CALCULATION OF DEPOSITION VELOCITY AND POSITION OF PARTICLE
Forces on an Indoor Aerosol
When an aerosol particle inside a room is considered, the airflow patterns and
the forces that act on the particle, based on its diameter, influence its behavior.
There are four major forces, which act on an aerosol particle inside a room either
individually or in a combination depending on the diameter range in which the
particle falls. These forces are
 Drag force
 Brownian force
 Lift force
 Gravitational force
Let us take a look at each of these forces; when and how they have an influence
on the behavior of the particles.
Drag force
Aerosols consist of two components, a gas or gas mixture, which most commonly
is air, and the particles suspended in it. The behavior of particles within the
aerosols depends to a large extent on the motion and the intrinsic properties of
the suspending gas. The motion of individual gas molecules affects particles
whose diameter is less than 0.1 m and hence the kinetic energy of gases is
useful in understanding the behavior of these particles. Larger particles can be
1
treated as being submersed in a fluid (the continuum regime). Intermediate
particles can be treated by the adjustments of the equations from the continuum
regime. This intermediate range is termed as a transition or slip regime. The
equations for various parameters that affect the drag force have been taken from
Paul [14].
Let us look at the concepts and parameters that affect gas and particle motion:
1. Reynolds number
When measuring an aerosol, it is important to understand what happens to an
aerosol in the environment on it’s way into the sensor of a measuring instrument.
Due to various external forces, the trajectories of the aerosol particles can
deviate from the gas flow. The flow pattern of the gas, whether smooth or
turbulent, is governed by the ratio of inertial force of the gas to the friction force of
the gas moving over the surface. This ratio is expressed as the Reynolds
number Re
Re 
 gVd Vd

.........................................................................(2.1)


The Reynolds number characterizes the flow and hence depends on the gas
density g and not on the particle density. At NTP i.e., 200C and 101 KPa (1 atm)
g = 1.192 * 10-3 g/cm3 and = 1.833 * 10-4 dynes/cm2
Re = 6.5 Vd …………………………………………………(2.2)
2
2. Gas density and Mach number
The density of a gas g is related to its temperature T and pressure P through the
equation of state
P   g RT 
 g R T
M
......................................................................(2.3)
When the gas moves at a high velocity relative to the acoustic velocity V g in that
gas, the gas becomes compressed. The degree of compression depends on
Mach number
Ma 
U
..........................................................................(2.4)
U sonic
where U is gas velocity in air and Usonic the sonic velocity at ambient temperature
is about 340 m/s.
3. Knudsen number
Large aerosol particles are bombarded constantly from all directions by a large
number of gas molecules. But when a particle is small, less than 1m in size, its
location in space may be affected by bombardment of individual gas molecules
and then its motion is no longer determined by continuum flow considerations,
but by gas kinetics. The average velocity of a molecule V is a function of its
molecular weight M and the gas temperature T. In air at normal temperature and
pressure, this molecular velocity Vr is 463 m/s. Using these air reference values,
the average velocity can be estimated for other gases and temperatures.
3
V  Vr T / Tr 
1/ 2
M r / M 1 / 2 ..............................................(2.5)
The mean free path  is the mean distance a molecule travels before colliding
with another molecule. In air at 200C and atmospheric pressure, the mean free
path r is 0.0665 m. Using these reference values  is determined for other
pressures and temperatures.
 110

 101.3  T  1 
293.15 ........................................................(2.6)
  r 



 P  293.15  1  110
T 

The gas properties for several gases at NTP i.e. 293.15 0K and 101.3Kpa is
given in Table 1.
Table 1: Properties of various gases at NTP
Gas
Dynamic viscosity
Density g
Mean free
 (P)
(10-3g/cm3)
path  (m)
Air
182.03
1.205
0.0665
Ar
222.92
1.662
0.0694
He
195.71
0.167
0.192
H2
87.99
0.835
0.123
CH4
109.77
0.668
0.0537
C2H6
92.49
1.264
0.0328
4
Iso-C4H10
74.33
2.431
0.0190
N2O
146.46
1.837
0.0433
CO2
146.73
1.842
0.0432
Source: Paul [14]
The Knudsen number Kn relates the gas molecular free path to the physical
dimension of the particle, usually the particle radius
Kn 
2
........................................................................(2.7)
dp
where dp is the average diameter of the particle.
Kn<<1 indicates continuum flow and Kn>>1 indicates free molecular flow. If Kn is
approximately in the range of 0.4-20, then the flow is referred to as transition or
slip flow regime.
If the particle is much smaller than the gas molecular mean free path (Kn>>1) it
can travel past an obstacle at a very small distance from the object since no gas
particle may impede it. If the particle is very large (Kn<<1), many gas collisions
occur near the surface and the particle is decelerated. When the Knudsen
number is of the order of unity, the particle may slip by the obstacle. When the
particle size is in this slip flow regime, it is convenient to assume that the particle
is still moving in a continuum gas flow. To accommodate the difference, a slip
correction factor Cc also known as the Cunningham slip correction factor is
introduced into the equations.

