aerosol physics and particle control. pm particle shape: -can be found in spherical, rectangular,...
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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 m to 100 m
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 System3 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 All sizes of particles in the air of concern
Inhalable (inspirable) 100 m ≥
Thoracic 10 m ≥
Respirable 4 m ≥
PM10 10 m ≥
PM2.5 2.5 m ≥
Inhaled Particle Deposition in Human Respiratory System
• Deposition of Inhaled Particles in the Human Respiratory Tract and Consequences for Regional Targeting in Respiratory Drug Delivery, Joachim Heyder
total
Head airways
Upper bronchial
Lower bronchial
Alveolar
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
• Particle Diameter
1. Equivalent volume diameter
2. Stokes diameter
3. 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.
20 m
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?
0
20
40
60
80
100
120
140
160
0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
Diameter (um) (not scaled)
Num
ber C
once
ntra
tion
per c
m3
0.001 0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
Range Number/cm30.001-0.01 1000.01-0.03 1500.03-0.08 200.08-0.10 300.10-0.30 1500.30-0.40 600.40-1.00 301.00-1.50 10
0.02 0.20
0
20
40
60
80
100
120
140
160
0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
Diameter (um) (not scaled)
Num
ber C
onc
ent
ratio
n pe
r cm
3
0.001 0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
Range Number/cm30.001-0.01 1000.01-0.03 1500.03-0.08 200.08-0.10 300.10-0.30 1500.30-0.40 600.40-1.00 301.00-1.50 10
0.02 0.20
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)
0
2000
4000
6000
8000
10000
12000
0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
Diameter (um) (not scaled)
Nu
mb
er
Co
nce
ntr
atio
n p
er
um
,cm
3
0.001 0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
The area of each rectangular gives the number of particles between Dp2 and Dp1 (Ni)
ΔDp = Dp2-Dp1 (μm)
Ni = ni ΔDpni = the value of the number size distribution function (#/μm/cm3)
ΔDp Diameter # /um/cm3 0.009 0.001-0.01 111110.02 0.01-0.03 75000.05 0.03-0.08 4000.02 0.08-0.10 15000.20 0.10-0.30 7500.10 0.30-0.40 6000.60 0.40-1.00 500.50 1.00-1.50 20
0
20
40
60
80
100
120
140
160
0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
Diameter (um) (not scaled)
Nu
mb
er
Co
nce
ntr
atio
n p
er
cm3
0.001 0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
0
2000
4000
6000
8000
10000
12000
0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
Diameter (um) (not scaled)
Nu
mb
er
Co
nce
ntr
atio
n p
er
um
,cm
3
0.001 0.01 0.03 0.08 0.1 0.3 0.4 1 1.5
• Infinitely small ΔDp dDp
Let n(Dp) denote the continous function of size distribution
n(Dp) dDp = Particle concentration with the diameters between Dp and Dp + dDp (#/cm3)
0
)( dDpDpnNn(
Dp)
(/u
m/c
m3)
Dp, um
dDp
dNDpn )(
dDp
Typical display of distribution function n(Dp)
Normalized Size Distributions
• Normalized size distribution ( ) can be obtained by:
)(Dpn
N
DpnDpn
)()(
The fraction of particles with diameters between Dp and Dp + dDp to the total number of particles in one cm3 air
Unit of normalized number distribution: μm-1
dDpDpn )(
Surface Area,Volume and Mass Distributions
• Surface area distribution ns(Dp)
• Mass distribution m(Dp)
)()( 2ppps DnDDn
ppp dDDnDS )(0
2
pppp
pppp
dDDnD
DnDDm
)(6
m
)(6
)(
0
3
3
Y: Characteristic function
*assuming particles are spherical
Y Y(Dp)
Number of particle (Np) 1
Length (Lp) Dp
Surface area (Sp) πDp2
Volume (Vp) 1/6πDp3
Mass (mp) 1/6πDp3ρp
Characteristic Functions for Particle Size Distributions (Y(Dp))*
Logaritmic Size Distributions• Since particles’ sizes vary over a very
large range, use of logaritmic scale for Dp is more appropriate.
