Aerosol Physics and Particle
Control
PM
Particle Shape:
-can be found in spherical, rectangular, fiber, or many other
irregular shape
-shape is important. Affects:
-particle behavior
-transportation
-control technology
-effect on the respirotary system (fiber shaped
particles particularly harmful to the lungs when they are
inhaled, since it is more difficult to remove once they are
settled or have clung to air ways.
PM
• Muallimköy ortam havasından toplanan partiküllere ait
SEM fotorafı
Particle Size
• The most important parameters since it affects:
– Behavior
– Transport
– Health effects
– Control technology selection
Very large range from 0.01 mm to 100 mm
A dust fall following a volcano contains large particles in
the range of millimeters that can settle down in a few
hours and smal particles (um range) can stay
airborne for months.
Human Respiratory System
3 major regions:
1. Head airways region
2. Tracheobronchial region
(thoracic)
3. Pulmonary or alveolar region
O2 –CO2 transfer take place in the
pulmonary region. For an adult total area
of this gas exchange region is 75 m2 and
total length of the pulmonary vessels is
about 2,000 km.
An adult person breathes about 10-20
times per minute and inhales about 10-20
liters of air/min.
Particle Size Categories
• Based on behavior in the human
respiratory system 3 categories can be
defined:
– Inhalable particles
– Thoracic particles
– Respirable particles
Particle Size Categories
Particles
Size Range
Total
Inhalable (inspirable)
All sizes of particles in
the air of concern
100 mm ≥
Thoracic
10 mm ≥
Respirable
4 mm ≥
PM10
10 mm ≥
PM2.5
2.5 mm ≥
Inhaled Particle Deposition in
Human Respiratory System
total
Head
airways
Upper bronchial
Lower bronchial
Alveolar
•
Deposition of Inhaled Particles in the Human Respiratory Tract and Consequences for
Regional Targeting in Respiratory Drug Delivery, Joachim Heyder
Size Distribution
• Particle Diameter
The highly irregular shape together with the
different densities depending on the
composition of the particle complicates its
size definition.
A particle’s size refers to its diameter
There are various definitions for the
diameter.
Size Distribution
•
1.
2.
3.
Particle Diameter
Equivalent volume diameter
Stokes diameter
Aerodynamic diameter
Particle Diameter
1. Equivalent volume diameter (de):
diameter of a sphere that would have the
same volume and density as the particle
Assume that following irregular shape
has a volume of V, de will be the diameter
of sphere whose volume equals to V.
Particle Diameter
2. Stokes diameter (ds): diameter of the
sphere that would have the same density
and settling velocity as the particle.
Particle Diameter
3. Aerodynamic diameter (da): diameter of
the sphere with a standard density
(1.000 kg/m3) that would have the same
settling velocity as the particle
Particle Diameter
de only standardizes the shape of the
particle by its equivalent spherical
volume
ds standardizes the settling velocity of the
particle but not the density
da standardizes both the settling velocity and
the particle density. Thus da is a
convenient variable to use to analyze
particle behavior and design of particle
control equipment
Particle Size Distribution
How to display particle size distribution?
160
Number Concentration per cm3
Range
Number/cm3
0.001-0.01
100
0.01-0.03
150
0.03-0.08
20
0.08-0.10
30
0.10-0.30
150
0.30-0.40
60
0.40-1.00
30
1.00-1.50
10
140
120
100
80
60
40
0.02
0.20
20
0
0.01 0.01
0.03 0.03
0.08 0.08
0.1 0.10.3 0.3 0.40.4 1 1
0.001
Diameter (um)
(not scaled)
1.5
1.5
160
Number Concentration per cm3
Range
Number/cm3
0.001-0.01
100
0.01-0.03
150
0.03-0.08
20
0.08-0.10
30
0.10-0.30
150
0.30-0.40
60
0.40-1.00
30
1.00-1.50
10
When shown as written in the table, we see that
number of particles with diameter between 0.01 0.03 um equals to particles with diameter between
0.1 and 0.3 um however, diameter range in the first
one is only 0.02 um while in the second is 0.2 um
(1000 times bigger than the first range)
140
120
100
80
60
40
0.02
0.20
20
0
0.1 0.10.3 0.3 0.40.4 1 1
0.08 0.08
0.03 0.03
0.01 0.01
0.001
(not scaled)
Diameter (um)
1.5
1.5
D
0.009
0.02
0.05
0.02
0.20
0.10
0.60
0.50
i
a
m
e
t
e
r
#
0.001-0.01
0.01-0.03
0.03-0.08
0.08-0.10
0.10-0.30
0.30-0.40
0.40-1.00
1.00-1.50
ni = the value
of the number
size
distribution
function
(#/μm/cm3)
/
u
m
/
c
m
3
11111
7500
400
1500
750
600
50
20
Number Concentration per um,cm3
ΔDp
12000
10000
8000
6000
The area of each
rectangular gives the
number of particles
between Dp2 and Dp1
(Ni)
Ni = ni ΔDp
4000
2000
0
0.01 0.01
0.03 0.03
0.08 0.08
0.1 0.10.3 0.30.4 0.4 1 1 1.51.5
0.001
Diameter (um) (not scaled)
ΔDp = Dp2-Dp1 (μm)
140
120
100
80
60
40
20
0
0.01 0.01
0.03 0.03
0.08 0.08
0.1 0.10.3 0.3 0.40.4 1 1
0.001
Diameter (um)
(not scaled)
1.5
1.5
Number Concentration per um,cm3
Number Concentration per cm3
160
12000
10000
8000
6000
4000
2000
0
0.01 0.01
0.03 0.03
0.08 0.08
0.1 0.10.3 0.30.4 0.4 1 1 1.51.5
0.001
Diameter (um) (not scaled)
Let n(Dp) denote the continous
function of size distribution
n(Dp) dDp = Particle concentration
with the diameters between Dp
and Dp + dDp (#/cm3)
n(Dp) (/um/cm3)
• Infinitely small ΔDp  dDp
dDp
Dp, um