 
C c  1  K n    exp 
 Kn


........................................................(2.8)

5
Using the Millikan cell apparatus ,  and  are constants determined taking
=6.73*10-8 m at 296.15K and 760 torr. The values for ,  and  for solid
particles and oil droplets are given in Table 2.
Based on the application of Modulated Dynamic Light Scattering (MDLS) method
[12] a range of different values of ,  and  were obtained which are given in
Table 3 and Table 4. The air pressure is maintained as 760 to 0.2 torr and the
Knudsen number ranges from 0.06 to 500.
Table 2: Values for ,  and  using Millikan Cell apparatus
Particle type



Solid particles
1.142
0.558
0.999
Oil droplets
1.207
0.440
0.596
Source : Paul [12]
Table 3: Values of ,  and  by direct computation and nonlinear fit ,  and 
by direct computation and nonlinear fit
Parametes
Direct computation
Nonlinear fit

1.2310
1.2291

0.4695
0.4711

1.1783
1.1705
Source : Paul [14]
6
Table 4: Values of , and  as calculated by different authors
Author
Mean free path



(m)
Knudsen and Weber (1911)
0.09417
0.772
0.40
1.63
Millikan (1923)
0.09417
0.864
0.29
1.25
Langmuir (1942)
0.133
0.62
0.22
1.25
Davies (1945)
0.066
1.257
0.40
1.10
DeMarcus and Thomas (1952)
0.0655
1.25
0.44
1.09
Reif (1958)
0.0652
1.26
0.45
1.08
Fuchs (1964)
0.0653
1.246
0.42
0.87
Dahneke (1972)
0.066
1.234
0.414
0.87
Allen and Raabe (1982)
0.0673
1.155
0.471
0.596
Allen and Raabe (1985)
0.0673
1.207
0.440
0.596
Hutchins and Harper (1995)
0.0673
1.231
0.4695
1.1783
For pressures other than atmospheric, the slip correction factor changes because
of the pressure dependence of mean free path in K n and the following can be
used for solid particles.
Cc  1 
15.39  7.518 exp( .0741Pd p )
Pd p
.........................................................(2.9)
where P is pressure in KPa and dp is particle diameter in m.
7
Cc equals 1 in continuum regime and becomes greater than 1 for decreasing
particle diameter in the transition regime.
4. Aerodynamic drag on particles
The externally applied forces on an aerosol particle are opposed and balanced
by the aerodynamic drag force. As an example, consider a skydiver. The air
resistance eventually balances the gravitational force pulling the skydiver
towards the earth. A particle’s drag force Fdrag relates the resistive pressure of
the gas to the velocity pressure and is determined by the relative motion between
the particle and the surrounding gas. When the particle dimensions are much
larger than the distance between the gas molecules, the surrounding gas can be
considered as a continuous fluid.
Under this condition, the drag force is given by
Fdrag 
 C d  gV 2 d p 2
8
.............................................................(2.10)
The aerodynamic drag force is related to the gas density and not the particle
density. The coefficient Cd relates the drag force to the velocity pressure. When
the inertial force pushing the gas aside, due to the velocity difference between
the gas and the particle, is smaller than the viscous resistance force, the drag
coefficient Cd is given by
Cd 
24
, Re p  0.1.....................................................................( 2.11)
Re p
8
Substituting equation (2.1) in equation (2.11)
Cd 
24
..................................................................................(2.12)
 gVd
Substituting this value of Cd in equation (2.10) we get
Fdrag  3Vd p ............................................................................(2.13)
This equation is also known as the Stokes law. The particle drag for shapes
other than spheres is usually difficult to predict theoretically. Therefore for
particles of other shapes, a dynamic shape factor , is introduced that relates the
motion of the particle under consideration to that of a spherical particle.
Fdrag  3Vd m ......................................................................(2.14)
where dm is mass equivalent diameter
The gases are not continuous fluids as indicated above, but consist of discrete
molecules. Therefore, when the particle size approaches the mean free path of
the gas molecular motion, we can apply a correction that takes the slip between
the gas and the particle into account. Thus, the Cunningham slip correction
factor Cc is introduced into equation (2.14)
Fdrag 
3Vd m
........................................................................(2.15)
Cc
Equation (2.15) assumes that the flow around the particles is laminar. For larger
Rep, empirical relationships for Cd have been developed to extend Stokes law.
9
For Rep above 0.1, Paul [14] gives the following equations.
Cd 
Cd 

