 , z0 e −QVn∆t
∆x∆y∆z
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Box Models – Air changes
Note the factor
e
− Qt / V
This models ‘Air changes’
The original ‘Q’ is diluted by a factor Q/V
Example 15 [air changes/(volume hr)]
15/3600 [air changes/volume second]
15V/3600 = Q, [air changes/second]
Example result
Concentration at (5,3.5, 1) due to 1000 particles at (4,3.5,1)
80
70
60
3
Conc. [particles/m ]
50
40
30
20
10
0
0
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1000
2000
3000
Time [sec]
4000
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6000
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Box Model Movie
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Box Models - Advection
Consider a outlet, with wind of s [m/s]
The length of the window is ∆x
The area of the cell is (∆x)2
Then the dilution is d = s ∆x / (∆x)2 = s/∆x
Thus, hold probability is exponential
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hA(∆t )=e-d∆t
(consider C’(t) = -d C(t), leads to
C(∆t)=C(0)e-d∆t, the concentration lowers
exponentially, thus also the hold probability)
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Box Models - Advection
The resulting equations for 2 dimensions
hA=e-d∆t
h = 1-4p = 1-4 D∆t/(∆x)2
Probability of hold incl. diffusion = e-(d∆t-ln(h)
We skip the detailed calculation !
Note that also air changes can now done
per cell.
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Box models - Combination
P(hold ) = h A h
Putting it all together
1
ln(h) + d∆t
P(downwind ) = 6
(1 − hA h)
ln(h) + d∆t
1
ln(h)
(1 − hA h)
P(other ) = 6
ln(h) + d∆t
A number m of walls?
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hm=h+mp
Change 1/6 to 1/(6-m)
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Box Models - Walls
Just set the right probabilities to zero
Example, a wall at index (2,3,4)
P((1,3,4)->(2,3,4)=0),…
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Box Models - Movie
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Box Models – At observer
9
8
7
6
5
4
3
2
1
0
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0
1000
2000
3000
4000
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6000
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Getting the dose
Integrate the cell of interest (e.g. observer,
outlet) over time
Take care of inhaled fraction
In our case:
Concentration*timestep*fraction
Then use the usual dose response curve
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A useful tool
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Part IV – Plume Models
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Plumes – Introduction
Modeling outdoor transport
Models developed in the area of pollution,
radioactive clouds, weather,…
But also very useful for pathogen
dispersal, spores, fungi, etc.
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Plumes - Example
Source: Slade et al. Meteorology and Atomic Energy, 1968.
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Plumes – Assumptions
start with the Convection Diffusion equation
Wind is in x-direction and constant
We have continuous emission
We have a steady state
Release at height ‘h’ at x=y=0
Plumes – Gaussian plume
equation
Concentration as a function of x,y,z
x is a part of the sigma’s
C( x, y, z) =
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Q
2πσ z σ yu
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e
y2
2σ y2
−
e
( z −h ) 2
2
2σ z
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Plumes - Dispersion
The dispersion parameters are dependent on the
‘Stability Class’, usually ‘Pasquill-Gifford’
Measures the turbulence in the atmosphere
Dependent on wind speed/cloud cover
Parameters also dependent on downwind
distance (x)
Get values from tables / graphs
any book on atmospheric dispersion
Or supplementary material
Stable (A) – Unstable (F)
(Figures from KNMI Netherlands)
Plumes – Limitations
Steady state assumed
Constant wind
Usually: short time scales
No barriers
Alternatives: Puff models, CFD models
We did not consider: deposition, plume rise,..
Part V - Exposure
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Exposure
Exposure
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Inhalation (primary route)
Ingestion
Dermal
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Inhalation
From concentration to dose
Example for adult male during light exercise
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Volume of inhaled air (1.25L )
Respiratory frequency ( 20 1/min )
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Deposition
Aerosols need to
deposit in ‘Alveoli’
Depends on
Exercise level
Nose/mouth breathing
Nose has a ‘filter’
and longer pathway
Size of aerosol
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Models for inhalation –
‘Dosimetrics’
E.g.
Rostami 2009, Inhalation Toxology 21
Hofmann 2009, Biomarkers 14
Price OT, et al. Multiple Path Particle Dosimetry
Model (MPPD V1.0): A Model for Human and Rat
Airway Particle Dosimetry. 2002.
Techniques include
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Fitting to in vivo studies (animal/man)
Modelling of respiratory tract
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Example ICRP Dosimetric model
International Commission on Radiological Protection,
publication 66, 1994
Do you need it? Is it implicit in the dose-response
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21-25
relation?
Exercise – Q fever
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Exercise
Suppose a goat has excreted 109
Coxiellae
The particles are about 10 µm
They are evenly emitted during about 10
hours.
It is a nice sunny day in juli, half cloudy,
with a wind of 4 m/s
You are at location downwind (2km),
crosswind (1km), resting for 1 hr.
Lets say 1 particle is risky, are you at risk?
Hints
Sketch the situation
Get the right sigma’s from the tables
Calculate the source’s nr. Of Coxiellas/sec.
Take care of units! (meters?, kilometers?)
Use the Gaussian plume formula
Solution
109 Cox/day = 109/(10*60*60)~ 28000 Cox/s
(Fig A22): Stability class = B
(Fig A21): σz~250
(Fig A20): σy~300
(x,y,z) = (2000,1000,1.8)
Formula gives 5.7x10-5 Cox/s ~ 0.2 Cox/h
Resp. fraction ~ 0.02, so dose ~ 0.004 Cox
(Or perhaps 0.002 if you count reflection)