N   n( Dp)dDp
0
dN
n( Dp) 
dDp
Typical display
of distribution
function n(Dp)
Normalized Size Distributions
• Normalized size distribution (n (Dp)) can be
obtained by:
n( Dp )
n ( Dp ) 
N
n ( Dp)dDp
Unit of normalized number distribution: μm-1
The fraction of particles with
diameters between Dp and
Dp + dDp to the total number
of particles in one cm3 air
Surface Area,Volume and Mass Distributions
• Surface area distribution ns(Dp)
ns (Dp )  Dp2n(Dp )

S    Dp2 n( Dp )dDp
0
• Mass distribution m(Dp)
m( D p ) 
m

6


6
D 3p  p n( D p )
3
D
 p  p n( Dp )dDp
0
Y:
Characteristic
function
Characteristic Functions for Particle Size
Distributions (Y(Dp))*
Y
Y(Dp)
Number of particle (Np)
1
Length (Lp)
Dp
Surface area (Sp)
πDp2
Volume (Vp)
1/6πDp3
Mass (mp)
1/6πDp3ρp
*assuming particles are spherical
Logaritmic Size Distributions
• Since particles’ sizes vary over a very
large range, use of logaritmic scale for Dp
is more appropriate.
Log Scale (logDp)
0.2
0.2
n(Dp) (/cmu3)
n(Dp) (/um/cmu3)
Linear Scale (Dp)
0.15
0.1
0.05
0
0
5
Dp (um)
10
0.15
0.1
0.05
0
0.01
0.1
1
logDp
10
Log Scale Size Distributions
Log Scale (logDp)

N   n( Dp)dDp
n(Dp) (/cmu3)
0.2
0
0.15
0.1

0.05
0
0.01
0.1
1
10
N   n(log Dp)d log Dp
logDp
 n(logDp )(logDp,2  log Dp,1 )
Number of particles with diameters
between logDp and logDp + dlogDp

Number
Surface
Volume
Seinfeld ve Pandis
Parameters of particle size distributions
1. Mean:
Averaged
diameter of
the sampled
particle
stream

D p ,mean
1
  D p n( D p )dDp
N0


D
pi
n( D pi )dDp
(number distribution)
N= Total number of
particles
1
N

D p ,mean 
D
pi m( D pi ) dDp
1
mT
(mass distribution)
mT = Total mass of
particles
2. Median:
taneciklerin
%50’sinin büyük,
%50’sinin küçük
olduğu çap
değeri
3. Mode:
The most
frequent
diameter
D p ,median
D p ,median 
Dp ,mod e
1
0 n( D p )dD p  2 N
 dn( Dp ) 