24 1  0.0916 Re p
Re p

24 1  0.158Re p
,0.1  R
2/3
,5  R
ep
ep
Re p
 5.......................................(2.16)
 1000.......................................(2.17)
Thus, using the appropriate form of the drag coefficient (equations 2.11, 2.16 or
2.17) and including the shape factor and slip coefficient we can calculate the
drag force calculated over a wide range of particles and conditions.
Fdrag 
Cd  g V 2 d p 2
8Cc
.......................................................(2.18)
Table 5: The shape factors for various types of compact particles
Dynamic shape factor, 
Shape
Sphere
1.0
Cluster of spheres
2-sphere chain
1.12
3-sphere chain
1.27
4-sphere chain
1.32
Glass fiber
1.71
Dusts
10
Bituminous coal
1.05-1.11
High-ash soft coal
1.95
Quartz
1.36-1.82
Sand
1.57
Talc
2.04
UO2
1.28
ThO2
0.99
Source : Paul [14]
Brownian Force
Suppose that a particle, about 0.001m in size, is suspended in a liquid at room
temperature. Then, its thermal motion may be observed, under a microscope, to
display a random zig-zagged path due to random thermal collisions with liquid
molecules. The collisions give rise to unbalanced forces, called Brownian forces,
acting on the particle. This random behavior of the particle is called Brownian
motion. This Brownian force causes an abrupt change of the particle velocity in
both magnitude and direction. Hence, the Brownian force is very important for
sub micron particles.
If a number of particles subject to Brownian motion are present in a given
medium and there is no preferred direction for the random oscillations, then over
a period of time the particles will tend to spread evenly throughout the medium.
Suppose if A and B are two adjacent regions, and at time t, A contains twice as
11
many particles as B, at that instant the probability of a particle leaving A to enter
B is twice as great as the probability that a particle will leave B to enter A.
Brownian force is modeled as a Guassian white noise random process as
described by Ahmadi and Smith [6]. The Brownian force is represented by n i (t).
where
ni (t )  Gi
S o
t
..................................................................(2.19)
and
So 
216T
...............................................................(2.20)
  d 5 S 2Cc
2
Lift Force
The Staffman’s lift force becomes important for particles that are not too small i.e.
particles between 2m and 4m in size with high shear rates. It is defined by
Ahmadi and Chen [3] as
Fl 
2 Kv1 / 2 d ij
Sd (d lk d kl )
(u j  u jp )
The deformation rate tensor dij is defined as being equal to
1
u i , j  u j ,i  . K is the
2
constant coefficient of Staffman’s lift force having a value of 2.594.
12
When distances very near to the wall are considered, the particle is assumed to
be primarily translating parallel to the wall. The lift force is mainly in the direction
perpendicular to the wall and is given by Ahmadi and Chen [3] as
27v 1 / 2  u j

Fl 
2 2 dS  h



1/ 2
u
j

 u jp J m
The lift forces in other directions are small and hence are neglected. The
Staffman’s expression for the lift forces is restricted to the condition that the
particle Reynolds number based on the particle-fluid slip velocity is much smaller
than the square root of the particle Reynolds number based on the shear field.
McLaughlin (1991,1993) evaluated the correction factor Jm and relaxed these
restrictions. When gravity is in the direction of flow, the lift force enhances the
particle deposition rate. For upward flow, however, the lift forces will move
particles away from the wall and reduces the deposition rate.
Gravitational Force
When indoor aerosol particles enter the room, the larger particles
(particle size > 4m) deposit on internal surfaces quickly under the influence of
gravity. For such particle deposition, gravitational force has a major role to play.
Gravitational force also has the influence on the deposition of the indoor particles
when the size of the particles is between 2m and 4m. When gravity is in the
same direction as the flow, particles move much faster than the flow due to
gravitational sedimentation effect. The gravitational force per unit mass is given
by Ahmadi and Chen [3] as
13
1