Linear Scale (Dp)
0
0.05
0.1
0.15
0.2
0 5 10
Dp (um)
n(D
p)
(/u
m/c
mu
3)
Log Scale (logDp)
0
0.05
0.1
0.15
0.2
0.01 0.1 1 10
logDp
n(D
p) (
/cm
u3)
Log Scale (logDp)
0
0.05
0.1
0.15
0.2
0.01 0.1 1 10
logDp
n(D
p) (
/cm
u3)
Log Scale Size Distributions
0
)( dDpDpnN
DpdDpnN log)(log
)log)(log(log 1,2, ppp DDDn
Number of particles with diameters between logDp and logDp + dlogDp
Seinfeld ve Pandis
Number
Surface
Volume
Parameters of particle size distributions
pppmeanp dDDnDN
D )(1
0
,
N
dDDnD ppipi )(1
T
ppipi
meanp m
dDDmDD
)(1
,
(number distribution)
N= Total number of particles
(mass distribution)
mT = Total mass of particles
1. Mean: Averaged diameter of the sampled particle stream
NdDDnDmedianpD
ppmedianp 2
1)(
,
0
,
0)(
mod,
mod,
epDp
pep dD
DdnD
2. Median: taneciklerin %50’sinin büyük, %50’sinin küçük olduğu çap değeri
3. Mode:
The most frequent diameter
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
1
)()( 2,
02
N
dDDnDD ppmeanpp
1
)(1
2,
N
nDD imeanppi
Standard Deviation= 2
Standard deviation: a measure of the distrubition width
Normal (Gaussian) Distribution
2
2
1exp
2
1)(
x
xf
2
,
2
1exp
2
1)(
meanppp DD
N
Dn
Normal Distribution Function
The mean of the population
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-.
Normal (Gaussian) Distribution
Linear Scale (Dp)
0
0.05
0.1
0.15
0.2
0 5 10
Dp (um)
n(D
p)
(/u
m/c
mu
3)
Are the particles in various air streams show a normal
distribution?
Lognormal Distribution
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.01 0.1 1 10
Diameter (um)
n(l
og
Dp
)
They mostly show a lognormal distribution
Log-Normal Dağılım
Dp,g=0.4μm
σg=2.5
Asıltı parçacıkların %68.2’si Dpg/σg ile Dpgσg arasındadır.
For lognormal distributions,
Dpg =Dp,median
Geometrik Ortalama ve Geometrik Standart Sapma
N
DnD
pii
gp1
,
lnexp
2/1
1
2,
1
))ln(lnexp
N
DDn gppii
g
Belirtilen Boyuttan Daha Küçük Olma Yüzdesi
Çap
, um
Tanecik Boyut Dağılımının Log-Olasılık Kağıtta Gösterimi
taneciklerin %50’sinin küçük olduğu çap
(Dp,50)
Belirtilen Boyuttan Daha Küçük Olma Yüzdesi
Çap
, um
Tanecik Boyut Dağılımının Log-Olasılık Kağıtta Gösterimi
Eğer dağılımdan hesaplanan noktalar bir doğru oluşturuyorsa dağılımın log-normal olduğu söylenebilir.
Tane sayısı, alanı,hacmi veya kütlesi için oluşturulabilir ve her dağılım için %50’sinin küçük olduğu çap bulunabilir.
Örnek: Dpg = 0.5 m ise, bu taneciklerin %50sinin bu çaptan küçük olması demektir. 100 tanecik varsa toplam , 50’sinin çapı 0.5 m’nin altındadır.
ÖrnekSize Interval Mean Size Interval Number of Particles0-0.02 0.01 200.02-0.04 0.03 1600.04-0.06 0.05 2640.06-0.08 0.07 2840.08-0.10 0.09 2760.10-0.12 0.11 2240.12-0.14 0.13 1700.14-0.16 0.15 1400.16-0.18 0.17 1100.18-0.20 0.19 900.20-0.28 0.24 1240.28-0.36 0.32 650.36-0.50 0.43 30
a) Plot the number size distribution n(Dp) and surface size distribution ns(Dp)a) Tane sayısı (n(Dp) ve yüzey alan boyut dağılımını çizin (ns(Dp). Dağılımları karşılaştırın.
b) Bu örneklenmiş asılı taneciklerin log-normal bir dağılım gösterdiği söylenebilir mi?
Size Interval Mean Size Interval Number of Particles n(Dp) ns(Dp)0-0.02 0.01 20 1000 0.3140.02-0.04 0.03 160 8000 22.6080.04-0.06 0.05 264 13200 103.620.06-0.08 0.07 284 14200 218.48120.08-0.10 0.09 276 13800 350.98920.10-0.12 0.11 224 11200 425.53280.12-0.14 0.13 170 8500 451.0610.14-0.16 0.15 140 7000 494.550.16-0.18 0.17 110 5500 499.1030.18-0.20 0.19 90 4500 510.0930.20-0.28 0.24 124 2480 448.54270.28-0.36 0.32 65 812.5 261.2480.36-0.50 0.43 30 272.7273 158.3416
Size Interval Mean Size Interval Number of Particles0-0.02 0.01 200.02-0.04 0.03 1600.04-0.06 0.05 2640.06-0.08 0.07 2840.08-0.10 0.09 2760.10-0.12 0.11 2240.12-0.14 0.13 1700.14-0.16 0.15 1400.16-0.18 0.17 1100.18-0.20 0.19 900.20-0.28 0.24 1240.28-0.36 0.32 650.36-0.50 0.43 30
a) Plot the number size distribution n(Dp) and surface size distribution ns(Dp)
n(Dp)
02000400060008000
10000120001400016000
0 0.1 0.2 0.3 0.4 0.5
Dp (um)
n(D
p)
(/u
m/c
m3
)
ns(Dp)
0
100
200
300
400
500
600
0 0.1 0.2 0.3 0.4 0.5
Dp (um)
ns
(Dp
) (
um
/cm
3)
Mean Size Interval Number of Particles Cumulative Number Cumulative Percentage0.01 20 20 0.010.03 160 180 0.090.05 264 444 0.230.07 284 728 0.370.09 276 1004 0.510.11 224 1228 0.630.13 170 1398 0.710.15 140 1538 0.790.17 110 1648 0.840.19 90 1738 0.890.24 124 1862 0.950.32 65 1927 0.980.43 30 1957 1.00
b) Bu dağılım log-normal mi?