0
 dD 
p

 Dp ,mod e
Beside mean values, it is important to know the how
distribution differs from these mean values.
For all three distribution shown above, the mean
diameters are the same while distribution width is
different.
Variance

 
2
 (D
 D p ,mean ) n( D p )dDp
2
p
0
N 1


2
(
D

D
)
 pi p,mean ni
1
Standard deviation: a
measure of the
distrubition width
Standard Deviation=
N 1
 
2
Normal (Gaussian) Distribution
Normal Distribution
Function
The mean of the
population
2

1
1 xm  
f ( x) 
exp 
 
 2
 2    
2

n( D p )
1
1  D p  D p ,mean  
 

exp 
N

 2
 2 
 
Normal (Gaussian) Distribution
Standard deviation () gives the characteristic width of
the symetric number size distribution. 68.2% of the
particles are between dp,mean – and dp,mean + , 84.1%
of it with diameters smaller than dpmean+ , and 15.9% of
it with diameters smaller than dpmean-.
Are the particles in various air
streams show a normal
distribution?
n(Dp) (/um/cmu3)
Linear Scale (Dp)
0.2
0.15
0.1
0.05
0
0
5
Dp (um)
10
Log-Normal Dağılım
They mostly show a lognormal distribution
Lognormal Distribution
Dp,g=0.4μm
0.35
σg=2.5
n(logDp)
0.3
0.25
0.2
Asıltı parçacıkların %68.2’si
Dpg/σg ile Dpgσg arasındadır.
0.15
0.1
0.05
0
0.01
For lognormal distributions,
0.1
1
Diameter (um)
10
Dpg =Dp,median
Geometrik Ortalama ve Geometrik
Standart Sapma
Dp,g
 

  ni ln D pi 

 exp 1
N




1/ 2
 

2 
  ni (ln D pi  ln D p , g ))  
 
 g  exp 1
 

N 1
 

 