G  1   g i
 S
Velocity Distribution
Doors and windows can be used to supply fresh air to rooms of a building but to
guarantee energy efficient air exchange rate mechanical ventilation is preferred.
For mechanical ventilation either displacement ventilation or mixing flow systems
can be used with supply flow rate reduced to meet hygienical requirements. For
comfort ventilation in rooms, vertical displacement flow systems have become
very popular. Wall mounted diffusers supply air directly into the occupied zone at
low velocity. The hot air released by the people entrains to the occupied zone
and creates an upward flow. To know the velocity distribution in a room, it
becomes important to examine the flow in front of an air terminal device and to
investigate whether this flow can be treated independent of parameters such as
room geometry, heat source location, location of exhaust opening etc. In order to
simplify the design procedure it is assumed that the flow depends only on certain
parameters such as type of diffuser, obstacles on floor, flow rate and Archimedes
number.
In our study it is assumed that the flow is radial flow from a single diffuser
wherein the air movement is not influenced by the sidewalls. Radial flow is the
flow close to the floor that has a virtual origin close to the diffuser. Radial flow at
the diffuser is common and is generated by blades in the diffuser in such a way
that the velocity obtains a radial distribution at the surface.
14
The velocity distribution near the floor is described by the decay of the maximum
velocity ux as a function of distance x. The velocity decay as given by Nielsen [7]
is
ux
h

uf
x  x0
where
h is the height of the diffuser and x0 is the distance to a virtual origin for selfpreserving flow, which in the study is assumed to be 1 feet.
The face velocity, uf, is given as the supply flow rate divided by the face area, af,
of the diffuser
The equation was further modified by Nielsen [8] to
ux
1
 K Dr
............................................................................(2.21)
q0
x  x0
where
K Dr 
ebm

 0  f (n)dn
0
e is a factor that represents the initial increase in flow rate due to entrainment in
the accelerating flow close to the opening.
bm is a factor which adjusts the flow in the direction  = 0 to the flow profile
generated by the diffuser.
15
KDr is independent of the distance x+ x0, but is a function of the Archimedes
number.
The velocity in most cases will be proportional to 1/(x+ x0) and the value of KDr is
adjusted to the situation to predict the velocity. A lot of experiments have been
conducted to establish a relationship between KDr and Archimedes number. KDr
is different for different products as it varies from 5 to 13 ms-1 at high Archimedes
number. From the experiments it was found that KDr increases with increasing
Archimedes number. The reason for this being that the gravity will accelerate the
vertical flow close to the opening and this will generate an air movement in a
relatively thin layer along the floor. The design of a diffuser is influenced by the
parameters in equation (2.21). A KDr value of 7 is obtained as given by Nielsen
[8],
when