Size Interval Mean Size Interval Number of Particles0-0.02 0.01 200.02-0.04 0.03 1600.04-0.06 0.05 2640.06-0.08 0.07 2840.08-0.10 0.09 2760.10-0.12 0.11 2240.12-0.14 0.13 1700.14-0.16 0.15 1400.16-0.18 0.17 1100.18-0.20 0.19 900.20-0.28 0.24 1240.28-0.36 0.32 650.36-0.50 0.43 30
a) Plot the number size distribution n(Dp) and surface size distribution ns(Dp)
Dpg=0.08
0.16
σg=Dp,84.1/Dp,50 = 2.0
Mean Size Interval Cumulative Percentage0.01 0.010.03 0.090.05 0.230.07 0.370.09 0.510.11 0.630.13 0.710.15 0.790.17 0.840.19 0.890.24 0.950.32 0.980.43 1.00
Size Interval Mean Size Interval Number of Particles0-0.02 0.01 200.02-0.04 0.03 1600.04-0.06 0.05 2640.06-0.08 0.07 2840.08-0.10 0.09 2760.10-0.12 0.11 2240.12-0.14 0.13 1700.14-0.16 0.15 1400.16-0.18 0.17 1100.18-0.20 0.19 900.20-0.28 0.24 1240.28-0.36 0.32 650.36-0.50 0.43 30
a) Plot the number size distribution n(Dp) and surface size distribution ns(Dp)
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
18
2pp
t
gdpv
Motion of Particles in a Fluid
Motion of Particles in a FluidNote: Continous FluidA 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
A continuum can be assumed if
31L
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.
d
Motion of Particles in a FluidNote: 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.
pn dK /
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/PT00 =1.340
Motion of Particles in a FluidRe>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/18=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/3vtCde
• 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 .
Nonspherical Particles and Dynamic Shape Factor
Shape Dynamic Shape Factor,x
Sphere 1.00
Cube 1.08
Fiber 1.06
Clustered Spheres
2 chain 1.12
3 chain 1.27
5 chain 1.35
10 chain 1.68
3 compact 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
• Where Cca and Cce are slip correction factors for da and de, respectively. p0=1000 kg/m3
• For nonspherical particles,da then:
• For spherical particles:
2/1
0
ca
cepea C
Cdd
1818
20
2
caaceept
gCdgCdv
2/1
0
ca
cppa C
Cdd
Aerodynamic Diameter• An irregular particle• de =10 m p = 3000 kg/m3
=1.3
• The aerodynamic equivalent sphere
• da=15.2 m
p=1000 kg/m3
vt:=0.0084 m/s vt:=0.0084 m/s
Aerodynamic Diameter
• Particles with dp > 3 m, the slip effect is negligible so:
2/1
0
p
ea dd
2/1
0
p
pa dd
for nonspherical particles, (de>3 m)
for spherical particles, (dp>3 m)
Aerodynamic Diameter
• For particles with diameter smaller than 3 m, slip effect must be considered.
52.2435.62
12/1
0
22 pceea Cdd
2/1
0
ca
cepea C
Cdd
2/1
0
ca
cppa C
Cdd
For non-spherical particles and 0.1 m<de<3 m
52.2435.62
12/1
0
22 pcpa Cdd
Aerodynamic diameterfor particles with diameter smaller than 0.1 m 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 MotionStopping 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 MotionNow 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:
dtdvm
vd
dtdvmvdF
p
pd
/3
)/(3
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.
)Re4082.0(tan04276.0Re 3/10
13/10 pp
g
pp xd
S
(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:
Since Re>1,
63.61081.1
510202.1Re
5
60
0
x
xxxVdap
)Re4082.0(tan04276.0Re 3/10
13/10 pp
g
pp xd
S
)63.40826.0(tan04276.063.62.1
10201000 3/113/16
xxx
S
S = 0.0046 m
Solution
• If the particle were assumed to be in Stokes region, then S
The error then would be
0.00614-0.0046/0.0046=33%
mxx
xxV
dS a 00614.0
1081.118
)1020(1000
18 5
26
0
20
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