Tanecik Boyut Dağılımının LogOlasılık Kağıtta Gösterimi
taneciklerin
%50’sinin küçük
olduğu çap
Çap, um
(Dp,50)
Belirtilen Boyuttan Daha
Küçük Olma Yüzdesi
Tanecik Boyut Dağılımının LogOlasılık Kağıtta Gösterimi
Eğer dağılımdan hesaplanan
noktalar bir doğru
oluşturuyorsa dağılımın lognormal olduğu söylenebilir.
Çap, um
Tane sayısı, alanı,hacmi veya
kütlesi için oluşturulabilir ve
her dağılım için %50’sinin
küçük olduğu çap bulunabilir.
Örnek: Dpg = 0.5 mm ise, bu
taneciklerin %50sinin bu
çaptan küçük olması
demektir. 100 tanecik varsa
toplam , 50’sinin çapı 0.5
mm’nin altındadır.
Belirtilen Boyuttan Daha
Küçük Olma Yüzdesi
Örnek
Size Interval
0-0.02
0.02-0.04
0.04-0.06
0.06-0.08
0.08-0.10
0.10-0.12
0.12-0.14
0.14-0.16
0.16-0.18
0.18-0.20
0.20-0.28
0.28-0.36
0.36-0.50
Mean Size Interval Number of Particles
0.01
20
0.03
160
0.05
264
0.07
284
0.09
276
0.11
224
0.13
170
0.15
140
0.17
110
0.19
90
0.24
124
0.32
65
0.43
30
Plot the
number
size distribution
n(Dp)boyut
and surface
size distribution
ns(Dp)
a)a) Tane
sayısı
(n(Dp)
ve yüzey alan
dağılımını
çizin
(ns(Dp). Dağılımları karşılaştırın.
b) Bu örneklenmiş asılı taneciklerin log-normal bir dağılım
gösterdiği söylenebilir mi?
Size
SizeInterval
Interval Mean
MeanSize
SizeInterval
Interval Number
NumberofofParticles
Particles n(Dp)
0-0.02
0.01
0-0.02
0.01
2020
1000
0.02-0.04
0.03
160
0.02-0.04
0.03
160
8000
0.04-0.06
0.05
264
0.04-0.06
0.05
264
13200
0.06-0.08
0.07
284
0.06-0.08
0.07
284
14200
0.08-0.10
0.09
276
0.08-0.10
0.09
276
13800
0.10-0.12
0.11
224
0.10-0.12
0.11
224
11200
0.12-0.14
0.13
170
0.12-0.14
0.13
170
8500
0.14-0.16
0.15
140
0.14-0.16
0.15
140
7000
0.16-0.18
0.17
110
0.16-0.18
0.17
110
5500
0.18-0.20
0.19
0.18-0.20
0.19
9090
4500
0.20-0.28
0.24
124
0.20-0.28
0.24
124
2480
0.28-0.36
0.32
0.28-0.36
0.32
6565
812.5
0.36-0.50
0.43
0.36-0.50
0.43
3030 272.7273
ns(Dp)
0.314
22.608
103.62
218.4812
350.9892
425.5328
451.061
494.55
499.103
510.093
448.5427
261.248
158.3416
a) Plot the number size distribution n(Dp) and surface size distribution ns(Dp)
ns(Dp)
600
16000
14000
12000
10000
8000
6000
4000
2000
0
ns(Dp) (um/cm3)
n(Dp) (/um/cm3)
n(Dp)
500
400
300
200
100
0
0
0.1
0.2
0.3
Dp (um)
0.4
0.5
0
0.1
0.2
0.3
Dp (um)
0.4
0.5
b) Bu dağılım log-normal mi?
SizeInterval
Interval Number
Mean Size
Interval Cumulative
Number of Particles
Mean Size
of Particles
Number Cumulative Percentage
0-0.02 0.01
0.01
20
2020
0.01
0.02-0.040.03
0.03
160
160
180
0.09
0.04-0.060.05
0.05
264
264
444
0.23
284
728
0.37
0.06-0.080.07
0.07
284
276
1004
0.51
0.08-0.100.09
0.09
276
224
1228
0.63
0.10-0.120.11
0.11
224
170
1398
0.71
0.12-0.140.13
0.13
170
140
1538
0.79
0.14-0.160.15
0.15
140
110
1648
0.84
0.16-0.180.17
0.17
110
90
173890
0.89
0.18-0.200.19
0.19
124
1862
0.95
0.20-0.280.24
0.24
124
65
192765
0.98
0.28-0.360.32
0.32
30
195730
1.00
0.36-0.500.43
0.43
a) Plot the number size distribution n(Dp) and surface size distribution ns(Dp)
SizeInterval
Interval Cumulative
Mean Size Interval
Number of Particles
Mean Size
Percentage
0-0.02 0.01
0.01 0.01
20
0.02-0.040.03
0.03 0.09
160
0.04-0.060.05
0.05 0.23
264
0.06-0.080.07
0.07 0.37
284
0.08-0.100.09
0.09 0.51
276
0.10-0.120.11
0.11 0.63
224
0.12-0.140.13
0.13 0.71
170
0.14-0.160.15
0.15 0.79
140
0.16-0.180.17
0.17 0.84
110
0.18-0.200.19
0.19 0.89
90
0.20-0.280.24
0.24 0.95
124
0.28-0.360.32
0.32 0.98
65
σ
=D
/Dp,50 = 2.0
g
p,84.130
0.36-0.500.43
0.43 1.00
0.16
a) Plot the number size distribution n(Dp) and
surface size distribution ns(Dp)
Dpg=0.08
Moudi Impactor for Mass Size
Distribution
Optical Particle Counter
Differential Mobility Analyzer
Aerodynamic Aerosol Sizer
(APS)
Motion of Particles in a Fluid
In all particle control technologies, particles are separated from
the surrounding fluid by the application of one ore more
forces:
-gravitational
-inertial
-centrifugal
-electrostatic
Those forces cause the accelarate the particles away from the
direction of the mean fluid flow, toward the direction of the
net force
The particles must then be collected and removed from the