 f (n)dn  1.1;   0.1m;
0
  ; e  2.5; f bm  1
0
If the value of bm is changed to 1.5 the KDr value changes to 11 as explained in
Nielsen [8]. However in order to obtain more information about the general air
movement, more measurements have to be taken and more experiments have to
be performed in this field.
Figure 1 shows the difference in the distribution of the indoor air contaminant in
the zones having displacement ventilation and mixed ventilation.
16
Zones having high gas concentration
Supply air inlet
Supply air inlet
Displacement Ventilation
Mixed Ventilation
Figure 1: Displacement Ventilation and Mixed Ventilation
Assumptions made in the development of the model
When an aerosol particle enters a room, it has an initial velocity uo because of
which it travels a distance xi. Based on the distance it travels within a specific
period of time, the velocity distribution in the room under displacement ventilation
is studied from which the deposition velocity of the aerosol particle is calculated.
A model has been formulated to study the aerosol dynamics based on the
diameter of the particles. In order to develop this model certain assumptions
have been made. They are
17
(a) No heat and mass transfer between the air and the particles.
(b) Rebound effect is neglected.
(c) When a particle reaches a wall it will stick to the surface.
(d) Negligible forces acting on particles of varying diameters are neglected.
(e) Aerosol size distribution is pretty constant.
(f) Aerosol particles are spherical in shape.
(g) Airflow is homogeneously and isotropically turbulent.
(h) There is no generation of particles indoors.
The size of the aerosol particle is taken as the prime criterion to develop the
model. Based on the size of the aerosols, the prominent forces acting are
considered and from that the various parameters such as distance traveled,
deposition velocity of the particle, velocity distribution etc. influencing aerosol
dynamics are calculated. Sections 2.8, 2.9, 2.10, 2.11 and 2.12 show the various
steps involved in the development of the model.
Equations for the deposition velocity and distance traveled when the
diameter of the particle is less than 0.01m (d<0.01m)
The basic governing equation as given by Chen and Ahmadi [3] gives the rate of
change of velocity as a function of the Brownian force per unit mass.
du ip
 ni (t )
dt
where
18
ni (t )  Gi
So 
S o
t
216T
 d 5 S 2 C c
2
and
S
particle density
gas density

 
C c  1  K n    exp 
 Kn




du ip
216T
 Gi
...................................................(2..22)
dt
d 5 S 2 Cc t
uip  Gi
216T
d 5 S 2 Cc
uip1  uip  Gi
t
216T
d 5 S 2Cc
.................................................(2.23)
t
....................... .(2.24)
Its known that
dxi
p
 ui .........................................................................(2.25)
dt
Substituting the value of uip from equation (2.24) in equation (2.25) we get
19
xi 1  xi
216KT
 uip  Gi
t
d 5 S 2 Cc
t ..........................................(2.26)
Rearranging the terms we get
xi 1  xi  uip t  Gi
216KT
( t ) 3 ..................................(2.27)
d 5 S 2 Cc
Further investigations have to be done to get the value of Gi.
Equations for the deposition velocity and distance traveled when
the diameter of the particle lies between 0.01m and 2m (0.01d<2)
duip
 ni (t )  Fd ................................................................(2.28)
dt
ni (t )  Gi
Fd 
S o
t
3C d Re
(u i  u ip )
2d (2 S  1)C c
2
where

 
C c  1  K n    exp 
 Kn




Based on the values of Re the values of Cd are calculated [14]
For (Re<0.1)
Cd 
24
Re
20
For (0.1  Re< 5)
Cd 
24(1  0.0916Re )
Re
For (5  Re< 1000)
Cd 
24(1  0.158Re
Re
2/3
)
During the calculation of drag force there is a term-involved ui, which is the initial
velocity of the particle. If the initial velocity of the particle is not known directly
but the face velocity and the flow rate is known then the initial velocity is
calculated from the velocity distribution of the particle.
 1 
ui

 K dr 
qo
x

x
0 

K dr 
ebm

 0  f (n)dn
0
Kdr is a function of the Archimedes number and is adjusted to the situation. It
varies from 5 to 13 ms-1.
e is a factor that represents initial increase in flow rate due to entrainment in the
accelerating flow close to the flow.
21
Bm is a factor that adjusts the flow in the direction  = 0 to the flow profile
generated by the diffuser.
 is the thickness of the profile
For particles in this size range the rate of change of deposition velocity is given
by
du ip
3C d Re

ui  u ip ....................................(2.29)
2
dt
2d (2S  1)Cc



3C R
duip   2 d e
ui  uip
 2d (2S  1)Cc

dt

Integrating the above equation we get
uip 
3C d Re
ui  uip t  C....................................(2.30)
2d (2S  1)Cc

2

At t = 0, uip = u0
uip 
3C d Re t
ui  uip  u 0 ....................................(2.31)
2d (2S  1)Cc

2

Arranging the terms the equation modifies to
uip 
3C d Re t
3C d Re
uip 
ui t  u 0
2
2d (2S  1)Cc
2d (2S  1)Cc
2
The equation for the calculation of uip now becomes
22