system to prevent ultimate re-entrainment into the fluid
Therefore we need to know the dynamics of particles in fluids
Motion of Particles in a Fluid
Drag Force
• FD = CDAppFv2r
– FD=Drag force, N
– CD=Drag coefficient
– Ap=Projected area of particle, m2
– pF=Density of fluid, kg/m3
– vr= relative velocity, m/s
Drag coefficient must be determined
experimentally since CD = f(particle shape and
the flow regime characterized by Reynolds
number)
Stokes’ Law
Motion of Particles in a Fluid
Motion of Particles in a Fluid
Motion of Particles in a Fluid
vt 
p p gd p2
18m
Motion of Particles in a Fluid
Motion of Particles in a Fluid
Note: Continous Fluid
A continuum is a region of spacee where charactheristic flow scales L are large enough
that properties like density and velocity can be assumed to vary smoothly and therefore
have point values.
Characteristic scale in diffusion or sedimentation for an aerosol particle is its diameter
d
A continuum can be assumed if
1
 L3
N
N: number of molecules per unit volume.
In water N is about 3.3x1028/m3, in air N is 2.5x1025 /m3. It is almost safe to
assume a continuum in liquids but not in gases.
Motion of Particles in a Fluid
Note: Continous Fluid
So does the aerosol particle sense a continuum or a series of discrete bombardements
by the air molecules? This can be judged by calculating Knudsen number.
Kn   / d p
d
When the Knudsen number is much less than unity (Kn<<1) a continuum can be
assumed.
Motion of Particles in a Fluid
Motion of Particles in a Fluid
dp>0.1 um
C=1+2.52/dp
Motion of Particles in a Fluid
If STP condition is not valid the value of  would change. STP: STP is 0 °C (32 °F
or 273 Kelvin) and 1 atm (101.335 kPa, 14.7 PSI, 760 mmHg)
Example:
At the top of a mountain, the atmospheric pressure is measured as 70 kPa, and T
as -20C. What would the error of the slip correction factor be for a 0.3 um particle if
the effect of P and T on  is ignored?
Solution:
Without considering the effect of T and P
C=1 + 2.52/dp = 1.55
When the effect of Pa nd T is taken into account:
= P0T/PT00 =1.340
Motion of Particles in a Fluid
Re>0.1
Motion of Particles in a Fluid
Motion of Particles in a Fluid
Motion of Particles in a Fluid
Example
• A grain of concrete dust particle is falling
down onto the floor through room air. The
particle diameter is 2 um and the particle
density is 2500 kg/m3. Assuming the room
air is still, determine the terminal settling
velocity of the particles. The room air is at
standard conditions.
Solution
• First calculate the slip correction factor for
the 2 um particle.
• At STP air viscosity=1.81x10-5 (Ns/m2), air
mean free path =0.066 um and air density
p=1.2 kg/m3
• C = 1 + 2.52/dp=1.08
• vt = ppdp2gC/18m=0.00026 m/s=0.26 mm/s
Nonspherical Particles and
Dynamic Shape Factor
• Most particle in practice are nonspherical. Particle
dynamic shape factor is defined as the ratio of the actual
resistance force of a nonspherical particle to the
resistance force of a spherical particle that has the same
equivalent volume diameter (de) and the same settling
velocity as the nonspherical particle.
• Shape factor = FD/3mvtCde
• Where FD is the actual drag force exerted on the
nonspherical particle and C is the slp correction factor for
de. The dynamic shape factor is always greater than 1
except for certain streamlined shapes .
Shape
Dynamic Shape Factor,x
Sphere
1.00
Cube
1.08
Nonspherical Fiber
Clustered Spheres
Particles and
2 chain
Dynamic Shape 3 chain
5 chain
Factor
10 chain
3 compact
1.06
1.12
1.27
1.35
1.68
1.15
Dust
Bituminous coal
1.08
Quartz
1.36
Sand
1.57
Aerodynamic Diameter
• In reality it is extremely difficult to measure or
calculate the equivalent volume diameters and
shape factors.
• Thus we need an equivalent diameter that can
be physically determined and can be used to
characterize the particle behavior
• Aerodynamic diameter, da, is defined as the
diameter of a sphere with unit density (1000
kg/m3) and the same settling velocity as the
particle concern.
Aerodynamic Diameter
 p de2 gCce 0 d a2 gCca
vt 