3C R
 2 d e

u
t

u
i
0
 2d (2 S  1)C

c
 ............................................( 2.32)
u ip  

3C d Re t 
1 

 2d 2 (2 S  1)C 
c 

We know that
dxi
 uip
dt
Therefore
Substituting the value of uip
dxi  u ip dt


3C R
 2 d e

u
t

u
i
0
 2d (2 S  1)C

c

 dt
dxi 

3C R t 
1  2 d e

 2d (2 S  1)C 
c 

In order to integrate the equation assume
3Cd Re
b
2d (2S  1)Cc
2
dxi 

ui bt  u0
.................................................................(2.33)
1  bt
ui bt
u
 0
1  bt 1  bt
Then
23
Let
1  bt  p
Differentiating the above equation
bdt  dp
dt 
t
1
dp
b
p 1
b
Substituting these values in equation (2.33)
dp
 p  1  1  dp
xi  bu i  
 u0 
 
bp
 b  p  b

bui 
b
2

1
u0
 1  p dp  b 
dp
p
Integrating the above equation with respect to p
xi 
bui 
2
 p  ln( p)  u0 ln( p)  c.......................................(2.34)
b
b
Substituting the values of b and p in equation (2.34)
24


3C R
 2 d e
u i 
2d (2 S  1)C c  
3C R t
1  2 d e
xi  
2 

  2d (2 S  1)C c
3C R
 2 d e

2
d
(
2
S

1
)
C
c 


2u 0 d 2 (2 S  1)C c 
3C R t
ln 1  2 d e
3C d Re
 2d (2 S  1)C c


3C R t
  ln 1  2 d e

 2d (2 S  1)C c




  C.............................................( 2.35)


At t = 0, xi=0 and ui =u0
which gives


3C R
 2 d e
u i 
2d (2S  1)C c 
C  
2
 3C d Re t 
 2

 2d (2S  1)C c 
Substituting the value of C in equation (2.35)


3C d Re
216T
 Gi


u
i
5 2
2

 

d
S
C

t
2
d
(
2
S

1
)
C
3C d Re t
c
c
 1 
xi  

2
2
 2d (2S  1)C c


3C R
 2 d e

 2d (2S  1)C c 


3C R t
  ln 1  2 d e

 2d (2S  1)C
c




3C d Re u i


2
2



2u d (2S  1)C c
3C R t
2d (2S  1)C c 
  
 0
ln 1  2 d e
.....................(2.36)
2
3C d Re
 
 2d (2S  1)C c   
3C d Re
 
  2
 
  2d (2S  1)C c  
25




Equations for deposition velocity and distance traveled when
the diameter of the particle lies between 2m and 4m (2 d < 4)
For particles in this range the Brownian force is negligible and hence is ignored.
du ip
 ni (t )  Fd  Fl  G
dt
Where
216T
d 5 S 2 C c t
ni (t )  Gi
The Brownian force is negligible for particles in this diameter range and hence is
neglected
27 1 / 2  u j

Fl 
2 2 dS  h
Fd 



1/ 2
u
j

 u jp J m
3C d Re
(u i  u ip )
2d (2 S  1)C c
2
 1
G  1   g i
 S
duip
3C d Re
27 1 / 2  u j
p


ui  ui 
dt
2d 2 (2S  1)Cc
2 2 dS  h


1/ 2



u
j

 1
 u jp J m  1   g i .........( 2.37)
 S
Integrating the above equation

3Cd Re
27 1 / 2  u j
p

du 
u  ui  2 
 2d 2 (2S  1)Cc i
2 dS  h

p
i


26
1/ 2



u
j
 1 
 u jp J m  1   g i dt
 S  

Integrating the above equation

3C R
27 1 / 2  u j
u  2 d e
ui  uip t  2 
 2d (2S  1)Cc
2 dS  h

p
i


1/ 2



u
j

 1
 u jp J m t  1   g i t  C ........(2.38)

 S


At t=0,uip = uo
Substituting these in the above equation we get C= uo
Substituting the value of C in equation (2.38) we get

3C R
27 1 / 2  u j
u  2 d e
ui  uip t  2 
 2d (2S  1)Cc
2 dS  h

p
i


1/ 2



u
j

 1
 u jp J m t  1   g i t  u0 .....(2.39)