18m
18m
• Where Cca and Cce are slip correction factors
for da and de, respectively. p0=1000 kg/m3
• For nonspherical particles,da then:
  p Cce 

d a  d e 
  0 Cca 
1/ 2
• For spherical particles:
  p Cc 

d a  d p 
  0Cca 
1/ 2
Aerodynamic Diameter
•
•


An irregular particle
de =10 mm
p = 3000 kg/m3
 =1.3
vt:=0.0084 m/s
• The aerodynamic
equivalent sphere
• da=15.2 mm
 p=1000 kg/m3
vt:=0.0084 m/s
Aerodynamic Diameter
• Particles with dp > 3 mm, the slip effect is
negligible so:
 p 

d a  d e 
 0  
1/ 2
 p 
d a  d p  
 0 
for nonspherical particles, (de>3 mm)
1/ 2
for spherical particles, (dp>3 mm)
Aerodynamic Diameter
• For particles with diameter smaller than 3 mm,
slip effect must be considered.
For non-spherical particles and 0.1 mm<de<3 mm
  p Cce 

d a  d e 
  0 Cca 
1/ 2
  p Cc 

d a  d p 
  0Cca 
1/ 2
1/ 2





1
p
2
2
  2.52 
d a   6.35  4d e Cce
2 
0  


1/ 2


p 
1 
2
2
  2.52 
d a   6.35  4d pCc
2 
0 


Aerodynamic diameterfor particles with diameter smaller than 0.1 mm can be solved
using iteration methods. In general those particles behave like gases for which
diffusion transport becomes more important than aerodynamic behavior.
Nonsteady State Particle Motion
settling velocity
Nonsteady State Particle Motion
Nonsteady State Particle Motion
Nonsteady State Particle Motion
Nonsteady State Particle Motion
Nonsteady State Particle Motion
Stopping distance of a particle in a gas flow: a simple means of explaining impaction,
answers how important a particle’s inertia is, relative to the viscous effects of the fluid
through which it moves
Collection of particles (when a flowing fluid approaches a stationary object such as a
fabric filter thread, a large water droplet, or a metal plate the fluid flow streamlines will
diverge around that object) on a stationary object occur with three different mechanisms
Impaction, interception, and diffusion:
Nonsteady State Particle Motion
Now assume a sphere in the Stokes regime is projected with an initial velocity of V0 into
a motionless fluid and ignore all but drag force:
Take equaiton 3 but assume no gravitational and buoyoncy force:
 Fd  3md p v  m(dv / dt)

3md p v
m
 dv / dt
Nonsteady State Particle Motion
Nonsteady State Particle Motion
Above equation is valid only for Stokes region. When Rep>1, the
stopping distance is shorter than is predicted by this equation since
drag force increases as the Rep increases. This increase in drag
force attenuates the velocity of the particle, and hence reduces the
stopping distance.
Since the Rep is proportional to the particle velocity and the drag
force is proportional to the V2, it is extremely difficult to obtain
stopping distance outside of Stokes region. Mercer (1973),
proposed an emprical equation to calculate S within %3 accuracy.
 pd p

S
Re1p/03  0.04276x tan1 (0.4082Re1p/03 )
g
(for 1<Rep0<1500)
Nonsteady State Particle Motion
If r’ is small, S (xstop) is also small. For instance if a 1 um particle with
unit density is projected at 10 m/s into air, it will stop after traveling 36
um
An impaction parameter NI can be defined as the ratio of the stopping
distance of a particle (based on upstream fluid velocity) to the
diameter of the stationary object, or
NI=xstop/d0
If NI is large most of the particles wil impact the ojbect. If NI is very
small most of the particles will follow the fluid flow around the object.
Nonsteady State Particle Motion
Example 2
• Calculate the stopping distance of a
particle with an aerodynamic diameter of
20 mm in still air. Assume the initial
velcoity of the particles is 5 m/s. What
would the error be if the particle motion
were assumed to be in Stokes region?
Solution
• The initial particle Reynolds number:
Re p 0
d aV0 1.2 x20x106 x5


 6.63
5

1.81x10
Since Re>1,
 pd p

S
Re1p/03  0.04276x tan1 (0.4082Re1p/03 )
g

1000x20x106
S
6.631/ 3  0.04276x tan1 (0.40826.631/ 3 )
1.2
S = 0.0046 m

Solution
• If the particle were assumed to be in Stokes
region, then S
0 d a2
1000x(20x106 ) 2
S
V0 
 0.00614m
5
18
18x1.81x10
The error then would be
0.00614-0.0046/0.0046=33%