 S


Arranging the terms the equation modifies to

3C d Re t
3C d Re
27 1 / 2  u j
p


u  2
ui   2
ui t 
2
2d (2 S  1)C c
2
d
(
2
S

1
)
C
2

dS
c
 h

p
i


1
 u j  u jp J m t  1   g i t  u 0 

 S






3C d Re
27 1 / 2  u j 
 1
p






u
t

u

u
J
t

1

g
t

u


i
j
j
m
i
0
2
 h 
 2d 2 (2S  1)C

S
2

dS


c


 ..........(2.40)
u ip  

3C R t 
1  2 d e

 2d (2S  1)C c 
Let
3C d Re
27 1 / 2  u j
 a; 2 
2d 2 (2S  1)Cc
2 dS  h
1/ 2



u
j

 1
 u jp J m  c; 1   g i  d
 S
27
We know that
dxi
 uip
dt
dxi
aui t  ct  dt  u o

dt
1  at
dxi  au i  c  d 
u0

t 
dt  1  at  1  at
Let
1  at  p
Differentiating the above equation
adt  dp
dt 
t
1
dp
a
p 1
a
xi  au i  c  d 
t
dt
dt  u 0 
1  at
1  at
1  dp
dp
 au i  c  d   

 u0 
  1  
2
p a
ap
a

 
dp
 p  1  1  dp
 au i  c  d  
 u0 
 
ap
 a  p  a
28

aui  c  d 
a
2
 p  ln( p)  u0 ln( p)  C......................................(2.41)
a

3C d Re
27 1 / 2  u j 
 1   
p



u

u

u
J

1   g i 
i
j
j
m
  2d 2 (2 S  1)C c
2 2 dS  h 
 S  
*

2


 3C d Re t 
 2

xi  

 2d (2 S  1)C c 







3c d Re t
3c d Re t
 1  




 ln 1  2

   2d 2 (2 S  1)C c 

2
d
(
2
S

1
)
C

c 




2u 0 d 2 (2 S  1)C c 
3C R t
ln 1  2 d e
3C d Re
 2d (2 S  1)C c


  C.......................................................(2.42)

At t=0,xi=0
Then

3C d Re
27 1 / 2  u j 
 1  
p



u

u

u
J

1   g i t 
i
j
j
m
 2d 2 (2S  1)C
2 2 dS  h 
 S 
c

C
..........................(2.43)
2
 3C d Re t 
 2

2
d
(
2
S

1
)
C
c 



Substituting the value of C in equation (2.42) we get
29

3C d Re
27 1 / 2  u j 
 1   
p



u

u

u
J

1   g i 
j
m
  2d 2 (2 S  1)C c i 2 2 dS  h  j
 S  
*

2


 3C d Re t 
 2

xi  

2
d
(
2
S

1
)
C
c






 

 
3c d Re t
3c d Re t
 1  

  ln 1  



  2d 2 (2S  1)C  
   2d 2 (2 S  1)C c  


c  

 
 


2u 0 d 2 (2 S  1)C c 
3C R t
ln 1  2 d e
3C d Re
 2d (2S  1)C c
 3C d ReU i
27 1 / 2  u 


 uj

 2d 2 (2 S  1)C c 2 2 dS  h 




3C R

 2 d e
 2d (2 S  1)C

c









 1 
 u jp J m  1   g i 
 S 

.............(2.44)
2










Equations for deposition velocity and the distance traveled when the
diameter of the particle is greater than 4m (d>4)
du ip
 Fd  G
dt
Fd 
3C d Re
(u i  u ip )
2d (2 S  1)C c
2
 1
G  1   g i
 S
30
du ip
3C d Re
 1

(u i  u ip )  1   g i ....................................(2.45)
2
dt
2d (2S  1)Cc
 S

3C R
 1 
duip   2 d e
(ui  uip )  1   g i dt
 S 
 2d (2S  1)Cc
Integrating the above equation we get
uip 
3C d Re
 1
(ui  uip )t  1   g i t  C.................................(2.46)
2d (2S  1)Cc
 S
2
At t=0, uip=uo
Substituting this in equation (3.24) we get C= uo
Equation (3.24) now becomes
uip 
3C d Re
 1
(ui  uip )t  1   g i t  u 0 .................................(2.47)
2d (2S  1)Cc
 S
2
3C d Re
 1
u i t  1   g i t  u 0
2d (2 S  1)C c
 S
u ip 
..............................................( 2.48)

3C d Re t 
1 

 2d 2 (2 S  1)C 
c 

2
dxi
p
 ui
dt
Substituting for uip
3C d Re
 1
u i t  1   g i t  u 0
dxi 2d (2 S  1)C c
 S

.
dt

3C d Re t 
1 

 2d 2 (2 S  1)C 
c 

2
31
3C d Re
 1
u i t  1   g i t  u 0
2d (2 S  1)C c
 S
dxi 
dt..............................................(2.49)

3C d Re t 
1  2

2
d
(
2
S

1
)
C
c 

2
In order to integrate let
3C d Re
 1
 a; 1   g i  b
2d (2S  1)Cc
 S
2
dxi aui t  bt  uo

dt
1  at
 aui  b t  uo 
dxi  
dt
1  at


Integrating the above equation
xi  au i  b 
t
dt
dt  u 0 
…………………………………….(2.50)
1  at
1  at
Let
1+at = p
a dt = dp
Which gives dt = dp/a
Substituting these in equation (2.50) we get
dp
 p  1  1  dp
 au i  b  
 u0 
 
p
 a  p  a
u
 au  b 
  i 2  p  ln( p)  0 ln( p)  c...........................................(2.51)
a
 a

Substituting the values of a, b and p into equation (2.51)
32
 3C R u
 1

d e i
 1   g i
2
 2d (2S  1)C  S 
c
xi  
2



3C d Re
 2


 2d (2S  1)C c 




 
3C R t
 1  2 d e
  2d (2S  1)C c


2u 0 d 2 (2 S  1)C c 
3C R t
ln 1  2 d e
3C d Re
 2d (2S  1)C c


3C R t
  ln 1  2 d e

 2d 2S  1C c

 



  C..................................................(2.52)


At t = 0, xi =0 and ui = u0
Substituting these values in equation (2.52) we get
 3C R u
 1

d e i
 1   g i
2
 2d (2 S  1)C  S 
c
C  
2



3C d Re
 2


 2d (2 S  1)C 
c 









Substituting the value of C in equation (2.52) we get
 3C R u
1


d e i
 1   g i
 2d 2 (2 S  1)C
S

c
xi  
2



3C R
 2 d e


 2d (2 S  1)C 
c 




 
3C d Re t
 1 
2
2d (2 S  1)C c
 




3C d Re t
  ln 1 
2


2d 2 S  1C c



3C Re
1


 1   g i
2
  2d (2 S  1)C c 
S
2u d (2 S  1)C c 
3C d Re t

 0
ln 1 
2

3C d Re
2d (2 S  1)C c  


3C R

 2 d e

 2d 2 S  1C 

c



2
33







............................(2.53)



Nomenclature
bm
Factor to adjust the flow
Cc
Stokes– Cunningham coefficient
Cd
Drag coefficient
d
Diameter of a particle (m)
e
Factor representing initial increase in flow
Fd
Drag force per unit mass (dynes)
Fl
Lift force per unit mass (dynes)
G
Gravitational force per unit mass (dynes)
Gi
Gaussian Random number
gi
Acceleration of body force (m2/s)
h
Height of diffuser (m)
Jm
McLaughlin’s correction factor
Kn
Knudsen’s number
KDr
Constant that depends on Archimedes number
L
Length (m)
Ma
Mach number
M
Gram molecular weight (gm/m3)
ni (t)
Brownian force per unit mass (dynes)
P
Pressure (dyne/cm2)
Ru
Universal gas constant (8.31 * 107 dyne cm/mole)
Re
Reynolds number
S
Ratio of particle density to gas density
34
Sc
Schmidt number
T
Temperature ( 0K)
t
Time step used during simulation (sec)
ui
Instantaneous gas velocity (m/sec)
u ip
Particle velocity in the ith direction (m/sec)
uj
Velocity component parallel to the wall (m/sec)
V
Volume of the room
vd
Deposition velocity (m/sec)
uo
Initial velocity of a particle (m/sec)
x
Distance traveled (m)
xo
Distance to a virtual origin (m)
Greek Symbols

Coefficient in slip correction coefficient

Coefficient in slip correction coefficient

Coefficient in slip correction coefficient

Boltzman’s constant (1.38*10-6 dyne cm/K)

Mean free path (m)

Dynamic viscosity (N/sm2)

Kinematic viscosity (m2/s)
g
Gas density (Kg/m3)
p
Particle density (Kg/m3)

Dynamic shape factor
35
36